I set up a usual photoelectric experiment in a laboratory that is movable. Let's say, when stationary, the metal I am working with has a large work function and the light I am shining cannot produce photons of sufficient energy. What if I now make the lab move at a relativistic speed and change nothing else about the experiment. Due to length contraction effects, will the energy of the photon now go up and photoelectric emission be observed? But that would mean physics is different in different inertial frames. What am I missing here? Thanks for any suggestions.

  • $\begingroup$ Do you co-move the light source with the rest of the experiment? $\endgroup$
    – A. P.
    Commented Dec 23, 2023 at 21:57
  • $\begingroup$ @A.P. I interpret "I set up a usual photoelectric experiment in a laboratory that is movable. /.../ What if I now make the lab move /.../" as that all of the lab equipment - light source and metal - move together. $\endgroup$
    – md2perpe
    Commented Dec 23, 2023 at 22:05
  • $\begingroup$ The lab you are working in is always considered stationary (you are moving with it). See rest frame. $\endgroup$
    – joseph h
    Commented Dec 23, 2023 at 22:07
  • $\begingroup$ Not sure why this has 3 downvotes, it's a decent question. It would help if you wrote down some four vectors in the lab frame, so we could then show you they are covariant. $\endgroup$
    – JEB
    Commented Dec 24, 2023 at 7:11
  • 1
    $\begingroup$ @JEB then we agree. This is an SR anti-pattern, and should be dealt with before attempting an answer. Any question that starts with "length contraction" as an initial premise, with no consideration of the Lorentz Transform, falls under the label "gamma slinging" to my way of thinking. $\endgroup$
    – m4r35n357
    Commented Dec 24, 2023 at 17:03

5 Answers 5


The meaning of saying, "The physics is the same in all inertial frame", means that the equation of motion does not change under Lorentz transformation (boost or rotation). But the hamiltonian is not conserved, hence the energy measured is different in every frame.

The work function is $\phi = h\nu_o$ is dependent on the nature of metal. The confusion here is with the doppler effect on the emitted light since if $\nu_1$ is observed in frame S, it can be $\nu_1'$ in frame S'. The case where the moving frame can observed higher frequency of light is when the movable stuff move towards the source since, $\nu = \nu_0 \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$. So Yes, it is possible to observed photoelectric emission.

In the moving frame S', one can observed photoelectric emission because in that frame, one can observed higher frequency of light emitted. In the stationary frame S, one can also observed photoelectric emission because with respect to that frame, the S' is moving at speed $v$ towards the stationary source which produce a light with frequency $\nu_1$,hence from the point of view of S, you are interacting with the light in an "altered frequency (greater) not equal to $\nu_1$ " because of your velocity $v$ towards the source.


For example, if the lab is in a galaxy so far away that it is receding from us, and is red-shifted, the photons of the experiment would be also red-shifted. The same for the H-spectra, but we don't say that the law of nature changes for different frames due to Doppler effect.

In this case, the photons are blue-shifted, but the reasoning is similar.


Let us consider a pulse of light, produced by a normal flashlight, said light pulse collides on a mirror that hangs on the front wall of a lab. The lab is moving in frame $F1$, and static in frame $F2$. (The flashlight is in the lab and co-moves with the lab)

In $F1$ light does work $F*d$ on the mirror, where d is the larger the faster the lab moves.

In $F2$ light does work $F*d$ on the mirror, where d is zero.

In $F1$ it is known that no part of the energy $F*d$ can be used to melt the mirror, because in $F2$ there is no such energy, but the melting of mirror would be noticed in $F2$.

It happens to be so that an object that moves distance $d$ while being pushed by force $F$, gains kinetic energy $F*d$. As that energy must be used to increase kinetic energy, it can not be used to anything else, like melting the object.

So everything works nicely in this case.


Kinetic energy does depend on the reference frame, whether it is for photons or massive particles. This is true even in classical physics. If you move towards a bullet, it impacts with more energy, if you move away it impacts with less.


Photons transport a quantum of energy from A to B at the speed of light. The amount of photon energy previously emitted by a subatomic particle depends on the prevailing gravitational potential at the point of emission. If the potential is higher than ours, the spectral lines are red-shifted, if lower, they are blue-shifted.

However, your test body is not very interested in where the photon comes from. The only thing that matters is its instantaneous energy content on impact AND the relative speed between the sample and the photon.

In case anyone has any objections, there is an example where the relative speed between the body and EM radiation is very relevant. In laser cooling of gases to achieve very low temperatures, the absorption of photons depends on the direction of movement of the molecules. If the particle moves towards the laser beam, the energy of the photon is sufficient to be absorbed.

What if I now make the lab move at a relativistic speed and change nothing else about the experiment.

Referring to the last paragraph, the photoelectric effect depends only on the energy of the photons used AND the velocity of your sample relative to the photon.

The energy of the photons used, however, depends on the relative velocity to your experimental setup. If both move together through space, nothing changes in the photoelectric effect. If both move away from each other, the source (in its rest system) must emit higher-energy photons. If they are moving towards each other, they must be lower-energy photons.


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.