# Relativistic derivation of energy for a photon inquiry

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.

First of all, doesn't 𝑃, by definition, still have 𝑚 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.

• It might be worth noticing that your explanation of momentum $(mc)^2=(E/c)^2-p^2$ relies on Susskind's result, and can't independently explain it. Sep 15, 2022 at 0:13
• While that is true I don’t see that the question was requesting an independent explanation. I actually would take this as a definition of mass rather than something that is in need of explanation
– Dale
Sep 15, 2022 at 0:34
• yeah i mean this doesnt really seem like an answer? you say p = E/c which is the exact same thing i have in my post. what do these depend on then if not mass? Sep 15, 2022 at 0:52

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 $$K=cp$$

• See this answer: “A lot of the confusion on this topic seems to arise from people assuming that $p=m\gamma v$ should be the definition of momentum.” Feb 17, 2023 at 5:19
• Alternatively, see this answer. Feb 17, 2023 at 5:22
• Ty @Ghoster, I don't remember anything of modern physics since when I was on that course a couple of years ago. I've lived with the wrong conviction that $p=\gamma m u$ for all this time but it looks like you've enlighten me. I also noticed that I was told about the distinction between rest mass and movement mass, but I found that this is now 'deprecated'
– tac
Feb 19, 2023 at 1:57