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This is a relativity-paradox which I can't resolve:

The distance between interference lines in the double-slit experiment is:

$$\ w = z \lambda/d$$

Where:

w: Distance between fringes

z: Distance from slits to screen.

$\lambda$: Wave length of light

$d$: Distance between slits.

Now, looking at the setup from a reference frame traveling in the direction of the distance $z$ between slit and screen and moving away from the slit in direction of the screen, the light gets redshifted and the distance $z$ gets Lorentz contracted:

$$\ w' = \frac{z}{\gamma} \gamma(1+\frac{v}{c}) \lambda/d =z(1+\frac{v}{c}) \lambda/d$$

This only holds for the first fringe or so, that aren't too far from the center, because the red-shift formula changes farther out. However, it's obvious that the distance between fringes has increased!

But vertical distances shouldn't change when boosting: I could place a photodetector at the first black fringe which beeps if it gets hit by a photon. In the reference frame at rest, it doesn't beep. But when looked at from the boosted reference frame, the fringes move and light hits the photodetector. It beeps. (The photodetector will not change its vertical position in a boosted frame - Lorentz contraction only happens along the axis of movement)

Where is my mistake?

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    $\begingroup$ I'm not sure I understand the setup. You define d as the distance between the slits but then it seems like you're saying that it's the distance between the slit and screen. Is the motion in a direction parallel to or perpendicular to the plane that contains the fringes? $\endgroup$
    – Noah
    Commented Aug 22, 2013 at 20:35
  • $\begingroup$ @Noah You're right, my mistake. I corrected it: The movement is perpendicular to the plane that contains the fringes. $\endgroup$
    – yippy_yay
    Commented Aug 22, 2013 at 20:51
  • $\begingroup$ One remark: special relativity doesn't need to be invoked. This problem occurs with classical velocity transformations as well (classical Doppler effect). $\endgroup$
    – fffred
    Commented Aug 23, 2013 at 0:49
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    $\begingroup$ @fffred A double slit experiment with water waves in a tank would be expected to alter the fringes, depending on whether the setup is moving through the water or standing still in it. But this isn't the case for the double split experiment with lightwaves which shouldn't be able to see an effect of the setup "moving through the ether". $\endgroup$
    – yippy_yay
    Commented Aug 23, 2013 at 11:21

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I've changed my answer substantially after further thought.

In the frame at rest with respect to the slits, interference is determined by how many wavelengths fit between one slit and the screen compared to how many wavelengths fit between the other slit and the screen. In the frame in motion with respect to the screen, three things are different:

  1. The wavelength is altered by the Doppler effect
  2. The screen-slit distance is altered by a Lorentz contraction, and
  3. The observer claims the light hitting the screen doesn't come from where the slits are at that instant but from where they were located a short time ago. The path length for the light is different than the instantaneous separation between the slit and screen.

In this frame the length $L$ that the light travels is given by

$L = \frac{z}{\gamma} - vt $

$L = \frac{z}{\gamma} - vL/c$

$L = \frac{z}{\gamma(1+\frac{v}{c})}$

This is the same factor by which the Doppler effect alters the wavelength, so observers in both reference frames agree on the number of wavelengths that lie along the path that the light travelled.

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