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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.
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 Quote by 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)
 Blog Entries: 2 Recognitions: Science Advisor 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.

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

 Quote by Bill_K 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> ?

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 How would you unambiguously define the ?
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.

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 Quote by dextercioby How would you unambiguously define the ?

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)
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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 ?

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