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I don't really have strong backgrounds studying quantum physics, but I did learn special and general relativity, and I have now a question how to get the momentum of photon.

For my understanding, Einstein "predicted" energy and momentum of photon through his research on photoelectric effects (1905) and Compton scattering (1916) by assuming photon as a "quantum" which contains the wave-particle duality. Those Einstein's predictions were actually "verified" by experiments.

However, some people are explaining the momentum of photon can be "derived" analytically from the special relativity by using $E^2 = (mc^2)^2 + (pc)^2$ with $m=0$.

But I can't derive the same conclusion because that famous equation comes from the "invariants" in space-time which is "interval (or displacement)" in Minkowski 4D space. In special relativity, if the "particle" is moving with the speed of light, the invariant (interval) must be zero. Considering four-velocity (derivative of "proper length" with respect to "proper time"), the invariant becomes the square of speed of light, and then the four-momentum is still defined by four-velocity multiplied by invariant mass $m_{inv}$. This should eventually give $0 = 0$ because $m_{inv} = m_{rest}$ in special realativity, not tell us $E^2 = (pc)^2$ when $m_{inv} = 0 $ .

I could not find any resources telling why/how we can use special relativity for "quantum" or "massless particle."

Can we really derive the momentum of photon from the special relativity? Do we have to use quantum physics, not special relativity to get the momentum of "massless quantum (or particle)?"

Qmechanic
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Yuu
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3 Answers3

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The four-momentum of any particle must be tangent to its world line, including photons. Since $ds^2=0$ for a photon, its four-momentum must also satisfy $P\cdot P=0$ to ensure that it is tangent along the photon's world line.

We must also note that the four-velocity of a photon cannot be defined, because the proper time experienced along a photon's world line is always zero. Thus, our definition for massive particles $P=mU$ will not work for light. However, the most basic definition of four-momentum is that it's a vector whose components contain the energy and momentum of the particle, like so:

$$P=\begin{bmatrix}E/c\\p_x\\p_y\\p_z\end{bmatrix}$$

For a photon moving in the $x-$direction, $p_y=p_z=0$. The inner product is:

$$P\cdot P=0$$ $$E^2-c^2p_x^2=0$$ $$p_x=E/c$$

Note that we have not imposed $m=0$ yet. In fact, we deduce that photons have zero mass because their world lines and four-momenta are null.

$$P\cdot P=m^2c^2=0$$

Chang Hexiang
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  • Thank you for your answers. Can you tell me what is the definition of your first component of four-momentum? I was thinking $P_{0} \equiv E/c = m_{inv} u_{0} = \gamma m_{inv} c$ and that will give us $0=0$. – Yuu May 14 '21 at 07:37
  • Ah yes, I forgot to mention that the four-velocity cannot be defined for photons. I will edit my answer. – Chang Hexiang May 14 '21 at 07:39
  • What I'm asking here is that, while the invariance becomes 0, the first components of the four-momentum is "still" defined by the four-velocity multiplied by invariant mass. Thus, all components should be 0 and that will give us $0 =0$. That means, the special relativity is analytically consistent for massless particle, but should not yield $E/c = P$. – Yuu May 14 '21 at 07:41
  • Cool! That is exactly what I want to know and expected! The special relativity should not be applied for massless "particle" and that's why Einstein "predicted" and we need quantum physics now – Yuu May 14 '21 at 07:42
  • Why do you say that the four-momentum is "still" defined by mU? This definition is simply not applicable for photons. – Chang Hexiang May 14 '21 at 07:43
  • I would say that this definition of four-momentum for light is still classified under special relativity. There are no general relativistic or quantum mechanical concepts used in the definition. – Chang Hexiang May 14 '21 at 07:44
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    Ah, you are right. I was saying the massless particle is still moving with velocity v, but if massless particle is moving with speed of light, the Lorentz factor becomes infinity and we can't have four-velocity. So, I was wrong, the four-velocity can't be defined for massless particle. – Yuu May 14 '21 at 07:50
  • Yep, you're right. Also, I want to add that it's a misconception that light having momentum was introduced by quantum mechanics or relativity. In fact, Maxwell's Theory of Electromagnetism already predicts momentum transfer via electromagnetic radiation, calculated using the Poynting Vector. – Chang Hexiang May 14 '21 at 07:53
  • Waves can transmit momentum as well, just as easily as they transmit energy, but yes I see your point. – Chang Hexiang May 14 '21 at 08:12
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The answer to the question in the title is: yes, we can. Special relativity states the energy momentum relation as $E^2=m^2+p^2$ in units where c=1. Setting m=0 gives $E=p$. The four momentum of a photon is $\hbar(\omega,\vec k)$.

In fact, $E=p$ can be derived from Maxwell's equations. It does not even require photons.

my2cts
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  • Actually, the special relativity did NOT state the energy $E$ is "invariant" in any inertial frame of reference. What is your definition of $E$? What I'm asking here is that is $E$ function of $m_{inv}$? – Yuu May 14 '21 at 10:13
  • @Takagaki Why this comment? I did not make such a statement. As for your second remark, it should be obvious that my definition of E is the same as Einstein's. – my2cts May 16 '21 at 16:11
  • Sorry, I could not follow why you said obvious. In 1905, one Einstein's paper "On the electrodynamics of moving bodies" mentioned $E=pc$ for "wave", which comes from Maxwell equations, not from special relativity. Also, his paper "Does the inertia of a body depend upon its energy-content?" in 1905 had $E=mc^{2}$ for electron, not massless photon. In 1935, his paper "Elementary Derivation of the Equivalence of Mass and Energy" gave $E=mc^{2}$ with E as the function of m. Can you give me any references where Einstein defined the "particle energy" E without using m? – Yuu May 17 '21 at 17:46
  • @YuTakagi My definition of E is totally mainstream, the same as Einstein's and everybody else's. Anyhow, E=pc can already be derived Maxwell's equations. – my2cts May 17 '21 at 17:58
  • For EM "waves," yes. That can be derived from Maxwell's equations, not special relativity but Electromagnetism. What I asked here is about photons "(quantum)", which has the particle-like aspect. I was thinking that Einstein used photon quantum "hypotheses" and that was "experimentally verified," not mathematically derived from special relativity. And, that experimentally verified facts can easily derive E=pc for particle-like photons. That's why I asked here such that can we really "derive" E=pc mathematically from special relativity. I didn't say E=pc is wrong. – Yuu May 17 '21 at 19:11
  • @YuTakagaki Experimental verification requires measuring radiation pressure and as you are looking for an experimental verification per photon, radiation pressure shot noise. I don't know that it has been done. – my2cts May 17 '21 at 22:18
  • The experiments I mentioned are photoelectric effects that yields $E=h\nu$ and Compton scattering that yields $p = \frac{h}{\lambda}$. Since $c=\nu \lambda$, $E=pc$ can be easily derived without special relativity. Also, as Chang showed above, it is consistent to special relativity. This means, $E=pc$ was not derived from special relativity but consistent and invariant, likewise Maxwell's equations are not derived from special relativity. That's why I asked here why people say $E=pc$ can be derived from special relativity? Can we derive $E=pc$ without using $E=h\nu$ and $p=\frac{h}{\lambda}$? – Yuu May 17 '21 at 22:40
  • It is still true that E=pc follows from special relativity. By your reasoning E=Mc$^2$ can be derived without special relativity. You can never derive an experimental result from a theory but you can predict it. – my2cts May 17 '21 at 22:48
  • Well, I did NOT ask whether $E=pc$ is correct. I said I believe $E=pc$ is experimentally confirmed and consistent to special relativity, not derived. Maxwell's equations also consistent in special relativity but no one said Maxwell's equations are derived from special relativity. As you said, if you agree with $E=pc$ comes from experiments, then why your answer was "yes," we can derive $E=pc$ from special relativity theory? Also, I'm sorry I could not follow what you mean by saying $E=mc^{2}$ can be derived without special relativity. How did you do that? – Yuu May 17 '21 at 23:01
  • Sorry, I just saw your modified answer and noticed that I misunderstand your comments. If the each components of four-momentum of photon is defined in different way from Einstein's original one by using the results of experiments, then I agree that we can derive $E=pc$ from special relativity. Sorry, my original question has ambiguity, and thanks for your help – Yuu May 17 '21 at 23:33
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When the particle is going at the speed of light, its line interval $ds^2 = 0$, but that does not mean that any relativistic invariant is zero.

Considering, as you did, the four-velocity $\underline{U}^2 = c^2$. The four-momentum is defined by $\underline{P} = m\underline{U} = 0$ if $m=0$, so $\underline{P}^2 = 0$. Since $P^\mu = \{E/c, \vec{p}\}$, you obtain $E^2 = c^2p^2$.

Anthony Guillen
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  • Thank you for answering my question, but I could not follow your last step. What is the definition of $\vec{p}$ in $P^{\mu}$? I thought $\vec{p} \equiv m\vec{u}$ where $\vec{u} = \frac{dx}{d\tau}$ where $d\tau$ is proper time – Yuu May 14 '21 at 07:01
  • Also, I thought the first component of the four-momentum should be defined as $p_{0} \equiv E/c = m_{inv} u_{0} = \gamma m_{inv} c$ where the first component of the four-velocity is $u_{0} = c \frac{dt}{d\tau} = \gamma c$ and $\gamma \equiv \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ – Yuu May 14 '21 at 07:08
  • I guess that Chang Hexiang answered these questions in the comments below his answer :) – Anthony Guillen May 14 '21 at 17:54
  • Yes, your are right. Thanks again for your help – Yuu May 17 '21 at 23:48
  • How do you define four-velocity $U$ for it to satisfy $U^2= c^2$? Four-velocity is conventionally defined as the derivative of the four-position w.r.t. the proper time, which implies $U = ( \gamma c, \gamma {\bf v})$, which is undefined for a particle at the speed of light. $U^2= c^2$ is in general the four-velocity of a particle in its rest frame, which again is undefined for a photon. – Albert May 30 '23 at 15:22