There is a difference between the energy of a particle and the momentum of a particle though.

When the speed is very fast (relativistic speeds), the formula:

Energy = sqrt( (momentum*c)2 + (mass*c2)2 )

must be used although it can be used at lower speeds also.

(Energy = kinetic energy + mass*c2) where c = speed of light.

So

E = Ekin+ mc2.

A charged particle's momentum is routinely measured by exploiting the fact that it bends in a magnetic field inversely proportional to the momentum.

You can bend the particle and measure how much it deflects with a known magnetic field and magnet length.

The kinetic energy can be measured by letting a particle (charged or neutral) slam into a piece of the appropriate material.

If there is enough of the material, the particle will stop leaving all its kinetic energy in the material.

If it is an antiparticle, it may annihilate and leave its rest mass energy (mc2) in the material as well.

One can solve the equation (c = 1) for the mass:

(Ekin + m)2 - p2 = m2

or

m2 - 2*m*Ekin + (p2 - Ekin2) = 0

So basically, in particle physics, one usually measures the energy and momentum and infers the mass.

Dear Al Timimi,

One cannot use Einstein's energy-momentum theorem E^2 - (cp)^2 = m(0)^2 c^4 on a lightspeed particle, but only on an atomic particle. This is because the E in the above equation is given by E = m c^2 = m(o) c^2 gamma, where the Lorentz factor gamma is infinity at lightspeed. If the mass of the particle, here denoted by m(0), is a constant, then the energy is infinite, and there is not enough energy in the whole Universe to make this come true. If we take it to be the constant mass of a photon of constant energy, then the photon would also have infinite energy which would prohibit from traveling at the speed of light. So a photon with constant mass when traveling at the speed of light, that is all photons normally experienced, cannot be validly said to obey Einstein's E = mc^2 or the associated energy-momentum theorem. Einstein's E = mc^2 does not apply to any particle that travels at the speed of light. I asked this question so the reader would be forced to come to this conclusion, which is the correct conclusion. Photons actually obey a different law, the law of photodynamics. Read my paper on Photodynamics, on this site, researchgate, under my name, to discover the actual equation which applies to the photon and all particles that travels at the speed of light. The bottom line is that any photon with a constant frequency, that is a constant mass, cannot obey Einstein's E = mc^2.

Take care,

Bob

Dear Robert,

You wrote " This is because the E in the above equation is given by E = m c^2 = m(o) c^2 gamma, where the Lorentz factor gamma is infinity at lightspeed. If the mass of the particle, here denoted by m(0), is a constant, then the energy is infinite, and there is not enough energy in the whole Universe to make this come true'

Please see how E =mc2 Has been derived at first place.... There is proper rigorous derivation ...Derivation started from work energy theorems dw =F. dr .... Where F = Mx acceleration .... Mass in this equation is a relativistic mass... Why you not look up the derivation .... You think that by some magic people has concluded E= M0 c2 ...No... There is no magic .... Equation is perfectly valid.... There are various other method through which same conclusion has been drawn...Take a look in attached paper. Stop your irrational thought...

Dear Gokaran,

I asked the question in that way so that people like you would conclude that Einstein's E = mc^2 does not apply to the photon. Congratulations if you came to the same conclusion, if indeed you did. Take a look at how I derived the Einstein energy momentum theorem in my paper Photodynamics. It is probably the easiest and quickest way of deriving it. The question shows that although the mass of the photon is constant, you cannot put that constant mass in the Einstein equation because it would lead to the photon having infinite energy. Again, this shows that the Einstein equation E = m(0) c^2 gamma, where m(0) is a constant mass does not apply to the photon . It must be replaced by the valid expression for the photon which reads E = m(0)c^2/gamma. Again, read photodynamics on this website, researchgate, under my name.

After you have read my paper, and you use the correct equation, tell me if I am irrational. By the way Einstein's equation is perfectly valid, but only for atomic particles whose relativistic masses goes like m = m(0)gamma, not m(0)/gamma like particles that go at the speed of light.

Take care,

Bob Martineau

Dear Gokaran,

Do you agree with my answer just above. Please answer.

By the way, when you talk, please use whole sentences so that the reader does not have to guess as to what you are saying.

Take care,

Bob Martineau