# Why massless particles must travel at c

1. Dec 6, 2007

### wtronic

1. The problem statement, all variables and given/known data

give a logical argument for why a particle with m0 = 0 with nonzero energy has to move with a speed equal to c.

2. Relevant equations

3. The attempt at a solution

I attempted using the momentum equation but I always get a multiplication by 0 which doesn't make sense to me.... please some help

thanks

2. Dec 6, 2007

### 2Tesla

You can relate p, E, and v without explicitly involving m. Do you know how?

3. Dec 6, 2007

### Mr.Brown

A particle with zero mass has to travel with c because his mass is independant of the reference frame (it´s zero in every reference frame) hence it has no rest energy in any reference frame ( you can´t transfer to one frame where it has energy and another where there is no energy because there is no rest energy).
But for the notion of particle to make any sense at all, having any manifestation in the real world, it has to contain some sort of energy and while this energy has to be the same in any reference frame it has to move with c.

4. Dec 6, 2007

### wtronic

2tesla, maybe since p = mv and E = mc^2 then I can write E = (c^2*p)/v.....? but this equation doesn't justify why it has to go like speed c...... because any v along with any p will give a value for energy..

am I wrong?

5. Dec 6, 2007

### wtronic

you seem to make sense, but I am guessing my professor is looking for something more mathematical..... I don't know, if anything I will try with that

6. Dec 6, 2007

### Mr.Brown

i guess the mathematical idea behind this is that $$p_{\mu}p^{\mu}=-M^2=0$$ hence if the mass is 0 the momentum and velocity are on the light cone which is one definition for speed c.

Last edited: Dec 6, 2007
7. Dec 6, 2007

### wtronic

hi Brown, I am wondering where did you get that equation..... and what mu means... thanks]

8. Dec 6, 2007

### 2Tesla

Just in case Mr.Brown's indices are unfamiliar to you, the equation he's referencing is:

$$E^2 = (pc)^2 + (mc^2)^2$$

So, if you combine this with m=0 and the v=c^2*p/E you discovered just now, you should find your answer.

P.S. It doesn't make a difference here, but for future reference, in relativity:

$$p = \gamma mv$$

and

$$E = \gamma mc^2$$

where

$$\gamma = (1 - (v/c)^2)^{-1/2}$$

Last edited: Dec 6, 2007