| Topic: |
Science > Physics |
| User: |
"Dr Photon" |
| Date: |
08 Aug 2005 07:54:24 AM |
| Object: |
Relativistic Young's slit experiment |
Imagine a Young's double slit expt where the screen is partially
transmitting and scatters, a piece of thin paper for example. You are
travelling towards this screen from the back at near the speed of light
so you see the red light used in the expt shifted to blue. It seems
that each slit has a chromatic diffraction, ie, straight through looks
blue, while off-axis a bit seems green, off axis further seems yellow.
First question: is this true?
Next: then there will be a colour shift across the fringes, the central
fringe will appear blue, the ones either side green, the next fringes
yellow. The spacing between fringes will be equal (as they are in the
expt frame).
What happens in the single photon case? If the emitter sends out a blue
photon (in my frame, red in his), this gets diffracted to the second
fringe (yellow), then scattered in my direction, I will see a yellow
photon? Or will the act of scattering mean I receive a blue photon
again, so all fringes appear blue?
I'm fairly sure the answer is blue so long as I'm far from the screen,
but any thoughts/references gratefully received.
br
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| User: "Bilge" |
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| Title: Re: Relativistic Young's slit experiment |
08 Aug 2005 11:57:57 PM |
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Dr Photon:
Imagine a Young's double slit expt where the screen is partially
transmitting and scatters, a piece of thin paper for example. You are
travelling towards this screen from the back at near the speed of light
so you see the red light used in the expt shifted to blue. It seems
that each slit has a chromatic diffraction, ie, straight through looks
blue, while off-axis a bit seems green, off axis further seems yellow.
First question: is this true?
Yes, but your reasons for why this is true aren't very clear.
Each of the red fringes acts as a _red_ source in the plane of the
screen. The doppler shift is given by:
\nu' = \nu \gamma(1 - \beta . n) = \gamma(1 - \beta cos(A))
where \beta = v/c, \gamma = 1/sqrt(1 - \beta^2) and n is a unit vector
along the line connecting you to the source (i.e., the direction along
which the light propagates).
If you are traveling along a path perpendicular to and toward the
central fringe, \beta . n = |\beta|, so the blue shift is a maximum.
The angle increases for each fringe away from the central fringe,
i.e., A goes from 0 -> \pi/2. However, note that even at \pi/2,
there is a doppler shift (i.e., transverse).
Next: then there will be a colour shift across the fringes, the central
fringe will appear blue, the ones either side green, the next fringes
yellow. The spacing between fringes will be equal (as they are in the
expt frame).
What happens in the single photon case? If the emitter sends out a blue
photon (in my frame, red in his), this gets diffracted to the second
fringe (yellow), then scattered in my direction, I will see a yellow
photon? Or will the act of scattering mean I receive a blue photon
again, so all fringes appear blue?
I'm fairly sure the answer is blue so long as I'm far from the screen,
but any thoughts/references gratefully received.
Treat the fringes as sources. Why should it matter if you replace
the fringe with a light source?
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| User: "Dr Photon" |
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| Title: Re: Relativistic Young's slit experiment |
09 Aug 2005 05:06:57 AM |
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Bilge wrote:
However, note that even at \pi/2,
there is a doppler shift (i.e., transverse).
yes! I think that is what I was trying to get a handle on (in a
round-about way!) - how to deal with the transverse doppler shift, and
the time dilation associated with it
http://www.kineticbooks.com/physics/17467/17506/sp.html
With the screen stationary wrt the slits, I will approach the screen
where the fringes are all red (in their rest frame), but will be
transverse doppler shifted wrt me. So they will have different colour
depending on their angle wrt my motion.
If I hold the screen in front of me at a fixed distance wrt me while
travelling towards the slits, then each point on the screen will see
the slits transverse doppler shifted, and so will still have different
colours. In this case, as there is no motion of the screen wrt me, then
they will be the colours I see.
phew, think I'm happy now.
thanks,
br
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| User: "Dr Photon" |
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| Title: Re: Relativistic Young's slit experiment |
09 Aug 2005 05:45:09 AM |
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Dr Photon wrote
phew, think I'm happy now.
just one last thing, which also got me started on this thread, about
modifications needed to Young's equation when describing a set-up
moving towards you at high velocity. Nothing revolutionary about that
I'm sure, if you length contract the distance from slits to screen,
while also length contracting the wavelength of light, the fringe
separation changes according to the plain formula lambda=y.d/ (m.L)
where y is the distance between m fringes, when it clearly shouldn't.
That was clear but how to account for the transverse doppler shift had
me temporarily confused.
br
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| User: "Ian Parker" |
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| Title: Re: Relativistic Young's slit experiment |
08 Aug 2005 08:05:38 AM |
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If you are going styraight towards the slit the light will look blue,
no real difficulty and no argument. If you are off axis the light will
still appear blue since your point of reference will be the source and
not the slits.
Looking at the problen from the point of view of conservation of energy
it is clear that all the photons are received with energy and momentum
set accourding to SOURCE. This is so since energy and momentum would no
longer be conserved.
Remember too that there will be time and distance dilation. The slits
will appear to be close in the moving frame.
Does this help to answer your question
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| User: "Dr Photon" |
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| Title: Re: Relativistic Young's slit experiment |
08 Aug 2005 08:26:32 AM |
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Ian Parker wrote:
If you are going styraight towards the slit the light will look blue,
no real difficulty and no argument. If you are off axis the light will
still appear blue since your point of reference will be the source and
not the slits.
Note that I am looking at light scattered from the screen, not at the
slits directly. Hence the light has travelled at high angle to reach a
lateral shift a few fringes to the side on the screen. Then it is
scattered back to the "straight through" direction. During the
high-angle path from the slits to the screen, the light will appear
colour shifted in my frame.
Looking at the problen from the point of view of conservation of energy
it is clear that all the photons are received with energy and momentum
set accourding to SOURCE. This is so since energy and momentum would no
longer be conserved.
Conservation of energy was what had me wondering. If (in my frame) the
photons start out "really" blue, then "really" become yellow, then are
scattered as either blue or yellow, what is the energy/momentum doing?
Remember too that there will be time and distance dilation. The slits
will appear to be close in the moving frame.
Closer to me yes (x-direction let's say), but they will appear the same
distance from eachother in either frame, as this is a lateral
direction. Time dilation also has my interest, ie, is the yellow "real"
and so can get scattered from the screen, or is it simply a Doppler
shift that can be shifted up or down depending on direction. I find
this interesting in the single photon regime.
Does this help to answer your question
getting there...
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| User: "Ian Parker" |
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| Title: Re: Relativistic Young's slit experiment |
08 Aug 2005 10:03:37 AM |
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Note that I am looking at light scattered from the screen, not at the
slits directly. Hence the light has travelled at high angle to reach a
lateral shift a few fringes to the side on the screen. Then it is
scattered back to the "straight through" direction. During the
high-angle path from the slits to the screen, the light will appear
colour shifted in my frame.
Relativity says that everyone is stationary in their own frame.
Therefore light that is scattered by a sheet of paper maintains its
color so long as you are moving with the paper. As I have said looking
from any point on the paper we are looking at the source not at the
slits.
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| User: "Dr Photon" |
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| Title: Re: Relativistic Young's slit experiment |
08 Aug 2005 12:31:16 PM |
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Ian Parker wrote:
Relativity says that everyone is stationary in their own frame.
Therefore light that is scattered by a sheet of paper maintains its
color so long as you are moving with the paper.
Therein lies my problem, you are approaching the paper at high
velocity.
Thinking about this from a purely Doppler point of view (as in sound
waves), I guess I have no problem. Each point on the screen re-radiates
at the same frequency as the next. If I am far from the screen, they
are all coming from pretty much the same direction (small angular
spread in front of me), so all get blue shifted by pretty much the same
amount. As I approach close to the screen, travelling on-axis wrt the
slits, then the off-axis angle to each fringe gets larger, and the
off-axis Doppler shift is not so great, so each fringe looks a
different colour (different sound frequency if using sound waves, less
blue shift). Once I haved passed over the screen, then the whole setup
red shifts as I recede from it. Once far from the screen on the far
side, everything is red-shifted by pretty much the same amount, so all
the fringes look red (infra-red).
However, with light I am just a little unsure if exactly the same thing
happens.
If I am travelling on-axis and am at the central fringe, then the light
is undoubtedly blue. If I were displaced a few fringes to the left, at
the screen, and still travelling parallel to the axis, the light would
appear yellow. My question I guess is how "real" is that? Does the
screen pick up this effect, and scatter it? Now I come to think about
it, it won't if it is stationary wrt the slits, but what if I am
holding the screen up ahead of me at a fixed distance as I travel
towards the slits? Will the outer fringes appear yellow?
As I have said looking
from any point on the paper we are looking at the source not at the
slits.
br
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