# Relativistic Mass/Energy

1. Sep 13, 2014

### Greatness

Pardon my elementary questions, but:
Why is it that photons can only travel at the speed of light?
I know because they have zero rest-mass it is only possible for them to travel at c, but is there some mathematical reasoning to this through the mass and momentum equivalence equation?

Lastly, is there some mathematical reasoning as to why anything with non-zero rest mass cannot reach light speed? I understand it would take infinite energy, but can that be reasoned mathematically, or is it solely physics?

Thanks, once again.

2. Sep 13, 2014

### ghwellsjr

Just two days ago, someone else asked very similar questions in this thread:

has relativistic mass gone out of fashion?

In that thread, I gave what I thought was a relevant answer in post #33. Why don't you have a look at it and see if it helps you?

3. Sep 13, 2014

### Staff: Mentor

That question is basically asking why light travels at the speed of light... it has to, because that is what "the speed of light" means.

However, you may be trying to ask a different question: "Why is the speed of light what it is?"
The answer is that light is electromagnetic waves, and the speed at which these waves propagate is determined by and can be calculated from the laws of electricity and magnetism which James Maxwell discovered in 1861 (google for "Maxwell's equations").

(as an aside.... except when you are studying the quantum mechanical properties of light, you should avoid thinking in terms of photons. They aren't what you think they are).

Last edited: Sep 13, 2014
4. Sep 13, 2014

### ChrisVer

photons travel at c everywhere...
light travels at c in the vacuum...
So the speed of light should refer at c??

5. Sep 13, 2014

### Staff: Mentor

The general relationship between (rest-)mass, energy and momentum is $E^2 = (pc)^2 + (mc^2)^2$. Therefore if m = 0, E = pc.

From the relationships between momentum and velocity, or between energy and velocity, and the relationship above, it's possible to show that
$$\frac{pc}{E} = \frac{v}{c}$$
which gives 1, i.e. v = c, for a photon.

6. Sep 13, 2014

### Greatness

Which relationships are you referring to?
Also, why in E = mc^2 is m sometimes referred to as either relativistic mass ( in the context of light? ), while at other times (and what I understand) as rest mass?

7. Sep 13, 2014

### Staff: Mentor

$$p = \gamma mv = \frac{mv}{\sqrt{1-v^2/c^2}}\\ E=\gamma mc^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$
When m is the "rest mass", E is the "rest energy", i.e. the energy that the particle has even when it is at rest: $E_0 = mc^2$. When m is the "relativistic mass", E is the total energy (rest energy plus kinetic energy): $E = E_0 + K = m_{rel}c^2 = \gamma mc^2$.

Last edited: Sep 13, 2014
8. Sep 13, 2014

### Staff: Mentor

The $E$ in $E=mc^2$ can be understood as either the total energy including the object's kinetic energy, in which case you would use the relativistic mass (and interpret $(\gamma-1)m_0$ as the kinetic energy); or as the energy in a frame in which the object is at rest so has no kinetic energy, in which case you would use $m_0$, the rest mass. Obviously the latter does not apply to light (no frame in which the light is at rest).

Note that the more modern $E^2=(m_0c^2)^2+(pc)^2$ reduces to $E=mc^2$ for a massive particle at rest ($p=0$) and to $E=pc$ for a massless particle. Thus, it's generally considered the more fundamental relationship these days.