Relativistic derivation of energy for a photon inquiry

Question

In Susskind's Special Relativity & Classical Field Theory, he presents the following argument for the energy of massless particles:

We know there is a relationship between the components of the velocity 4-vector as follows: $$(U^0)^2 - (U^x)^2 - (U^y)^2 - (U^z)^2 =1$$ $$m^2(U^0)^2 - m^2(U^x)^2 - m^2(U^y)^2 - m^2(U^z)^2 =m^2$$ $$E^2 - P^2 = m^2$$ $$E = \sqrt{P^2 c^2. + m^2c^4}$$ $$E = c |P|$$

Now, I ave two questions about this argument. First of all, doesn't $P$, by definition, still have $m$ in it? Therefore, this would be saying the energy is zero? Also, my understanding of Susskind's derivation of all of this is to start with the notion that there is a particle with a maximum speed and to use Lagrangian mechanics and the invariance of 4-vectors to build up to this point. At what point does he make an assumption that is inconsistent with quantum mechanics? It seems to be the invariance of 4-vectors and lagrangian mechanics should hold consistently with quantum mechanics, but I am not entirely sure. I would expect, based on quantum mechanics, the energy of a particle to depend on more than the velocity and mass (to depend on the wavelength/frequency), so I want to know where this theory Susskind is building up breaks down in the quanutm picture.

Answer 1

First of all, doesn't , by definition, still have in it?

No. In Newtonian physics momentum is defined as $p=mv$, but that is not the definition in relativistic physics. In relativistic physics it is $m^2 c^2=E^2/c^2-p^2$. For a massive object this reduces to $p=mv$ for $v\ll c$. But for massless objects we get $p=E/c$.

At what point does he make an assumption that is inconsistent with quantum mechanics?

It doesn’t, but when you get to QFT you usually need to consider the stress-energy tensor, not just the four momentum. The four-momentum isn’t inconsistent, it is just less useful.

Answer 2

No, not necessary $P$ involves mass because is the four-vector definition. Since $\vec{P} = (E/c, \{\gamma(u)m\vec{u}\})$, if $m=0$ you get $P = E/c$ (*).

And regarding your second question, I think you are asking for the difference between the two mechanics. Non-quantum mechanic (both Einstein and Newton mechanic) denies in its theory and equations (implicitly) that the Uncertainty Principle holds (recall Einstein's famous quote). And it can be thought as a case of the quantic mechanics for huge mass particles. If you think on a macroscopic situation, say for example a particle of $100g$ that travelled a distance of $15m$ at $2m/s$. Through Uncertanity Principle one discovers that:

$$\Delta p \geq \frac{\bar{h}}{2 \Delta x} = 3.5E{-36} \: kgm/s$$ $$\implies \Delta v = \frac{\Delta p}{m} \geq3.5E{-35} \: m/s << v$$

Obviously this doesn't proves anything (it's a lower bound for $\Delta v$) but gives you an idea of why in non-quantic mechanic is useless to draw on that principle.

(*) PS: While reading this, I think however that I found a simpler (perhaps more informal) argument for a definition of $\vec{p}$ for massless particles. Starting for the fundamental relationship of kinectic energy to momentum $dK = \vec{u}\cdot d\vec{p}$ , if we assume that $K$ and $p$ are functions of $u$ , since on a photon is fixed $u=c$, then $K$ and $p$ are fixed too. If at any instant $\vec{u}$ and $d\vec{p}$ were not parallel, this could only mean that the particle is accelerated, then $K$ is not constant. So it must be $dK=cdp$, and since no speed means no kinectic energy. integrating both sides gives $K=cp$