Watching young's double slit experiment in moving frame

[Sourabh N]

Suppose an observer is moving at some constant velocity(<< c) and observing the young's double slit experiment. What changes will it observe compared to an observer at rest? This is what i could make out :
1) The wavelength of incoming light will change (doppler effect)
2) By the intuition that interference pattern will be the same, the moving slit somehow cancels the doppler effect, hence same pattern for the both the observers.

also how will things change if its an electron beam.

 

[Sourabh N]

anyone?

 

[cesiumfrog]

The space between slit and screen will be contracted.

 

[atyy]

The double slit formula doesn't depend on frequency, only on wavelength. Newtonian physics would predict that the wavelength is constant, so the speed of light must change to compensate for the frequency change. So Newtonian physics makes the intuitively correct prediction, as long as you don't measure the speed of light.

I have no idea what the experimental result is, since the real world is special relativistic.

 

[i_island0]

As far as i know, wavelength changes only when the source moves (if v<<c). Motion of observer doesn't change the wavelength. Thus, no change in fringe pattern will be observed.

 

[Sourabh N]

So, using doppler effect for wavelength ( $$\lambda'$$ = $$\sqrt{\frac{1+\beta }{1-\beta}}$$ $$\lambda$$)
and length contraction for the space between slit and screen(D' = $$\frac{D}{\sqrt{1-\beta^2}}$$),

i got w' = (1 + $$\beta$$) w
where w is the fringe width for the stationary observer and w' for the moving chap.

Here the effects don't cancel out, please correct me.

 

[i_island0]

your point is correct only iff we take 'v' to be comparable to 'c'. But in your question in the beginning you have mentioned that v<<c

 

[Sourabh N]

I agree that in non-relativistic case, the pattern won't change. But in the relativistic case, let's say I mark the peaks (seen from the stationary frame), and then I perform this experiment, will the peaks come closer relative to the marks? The formula suggests it to be so.

 

[cesiumfrog]

The Doppler shift varies directionally.

 

[bernhard.rothenstein]

Suppose an observer is moving at some constant velocity(<< c) and observing the young's double slit experiment. What changes will it observe compared to an observer at rest? This is what i could make out :
1) The wavelength of incoming light will change (doppler effect)
2) By the intuition that interference pattern will be the same, the moving slit somehow cancels the doppler effect, hence same pattern for the both the observers.

also how will things change if its an electron beam.

I think that you could mention how is located the line that joins the two slits. If it is normal to the direction of relative motion the the distance between them is unchanged and we deal with a longitudinal Doppler shift of the two incident rays. Other scenarios are also possible which are more complicated.

 

[atyy]

The derivation of the positions of the interference maxima requires that the path difference to an interference maximum be some integral multiple of 2*pi (or something like that). The path differences to different orders of maxima contain different amounts of components parallel and transverse to the direction of motion. Perhaps this should be taken into account.

 

[atyy]

Looks like the non-central maxima will also involve a mix of transverse and longitudinal Doppler effects?

 

[Dale]

In any experiment involving waves what is measured is the phase of the waves at a given point, so the phase must be a Lorentz invariant. In the case of the double-slit you get a bright band when the waves from the two slits are in phase and a dark band when they are out of phase. The phase is given by: φ = ωt-k.x but since (t,x) is a four-vector the phase can also be written φ = (ω,k).(t,x). Since φ is a Lorentz invariant, and since (t,x) is a four-vector, then (ω,k) is manifestly also a four-vector and is known as the http://en.wikipedia.org/wiki/Wave_vector" .