Is it Valid to Derive $E = pc$ From the Energy-Momentum Relationship for Photons?

Question

Given a particle with mass $m$ moving at velocity $v$, total energy is:

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

Note I am not using the relativistic - rest mass convention, as I was taught to think in terms of rest - total energy instead. "Relativistic mass" would be represented as $\gamma m$ , where the mass of the particle is being changed by a factor of $\gamma$ depending on it's relative velocity $v$, where gamma is equal to,

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

To obtain momentum of Particle, we expand the original equation to account for relativistic momentum:

$$E^2 = (\gamma mvc)^2 + (mc^2)^2$$

If the above equation is an accurate representation of the energy-momentum relationship, how does my professor use this equation to derive $E=pc$ for a photon? He says for a massless particle (photon),

$$E^2 = (pc)^2 + ((0)c^2)^2$$

therefore,

$$E = pc$$

But he appears to me to neglect the fact that you can expand the $p$ variable into relativistic momentum, which is a function of mass, gamma, and velocity. Was his move acceptable? If so , why? How can you make one mass 0 but not the other?

Answer 1

Your equation involving $\gamma m$ is useless for photons because $\gamma$ is infinite and $m$ is zero. That product is indeterminate.

Your professor is correct.

Answer 2

I think the existing answers are missing the essential point here. The main point is that we can construct a 4-vector with components given by quantities called $E/c$ and $\bf p$. We then try to interpret what those quantities are. We can either deduce (by symmetry arguments) that they are conserved in an isolated system, or else we propose that at least one of them is conserved, and then we can deduce that the other is conserved also. (The argument is that if $P$ is 4-momentum and we define $Q = P_{\rm after} - P_{\rm before}$ and then assert that one component of $Q$ is zero in all frames then the whole of $Q$ must be zero; the technical term for this is "the zero component lemma" but it is easy to prove).

Ok, so far so good. Since $E/c$ and $\bf p$ make a 4-vector, we can deduce that the quantity $$ E^2 - p^2 c^2 $$ is Lorentz invariant. Also, one finds that the notion of conservation does not make physical sense for spacelike 4-vectors, so we have that the energy-momentum 4-vector is either null or timelike. It follows that the invariant quantity is either zero or positive. So we can associate it with a non-negative quantity, which for the sake of convenience we write this way: $$ E^2 - p^2 c^2 = m^2 c^4 . $$ Here, $m$ is Lorentz-invariant. Clearly if $m=0$ then $E=pc$ and otherwise $E > pc$. The factors of $c$ are introduced so that in the limit of low velocities, $E$ shall equal a constant plus the Newtonian kinetic energy $(1/2) m v^2$ (and then also in that limit one will find $p = m v$.)

The main point in all the above is that we have not needed to introduce the Lorentz factor $\gamma$. That factor only comes in when we are considering the case $m \ne 0$. In that case we can invoke a 4-velocity $U$ and furthermore associate an inertial frame with that 4-velocity (this can only be done when $U$ is non-null, i.e. not associated with motion at the maximum speed). In this case the 4-momentum can be written $$ P = \frac{d U}{d\tau} = \frac{dU}{dt} \frac{dt}{d\tau} = \frac{dU}{dt} \gamma $$ where the last step used $dt/d\tau = \gamma$ where $t$ is time in some given inertial frame, $\tau$ is proper time along the worldline of a particle with 4-momentum $P$. This leads to the results $$ E = \gamma m c^2, \qquad {\bf p} = \gamma m {\bf v}. $$

So you see the reasoning is this way round: first we construct the 4-vector (calling its components $E/c$ and $\bf p$), then the invariant, and then the use of $\gamma$ when considering the case $m > 0$.

Answer 3

...He says for a massless particle (photon),

$$E=pc$$

But he seems to me to neglect that you can expand the $p$ variable into relativistic momentum, which is a function of mass, gamma, and velocity. Was his move acceptable? If so, why? How can you make one mass 0 but not the other?"

As you explained, the relation $E=pc$ is obtained by replacing $m=0$ in the general relation. So bringing back $m$ into the equation to express $p$ is not a consistent procedure. One must find another way to introduce either $E$ or $p$ for a massless particle.

In this case, the energy is obtained by the Planck-Einstein relation $E=hf$, which also determines $p$ using $E=cp$, giving the de Broglie equation.

By the way, if we use the relativist momentum $p=m\gamma v$ and put $m=0$ we get $p=0$ unless we put $v=c$ which leaves the equation undetermined, so with a possible value $p\neq0$. This shows that a mass zero particle must move at the speed of light.