On why massless particles move at the speed of light

[PAllen]

On "why massless particles move at the speed of light"

It has come up a few times whether you can derive that massless particles must go the speed of light, strictly using SR. Bcrowell proposed a way that some argued against. I have a different tack for consideration.

I recently derived that, for any particle with E0 (rest energy) > 0, the following is true:

(1 - v^2/c^2) = 1 - KE / (KE + E0)

From this you can say:

1) limit as E0 -> 0 leads to v=c
2) If there is to be such a thing as a particle with E0=0 and nonzero KE, and if it should be consistent with the SR, then this relation directly requires that v=c.

 

[PAllen]

It has come up a few times whether you can derive that massless particles must go the speed of light, strictly using SR. Bcrowell proposed a way that some argued against. I have a different tack for consideration.

I recently derived that, for any particle with E0 (rest energy) > 0, the following is true:

(1 - v^2/c^2) = 1 - KE / (KE + E0)

From this you can say:

1) limit as E0 -> 0 leads to v=c
2) If there is to be such a thing as a particle with E0=0 and nonzero KE, and if it should be consistent with the SR, then this relation directly requires that v=c.

That should be:

sqrt(1 - v^2/c^2) = 1 - KE / (KE + E0)

 

[Bill_K]

The rest mass of a particle is defined to be the norm of its momentum 4-vector. There are only two kinds of vectors that have zero norm: null vectors and the zero vector. So a particle which is massless either travels at the speed of light or carries no energy and momentum at all.

 

[dextercioby]

The rest mass of a particle is defined to be the norm of its momentum 4-vector. [...]

How would you unambiguously define the <momentum 4-vector> ?

 

[Bill_K]

How would you unambiguously define the <momentum 4-vector> ?

It doesn't have to be unambiguous! The momentum vector for a particle is P = (α, αv) for some α, I don't care what. But so long as the norm is zero, P·P = 0, then v = c for any α, which is all we wanted to show. You guys are making this way too hard.

 

[pervect]

How would you unambiguously define the <momentum 4-vector> ?

Let's start with one spatial dimension. Then

given the Lagrangian L(x, v) where v = dx/dt

define

$$p = \frac{\partial L}{\partial v}$$

This serves as an adequate definition of momentum.

A Lagrangian is defined as "good" if the equations of motion

$$\frac{dp}{dt} = \frac{\partial L}{\partial x}$$

are correct, i.e. match observation. And E would be given by

E = p*v - L (more generally the sum over i of p_i *v_i - L when you have more than one dimension)

 

[dextercioby]

The point I was trying to make is that postulating the existence of rest/invariant mass, assuming it nonzero and the form of the Lagrangian of a free particle of rest mass m_0 leads to P^2 = m^2. Point: P^2 = m^2 is a consequence, not an assumption/definition/axiom.

Assume it in reverse, then what are the Lagrangian & Hamiltonian formulations of the theory with P^2 =0, thus with invariant mass = 0 ?

 

[PAllen]

The rest mass of a particle is defined to be the norm of its momentum 4-vector. There are only two kinds of vectors that have zero norm: null vectors and the zero vector. So a particle which is massless either travels at the speed of light or carries no energy and momentum at all.

The aim is was to justify this with the fewest assumptions and without apparent division by zero that could be criticized by a skeptical newbie. Playing skeptical newbie, all you've got from this is:

E^2=P^2 c^2

So now newbie asks, isn't P = m gamma v = 0 if m=0? Contradiction (since E assumed nonzero)?
So now the problem has been transposed to showing why P=E/c implies speed is c.

[Edit: Of course, one way to do this is to write the general:

P = Ev/c^2 ; then the requirement that P = E/c implies v=c. But with this addition I don't see much superiority to this approach. ]