Classical aberration of plane waves

Discussion in 'Physics & Math' started by tsmid, Apr 8, 2017.

  1. tsmid Registered Senior Member

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    We know that the path of a particle moving along the axis of one reference frame is inclined by some angle against this axis in a reference frame moving relatively to the first one with a certain speed transversely to the particle velocity. This is simply a result of the coordinate transformation involved.
    However, I was wondering recently how the aberration can be understood in case of waves. Consider the following scenario

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    We create a locally plane parallel wave-front in a stationary medium (v=0) which propagates vertically upward with speed c (I indicated 4 positions here). Now we put the medium in motion with a velocity c to the right. The vertical speed c will obviously be unaffected by this, but the horizontal position of the wave front will be progressively shifted to the right. However, for small velocities v, this should be unobservable: I have schematically indicated the tube of a detector (e.g. telescope) in red, and it is obvious that the wavefront progresses through the detector exactly identically in both cases, because in either case it is exactly horizontal and travels upwards with speed c. So how can we understand the aberration here?

    Thomas
     
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  3. Q-reeus Valued Senior Member

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    Suggestion: actually specify ALL relevant factors. For instance, I have no idea what kind of wave is being generated in what kind of medium. EM, acoustic, in vacuum, air, water, solid matter, - what?
    Is the observer outside of or immersed in said medium? If not, what is the nature of the medium of the observer? I could guess air or vacuum, but that's just a guess.
    What can be said without further info is that your assertion "The vertical speed c will obviously be unaffected by this" is at best a 1st order approximation that ignores SR. Which may be ok if acoustic waves are being considered, but will fail miserably if light or similar EM waves are implied, and v (you made a typo and used c first up), is an appreciable fraction of c.

    Have you not first read through e.g. the Wikipedia article on the topic?: https://en.wikipedia.org/wiki/Aberration_of_light
     
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  5. tsmid Registered Senior Member

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    Hi Q-reeus,

    Thanks for your reply. It doesn't matter what kind of medium it is. As indicated in the title, I am referring in the first place to classical waves though. For the sake of the argument, assume for instance you dip the edge of a long board into a reservoir of resting water, causing a wave propagation to travel away with speed c from you. Then do the same with now the water running with speed v to the right. As illustrated in the above diagram, the localized disturbance will become progressively displaced to the right, but the wavefront would still be aligned in the same direction (and thus by definition the propagation direction of the wave as well).

    Thomas
     
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  7. Q-reeus Valued Senior Member

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    In that example of surface water waves SR is not relevant. For a point source, with v < c, it's obvious the wavefront will no longer be circular about the source but as shown here: http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/dopp.html
    In that case the velocity component of a wavefront radial to the source and normal to v will remain c, but not in any other direction as there is simple addition of velocities applying i.e. v + c(θ).
    A continuous linear array of such point sources with equal phase would generate a uniform wavefront parallel to the source array axis and propagating with a radial velocity component |c(r)| = c, whether or not v is non-zero. Instantaneous velocity of the wavefront is obviously the simple vector sum of v + c(r).
    Assuming awareness of the Fizeau experiment, you will know the situation is quite different for light propagating in a moving medium: https://en.wikipedia.org/wiki/Fizeau_experiment
     
  8. rpenner Fully Wired Staff Member

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    If f(t,x) is a wave that keeps its shape when moving with velocity v, then f(T, X) = f(T+t, X+vt). So f(t,x) = f(vt-x). If f(t,x) is a periodic wave then it repeats in space and time so f(T,X) = f(T + n P, X + m L) for any integers m,n and a certain period P and a certain wavelength displacement L. Putting these together, we have v= L/P or f( (L/P) t - x ) = f( (L/P) t - x - n L).

    Sine and Cosine functions have periodicity of 2π as does the exponential of an imaginary quantity, so we have ω = 2 π/P, k =L 2 π/L^2, defining phase frequency, ω, and wave vector, k. Exp( i ( ωt - k • x)) = Exp( 2πi ( t/P - L•x/L^2)) describes a simple plane wave (filling all space) and moving with velocity v as before.

    So now that you can model any wave, how does it transform under a Lorentz transform?

    Simply solve v t - x = v' t' - x' for v' in light of the proper Lorentz transformation. If you want to be fancier you could solve for P and L or ω and k but the v' you compute from those pairs is the same as for v' computed directly and is the same for waves and particles.
     
  9. tsmid Registered Senior Member

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    I don't know why you think the Doppler effect would apply here. We consider a wave propagation perpendicular to the flow velocity of the medium. Assume you create a water wave from a river bank (a locally plane wave with the wave front parallel to the river bank). Does the time it takes the wave to reach the other bank depend on the flow velocity of the river or not, and is the wavefront still parallel to the river bank or not?

    Thomas
     
    Last edited: Apr 9, 2017
  10. rpenner Fully Wired Staff Member

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    Perpendicular in which reference frame? The speed of the wave is constant relative to the inertial frame where the water is at rest. So necessarily a wave which crosses a channel of water width L where the speed of wave in still water is u and the speed of the water is v is moving "upstream" at an angle in the water's rest frame. Galilean relativity gives the relation:
    t = L / sqrt ( u^2 - v^2 ) so that in water's rest frame the wave moves L v / sqrt ( u^2 - v^2 ) upstream and L cross-channel. This movement of v upstream and sqrt ( u^2 - v^2 ) perpendicular to upstream give a speed of u at an angle of arctan (v/sqrt ( u^2 - v^2 ) ) in the water's rest frame and a speed of sqrt( u^2 - v^2 ) and angle of 0 in the frame where the water moves at speed v. If v^2 >= u^2, the flow is supersonic and the wave can't make it.
     
    Last edited: Apr 10, 2017
  11. rpenner Fully Wired Staff Member

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    The differences in Special Relativity are 1) there is no medium in the vacuum, and 2) the speed of light in vacuum is the same for all inertial frames, 3) the elapsed time is not required to be the same between two events for all observers.

    So Galilean relativity suggests the angle of aberration for a sideways-moving telescope should be arctan (v/sqrt ( u^2 - v^2 ) ), but the speed of propagation should be sqrt ( u^2 - v^2 ) so setting that equal to c, we get arctan(v/c).
    But Special Relatibity says the angle of aberration for a sideways-moving telescope should be arctan (v/sqrt ( c^2 - v^2 ) ), but the speed of propagation should be c so setting that equal to c, we get arcsin(v/c).

    The Taylor series for arctan(v/c) (in radians) starts v/c - (1/3) v^3/c^3 + ...

    The Taylor series for arctan(v/c) (in radians) starts v/c + (1/6) v^3/c^3 + ...

    So the leading term in the difference between these is (1/2) v^3/c^3.

    So for v/c ≈ 10^-4, this predicts a difference of angle of 5×10^-13 radians or 10^-7 arcseconds.

    Clearly this is not the best angle to test between the theories. At an angle of 45 degrees, instead of 90, the difference between the aberration formulas is as much as 2.5×10^-9 radians or 0.5 milliarcseconds. That's at the limit of human positioning technology, I believe, by looking at Hipparcos data.

    Indeed, that's been done http://adsabs.harvard.edu/full/1997ESASP.402...49F
     
    Last edited: Apr 10, 2017
  12. tsmid Registered Senior Member

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    In the reference frame of the river bank obviously (in the water's reference frame there is no flow velocity). And my question was whether in this reference frame the orientation of the wave front depends on the flow velocity or not.

    In the water's rest frame there is no flow, so how should the wave be affected by it?
    In the water's rest frame, the propagation direction is once and for all set by the initial disturbance of the water molecules, and if that happens to be a line of simultaneous disturbances at one river bank, then the resulting plane wave should reach the opposite river bank after time t=L/u (i.e. independent of v). The only thing that would depend on v is the location where the wave would run ashore, but the wavefront would still be parallel to the shore (assuming a straight river obviously).

    Thomas
     
  13. Neddy Bate Valued Senior Member

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    Isn't this just another example of "relativity of simultaneity" affecting the way the two reference frames record the simultaneity of events?
     
  14. Q-reeus Valued Senior Member

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    As per #4 there is aberration for a point source but not for a (very long) line source in your water wave example which safely ignores relativistic corrections. So your above reasoning holds. If SR matters then as hinted at in #10, simultaneous signal propagation of a line source in the water rest frame will NOT translate into simultaneous propagation in the 'riverbank' frame, and vice versa. Hence there will be aberration i.e. wavefront at an angle even for a line source. Which is a necessary consequence of the constancy of (in vacuo) c in any frame. So then just work from that given in #5.
     
  15. Q-reeus Valued Senior Member

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    Erratum: In #11 'riverbank frame' should have read 'river frame'. In the riverbank frame, water wavefront remains parallel to the line source (assumed parallel to both river banks) regardless of relativistic considerations and there is no progressive phase differential along the line source in that frame.
     
  16. tsmid Registered Senior Member

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    As emphasized earlier, I am only interested in the classical (i.e. Galilean) aberration theory here, not any relativistic effects (which would only differ in second order anyway).
     
  17. tsmid Registered Senior Member

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    I am not assuming the constancy of c here. I am only considering classical aberration i.e. Galilean velocity addition.
     
  18. tsmid Registered Senior Member

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    I am indeed only interested in the situation in the riverbank frame. And a line source parallel to the riverbank should causes a water wave front also parallel to the riverbank, regardless of the flow of the water. Otherwise we would end up with a situation like in the right part of the diagram below, which would imply that the water near the points A and B on the riverbank is disturbed at different times, in contradiction to the assumption. I can't see that anything but the situation shown in the left part of the diagram should result, that is a shear of the wave front along the flow velocity but not a tilt (aberration). Even that shear would not be observable for an ideal plane wave i.e. an infinite long wave front.

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  19. Neddy Bate Valued Senior Member

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    Okay then, that makes things a bit easier.

    When v>0, the reference frame in which your medium is stationary would record your red detector tube as moving to the left with velocity -v. So that reference frame must record the velocity of the waves inside the tube as having that same leftward horizontal component. The vertical component of those waves' velocity would be c in the vertical direction.

    The resultant velocity of the waves (combining both the horizontal and vertical components) would then be something greater than c, with its direction pointed on an angle in the same direction that the red detector tube is moving. Thus the waves are suffering aberration, which is that angle I just mentioned, when recorded from the reference frame of the medium, (even though the wave fronts inside the tube remain perpendicular to the walls of the tube).

    Note that the aberration in the above case would be caused by the motion of the detector, not the motion of the medium. The opposite would be the case when recorded from the reference frame of the detector.
     
    Last edited: Apr 11, 2017
  20. Neddy Bate Valued Senior Member

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    To make things easier to understand, you could leave the velocity of the medium v=0, and just consider a reference frame which moves horizontally past the whole arrangement at velocity v>0.
     
  21. Q-reeus Valued Senior Member

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    Sure the left illustration has it right. Considered as an infinite and infinitesimally spaced linear array of point sources, the individual abberations from each point source cancel each other overall. A point source is more interesting because one considers all propagation directions about the source, and it can go from that Doppler pattern linked to in #4, to a Mach cone for v > c. Even for the latter, a line source still has the wavefront parallel to both banks.
    Of course all this assumes laminar flow always applies but in practice, turbulence will likely rear it's ugly head especially for point source and v > c case.
     
  22. tsmid Registered Senior Member

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    I agree basically with that, but you can't call the resulting lateral displacement of the wavefront an 'aberration'. If it would be an aberration, you could restore the original situation by tilting the detector by the corresponding angle, but then the wavefronts would here not be perpendicular to the walls of the tube anymore (as illustrated in the right part of the below diagram). They would only stay perpendicular if you leave the tube in the original vertical position. Otherwise the phase of the wave would vary across the cross section of the tube and the observed object would not be in the center of the field of view anymore.

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  23. Neddy Bate Valued Senior Member

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    If your plane waves were only about as wide as the red detector tube, (for example if the waves first passed through an aperture), then, for the case of v>0, the waves would run into the side wall of the vertical tube, and might never successfully reach the far end of the tube, (neglecting reflection of the waves off the tube wall). The wavefronts would still be perpendicular to the side walls of the vertical tube, but the waves would be absorbed before reaching the far end.

    In that case, in order for the waves to reach the far end of the tube, the tube would have to be oriented on an angle, and this would demonstrate the aberration. This would be so even though the wavefronts would no longer be perpendicular to the walls of the tube anymore. So, in a way, your arrangement is 'hiding' the aberration by having the plane waves much wider than the detector tube.
     
    Last edited: Apr 13, 2017

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