I was wondering as how would appear the interference pattern of a double-slit experiment moving at a relativistic speed v, 1) in the case of light and, 2) in the case wave matter (i.e. electrons for instance) as seen in comparison between rest and moving reference frames (respectively S and S’).
More precisely, I am considering the spacing between fringes and their nature. In terms of the spatial distribution of the individual speckles observed on the screen, my first impression was that it must be exactly of the same nature, i.e. fringes being centered at same screen positions (the y axis being transversal to the relative movement). Doing the usual simple math (see details below) to obtain fringes positions and spacing, and using the longitudinal length contraction L’ = L / G(v) (where G(v) is the usual gamma factor G(v) = 1/(1 – (v/c)**2)**0,5) gives me finally an equation relating the respective wavelengths in S’ and S :
Lambda’ = G(v) Lambda
And then, I have the following two questions:
1) In the case of light (with Lambda = c/f), this would imply that f’ = f / G(v) (see details below). Then is it true that a red shift must always be taken into account due to a Doppler effect acting only transversally?
2) Since for a matter wave, the de Broglie wavelength is Lambda = h/p, the observed pattern spacing would be maintained if p’ = p / G(v) (see details below). But I doubt seriously that there is such a simple relativistic transformation between the respective linear momenta of rest and moving frame (the correct composition of momenta is different). What are the errors then?
Thanks for your patience. I would clearly appreciate the help of someone.
Details of the simple math:
Starting in the frame S with the walk difference delta(m) for a maximum of intensity observed at fringe of order m on the screen (along the y axis):
delta(m) = r2 – r1 = m Lambda and d/L -> epsilon
=> y(m)/L = tan theta -> sin theta = delta(m) / d = m Lambda / d
=> position of fringe m: y(m) = m Lambda L / d
and spacing : Dy = y(m+1) – y(m) = Lambda L / d
The same shall be true in the frame S’: position of fringe m: y’(m) = m Lambda’ L’ / d and spacing Dy’ = Lambda’ L’ / d. Then, the length contraction, together with the requirement that the spacing must remain the same give:
L’ = L / G(v) with Dy = Dy’ => Lambda L / d = Lambda’ L’ / d = Lambda’ L / (G(v) d) => Lambda’ = G(v) Lambda
1) In the case of light having a constant speed
c = Lambda f = Lambda’ f’ = G(v) Lambda f’ => f’ = f / G(v).
which is the transversal Doppler Effect with the resulting red shift of the fringe colors. This seems a priori reasonable since the moving experimental setup is observed transversally as it moves (from left to right), thus, there is no need to have a longitudinal Doppler Effect. However, this appears somehow in contradiction with the fact that the walk difference delta’(m) = m Lambda’ is drawn along a diagonal (at angle theta’) and has also a longitudinal component. What does that mean in terms of possible x and y components of (longitudinal and transversal) Doppler effects? What is missing or wrong here (either in the calculation or the interpretation)?
2) In the case of wave matter (electrons) having the de Broglie wavelength Lambda = h/p, things become more complicated since one should suppose for the electron beam a speed u in the reference frame S and a speed w in S’ being related through the relative speed v of S’ by the subtractive relativistic composition of speeds (with u > v):
w = (u – v)/(1 – uv/(c**2)) with the well-known geometric identity
w G(w) = (u – v) G(u)G(v)
Then, the relation between wavelengths leads to a relation between p = p(u) = G(u) m0 u and p’ = p(w) = G(w) m0 w (where m0 stands for the electron rest mass and G for the relativistic gamma factor):
Lambda’ = G(v) Lambda => h/p(w) = G(v) h/p(u) => p(w) = p(u)/G(v)
which is surprising and possibly too simple to be true. Indeed, using the geometric identity w G(w) = (u – v) G(u)G(v), one can see that the correct composition of momenta for p’ = p(w) should read as :
p(w) = G(w) m0 w = G(w) m0 (u – v) G(u)G(v)/G(w) = m0 (u – v) G(u)G(v)
= G(v) G(u) m0 u – G(u) G(v) m0 v = G(v) p(u) – G(u) p(v)
And again, as it was the case for the light situation, one should observe that these momenta are not necessarily longitudinal since they are related to the walk difference delta’(m) = m Lambda’ which is drawn along a diagonal (at angle theta’). So, what is wrong in this logic and in these probably too simple calculations?
Thanks for your help!
Wilj
Best Answer
First of all, light waves and matter waves may be treated together, using the same maths, because the waves associated with light and the waves associated with matter are fundamentally the same thing.
Second, all the waves before they interfere and after they interfere may be written in terms of the probability current $j^\mu (x,y,z,t)$, and its transformation from one frame to another is obtained by a simple Lorentz transformation of the coordinates $(t,x,y,z)$. In the non-relativistic limit, the Galilean transformation will work as a good approximation of the Lorentz transformation.
Third, whenever a wave – locally or globally – has a well-defined energy-momentum vector of the corresponding particle (whether it's a photon, electron, or something else), then the wave has a simple plane-wave form $$ \psi = \exp (ip^\mu x_\mu) $$ At least the frequencies and wavelengths are determined by this simple complex exponential. Note that $x_\mu$ and $p^\mu$ transform according to the Lorentz transformation, in the same way, and their inner product – the argument of the exponential – is therefore invariant under Lorentz transformations which is why the phase above is invariant, too. One doesn't need to make "explicit" checks; it is clear that the predictions for the locations of the interference minima and maxima will be the same in all reference frames.
Now, if the momentum in the direction between "slit and the photographic plate" which I will call the $z$-direction is $p_z$ and the total energy is $E$ for the interfering particle, then $(E,p_z)$ is transformed by the standard Lorentz transformation when boosted by the velocity $v$. This transformation of $(E,p_z)$ to $(E',p'_z)$ is equivalent to the calculation of the velocity $V$ of the (wave-represented) particle $$ V = \frac{p_z c^2}{E} $$ (the denominator is the total energy so in the non-relativistic limit, it has to be replaced by $m_0 c^2$ and not by $mv^2/2$) and composing this $V$ with the $v$ by the relativistic composition formula, $$V'=(V+v)/(1+Vv/c^2)$$ For photons and other particles moving at the speed of light, the speed is $V=c$ and we get the "new" speed as $V'=c$ again. That's not too informative because for photons, the frequency or energy is not encoded in the speed (the latter is always $c$).
The transformation of $E,p_z$ – that obey $E=|p_z c|$ for $z$-directed photons – under the Lorentz transformation given by the motion at speed $v$ is given by the Doppler shift – both the momentum and the frequency get rescaled by $$ \sqrt{ \frac{c+v}{c-v} }$$ The wavelength in the $z$-direction gets contracted by this factor. However, the distance between the slits and the plate gets contracted as well, and the time gets shortened because the plate is moving "against" the photons. These three modifications are consistent with each other because the number of wavelengths between the slit and the plate gets multiplied by $$ \frac{ {\sqrt{1-v^2/c^2}} }{ \sqrt{ \frac{c+v}{c-v} } } = 1-\frac vc$$ which is exactly the factor by which the time spent between the slit and the plate will shorten because the plate is moving against the photons.
Let me emphasize once again that if we only change the speed in the $z$ direction, the transverse momentum components $p_x,p_y$ and the corresponding components of the wave vector $k_x,k_y$ will remain unchanged. That's true for relativity and non-relativistic physics, massless as well as massive particles.