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.


2 Answers 2


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – RC_23
    Sep 15, 2022 at 0:13
  • $\begingroup$ 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 $\endgroup$
    – Dale
    Sep 15, 2022 at 0:34
  • $\begingroup$ 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? $\endgroup$ 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 $K=cp$

  • $\begingroup$ 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.” $\endgroup$
    – Ghoster
    Feb 17, 2023 at 5:19
  • $\begingroup$ Alternatively, see this answer. $\endgroup$
    – Ghoster
    Feb 17, 2023 at 5:22
  • $\begingroup$ 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' $\endgroup$
    – tac
    Feb 19, 2023 at 1:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.