I was thinking about the Photoelectric Effect and Blue-Shifting when I came up with a thought experiment that I couldn't think of an answer for. The thought experiment is as follows:

A metal plate is set up to measure the photoelectric effect. Travelling toward the metal plate is a space-ship travelling at near C. The space-ship is emitting light that is of a frequency slightly below the threshold frequency of the metal plate.

From what I understand, from the metal plates frame of reference the light emitted by the space-ship would undergo Blue-Shifting and hence have a higher measured frequency, exceeding the threshold frequency of the metal plate and causing a current to be measured.

However from the space-ship's frame of reference the light would not undergo Blue-Shifting and thus would be below the threshold frequency of the metal plate, thus a current would not be measured.

Could someone please explain where I've gone wrong, and what would actually happen in this situation.

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    $\begingroup$ From the spaceship's point of view either the metal's energy threshold for releasing a charged particle would change, or there would have to be a change of electromagnetic boundary conditions between the metal surface and the vacuum. I am not making this an answer, because I don't feel like analyzing the situation in microscopic detail, but it can't be any other way. $\endgroup$
    – CuriousOne
    Aug 28, 2014 at 4:29
  • $\begingroup$ Check out en.wikipedia.org/wiki/M%C3%B6ssbauer_effect for a similar situation. $\endgroup$ Aug 28, 2014 at 11:30
  • $\begingroup$ As I said in my answer - it only makes sense to define the work function in the rest frame of the plate. The threshold frequency will be Doppler shifted according to the standard transformations. In other words you could use the fact (in your thought experiment) that a photoelectric current is generated to tell you that the relative velocity of the plate and spaceship exceeds a certain value. $\endgroup$
    – ProfRob
    Sep 18, 2014 at 7:09
  • $\begingroup$ You may also want to check out Doppler cooling en.wikipedia.org/wiki/Laser_cooling#Doppler_cooling $\endgroup$
    – suresh
    Sep 18, 2014 at 7:16
  • $\begingroup$ I think CuriousOne is right: after all when we study quantum mechanical treatment of the photelettric effect, for example in order to determine the threshold, we always put ourselves in the atomic nucleus frame of reference. $\endgroup$ Sep 22, 2014 at 22:52

1 Answer 1


I believe CuriousOne is correct, however it does not make any sense to define the threshold frequency for the photoelectric effect in anything but the rest frame of the metal.

From the rest frame of the spaceship, the metal plate is rushing towards it and the apparent threshold frequency is lowered, but an occupant of the spaceship should realise that this is just because the light they are emitting is blueshifted to a higher frequency in the rest frame of the metal.

An analogous situation in astrophysics would be the absorption of the light from a high redshift quasar by clouds of intervening gas at a variety of intermediate distances and redshifts. Each of these clouds is travelling at a different velocity with respect to the light from the quasar and absorbs light due to electronic transitions in gas atoms at apparently different wavelengths in our rest frame, but at the similar wavelengths in their own respective rest frames.

In X-ray astronomy, the redshift of the photoelectric absorption edges of various metal ions is frequently taken into account for extragalactic sources.

So, quantitatively: In the rest frame of the metal, nothing changes.

From the rest frame of the spaceship - if we define $v$ as the relative velocity of the spaceship and metal plate ($v$ is positive when they are moving towards each other), and we define $W'$ as an "effective observed work function" as judged from the space ship (compared with $W$ in the rest frame of the plate), and $\nu'$ as the corresponding threshold photoelectric frequency. Then:

$$\nu' = \nu \left(\frac{c-v}{c+v}\right)^{1/2}$$ and $$W' = W \left(\frac{c-v}{c+v}\right)^{1/2}$$

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    $\begingroup$ Very nice. The concept of an apparent work function is interesting. I wonder if one could come up with a good explanation. Maybe the greater mass of the electrons changes the momentum exchanged during the interaction? $\endgroup$
    – Floris
    Sep 22, 2014 at 23:31

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