I thought a scientific calculation can only become fact if it remains true forwards and backwards? If this is the case, then can Einstein's E=MC2 also be written as M=EC2? Or is it M=E/C2?
Seems right to be able to say Mass=Energy times the speed of light. But, why does the speed of light have to be squared? Isn't that implying something faster than the speed of light?
I am perfectly capable of making up my own mind. Right?

that'd be m = E/c^2 (after dividing both sides by c^2).
without deriving the equation, you can still check whether the units are right: basic units: mass : kilograms (kg) distance: meters (m) time: seconds (s) derived units: speed = distance/time : m/s acceleration = speed/time : m/s^2 energy = mass * acceleration * distance : kg*m^2/s^2 the last equation might not be familiar, but it's the expression for the energy of motion. if a mass is accelerated at a constant rate over a distance, multiplying all 3 values together tells you how much energy was added by the acceleration. In the same way that you can only get an acceleration by dividing distance by time squared, I hope you can see, to get an energy from a mass and a speed, you have to multiply mass*speed^2. as it happens if you use the speed of light, the result is the energy inherent in a massive object when it's at rest. since in our everyday experience speeds are typically much less than that of light, that means that most of the energy in the stuff we see is present as mass, not motion. I hope that helps a bit... clear enough?
<Insert Einstein Quote Here>

...if a mass is accelerated at a constant rate over a distance, multiplying all 3 values together tells you how much energy was added by the acceleration. Is that equation (E=MC^2) 'THE' theory of relativity? I just don't know what it has to do with the examples used to explain the theory. Such as, two people, each at opposite ends of a train car and jumping up at the same time, while someone standing outside of the moving train sees one man jumping before the other. Also, how did they get the atom bomb from this equation?
I am perfectly capable of making up my own mind. Right?

no, it's just one result of the theory. the core principles are 1) the speed of light is constant for all observers 2) speeds (other than that of light) are relative  no absolute frame of reference you might think that those principles are contradictory. Einstein however showed that they're not, which was pretty cool. all that weird stuff like time dilation and length contraction happens because the above principles apply simultaneously. each observer can mark out their own coordinate system for spacetime (x,y,z,t). according to principle 2, there's no special coordinate system that everyone can use, but what relativity does do is allow you to transform between coordinate systems if you know the relative velocity of the observers involved. in mathematical terms, special relativity is essentially a coordinate transformation. the examples with trains and whatnot are intended to show the strange consequences of the theory without having to do any algebra. Apparently this is necessary because everyone hates algebra. that would be a bit oversimplified. development of the atom bomb had as much to do with experimental advances like the discovery of radioactivity and atomic structure, and the theory of quantum mechanics. in nuclear reactions less than 1% of the mass is released as energy. the same thing happens in ordinary chemical reactions, it's just that the mass loss is too small to measure. so E=mc^2 isn't really telling you anything fundamental about nuclear reactions.
<Insert Einstein Quote Here>

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Dark_Equations.docx
Equation of Dark Matter M = ec2 (2 is exponent) is the inverse function of the Einstein's formula (E = mc2). Hence, it can be the massenergy equivalence of dark matter. Although this is an educated guess, it gives a reasonable answer to the big question called dark matter. The theory is a simple and straightforward approach to the dark matter problem. It has a mathematical basis. Any onetoone function has an inverse function. The equation E = mc2 is a onetoone function. Thus, it has an inverse function. The inverse function of E = mc2 is M = ec2. Dark photon travels at the speed of c seconds per meter (not c m/s). Based on the equation M = ec2 , energy e converts to a notable mass M. This explains dark matter's great mass. M = ec2 , where e is relativistic energy, can be written as: M = (gamma) e0 c2 where e0 is rest energy. Thus, M = e c2 is the massenergy equivalence of dark matter from our point of view as observers made of ordinary matter. The gamma is dark transformation factor from our viewpoint; its value is as follows: Transformation Factor of Dark Matter: gamma = 1 / [ 1  ( v2 / c2 ) ] 1/2 Both 2 are exponents. 1/2 is exponent too, which lies above the parenthesis. Both v and c are in terms of s/m. v is relative speed of a dark matter object or particle, and c is almost equal to 3 x 10 (8) s/m where 8 is exponent. The domain of v is the closed interval [ 0 s/m , c s/m ]. [It's hard to write equations here because there is no suitable tool to write them, so exponents come in front of the variables; they must lie above the variables.] The equation implies that dark matter in its own perspective has a three dimensional time and a one dimensional space. However, dark matter can be detected indirectly by observers who live in ordinary spacetime. Also, dark matter has computational aspects because mathematics is able to demonstrate an inverse spacetime function. Telling that we are not familiar with such a spacetime in our daily activities is not a plausible reason to deny its possible existence. Another reason that causes the inverse spacetime theory of dark matter to be rejected promptly and sometimes even angrily without any deliberation, as I have experience about that, is the impact of the current massenergy equivalence which is considered as a universal, absolute and everywhere working equation by almost all scholars and physicists. However, the history of science has falsified many ideas that were regarded as universal and absolute knowledge in the past. The equation E = mc2, i.e. the current massenergy equivalence, is just true for ordinary matter and ordinary antimatter. It is not true for dark matter. The reason is that dark matter is not baryonic matter to be determined based on the current massenergy equivalence. The inverse function of E = ec2 improves our understanding of universe. There is no physical or mathematical contradiction about that. The current physics based on E = mc2 cannot explain how a neutrino changes its flavor possessing a different mass. For describing Higgs mechanism, particle physicists have to assume a sequence of massive particles which their number will go toward infinity. They also require a lot of complex assumptions and explanations to convince themselves. No, no. It really doesn't work because there is a simple and better way that is consistent with reality. This is the essence of Occam's razor. It is possible to explain motion of all planets in the Solar System by considering the Earth as the center of the System. However, it needs a lot of complex assumptions, and finally the Earth is not the center of Solar System. Once it was a taboo to talk about dark photon because it doesn't fit the current massenergy equivalence or it cannot be defined by the Standard Model. Now, what happened that some physicists have started to search for it experimentally though there are no theoretical tools to describe it within the Standard Model? Banning ideas just because they don't fit this or that model is actually an action against the development of science. The most challenging part of the new theory was to find the correct equation of momentum in dark matter from our point of view. It took a long time to find it. After that, workmass theorem of dark matter in our viewpoint was a difficult task. The idea that dark photon in its own perspective traverses time per unit space was the initial thing that came to the mind while I was studying nonEuclidean geometry. Then, I realized that the idea corresponds to the inverse function of E = mc2. The strange behavior of neutrinos pushed me to think that neutrino cannot be an ordinary matter particle. Pauli proposed neutrino to explain incompatibility of beta decay to law of conservation of energy though Bohr said that beta decay didn't obey the law. Adding neutrino to the equation of beta decay has raised a new problem that scientists equipped with the current massenergy equivalence cannot still resolve it in a perfect way. It is a mass problem leading to search for super massive particles with unknown origin. They just pretend to know origin. However, the origin is standing out there while teasing them. That's dark matter. The two W bosons were proposed to explain neutrino absorption and emission. The Z boson explains elastic scattering of neutrinos from matter. Then, Higgs field and its particle, i.e. Higgs boson, were proposed to give mass to those particles. All of these particles have been discovered but it doesn't imply that the explanation presented by the Standard Model is true. The question is that what does give mass to Higgs boson? Thus, the Model needs to suggest some new particles. However, is there any end to these suggestions? There is really no end: Particle after particle, without having suitable equations or formulas to predict the exact values of their masses. Why do W bosons, Z boson, and Higgs boson have very short lifetimes? The Standard Model also had to claim that Higgs field fills everywhere. Such a claim is not acceptable. What can be an elegant solution to the mass problem in both particle physics and astrophysics? Neutrinos, W bosons, Z boson, and Higgs boson are all dark matter particles. The inverse spacetime theory of dark matter depicts a symmetry between variables of ordinary matter and dark matter. The symmetry is not fake, it originates from properties of spacetime. Ordinary particles and dark particles appear together in all equations. They are not separate events. Dark matter prepares mass for ordinary matter, and ordinary matter prepares energy for dark matter. The theory explains creation of neutrino, flavor change of neutrinos, the mass origin of W bosons, Z boson, Higgs boson, and any supper massive particle with a short lifetime. It formulates great mass of dark matter and its effects in astrophysics. All of these are done with the least assumptions and simple equations. The theory can also be called inverse function theory of dark matter. E = mc2 means that 1) mass can convert to energy, and 2) energy is locked within a mass. However, M = ec2 means that 1) energy can convert to mass, and 2) mass is locked within an exotic energy. The main point is to think of mass quantization. However, mass quantization is not possible for ordinary matter. We have energy quantization in ordinary matter. To quantize mass, it requires to assume an inverse spacetime, and this leads to dark matter. Thus, we can quantize mass for dark matter. This implies that there is dark photon, and we need to know equation of dark photon mass. The equation of mass is M = h f where h is numerically equal to the Planck's constant, and f is frequency of dark photon. Dark photon doesn't have energy at least theoretically but it has mass which depends on its frequency. Therefore, the formula of the mysterious dark matter in our viewpoint is M = (gamma) e0 c2. This is the key equation to explain all aspects of dark matter (DM). **** References I have used the references below in my research. However, the assumptions, arguments, computations, and equations I presented for dark matter and dark energy are my own work, and cannot be found in any books, academic papers or articles. 1) Wikipedia. 2) Nicolson, Iain. (2007). Dark Side of Universe. 3) Close, Frank. (2010). Neutrino. 4) Randall D. Knight. (2013). Physics. Third edition. 5) Salas, Hille, Etgen. (2007). Calculus. Tenth edition. **** Fundamental Argument of the Inverse SpaceTime Theory of Dark Matter Why cannot we see dark matter? If dark matter is in spacetime and it has mass, we should be able to see it. Thus, what causes it to be invisible? There are three dimensions of space and one dimension of time. Anything we see is located in a 3D space and 1D time. Is the distinction between the dimensions of spacetime real? If it is, then nature is made of an absolute distinction of dimensions, i.e. only a 3D space and a 1D time. Since dark matter is in the spacetime, we should be able to see it, otherwise, there is something wrong here. What is that? The distinction between dimensions of spacetime is not real. The distinction is just a product of our sensory system. That's the work of our brain. The nature makes no distinction between the four dimensions of spacetime. Nature doesn't care if our perception implies that space is three dimensional and time is one dimensional. She can use three of the four dimensions as time and one of them as space. If our sensory system cannot see or perceive such a thing, we cannot deny its existence or blame the nature for the way she applies the four dimensions. In brief, there are 4 dimensions in spacetime but the distinction between the dimensions depends on the nature's application of dimensions. She uses 3 dimensions as space and 1 dimension as time to create ordinary matter. However, she uses 3 dimensions as time and 1 dimension as space to create dark matter. Since we are made of ordinary matter, we can see just ordinary matter. If we were made of dark matter, we could see just dark matter. Suppose that an object travels 2 meters in 1 second. We depict a coordinate system with the horizontal axis as time and the vertical axis as position. We can show the object as a point with coordinates ( 1 s , 2 m ). The speed of the object is 2 m/s. We call the point ( 1 s , 2 m ) as point A. If we exchange just the numbers in ( 1 s , 2 m ), but not the units, we will obtain: ( 2 s , 1 m ). What means the new point? It implies that an object travels 1 meter in 2 seconds, and its speed is ( 1 / 2 ) m/s. The new point is symmetric to the first point with respect to the line y = x . The two points are symmetric with respect to the line y = x . We call the point ( 2 s , 1 m ) as the point B. Now, we exchange the axes, i.e. considering the horizontal axis as position, and the vertical axis as time. What is the meaning of the point ( 2 m , 1 s ) in the new coordinate system? Is the same as the point A in the first coordinate system? No, they are not the same. We call the point ( 2 m , 1 s ) in the second coordinate system as the point C. The point ( 2 m , 1 s ) in the second coordinate system implies that an object travels 1 second in 2 meters, and its speed is ( 1 / 2 ) s/m! However, this is meaningless for us because our sensory system cannot perceive such an object. It makes no sense based on the ordinary physics too. Thus, we just exchanged the axes and confronted with a puzzle. What happened here? The point ( 2 m , 1 s ) in the second coordinate system stands at the same point as the point ( 2 s , 1 m ) in the first coordinate system. That is if we put the second coordinate system over the first coordinate system and ignore what the axes show as position or time, the points B and C lie on each other. What means this? We cannot perceive what the point C in the second coordinate system means, but we can perceive the point B. The second coordinate system was created by exchanging the position and time axes of the first coordinate system. The points C and B have different meanings. There is no way to think of point C in the current physics. Look at the three points: A ( 1 s , 2 m ) ; B ( 2 s , 1 m ) ; C ( 2 m , 1 s ). The points A and B are in a familiar coordinate system. The points A and B are mathematically the inverse functions of each other. They have physical meanings because they are in a coordinate system that makes sense to our sensory system. The point C is numerically the same as the point B, but by considering the units they are very different. The point C is just numerically the inverse function of the point A, but if we consider the units they are very different. The point C is obtained if we exchange both numbers and units in A ( 1 s, 2 m ). How can we interpret the point C in the second coordinate system where the horizontal axis is position and the vertical axis is time. We have no sensory tools to detect it because our sensory system cannot perceive such a thing. The point C is in the case that time is three dimensional and space is one dimensional. This is the inverse of what we know as spacetime or what we can perceive as spacetime. If the distinction between a 3D space and a 1D time is just a product of our sensory system, then nature uses three of the four dimensions as time and creates dark matter which cannot be seen by mankind. This assumption allows us to determine dark photon and quantize mass. If there is something as point C in the nature, we can physically analyse it by considering an inverse spacetime consisting of a 3D time and 1D space. We can use inverse functions to find the equations that demonstrate behavior of objects like the point C. Finally, we can say that an object that moves like the object C in the second coordinate system is a dark matter. The equation of motion of the point A is d = 2 t, where d is distance, and t is time. The constant 2 shows the speed of motion. We have: v (A) = 2 m/s. The equation of motion of the point B can be derived in this way: d = v t works for the point A. Since the point B satisfies an equation that is the inverse function of the equation of the point A, we can exchange the variables in d = v t and solving for d. Thus, we have t = v d and then d = t / v = (1 / v) t. In the equation d = (1 / v) t, we can replace 1 / v with v' as the speed of the point B. Thus, v' = 1 / v. Here, we should be careful about the units. v in (1 / v) is just numerically equal to the v, i.e. we ignore its units. Thus, we can write v' = (1 / v) m/s, and so d = v' t. Since v = 2 just numerically, we have v' = (1 / 2) m/s, and so d = (1 / 2) t is the equation of motion of the point B. Hence, v (B) = (1 / 2) m/s = 0.5 m/s. The equation of motion of the point C is based on t = v d, where we depict vertical axis for time t, and horizontal axis for position d. Thus, the unit of speed will be seconds per meter or s/m. The equation of the point C is t = (1 / 2) d. The speed of the point C is v (C) = (1 / 2) s/m = 0.5 s/m. The object moves in a 3D time and 1D space. **** Kinetic Mass of Dark Matter The relation between force and acceleration in dark matter from our point of view is F = e a. Thus, Newton's first and second laws of motion are not true in dark matter from our viewpoint. The relation between position x and time t in DM from our perspective is v = dx / dt. The relation between v and acceleration a in dark matter from our perspective is a = dv / dt . Dark matter has kinetic mass, not kinetic energy. The work W done by a net force F, parallel to the direction of motion, on a dark matter particle through a distance x from our point of view is as follows: WorkMass Theorem in Dark Matter: W = Integral of (F) dx from (x i) to (x f) = delta( Mk ) from Our Viewpoint Here is the proof: W = Integral of (F) dx = Integral of ( e a ) dx = ( e ) Integral of ( dv / dt ) ( dx ) = ( e ) Integral of ( dx / dt ) ( dv ) = ( e ) Integral of ( v dv ) = e [ ( 1 / 2 ) v2 ] [ from ( v i ) to ( v f ) = ( 1 / 2 ) e ( v f ) 2  ( 1 / 2 ) e ( v i ) 2 = delta( Mk ) where Mk is kinetic mass of dark matter particle from our point of view. ( v i ) and ( v f ) are initial and final speeds. We can conclude: Kinetic Mass of Dark Matter: Mk = Integral of v d p from Our Viewpoint where p is momentum of dark particle from our viewpoint as p = e v . The integral results in: Kinetic Mass of Dark Matter: Mk = ( 1 / 2 ) e v 2 (second 2 is exponent) from Our Viewpoint It can be proved that the equation F dt = d p is true in dark matter from our point of view, where F is force and p is momentum. Therefore, we have: Impulse  Momentum Theorem: J = Integral of F dt from ( t i ) to ( t f ) = ( p f )  ( p i ) of Dark Matter from our Viewpoint where the subscripts i and f represent initial and final conditions. J represents impulse. The equation implies that the law of conservation of momentum in dark matter from our point of view is different than the law in ordinary matter because p = e v. The Law of Conservation of Momentum in Dark Matter Consider a system having so many dark matter particles only. The particles can be microscopic or macroscopic; it makes no difference. If a particle has speed v, its momentum is p = e v from our viewpoint. The total momentum of the system is: P = p 1 + p 2 + p 3 + p 4 + ... where 1, 2, 3, ... are subscripts. The rate of change of P of a dark system is equal to the net force applied to the system: d P / dt = F (net) Since there is no interaction between dark matter particles from our point of view, the net force exerted on the system is just due to external forces. If the net external force i.e. F (net) is zero, it means that the system is an isolated dark system. In this case, there are no external forces or they add to zero. Thus, in an isolated dark system we have: Isolated Dark Matter System: d P / dt = 0 from Our Viewpoint It means that the total P remains constant. Therefore, the total P of an isolated dark system is constant: Any interaction within the system doesn't change its total momentum. This is the law of conservation of momentum in dark matter from our point of view; a fundamental law. The Dark Transformation Factor The dark transformation factor in our perspective as observers made of ordinary matter is gamma = 1 / [ 1  ( v2 / c2 ) ] 1/2 where all 2 and 1/2 are exponents. The both v and c are in terms of seconds per meter. Thus, gamma is a dimensionless factor. Now, I show how we can derive it. This is a fundamental step to decode the mystery of dark matter. Suppose that the equation of motion of an ordinary object is x = 4 t. Thus, the object travels 20 meters within 5 seconds. It can be shown as ( 5 s, 20 m ). The inverse function of the equation is t = 4 x which is the equation of motion of a dark object. Thus, the speed of dark object is 4 s/m in its own frame of reference. The dark object travels 4 seconds within 1 meter. It can be shown as ( 4 s, 1 m ). The maximum value of speed for a dark particle is c s/m. Suppose that there are two inertial reference frames R and R' in dark matter. The origins of the frames coincide at x = x' = 0 , and then the frame R' moves at speed of v s/m with respect to the frame R. A beam of dark light is emitted from the origins ( t = t' = 0 ) at the distance they coincide ( x = x' = 0 ). We call it the event 1. The dark light hits a detector. We call it the event 2. The coordinates of the event 2 are (t, x) in frame R and (t', x') in frame R'. The dark light travels at speed c s/m in the both reference frames. Since a new transformation should agree the dark transformation, we can write the equations below: t = (gamma) ( t' + v x' ) t' = (gamma) ( t  v x ) The positions of event 2 are t = c x in R and t' = c x' in R'. Thus, we have: t = c x ; t' = c x' Therefore, c x = (gamma) ( c x' + v x' ) = (gamma) ( c + v ) x' c x' = (gamma) ( c x  v x) = (gamma) ( c  v ) x We find x' from the second equation and use it in the first one: x' = (gamma) [ 1  ( v / c ) ] x x = (gamma) (gamma) [ 1 + ( v / c ) ] [ 1  ( v / c ) ] x We simplify the second equation: 1 / (gamma) 2 = [ 1  ( v2 / c2 ) ] (gamma) 2 = 1 / [ 1  ( v2 / c2 ) ] where all 2 are exponents. Thus, we have: gamma = 1 / [ 1  ( v2 / c2 ) ] 1/2 That's the dark transformation factor. v is the speed of the dark frame R' with respect to the dark frame R. Thus, the domain of v is the closed interval [ 0 s/m , c s/m ]. MassEnergy Equivalence of Dark Matter from Our Perspective We can derive massenergy equivalence of dark matter from our perspective as observers made of ordinary matter. We have p = (gamma) e v in our perspective, where gamma = 1 / [ 1  ( v2 / c2 ) ] 1/2 . From the workmass theorem, we can write: Mk = Integral of v d p = v p  Integral of ( p d v ) = gamma ( v2 e )  Integral of gamma ( v e ) d v = gamma ( v2 e )  ( e ) Integral of [ v / ( 1  v2 / c2 ) 1/2 ] d v = gamma ( v2 e )  ( e ) Integral of [ v c / ( c2  v2 ) 1/2 ] d v = gamma ( v2 e )  ( e c ) Integral of [ v / ( c2  v2 ) 1/2 ] d v where all 2 and 1/2 are exponents. Since Integral of [ v / ( c2  v2 ) 1/2 ] d v =  ( c2  v2 ) 1/2 + C , we have: Mk = gamma ( v2 e ) + ( e c ) [ ( c2  v2 ) 1/2 ]  ( M0 ) We can write: Mk = gamma ( v2 e ) + ( e c2 ) [ 1  ( v2 / c2 ) ] 1/2  ( M0 ) Since [ 1  ( v2 / c2 ) ] 1/2 = 1 / gamma , we have: Mk = gamma ( v2 e ) + ( e c2 ) ( 1 / gamma )  ( M0 ) Mk = gamma ( e ) [ v2 + ( c2 / (gamma) 2 ) ]  ( M0 ) After simplifying, we have: Mk = (gamma) e c2  M0 That's the relativistic kinetic mass of dark matter from our perspective. Thus, we have: M = Mk + M0 M = gamma ( e c2 ) That's the massenergy equivalence of dark matter. MassMomentum Equation of Dark Matter We got M = gamma ( e c2 ), so: M2 = (gamma) 2 e2 c4 M2 = e2 c4 / [ 1  ( v2 / c2 ) ] M2  ( M2 v2 / c2 ) = e2 c4 M2 = e2 c4 + ( M2 v2 / c2 ) Since M = (gamma) e c2 , we replace M in the right side with (gamma) e c2 and simplify: M2 = e2 c4 + [ (gamma) 2 e2 v2 c2 ] Since p = (gamma) e v , we have p2 = (gamma) 2 e2 v2 , we have: M2 = e2 c4 + p2 c2 That's the massmomentum equation of dark matter. Dark Photon Dark photon has no energy i.e. e = 0. However, it has mass and momentum. Speed of dark photon is c s/m. Since v = c s/m, it results in ( 1 / gamma ) = 0. We take e = 0 in the massmomentum equation above and find mass of a dark photon. Thus: Mass of Dark Photon: M = p c where p is momentum of a dark photon. For an ordinary photon, we have: Ordinary Photon: ( m , E ) = ( 0 kg , hf Joule ) We consider mass and energy of a dark photon as an inverse function of mass and energy of an ordinary photon. Thus, for a dark photon we have: Dark Photon: ( m , E ) = ( h f kg , 0 Joule ) It implies that energy of dark photon is zero, and its mass is h f where h is numerically equal to the Planck's constant. Mass of a dark photon is M = h f where h is numerically equal to the Planck's constant, and f is frequency. Since c = f (lambda) for a dark photon, where lambda is wavelength in terms of seconds and f is frequency in terms of 1/meter, we have M = h c / (lambda) as equation of dark photon mass. By taking the two mass equations equal, we obtain: Momentum of Dark Photon: p = h / (lambda) I also write the equation: Mass of Dark Photon: M = h f Dark photon is a wave of mass. **** Gravitational Force If we suppose that a dark matter has a low speed or no speed, its rest mass is M = e c2. Thus, the gravitational force between a dark matter (DM) object and an ordinary matter (OM) object based on the Newton's law of gravity is as follows: Gravitational Force between Dark Matter and Ordinary Matter: F = G m e c 2 / r 2 (both 2 are exponents) where e is equal to energy of dark matter object in terms of joule, m mass of ordinary matter object, c is numerically equal to the speed of light, r is distance between the objects, and G is the gravitational constant. The formula explains galaxy rotation curves: G m e c 2 / r 2 = m ( v 2 / r ) Thus, v 2 = G e c 2 / r v 2 = G M / r where v is speed of a star with mass m rotating very far from center of a galaxy. The dark matter is in the center of galaxy with a rest mass of M = e c2 . Gravitational Potential Mass of Dark Matter Dark matter objects have gravitational potential mass from our point of view, not gravitational potential energy. Suppose that there is a a dark matter object near to the surface of the Earth. The gravity that the Earth exerts on it is: F = G M e c 2 / R2 (2 is exponent) where R is the radius of the Earth, M mass of the Earth, G is gravitational constant, and e is energy of dark object in terms of joule. We have F = e a , thus: G M e c 2 / R2 = e a It gives: a = G M c 2 / R2 We can calculate the gravitational potential mass of the dark object as follows: delta( Mk ) =  delta( Um ) Integral of F dy = delta( Mk ) Thus: Integral of F dy =  ( Umf  Umi ) where (Um) f is final gravitational potential mass, (Um) i is initial gravitational potential mass, F is force of gravity, y is position of the object. (m in Um is subscript.) It results in: d( Um ) =  F dy Since we have F =  e a , where the negative sign implies that the force of gravity points downward, we can write: d( Um ) = e a dy By integrating both sides, it gives: Um = e a y Therefore, Gravitational Potential Mass (Um) of a Dark Matter Object: Um = e a y Near to the Earth Surface where e is numerically equal to the energy of dark matter object in terms of joule, a is free fall acceleration of dark matter, y is position of the object. The law of conservation of mechanical mass implies that sum of kinetic mass and gravitational potential mass, of a dark matter object near to the Earth surface, remains constant at any point. As a reminder, the gravitational potential energy of an ordinary matter with mass m near to the Earth surface is equal to (m g y). We can write: Mechanical Mass of Dark Matter: M mech = Mk + Um The mechanical mass of dark matter object is conserved in the case mentioned. Law of Conservation of Mechanical Mass in Dark Matter We can use the law of conservation of mechanical mass to determine speed of a dark object tossed near to the Earth surface: Law of Conservation: ( Mk ) i + ( Um ) i = ( Mk ) f + ( Um ) f of Mechanical Mass in Dark Matter where the subscripts i and f represent initial and final kinetic mass or gravitational potential mass. Thus, ( 1 / 2 ) e ( v i ) 2 + ( e a h i ) = ( 1 / 2 ) e ( v f ) 2 + ( e a h f ) All e cancel out, so: ( 1 / 2 ) ( v i ) 2 + ( a h i ) = ( 1 / 2 ) ( v f ) 2 + ( a h f ) Multiplying both sides by 2, we have: ( v i ) 2 + ( 2 a h i ) = ( v f ) 2 + ( 2 a h f ) Thus, ( v f ) 2 = ( v i ) 2  2 a ( h f  h i ) where h i and h f represent initial and final height, the both 2 after ( v f ) and ( v i ) are exponents, and ( a ) is a positive number. We can rewrite the equation as: ( v f ) 2 = ( v i ) 2  2 a ( delta h ) **** Toward a New Physics: Dark Physics The universality of the current massenergy equivalence, i.e. E = mc2 , has been broken since long time ago, but nobody pays attention. There are at least two great evidences for the claim: 1) the existence of dark matter; 2) neutrino oscillation. The current massenergy equivalence is just true for ordinary matter and ordinary antimatter, not for dark matter. **** The Nature of Neutrino Oscillation Neutrino has oscillation in observations. That means one flavor of neutrino changes to another. The Standard Model of particle physics implies that there are three flavors of neutrino, and neutrino is a lepton like electron. However, neutrino is not an ordinary lepton. Why? Because if neutrino was an ordinary lepton, electron might have oscillations too. But electron shows no oscillation. We can stop a beam of electron with a small piece of aluminium. However, neutrino easily passes through the Earth and even the Sun. We cannot stop neutrino with a block of lead 1,000,000 kilometers thick! Now, do you still believe that neutrino is an ordinary lepton like electron as the Standard Model considers? Ordinary matter (OM) particles never behave like neutrino. Neutrino is not an ordinary matter. Neutrino is a dark (matter) particle. The equation E = mc2 is not true for neutrino because it is a dark particle. Neutrino has rest energy (e0) in our viewpoint, which leads to its rest mass (M0) in accord to M0 = e0 c2. Neutrino has kinetic mass too. On the other hand, dark photon has no energy (e0 = 0); it has only mass. Neutrino is made of an exotic energy. This energy is different than any known energy because mass (M) is locked within the energy (e) in accord to the equation M = ec2. Thus, mass has been locked within a neutrino in our viewpoint. As a result, neutrino has no mass in the sense that an object made of ordinary matter has. Neutrino has rest energy (e0) and so rest mass (M0) which is equal to e0 c2. Neutrino is a dark lepton. Thus, for example, an electron neutrino is a dark electron, and an electron antineutrino is a dark positron. To explain neutrino oscillation, the current physics erroneously assumes neutrino as an ordinary matter particle, and then presents a mathematical matrix for the probability of the three flavors of neutrino to convert to each other. However, I present an explanation of neutrino oscillations below based on emission or absorption of dark photon. Neutrino Mass levels: Absorption Spectrum Electron neutrino is a dark electron. When an ordinary electron makes transitions between two energy levels in an atom, its energy changes depending on the emission or absorption of an ordinary photon. A similar phenomenon happens to a dark electron but the difference is that a dark electron makes transitions between two mass levels in a dark atom. This is due to the inverse function property. We observe dark electron as an electron neutrino. The observed mass oscillations are due to an electron neutrino making transitions between two mass levels in a dark atom. Electron neutrino has a set of dark quantum states, each with a distinct mass. We can roughly imagine them as orbits. When an electron neutrino absorbs a dark photon, it makes a transition, or jump, from a lower mass state to a higher mass state, and so appears as a muon neutrino or tau neutrino. The mass of the absorbed dark photon is equal to the mass difference between the two mass levels. Thus, muon neutrino and tau neutrino are just neutrinos in higher mass levels. What is called neutrino oscillation is just mass absorption spectrum of a dark atom. The relation between the wavelength, lambda, of a dark photon and its mass M is as follows: Wavelength of Dark Photon: lambda = hc / M where h is just numerically equal to the Planck's constant. The mass of a dark photon can also be obtained as: Mass of Dark Photon: M = h f where f is frequency, and h is just numerically equal to the Planck's constant. By using M from the second equation in the first equation, we find the relation between frequency and wavelength of a dark photon: Frequency and Wavelength of Dark photon: f (lambda) = c Compton wavelength of a Dark Matter Particle Compton wavelength of a dark matter particle in our viewpoint is given by the equation: Compton Wavelength of a Dark Matter Particle: lambda = h / e c where e is the dark particle's rest energy, and c and h are numerically equal to the speed of light and the Planck's constant respectively. The equation comes from the fact that mass of a dark photon in our viewpoint is M = h f = h c / lambda = e c2 which after simplifying gives the equation. So far, so good. **** Neutrino decay It is possible to think of neutrino decay. For example, A muon neutrino can decay to an electron neutrino, a positron, and a muon. It can be written as: muon neutrino > electron neutrino + positron + muon The exotic energy of the muon neutrino converts to a great mass based on M = ec2 , which produces a muon, a positron, and the remaining mass is locked within the electron neutrino. **** Neutrino Speed Neutrino speed is very close to the light speed, i.e. a little bit less than c m/s. However, as the transformation factor, i.e. gamma = 1 / [ 1  ( v2 / c2 ) ] 1 / 2 implies the greatest lower bound of speed in dark matter is (1 / c) m/s. **** Dark Energy The standard model of cosmology, the LambdaCDM model, says that there is an unknown form of energy called dark energy (DE) that accelerates the expansion of the universe. It makes up about 68 percent of the massenergy of universe; the remaining 32 percent is matter which includes dark matter and ordinary matter. The evidence is the faintness of distant Type Ia supernovae. The model implies that when the universe was 380,000 years old (the starting point is Big Bang) i.e. 13.7 billion years ago, dark energy was negligible, but dark matter (DM) had made up about 63 percent of the total massenergy. Then, dark energy (DE) started to grow and expanded the universe. As a result, DE constitutes about 68 percent of massenergy today, while DM decreased to be about 26.5 percent, and ordinary matter makes up less than 4.9 percent now. The model indicates that during the first 8 billion years after Big Bang, the expansion of universe was slowing down. Then around 56 billion years ago, it changed to acceleration because DE dominated the two sorts of matter. If you ask what DE is, the model gives you a bunch of problematic and somewhat paradoxical answers which I will explain them. Since the current physics believes that the expansion of universe is accelerating, it requires something to provide a repulsive effect that drives the acceleration. That thing is called DE. It has a negative pressure that makes a repulsive form of gravity. What is the origin of the idea? In 1917, Einstein added a term called cosmological constant to his equations of general relativity to derive a static universe as the current belief at that time. If cosmological constant is positive, it causes repulsion; if it is negative, it causes attraction. Now, the cosmological constant is considered as the vacuum energy in the current physics. However, quantum physics gives a value for the vacuum energy density that is greater than the value of cosmological constant by a factor of about 10 (120) [120 is exponent] ! While the cosmological constant is estimated to be 10 (9) Joules per cubic meter (9 is exponent), the vacuum energy is calculated to be 10 (113) Joules per cubic meter (113 is exponent). This is the worst prediction in physics. To account for the huge amount of discrepancy, the model says that something unknown cancels out the great extra value of the vacuum energy to make it equal to the very small value of the cosmological constant. Thus, the model assumed the value of cosmological constant in a way that 1) it had been extremely small in the early universe and 2) it would be the dominant component of universe today. The assumptions towards a right modification to meet the model's requirement is called the finetuning problem. Since the model suggests that the start of accelerating expansion of universe is relatively close to the recent time, it means that we live in a special time in the history of the universe in which the acceleration began just around 56 billion years ago. It is similar to the idea that human being had in the past and considered the Earth as the center of the universe. Why did the accelerating expansion begin to occur very close to the present? This is called the coincidence problem. None of those problems has a convincing solution. Although some cosmologists try hard to explain dark energy in an evolving way, instead of cosmological constant which remains constant over time, by using quintessence or phantom energy, their assumptions suffer the mentioned problems too. **** Physicists have considered many possible candidates for dark matter (DM) except the possibility of the inverse function of the equation E = mc2 , which is symmetric to the graph of the equation with respect to the line y = x. Their candidates and all efforts to detect DM particles have failed so far, and will continue to fail for ever unless they revise their ideas. The reason is that they approach DM by using ordinary matter (OM) formulas, but DM is a nonbaryonic matter. As a better definition, ordinary matter is anything that obeys the equation E = mc2, while dark matter is anything that obeys the equation M = ec2 , where both 2 are exponents. Since DM neither emits nor absorbs significant amounts of electromagnetic radiation, we cannot see it directly. If we could travel to a region of the universe where is assumed to have rich amounts of DM, we would observe nothing there at all except the gravitational effects of DM. If we could convert ourselves to DM or we were made of DM, we would not observe ordinary matter any more which constitutes luminous galaxies and stars; we could see DM only. **** The Nature of Dark Matter Suppose that an observer moves along with an object or particle. In the case, anything made of ordinary matter has a mass (except ordinary photon), and energy is locked within the mass. The ratio of this energy to the mass (E/m) is equal to c2 (2 is exponent). In contrast, anything made of dark matter has an energy (except dark photon), and mass is locked within the energy. The ratio of this mass to the energy (M/e) is equal to c2. These two fundamental rules are based on the mathematical symmetry between massenergy equivalences of ordinary matter and dark matter. It can be stated in other words by considering the observer above: Ordinary matter is a kind of matter in which energy has been locked within a mass; however, dark matter is a kind of matter in which mass has been locked within an energy. You may ask how it is possible. The answer is out there: watch neutrino as a dark particle. Therefore, the physics of DM, or dark physics, may be based on the assumption above. In some circumstances such as fission events in dark matter particles, part of the energy of DM particle converts to mass in accord to the equation M = ec2 (2 is exponent), and so a huge amount of mass appears. In the late 19th century, scientists discovered that radioactive elements have beta decay which seemingly produces a proton and an electron. Then, they found that a small amount of energy is missed. Neils Bohr said that energy is not conserved in beta decay. However, Pauli suggested a new invisible particle that carries the missing energy to balance the account. Fermi called it neutrino. The particle that Pauli proposed is actually an antineutrino. Although nobody thinks that it is a dark particle so far, I have an idea: When a beta decay occurs, there is both a losing mass of ordinary side of neutron, which changes neutron to a proton, and a losing exotic energy of dark side of neutron, which converts to mass in accord to M = ec2 and after giving mass to an electron then appears as an electron antineutrino. The two events balance the account in accordance to the assumed symmetry between the two massenergy equivalences. Hence, the known antineutrino is due to a dark interaction. The creation of electron antineutrino conserves exotic energy because the disappeared exotic energy makes a dark particle, i.e. electron antineutrino. Dark matter is an exotic energy. If a dark matter particle loses an amount of energy, the energy converts to mass. The mass can appear as dark photons as in dark gamma decay, or can be locked within a dark particle such as an electron antineutrino in dark beta decay. Thus, a neutrino or antineutrino has no mass in the sense that the word mass is used for ordinary matter objects or particles. Based on the modern physics, when beta decay occurs, one of the down quarks of a neutron (udd) turns to an up quark and a W (negative) boson. The process changes the neutron to proton (uud). Then, W boson decays to an electron and an electron antineutrino. The mass of W boson is too much; almost 80.41 GeV/c2, i.e. about 86 times massive than proton with a mass almost equal to 0.938 GeV/c2. W (positive or negative) bosons are force carriers of weak interaction that explains beta decay and interactions between neutrino and electron. Since W bosons and Z boson are very heavy particles, an unusual scalar field called Higgs field assumed to give them mass. Higgs boson, with a mass almost 125 GeV/c2, is assumed as the quantum of Higgs field. I add that the mass of Z boson is almost 91.187 GeV/c2 , i.e. about 97 times massive than proton. A protonantiproton interaction at energy of 630 GeV was used by CERN in 1978 to detect W and Z bosons. It is obvious that proton and antiproton with this great kinetic energy can produce massive particles, for their kinetic energy converts to massive particles. The question is that whether W and Z bosons can also be detected in low energy conditions. No, they cannot. Thus, the modern physics presents a new problem associated with the origin of the mass of these massive particles produced just by high energy collisions and claimed to have role in weak interactions. Why do I mention all these? Because the mass account is not balanced in the explanation associated with weak interactions. A down quark results in an up quark and a W boson while the latter particle is much more massive than the neutron (with a mass almost equal to 0.939 GeV/c2 ). When an electron and a neutrino interact, they exchange a W boson which is extremely massive than both. What is the origin of W boson mass in low energy conditions? Where do the mass of it and even the mass of Higgs boson come from? How are these great masses generated in low energy circumstances? Detecting them in high energy protonantiproton or protonproton collisions is not an enough evidence to connect them to beta decay or weak interactions. Consequently, the argument is that if the two W bosons, Z boson and Higgs bosons are really intermediate particles in weak interactions, they cannot be ordinary matter particles. They are all dark particles, for their masses are not acceptable based on the ordinary massenergy equivalence. The bottom line is that we don't need to assume an odd field called Higgs field to explain the mass of the heavy bosons in the GlashowWeinbergSalam theory of weak interaction. There is a simple and elegant way to solve the mass problem. Occam's razor states that the fewer are assumptions, the hypothesis is better. We don't need the complex Higgs field and particles that may arise from it. I remove all with Occam's razor presenting a simpler way, for the nature is simple. Let's review again: One of the down quarks of neutron turns to an up quark and results in a W negative boson. There is no Higgs field or something like that to involve here. Also, it is absolutely impossible to create such a huge mass based on the neutron mass or even its mc2. However, the neutron losses a tiny amount of an exotic energy which goes to the formula M = ec2 creating the great mass of W negative boson. This kind of energy is completely different than any sort of known energy. An electron immediately takes a tiny amount of the W boson's mass. Then, part of the remaining mass of the W boson is locked within an electron antineutino in accord to M = ec2, and the other part gives kinetic mass to the antineutrino. Hence, the weak interaction is mediated by a dark interaction. The W boson is about 15736 times massive than electron. The electron's mass is almost 0.511 MeV/c2. Since neutron losses a small amount of mass to change to proton, the missing mass converts to energy based on E = mc2, and the energy acts as kinetic energy for the electron. What did it happen here? An ordinary matter particle called neutron decayed to two ordinary matter particles called proton and electron, and produced a dark particle called electron antineutrino. The process was mediated by the huge mass of a dark particle called W negative boson. Both the mass and energy accounts are balance, and it is not necessary to resort to the complicated Higgs field hypothesis. Thanks to the Occam's razor that works here. If we accept the complicated solution that the Standard Model of particle physics proposes, we are going to make a zoo of massive heavy particles with extremely tiny life times less than a zeptosecond, i.e. 10 (21) second where 21 is exponent. Please, how far can this mass game go? In fact, there is a simple model or computational hypothesis that unifies and calculates all these masses. On the other hand, the problem associated with the idea that neutrino is an ordinary particle is that the modern physics cannot measure its mass. The reason is that neutrino has no mass in the sense that any ordinary object or particle has. Neutrino and antineutrino are dark particles. Therefore, mass has just been locked within a neutrino or antineutrino. Actually, antineutrinos are antidark matter particles. Antidark matter can easily be defined by the equation M2 = e2 c4 (all numbers are exponents). However, it is first better to understand the dark matter' nature, then define antidark matter. So far, so good. The below is a summary of my hypothesis of beta decay to make sure that the whole process is clear: 1) a neutron (udd) losses a tiny amount of its mass and turns to a proton (uud). This mass (m) goes to the equation E = mc2 and converts to energy (E); 2) at the same time, the neutron losses a tiny amount of an exotic energy. This energy (e) goes to the equation M = ec2 and converts to a notable mass (M). This mass is the mass of W boson which has a negative charge to balance with the proton's positive charge. In other words, the exotic energy (e) produces a W boson by converting to the mass (M); 3) thus, the mass of W boson comes from the neutron, not from outside such as a Higgs field; 4) part of the mass of W boson gives mass to an electron which takes the negative charge of the W boson too; 5) the remaining mass of the W boson goes to the equation M2 = e2c4 (all numbers are exponents) and converts to an exotic energy (e), which is equivalent to say that an electron antineutrino is created; 6) the energy (E) in the section (1), which comes from the missing mass (m) of the neutron, gives kinetic energy to the electron; 7) since the origin of the both W boson and the electron antineutrino is the equation M = ec2 as explained, they are dark particles. The life time of W boson is extremely short. Thus, the weak interaction in beta decay is mediated by a dark interaction. That's it. Numerical Analysis of Beta Decay: Still Against the Current Physics I analyse beta decay numerically to show that how the current physics is still wrong about it. We have: neutron > proton + electron + electron antineutrino The range of kinetic energy of the electron is 0 to around 1 MeV. However, the maximum kinetic energy of the electron can be a few tens of MeV. The mass of the neutron is about 939.565 MeV/c2. The mass of the proton is about 938.272 MeV/c2. Thus, the neutron's loss of mass is about 1.293 Mev/c2. This mass gives 1.293 MeV energy based on the equation E = mc2. Suppose that the electron's kinetic energy is 1 Mev. Thus, the electron's kinetic energy is provided by part of 1.293 MeV. The remaining energy is equal to 0.293 MeV. This energy gives a mass just equal to 0.293 MeV/c2. However, the electron rest mass is about 0.511 MeV which is greater than 0.293 MeV/c2. So, how can the electon be created? If neutrino has a mass as the Standard Model emphasizes on it, the mass problem will be worse. Also, if the kinetic energy of the electron is a few tens of MeV, the mass problem will be worse. On the other hand, we have: 1.293 MeV/c2  0.511 MeV/c2 = 0.782 MeV/c2 which is equivalent to the energy value of 0.782 MeV based on E = mc2. Therefore, if the kinetic energy of the electron is less than 0.782 MeV, where does the remaining energy go? This question is answered by assuming an electron antineutrino which takes the additional energy in this case. However, if the kinetic energy of the electron is equal to 0.782 MeV, no energy or mass remains for an electron antineutrino to be created. In spite of this fact, an electron antineutrino appears there. To solve either the mass or energy problem, the GlashowWeinbergSalam theory suggested W negative boson. Thus, we can write: neutron > proton + W W > e + electron antineutrino Hence, neutron > proton + electron + electron antineutrino However, the mass of the W boson is about 80.41 GeV/c2, i.e. about 86 times massive than neutron. When you ask how neutron gives a particle that is about 86 times massive than itself, the Standard Model introduces a scalar field called Higgs field, which its quantum is Higgs boson with a great mass about 125 Gev/c2, and saying that Higgs field gives mass to the W boson via the Higgs mechanism. When you ask how the Higgs mechanism prepares for that, the Model suggests the Goldstone bosons for the explanation. When you ask where the Goldstone bosons come from, the Model has no answer but yet can go ahead with another new massive particle. When you ask where the Higgs field is, the Model says that it permeates all space. There are 6 essential problems in the Standard Model associated with beta decay: 1) W boson, Higgs boson, and Z boson are found in high energy experiments but beta decay is not a high energy phenomenon. 2) Although Higgs boson was found in high energy protonantiproton collisions, its mass cannot be calculated. The mass of the Higgs boson is just an experimental finding. It cannot be shown mathematically. 3) Assuming a field outside the neutron for explaining beta decay is not reasonable. It means to involve something beyond the neutron or the equation of beta decay for explaining beta decay. 4) Higgs boson is not an evidence for the existence of the Higgs field. In fact, there is no evidence for Higgs field. 5) Such a successive proposing of far massive particles backing by a high energy physics for describing the process of low energy beta decay can continue infinitely. This fact strongly implies that the explanation is wrong. 6) There is no evidence that a W boson with just an established mass is involved in beta decay. It is possible to think of particles with masses a few GeV/c2 above or below the mass of the W boson that can mediate beta decay depending on the kinetic energy of the beta particle, i.e. electron. The approach of the Standard Model is similar to the approach of people in the past toward the standing point of the Earth in sky. People believed that the Earth is standing on four elephants. When you ask what do the elephants stand on? They answer that the elephants lie on a tortoise. Then, when you ask what does the tortoise stand on? Yet they can continue in the same way. Let's rephrase the problems facing in beta decay because it has some profound consequences: A) If the kinetic energy of the electron in beta decay is less than 0.782 MeV, there will be an excess of energy. Where does it go? Pauli suggested that an unknown particle takes this energy. The particle is called an electron antineutrino. B) If the kinetic energy of the electron in beta decay is equal to 0.782 MeV, no energy or mass will be available for an electron antineutrino to be created. C) If the kinetic energy of the electron in beta decay is more than 0.782 MeV, there will be a mass deficit for both the electron and the electron antineutrino to be created. Thus, neither electron nor electron antineutrino will be there. To solve the mass deficit problems in B and C, W boson was assumed to prepare the required mass. It is assumed that W boson gives mass and energy required for the creation of both electron and electron antineutrino. Then, a chain of far massive particles is presented to account for the origin of the W boson. First of all, the conclusion is that if the kinetic energy of the electron is less than 0.782 MeV, there is absolutely no need for the involvement of the W boson, because the energy and mass accounts are in a balance. This is in accord to the E = mc2, or the current massenergy equivalence. This is when electron antineutrino is considered as an ordinary particle with a rest mass. Thus, the problem begins with an electron having a kinetic energy equal to or more than 0.782 MeV. By introducing W boson for solving the problems B and C above, the Standard Model has to accept that whether or not the electron's kinetic mass is more than 0.782 MeV, W boson is involved in beta decay. However, this is not reasonable. I give a numerical example. Suppose that the kinetic energy of the electron is 0.645 MeV, i.e. less than 0.782 MeV. As a result, it remains 0.137 MeV for both the energy and the rest mass of the electron antineutrino, which each of them is determined by the equation E = mc2 in accord to the Standard Model that assumes electron antineutrino as an ordinary particle with a rest mass. Thus, there is no need to involve W boson in this case. However, the Standarl Model cannot say that sometimes W boson is not required. Now, by entering a W boson, since neither electron nor electron antineutrino needs the W boson, what happens to the great mass of the W boson, i.e. about 80.41 GeV/c2? The Model has to accept that this mass converts to energy based on E = mc2 and gives the energy to the electron antineutrino. It means that the W boson gives about 80.41 GeV energy to the electron antineutrino. Therefore, the overall energy available for the electron antineutrino will be about 80.41137 GeV, i.e. the sum of (0.00137 GeV + 80.41 GeV). The result is that the value of the energy available for the electron antineutrino is a huge amount of about 80.41 GeV rounded off to four significant figures. However, there is not such a thing at all. Thus, the Standard Model faces a great paradox in the case that the electron's kinetic energy in beta decay is less than the critical value of about 0.782 MeV. Someone may argue that W boson gives mass and energy to an electron with a kinetic energy of 0.782 MeV. Let's see what happens. The rest mass of the electron is about 0.511 MeV/c2 or 0.00511 GeV/c2. So, we have: 80.41  0.00511 = 80.40489 GeV/c2. Thus, the remaining mass of the W boson is about 80.40 GeV/c2 rounded off to four significant figures. This is equivalent to the energy of 80.40 GeV based on E = mc2. Therefore, an energy equal to 80.40 GeV will be available for the electron antineutrino. Since the rest mass of the electron antineutrino is very small based on the Standard Model, the notable part of the value 80.40 GeV will be the kinetic energy of the electron antineutrino. Also, the value of 1.293 MeV or 0.001293 GeV from the mass difference between neutron and proton should be added to 80.40 GeV. Hence, the kinetic energy of the electron antineutrino will be about 80.401293 GeV or 80.40 GeV with four significant figures. However, there is not such a thing in beta decay. Consequently, the paradox exists again. (I'll be back to complete this part.) Another example: We know that chlorine37 interacts with an electron neutrino and changes to argon37 and an electron: Cl37 + electron neutrino > Ar 37 + e The reaction is used to detect solar neutrinos. Cl37 consists of 20 neutrons and 17 protons. Argon37 consists of 19 neutrons and 18 protons. Thus, one of the chlorine's neutrons changes to a proton and results in argon37. The numbers state it clearly. The underlying process is as follows (n is neutron, p is proton, and e is electron): n + electron neutrino > p+ + e We can explain it based on the hypothesis above. I can say there is something more in the result side though the Standard Model can never predict it. What is that? There are also two dark photons in the result side. The real process is as follows: 1) an electron neutrino hits nucleus within an atom of Cl37 and causes a neutron to react in this way: n > p+ + W 2) W > e + electron antineutrino Here, someone may suggest ( W + electron neutrino > e ) and conclude that electron antineutrino is not produced. However, their suggested reaction cannot be explained. 3) electron neutrino + electron antineutrino > two dark photons It means that electron neutrino and electron antineutrino annihilate each other and produce two dark photons. 4) thus, n + electron neutrino > p+ + e + 2 dark photons 5) This is against the Standard Model which is completely wrong about neutrinos. Each dark photon of the reaction has a mass which is almost equal to (mass of the W boson  rest mass of the electron), i.e. about (80.41  0.00511) = 79.99 GeV/c2 with 4 significant figures. The rest mass of the electron is about 0.00511 GeV/c2. Dark photon has no energy, or strictly speaking no exotic energy which is associated with dark matter. How did I get the value of the dark photon mass? The mass of the dark photon of the reaction is exactly equal to the mass (M0) of the rest exotic energy (e0) of the antineutrino or neutrino. We have M = (gamma) e0c2 = (gamma) M0 for a neutrino or antineutrino. Since gamma is about 1, we can say that M is almost equal to M0 (or e0c2). Because of the (W boson > e + electron antineutrino), we have M = (mass of the W boson  rest mass of the electron), which is almost equal to M0, i.e. the value we needed. If two particles are detected in the reaction above, each having a mass about 79.99 GeV/c2, it will be a great proof for the mentioned hypothesis. Mass (M) of a dark photon varies in different situations, and it depends on dark photon's frequency (f): M = h f , where h is just numerically equal to the Planck constant, i.e. its unit is different. I'll discuss the formula later. However, the Standard Model cannot define dark photon. Who wins? Dark beta decay In dark beta decay, a dark neutron decays to a dark proton, a dark electron, and an electron antineutrino in its own frame of reference. A dark electron appears as an electron neutrino with respect to us, and an electron antineutrino appears as a positron with respect to us. Notice that in writing the products of dark beta decay, I didn't use the word dark before electron antineutrino. Think of the reason. What does a dark proton appear with respect to us? Maybe it appears as a particle without electric charge with respect to us, say dp (abbreviation for dark proton). What does a dark neutron appear to us? Maybe it appears as a particle with a positive charge with respect to us in order to balance with the positron's charge, say dn+ (abbreviation for dark neutron). Thus, what we will detect, as observers made of ordinary matter, is as follows: dn+ > dp + electron neutrino + ordinary positron (e+) The charges are conserved in the equation. We detect 3 dark particles and just one ordinary particle, i.e. positron. Therefore, 1) a dark neutron has a positive charge in our point of view but it has no charge in its own reference frame; 2) a dark proton has no charge in our point of view but it has a positive charge in its own reference frame. We also realize that any electron antineutrino produced in dark matter will appear as a positron with respect to us. Since I am talking about two different realms, dark matter and ordinary matter, the changes in the appearance of comparable particles in both dark and ordinary side are just a relativistic issue. I call the process that was used above for finding the particles that we observe from the reactions that occur in dark matter as translation of dark reactions to observable reactions, or in short translation. So, we can write: In dark matter (dark beta decay): dark neutron > dark proton + dark electron + electron antineutrino Our observation: dn+ > dp + electron neutrino + ordinary positron (e+) Based on our observation of dark beta decay, we can predict the following equations to be detectable by us: dark neutron (dn+) + electron > dark proton (dp) + electron neutrino dark neutron (dn+) + electron antineutrino > dark proton (dp) + positron dark proton (dp) + electron neutrino > dark neutron (dn+) + electron dark proton (dp) + positron > dark neutron (dn+) + electron antineutrino dark neutron (dn+) + dark antiproton > electron neutrino + positron If a dark electron collides with a dark positron in dark matter, what is the translation of the event as we observe it? It will be: In dark matter: dark electron + dark positron > 2 dark photons Our observation: electron neutrino + electron antineutrino > 2 dark photons The rules of translation of dark matter reactions to our ordinary observations 1) Any particle with charge is translated to a particle with no charge. 2) Any particle without charge is translated to a particle with charge. 3) Photon comes the same. 4) Dark electron is translated to electron neutrino. (Thus, e.g. dark muon to muon neutrino) 5) Dark positron is translated to electron antineutrino.(Thus, e.g. dark antimuon to muon antineutrino) 6) Electron neutrino is translated to ordinary electron. (Thus, e.g. muon neutrino to ordinary muon) 7) Electron antineutrino is translated to ordinary positron. (Thus, e.g. muon antineutrino to ordinary antimuon) Suppose that a given equation in dark matter is as follows: In dark matter: dark neutron + electron neutrino > dark proton + dark electron + 2 ordinary photons What is the reaction as we observe it? The equation shows that some interactions in dark matter, in their own perspective, can produce ordinary photons: Our observation: dn+ + electron > dp + electron neutrino + 2 ordinary photons If there is a reaction in dark matter as: In DM: electron antineutrino + dark proton > dark neutron + dark positron + 2 ordinary photons What is the reaction with respect to us as we detect it? The answer is: Our observation: ordinary positron + dp > dn+ + electron antineutrino + 2 ordinary photons Why do particles change in our detection? Why are there the rules of translation? The answer lies in the relativistic relations between dark matter and ordinary matter. Any observation depends on the viewpoint of an observer. I can explain more by using mathematics, but actually my experience indicates that it is confusing not only for most of people but even for physicists. The reason is that individuals are so accustomed with the established applications of physical concepts and formulas that they can hardly visualize something beyond those applications. As a result, I decided to depict the answer within a conversation. It really works. David is a scientist on the Earth doing research on dark matter. Susan is another scientist living somewhere in the universe which is considered a dark side from the viewpoint of David. They can somehow communicate to each other. Below is a part of their conversation. David: Hi Susan. How are doing today? Susan: Fine, thanks. What's up there? David: I think you are made of dark matter; something that I cannot directly see it. Susan: Telling that I am made of dark matter originates from your point of view. David: Do you mean that I am wrong? Susan: No David. It is just a relativistic perspective. From my point of view, you are made of something that I consider it as dark matter, because I am not familiar with that kind of matter in a direct way. I know the matter around me but it is not dark at all. I can see and touch it. David: Good point. I can understand. I calculated the value of light speed in your world to be c meters per second, where c is equal to 299792458. Am I right? Susan: Well, again your calculation is just based on your perspective. The speed of light as I measure here is c seconds per meter. However, since the arrangement of space and time in your world is the inverse of what is the case in my world, you calculate the light speed of my world based on the arrangement of spacetime in your world. David: What? I cannot figure out. Susan: When you want to draw a position versus time graph in your world, you choose the horizontal axis as time and the vertical axis as position. In this way, you obtain a straight line for the motion of an object in your world. The slope of the line gives the velocity and so speed of the object, which is equal to v m/s. Then, you try to find another straight line as an inverse function of the first equation. The second line is symmetric to the first line with respect to the line y = x , and vice versa. They are inverse functions of each other. In fact, you consider time as an independent variable and position as a dependent variable. However, I draw my graph for the same job, i.e. to find velocity, in a different way. David: How do you draw a position versus time graph? Susan: Since time is considered to be an independent variable in your world, you choose the horizontal axis to show time values. However, since space is considered to be an independent variable in my world, I choose the horizontal axis to show space values and the vertical axis to show time values. As a result, the unit of velocity or speed is seconds per meter in my world. This is due to the nature of the world I live in, and so the way I apply to draw a time versus position graph. David: How can time be a dependent variable in your world? Susan: There are at least 4 dimensions. What are they? David: 3 dimensions of space, and 1 dimension of time. Susan: That's true just for you, not for me at all. David: What? Susan: There are 3 dimensions of time and 1 dimension of space in my world. David: I cannot believe it. Susan: Actually, nature has no distinction between the four dimensions. Nature works based on her own requirement. If nature needs to use 3 of the 4 dimensions as space in a situation, she will do it. And if she needs to use 3 of the 4 dimensions as time in another situation, she will not hesitate to do so. It is just your specific sensory system that perceives 3 dimensions as space and 1 dimension as time always. In contrast, it is just my specific sensory system that perceives 3 dimensions as time and 1 dimension as space always. But nature doesn't care about our perceptions. I'll complete this part later. Numerical Analysis of W boson Mass We have: neutron > proton + W boson or n > p+ + W  . I write the equation as: n > p+ + exotic energy , exotic energy > W It means that neutron releases an amount of exotic energy in beta decay, and then the exotic energy produces W boson with a mass about 80.385 GeV/c2. Since W boson mass is about 86 times bigger than neutron mass, neutron cannot produce such a great mass based on E = mc2. Thus, there must be another way for neutron. I delete Higgs field or any other assumptions that originate from outside of neutron. Neutron itself should be able to handle the equation; a reasonable idea. Nature always applies simple ways. How much is the value of the exotic energy released? I convert the mass of W boson to kg. Since 1 eV/c2 is about 1.7827 x 10(36) kg, 36 is the exponent, the mass of W boson is about 1.4330 x 10(25) kg where 27 is exponent. Now, we have M = 1.4330 x 10(25) kg. Using the equation M = ec2, we can find the value of e or exotic energy: c2 is about 2.99(2) x 10(16) where 2 and 16 are exponents. Thus, the value of e is M/c2, which is about 1.6029 x 10(42) joule rounded off to five significant figures, where 42 is exponent. The exotic energy 1.6029 x 10(42) J goes to the equation M = ec2 and produces the mass of W boson. It means that the W boson's mass is locked within the released exotic energy. We can convert it to electron volt. One electron volt is about 1.6022 x 10(19) joule where 19 is exponent. Thus, the value of the exotic energy will be about 1.0004 x 10(23) eV where 23 is exponent. Hence, we have: Exotic Energy Released by a Neutron = about 1.6029 x 10(42) Joule = about 1.0004 x 10(23) electron Volt in Beta Decay The exotic energy is about 0.0100 zepto eV, i.e. ten trillionth of an electron volt; extremely small. **** Again Dark Energy After a lot of arguments for dark matter, is there any clue to the nature of dark energy, another big question facing the current physics, in our findings? Maybe. Because I have introduced dark energy in a section above, I will not repeat it again. Rather, I will try to make a connection between what I wrote about dark matter and what is called dark energy. Scientists revealed that about 13.7 billion years ago, universe consisted of 12% ordinary atoms, 15% ordinary photons, 10% neutrinos, and 63% dark matter. However, today universe consists of 4.9% ordinary atoms, 26.8% dark matter, and 68.3% dark energy. Why did 36.2% of dark matter disappear? Why did 7.1% of ordinary atoms disappear too? If neutrinos are dark matter particles, then 46.2% of dark matter disappeared. If we consider ordinary photons and atoms together as ordinary matter, then 22.1% of ordinary matter disappeared. Thus, we have: 46.2% + 22.1% = 68.3% which is the dark energy today. Therefore, the disappeared dark matter and ordinary matter are equal to the present dark energy. It means that both dark matter and ordinary matter have contributed to the creation of dark energy. This is possible only if ordinary matter, dark matter and dark energy are somehow hidden within each other. Also, there should be some conditions that one of the dark things changes to or appears as the other. **** Dark Matter Waves and Mass Quantization The massmomentum equation of dark matter above implies that momentum of a dark photon is p = M / c. Thus, we have M = p c for a dark photon. Since M = h f is true for a dark photon, we conclude that p c = h f . Also, we have c = f (lambda). Hence, we can write p = h f / c or p = h / (lambda) which is true for a dark photon. The equation p = h / (lambda) is true for any dark particle too. For a standing dark wave in a onedimensional box of length L, we have: lambda = 2 L / n . Since lambda = h / p , by equating these two equations, we obtain: 2 L / n = h / p . We discussed before that p = e v is true for a dark matter particle. Thus, we can write 2 L / n = h / e v which gives: (v)n = n h / 2 L e where the first n is a subscript, and n can be 1, 2, 3, ... The equation shows speed of a dark particle in a standing wave with n antinodes within the box. Since equation of kinetic mass is Mk = ( 1 / 2 ) e v2 where k is subscript, and the second 2 is exponent, we can use (v)n in the kinetic mass equation to obtain: (M)n = n 2 h 2 / 8 e L 2 where the first n is a subscript ( n = 1, 2, 3, ...), and all 2 are exponents. The equation shows the mass of dark particle. e stands for energy of a dark particle. Thus, a dark particle in the box can only have certain masses, or in other words, its mass is quantized. These masses are mass levels. The fundamental quantum of mass is: (M)1 = h 2 / 8 e L 2 All of these equations are from our point of view. The equation of (M)n can be applied to neutrinos to measure their mass. **** Dark Wave Function: Law of Dark Quantum Mechanics A wavelike function is: psi(x) = (psi)0 sin( 2 pi x / lambda ) where 0 is a subscript, and lambda is wavelength. The second derivative of psi(x) is: d2 psi / dx2 =  (2 pi) 2 psi(x) / (lambda) 2 where the both 2 after (2 pi) and (lambda) are exponents. For any dark particle, we have: lambda = h / p . Since p = e v = [ 2 e (Mk) ] 1/2 where 1/2 is exponent, and Mk = ( 1 / 2 ) e v2 is kinetic mass, we can substitute it in the lambda equation: lambda = h / [ 2 e (Mk) ] 1/2 where 1/2 is exponent. We square the lambda and substitute in the pesi equation: d2 psi / dx2 =  2 e ( Mk ) psi(x) / ( h bar ) 2 where ( h bar ) = h / (2 pi). ( h bar ) represents the reduced Planck's constant. The law of conservation of mechanical mass in dark matter, as it was presented in a section above, implies that: M mech = Mk + U So, Mk = M  U(x) where M is total mechanical mass of dark particle, and U(x) is potential energy of the dark particle as a function of position x. By substituting it in the psi equation, we have: d2 psi / dx2 =  2 e [ M  U(x) ] psi(x) / ( h bar ) 2 where the number 2 after ( h bar ) is an exponent, and e represents energy of a dark particle. That's the dark wave equation of dark matter particles. It is the law of dark quantum mechanics. That's a great triumph. Similar steps was accomplished by Schrodinger in 1925 to find Schrodinger equation of ordinary particles. If U(x) is zero as for the box in the previous section, then we can write: d2 psi / dx2 =  2 e M psi(x) / ( h bar ) 2 By solving this equation for the allowed masses, we will obtain: (M)n = n 2 h 2 / 8 e L 2 which is the same result as shown in the mass quantization. **** Ordinary Charge vs. Dark Charge Suppose that an ordinary particle with charge q is a distance r apart from a dark particle with charge q'. By ignoring signs of charges, the magnitude of electric force between the two charges is: F = K' q q' / r 2 where K is electrostatic constant between ordinary charge and dark charge, and 2 is exponent. The dark charge experiences a force by the electric field of the ordinary charge, that is: F ( on q' ) = q' / E where E is the electric field of the ordinary charge q. Thus, we have: E = r 2 / K' q The ordinary charge experiences a force by the electric field of the dark charge, that is: F ( on q ) = q E' where E' is the electric field of the dark charge q'. Thus, we have: E' = K' q' / r 2 K' is far less than the electrostatic constant K (which works for ordinary charges) of the Coulomb's law. By multiplying E and E', we obtain: E E' = q' / q **** It can be shown that ordinary electromagnetic fields create dark electromagnetic fields and vice versa. This implies that ordinary matter creates dark matter and vice versa. It makes sense because otherwise there will be no massenergy conservation associated with both ordinary and dark massenergy. The proof needs a lot of new equations to be introduced as we go ahead to formulate them. Ordinary matter prepares energy to dark matter interactions, and dark matter prepares mass to ordinary matter interactions. Thus, there are energyproducing fields by ordinary matter and massproducing fields by dark matter. **** Electric Potential Mass Suppose that there is a parallelplate capacitor made of ordinary matter with positive plate in the right side and negative plate in the left side. We consider an axis s that its zero starts from the negative plate, i.e. s = 0 at the negative plate. A positive dark charge q' moves from the positive plate to the negative plate. We want to calculate the change in its inverse of electric potential mass. Ordinary charges in the similar case has a change in electric potential energy but dark charges has a change in inverse of electric potential mass. The uniform electric field E of the capacitor exerts a constant force F ( on q' ) = q' / E on the dark charge, so the work done on the dark charge is: W elec = F ( Delta r ) where ( Delta r ) is displacement vector. We have: Delta r =  sf  si  = si  sf where i and f are subscribes indicating the initial and final position of the dark charge on the axis s. Thus, W elec = F ( si  sf ) = q' ( si  sf ) / E Since Delta ( U elec ) =  W elec, we have: Delta ( U elec ) = q' ( sf  si ) / E = ( q' sf / E )  ( q' si / E ) where ( U elec ) is electric potential mass, and Delta ( U elec ) is the change of it, which is physically important. The last equation can be used in law of conservation of mass in dark charge interactions. **** Mass of a Dark Photon Evaluate the mass and wavelength of a red dark photon, given that the wavelength of a red ordinary photon is 650 nm? Momentum of an ordinary photon is p = h / lambda . Momentum of a dark photon is p' = h / ( lambda ) '. Since momentum of a dark photon is numerically equal to momentum of a correspondent ordinary photon, by equating the two equations we find: p' = p , so ( lambda ) ' = lambda Thus, the wavelength of a red dark photon is: 650 (10)(9) m = 650 ns, where 9 is exponent. Hence, the wavelength of a red dark photon is 650 nanoseconds. To find the mass of dark red photon, we need frequency of the dark red photon. We know that: c = ( lambda ) f for ordinary photon c = ( lambda ) ' f ' for dark photon By equating the two equations, we obtain: ( lambda ) f = ( lambda ) ' f ' Since ( lambda ) ' = lambda , we obtain: f ' = f Thus, we can numerically know frequency of a dark photon by knowing the frequency of its correspondent ordinary photon. We use c = ( lambda ) f to find the frequency of the red ordinary photon: f = 3 x 10 (8) / 650 x 10 (9) = 4.62 x 10 (14) Hz where 8 and 9 are exponents. Thus, f ' = 4.62 x 10 (14) 1 / meters. Since M = h f ' and f ' = f , we conclude: M = h f ' = h f The equation means that mass of a dark photon is equal to h f where f is the frequency of its correspondent ordinary photon. Therefore, we can find the mass of red dark photon as: M = 6.63 x 10 (34) x 4.62 x 10 (14) = 3.06 x 10 (19) kg where 34, 14, and 19 are exponents. By converting the mass in kg to neV.c2 where 1 neV.c2 as a unit of mass in dark matter is almost equal to 14.40 x 10 (12) kg, where 12 is exponent, we obtain: M = 2.13 x 10 ( 8) neV.c2 where  8 is exponent. Thus, the mass of the red dark photon is almost 3.06 x 10 (19) kg or 2.13 x 10 ( 8) nanoeV.c2. I moved the last part, namely The Dark Transformation Factor, to the above. **** Dark Photon Space and time are meaningless for a photon. As a result, speed is also meaningless for a photon. The time that it takes a photon to travel a distance of 3 x 10 (8) meters is zero for the photon (8 is exponent); thus the distance is also zero for it. Therefore, if it is said that a photon travels 3 x 10 (8) seconds in 1 meter, it makes no difference for the photon. However, it makes a difference for us. If a photon lies on an ordinary spacetime, it is an ordinary photon. If a photon lies on a dark spacetime, it is a dark photon. Photon is not a product of space and time. From the viewpoint of a photon, saying that the speed of photon is c s/m is as meaningless as saying that the speed of photon is c m/s. However, if we suppose that photon can also travel at speed of c s/m, we can explain dark matter. When a photon is within an ordinary spacetime, its speed is c m/s, but when it is within a dark spacetime, its speed is c s/m. Both of these statements are meaningless for a photon, but they are meaningful for us because we can explain our observations of ordinary and dark objects. **** Dark Neutron What are the values of energy and rest mass of a dark neutron? Dark neutron has no rest energy: Its energy is constant. However, dark neutron has a rest mass equal to 150.5349738 picokilogram, which is equal to 10.4540751 neV.c2 (or nano electronvolt.c2) where 2 is exponent. neV.c2 is the unit of mass of dark matter. That means if a dark neutron converts to mass, a mass of 150.5349738 pkg will be released, which is 8.98755 x 10(16) times or almost 9.0 x 10(16) times the mass of an ordinary neutron, where 16 is exponent. As it is obvious, 8.98755 x 10 (16) is almost equal to the square of c. Therefore, the energy of a dark neutron is 10.4540751 neV, or 150.5349738 pkg / c2, or 1.008664916u where u is the unit of energy of dark matter equal to 1 u = 1.660539040 x 10 (27) J = 149.24181 pkg / c2, where 27 is exponent. pkg / c2 is the unit of energy of dark matter. The values are summarized below: Energy of Dark Neutron = 10.4540751 neV = 150.5349738 pkg / c2 = 1.008664916 u 1 u = 1.660539040 x 10 (27) J = 149.24181 pkg / c2 Rest Mass of Dark Neutron = 150.5349738 pkg = 10.4540751 neV.c2 1 neV = 14.39964534 pkg / c2 1 neV.c2 = 14.39964534 pkg These are so exact values. **** Dark Electron The values of energy and rest mass of a dark electron are as follows: Energy of Dark Electron = 5.685630065 peV = 0.08187105647 pkg / c2 = 81.87105647 fkg / c2 = 5.48579895 x 10 (4) u Rest Mass of Dark Electron = 81.87105647 x 10 (3) pkg = 81.87105647 fkg = 5.685630065 peV.c2 These are so exact values. Therefore, if a dark electron converts to mass, a mass of 81.87105647 femtokilogram will be released, which is 8.98755 x 10 (16) times or almost 9.0 x 10 (16) times the mass of an ordinary electron, where 16 is exponent. As it is obvious, 8.98755 x 10 (16) is almost equal to the square of c. **** Dark Proton The values of energy and rest mass of a dark proton are as follows: Energy of Dark Proton = 10.43968485 neV = 150.3277593 pkg / c2 = 1.007276467 u Rest Mass of Dark Proton = 150.3277593 pkg = 10.43968485 neV.c2 These are so exact values. Therefore, if a dark proton converts to mass, a mass of 150.3277593 pkg will be released, which is 8.98755 x 10 (16) times or almost 9.0 x 10 (16) times the mass of an ordinary proton, where 16 is exponent. As it is obvious, 8.98755 x 10 (16) is almost equal to the square of c. **** Dark Beta(minus) Decay We have: Dark Neutron > Dark Proton + Dark Electron + Dark Electron AntiNeutrino How much mass is released in the process? The mass of decay is: Q = Rest mass of dark neutron  (Rest mass of dark proton + Rest mass of Dark electron) or: Q = Rest mass of dark neutron  Rest mass of dark proton  Rest mass of Dark electron Using the values presented in the last three parts above: Q = (150.5350  150.3278  0.0819) pkg = 0.1253 pkg = 125.3 fkg Thus, the value of the released mass is almost 125.3 femtokilogram. This mass is shared between dark electron and dark electron antineutrino as kinetic mass. **** Units neV is the unit of energy. pkg/c2 is the unit of energy of dark matter too. We have: 1 neV = 14.39964534 pkg/c2 If 1 neV of energy in dark matter converts to mass, how much mass is released? The answer is that 14.39964534 pkg mass is released. eV.c2 is the unit of mass of dark matter. We have: 1 eV.c2 = 14.39964534 x 10(3) kg = 14.39964534 g where 3 is exponent. That means 1 eV.c2 of mass in dark matter is equal to 14.39964534 gram or almost 14.400 gram. **** Solution of Galaxy Rotation Curves We assume that there is dark matter at center of a galaxy and as it converts to mass, the mass makes gravitational force on a star far away from the center of galaxy. The equation of the force is: F = G m e c2 / r2 where both 2 are exponents, and m is the mass of star. ec2 , which is equal to M, is the rest mass of the dark matter. Therefore, the speed of the star is: v = (G e c2 / r ) (1/2) where (1/2) is exponent. Thus, the speed of star doesn't depend on the star mass. These two equations solve the problem of galaxy rotation curves. I will show this by giving an example. Suppose that there is a star with a mass equal to the mass of the Sun, i.e. 1.989 x 10 (30) kg, located at 20 kpc away from center of a galaxy. 1 kpc is 3261.56 light years. 1 light year is 9.461 x 10 (15) meters. G is numerically equal to 6.6742 x 10 (11). We suppose that there is a dark matter object at the center of the galaxy with an exotic energy of 1.18 x 10 (25) J, and the whole exotic energy converts to mass. Thus, the rest mass of the dark matter is 1.062 x 10 (42) kg where 42 is exponent. With these values, we can calculate both the gravitational force on the star and the speed of star. The gravitational force is equal to almost 3.70 x 10 (20) N where 20 is exponent. The amount of the force is great. The speed of star is equal to almost 3.39 x 10 (5) m/s or 339 km/s. The speed is reasonable based on the galaxy rotation curve data. If the same star is located at 30 kpc from the center of galaxy, what are the values of the gravitational force and the speed of star? The gravitational force is equal to almost 1.65 x 10 (20) N. The speed of star is equal to almost 2.77 x 10 (5) m/s or 277 km/s. If the same star is located at 40 kpc from the center of galaxy, what are the values of the gravitational force and the speed of star? The gravitational force is equal to almost 9.25 x 10 (19) N. The speed of star is equal to almost 2.40 x 10 (5) m/s or 240 km/s. This matches the real data of the galaxy rotation curve. If the same star is located at 50 kpc from the center of galaxy, what are the values of the gravitational force and the speed of star? The gravitational force is equal to almost 5.92 x 10 (19) N. The speed of star is equal to almost 2.14 x 10 (5) m/s or 214 km/s. The values of speed for the distances 20 kpc, 30 kpc, 40 kpc, 50 kpc from the center of galaxy are 339 km/s, 277 km/s, 240 km/s, 214 km/s respectively. That's a very good approximation of the real data of galaxy rotation curve. Thus, the equations above work. **** Ordinary Matter and Dark Matter. Is There Dark Energy? There is no dark energy. Why? The dimensions of spacetime are determined by the sort of matter that lies on spacetime. If an ordinary matter lies on spacetime, then there are 3 dimensions of space and 1 dimension of time which passes. If an exotic energy of dark matter lies on spacetime, then there are 3 dimensions of time and 1 dimension of space which expands. The expansion of space is analogous to the passage of time. Passage of time is an inseparable part of an ordinary matter; Expansion of space is an inseparable part of a dark matter. We can see an ordinary object in 3dimensional space but time, as the fourth dimension for that, passes. In contrast, we cannot see exotic energy of dark matter because it lies on a 3dimensional time, however, we can observe the expansion of space which works as the fourth dimension for dark matter. **** Fundamental Equations of Dark Matter The fundamental equations of dark matter are as follows: v = dt / dx (unit of v: s/m ) ; a = dv / dx ; F = e a ; p = e v ; Integral ( F dx ) = delta p = e ( delta v ) Integral ( F . dt ) = delta (M k) = e v 2 / 2 where the first 2 is exponent, and (M k) is kinetic mass; M = (gamma) e c2 ; M0 = e c2 ; M 2 = M0 2 + p2 c2 ; M k = ( gamma  1 ) e c2 where (M k) is kinetic mass; gamma = 1 / [ 1  ( v2 / c2 ) ] 1/2 where the both 2 and 1/2 are exponents; beta x M = p c ; M = h f ; c = f x lambda ; p = h / lambda ; Unit of h: kg.meters ; Unit of f or frequency: meters (1) where 1 is exponent; Unit of lambda or wavelength: seconds. Yes; this is dark matter, not ordinary matter, and those are the gamechanging equations. **** The All Idea at a Glance I give a summary of what I assumed to approach the problem of dark matter: 1) I assumed that there is a 3D time and 1D space overlapping the known 3D space and 1D time. Thus, dark matter particles exist in a 3D time and 1D space. Dark matter interacts with ordinary matter through gravitational force. 2) Consequently, a dark photon travels within 3 dimensional time and 1 dimensional space. Thus, the speed of dark photon is c seconds per meter. Since space and time are meaningless from the viewpoint of both ordinary and dark photon, the assumption is reasonable. 3) The particles in the assumed space and time are made of energy with the exception of dark photon which has no energy and is made of mass. Thus, Dark photon has mass, but it has no energy. Since the assumed spacetime is the inverse of the familiar spacetime that we live in, the massenergy equivalence of dark matter is the inverse function of the massenergy equivalence of ordinary matter. If the energy of any dark particle converts to mass, a computable amount of mass will be released. Mass in dark matter acts the same as energy in ordinary matter. That's why we cannot see any mass associated with dark matter. Thus, for example, we have kinetic mass and thermal mass in dark spacetime. Any kind of dark mass interacts with ordinary mass through gravitational force. 4) Therefore, I wrote the equations governing the assumed spacetime and calculated the energies and rest masses of dark electron, dark proton, and dark neutron as follows: Energy of Dark Electron = 5.685630065 peV Rest Mass of Dark Electron = 81.87105647 fkg Energy of Dark Proton = 10.43968485 neV Rest Mass of Dark Proton = 150.3277593 pkg Energy of Dark Neutron = 10.4540751 neV Rest Mass of Dark Neutron = 150.5349738 pkg The rest masses of these dark particles are released when their energies are converted to mass. 5) The new theory easily resolves the enigma of galaxy rotation curves as explained above. This is a bold approach for solving the dark matter problem. **** Equation of Dark Energy Now can we write down the equation of dark energy, another big problem? Yes, we can! We have ordinary matter and ordinary antimatter. On the other hand, we have dark matter, but what about antidark matter? Let's do a scaling: Proved: Ordinary matter: E , m Ordinary antimatter: E' , m' Assumed: Dark matter: M , e AntiDark matter: M' , e' Scaling: E . E' . e . e' <> M . M' . m . m' I used the sign <> for showing the proportionality. Since E' =  E m' = m M' =  M e' = e So  (E2) (e2) <>  (M2) (m2) where all 2 are exponents; Rearrange (E/m)2 <> (M/e)2 Or c4 = c4 where 4 is exponent. Thus, we have: E . E' . e . e' = M . M' . m . m' Hence, the scaling gives a reasonable result. The equation above is so fundamental. However, suppose that someone makes another scaling as follows: E . E' . e . e' . e'' . e''' <> M . M' . m . m' . M'' . M''' where Dark Energy: M'' , e'' AntiDark Energy: M''' , e''' since M''' =  M'' e''' = e'' so c4 . e'' . e''' <> c4 . M'' . M''' (e'')2 <>  (M'')2 But there is no equation to account for the proportionality above because they have already used the two possibilities, i.e. E = mc2 and M = ec2. If e'' = M'' c2, it is ordinary matter, and if M'' = ec2, it is dark matter. Also, e'' = M'' cannot be true, because mass and energy cannot be equal without an appropriate proportionality constant factor. Thus, they will end up with a paradox. As a result, dark energy is just antidark matter for which the first scaling works. We can conclude that the equation of dark energy is the same as the equation of dark matter but with a negative mass M: Equation of Dark Energy: M2 = e2 c4 + p2 c2 where M < 0 or Dark Energy = AntiDark Matter All numbers in the equation are exponents. The negative mass M means that the charges of antidark proton and antidark electron are opposite the charges of dark proton and dark electron respectively. I should emphasize that there is no dark energy photon because dark energy is just antidark matter, and so it cannot have dark energy photon. Similarly, there is no photon in ordinary antimatter. **** A Correction to the Current Mass Density of Universe I just write the equations that I got. I will explain them later. m: mass of ordinary matter M: mass of dark matter M': mass of dark energy F: Gravitational force of dark matter on ordinary matter F': AntiGravitational force of dark energy on ordinary matter F net = F'  F = (G M' m / x 2)  (G M m / x 2) F net = (G m / x 2) (M'  M) Work = W (initial > final) = Integral (F net) (from x i to x f) Delta(Potential energy) =  W (i > f) Delta U = U at infinity  U at r =  Integral (G m / x 2) (M'  M) dx (from infinity to r) =  (G m)(M'  M) Integral (1 / x 2) dx (from infinity to r) =  (G m)(M'  M) [1 / x] (from infinity to r) =  (G m)(M'  M) (1 / r) Delta U = U at infinity  U at r =  (G m)(M'  M) (1 / r) U at infinity = 0 >  U at r =  (G m)(M'  M) (1 / r) Potential Energy of All Universe: U at r = (G m)(M'  M) (1 / r) (Unbound System) The last equation is so fundamental and changes the longlasting equations of astrophysics. U at r is the gravitational potential of all universe at distance r. Since U is positive, the system is unbound. Thus, ordinary matter is influenced by a net antigravitational force exerted by dark energy. This means that the universe both expands and the expansion speeds up. The expansion can be stated with or without an initial velocity. To calculate initial and final velocities at a distance r, we have: (1/2) m v2 + (G m)(M'  M) (1 / r) = (1/2) m v'2 + 0 Initial and Final Velocities of Total mass of Ordinary Matter: v' 2  v 2 = (2 G / r)(M'  M) Due to the Combined Gravitational and AntiGravitational Forces of Dark Matter and Dark Energy Respectively where both 2 are exponents. v and v' are initial and final velocities respectively. m is total mass of ordinary matter. M' and M are total masses of dark energy and dark matter respectively. 0 is the gravitational potential energy at infinity. Now I relate the last equation to the mass densities of dark energy and dark matter as follows: If v = 0 or v is far less than v', then: v' 2 = (2 G / r)(M'  M) Thus: v' = Expansion Velocity at the Present Time After multiplying both numerator and denominator by (4/3) pi r 2 : v' 2 = (8/3) (pi G r 2) [(M' / V)  (M / V)] and simplifying where V is the volume at distance r. Therefore, we obtain: v' 2 = (8/3) (pi G r 2) (Density of DE  Density of DM) where 2 is an exponent. DE and DM stand for dark energy and dark matter respectively. Since: v' 2 / r 2 = H 2 (All 2 are exponents.) H 2 = (8/3) (pi G r 2) (Density of DE  Density of DM) Or: Mass Density of DE  Mass Density of DM = 3 H 2 / 8 pi G [correct] where 2 is exponent. H is the Hubble constant: H = 70 km/s per megaparsec = 2.3 x 10(18) 1/s, where 18 is exponent. However, the current physics says that: Critical Mass Density of Universe = 3 H 2 / 8 pi G [wrong] which is wrong because they use the equation of gravitational potential energy to find the escape velocity of the universe, but they don't consider the antigravitational force of dark energy which makes the system unbound and has a different equation for the potential energy as I derived it above. I should add that if someone argues for the equation below: (1/2) (m + M) v2 + (G M')(m + M) (1 / r) = (1/2) (m + M) v'2 + 0 [wrong] v' 2  v 2 = 2 G M' / r [wrong] which means mass of dark energy M' exerts antigravitational force on both ordinary matter m and dark matter M, it cannot give the correct recession velocity observed by astronomers, because dark matter exerts a gravitational force on ordinary matter which competes with the antigravitational force of dark energy. We have to show the gravitational effect of dark matter in our equation for having the observed velocity. **** Two Fundamental Approaches to Dark Matter There are two approaches for writing dark matter equations. These approaches are equivalent but each have their specific advantages. 1) If we exchange space and time, then the physical constants in dark matter will have the same values of the physical constants in ordinary matter but the units will be different. Advantage: The approach makes it easy to write dark matter equations. I call this method natural approach. 2) The physical constants in dark matter are inverse of the physical constants in ordinary matter but the units are the same. Advantage: The approach makes more sense for people because it presents dark matter in our familiar 3D space and 1D time. However, finding dark matter equations in this way is difficult. I call this method popular approach. I prefer the natural approach for finding dark matter equations. However, sometimes we need the popular approach to predict the behavior of dark matter in a better way. I give an example: 1) Dark photon based on natural approach: M = pc ; p = h / lambda ; c = (lambda) f ; 2) Dark photon based on popular approach: M = pc ; p = 1 / ( lambda . h ) ; c = 1 / ( lambda . f ) M = ( c 2 / h ) f ; p = c f / h **** The File A file "Dark Equations" was uploaded at top of the post, which explains the theory clearly. * File update on April 30, 2017. * File update on May 13, 2017. * File update on May 25, 2017. * File update on May 27, 2017. * File update on June 19, 2017. * File update on August 9, 2017. Matrix transformations of dark matter, dark energy, and antimatter is shown in the file. I'll be back. 
In reply to this post by Alien Observer
to alien observer..... you are relatively way too smart lol have a warp speed day :) 
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I wish Alien was back here on iwonder... this is an old post... he hasn't been around for a longggg time.... ): but.. he is smart though ... 
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In reply to this post by parallel400
Bumping the post up for you mardebaran2005 ..Hope my email helped you fix the problem you had on here.

In reply to this post by *StarShine*
Alien Observer once posted that I almost made him cry.

In reply to this post by parallel400
my brain is **warped lol 
In reply to this post by Alien Observer
"Also, how did they get the atom bomb from this equation? "
Since the equation shows how much energy is contained within a massive object(particle) it just focuses the attention to getting that energy out of that object(for obvious reasons). That energy is released by splitting an atom's nucleus. Untill the Uranium(it has unstable nucleus which results in radioactivity) came into the picture the energy that has been taken out by splitting other atoms was the same amount that was wasted for splitting it. The problem with Uranium is that it's unstable, so once the splitting starts it becomes uncontrolled and a chain reaction starts whitch releases that tremendous amount of energy in a split second, which ofcourse due to the destructive nature of man led to the discovery of the bomb. http://chemwiki.ucdavis.edu/@api/deki/files/16111/20.7.jpg 
In reply to this post by Alien Observer
How did E=mc^2 excatly come to frution, and become so widely accepted (besides Planx urging everyone to take Einstein seriously)?
Since even the remote idea of dark matter/energy did not exist during Einstein's time, how is it that E=m (energy=mass) was accepted along with E=mc^2 (energy=mass*celeritas^2). Since according to logic and common sense it would be similar to saying something like 1=1*10^2 or 1=100. 
Nvm
I realized that the formula m=Ec^2 is acctually just the original way Einstein had written it, also E=m did not acctually exist it was just the idea that used up energy was converted to the mass of an object. And to partially correct parrallel400 it does not necessarily apply only to dark matter/energy, but mass. 
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