[math-fun] Reflection from a moving mirror
This is a "test your physics intuition" puzzle. When light reflects from a stationary mirror, we know that (1) the angle of reflection equals the angle of incidence, and (2) the frequency of the reflected light equals the frequency of the incident light, i.e. there is no Doppler shift. Does any of this change if the mirror is moving with a velocity parallel to its surface? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
My intuition says that it will. However, suppose that the mirror is sliding in the +y direction, and we shoot a photon at it at an angle from the normal (x direction); if we let f be the frequency of the light, traveling at (f,f*x,f*y,0) (in the +x,+y direction, with x^2 + y^2 = 1 and the incident angle having tangent y/x) in the stationary reference frame, and consider the mirror to be moving with velocity vector (gamma, 0, gamma*v, 0) (in the +y direction), and *assume* that the mirror is like a rocket that absorbs a photon and reemits it with the same (rocket) observed frequency, with the same (rocket observed) angle, but reflected back (-x direction) we get f_hat (frequency observed by the mirror as it absorbs the incoming photon) = gamma*f - gamma*v*f*y (which I think agrees with the usual Doppler shift result), and we get a different measured incident angle. But, if we reemit back in the -x, +y direction with the same observed values for the mirror, I think that everything cancels to give the same (reflected) angle and initial frequency in the stationary frame. So...I think that, although the mirror observes a different frequency and incident angle, it reflects in such a way to cancel out the relativistic effects. I am not, however, confident that I knocked this off correctly. Bill -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Tuesday, January 09, 2007 8:59 AM To: math-fun@mailman.xmission.com Subject: [math-fun] Reflection from a moving mirror This is a "test your physics intuition" puzzle. When light reflects from a stationary mirror, we know that (1) the angle of reflection equals the angle of incidence, and (2) the frequency of the reflected light equals the frequency of the incident light, i.e. there is no Doppler shift. Does any of this change if the mirror is moving with a velocity parallel to its surface? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
i think that this is right, what bill wrote below In the mirror's frame, the photon's frequency is transverse doppler shifted, but i think if you work only in the stationary frame everything cancels on the return under those assumptions might be fun to work out the angle you should shoot a photon toward a moving perpendicular mirror so that there's no transverse doppler shift observed in the mirror's frame when it arrives shooting the photon at that angle, it won't come back to you, but when you call the guy riding on top of the mirror on the phone, you'll agree with him on the frequency you sent it ...i think... corrections welcomed On 1/9/07, Cordwell, William R <wrcordw@sandia.gov> wrote:
My intuition says that it will. However, suppose that the mirror is sliding in the +y direction, and we shoot a photon at it at an angle from the normal (x direction); if we let f be the frequency of the light, traveling at (f,f*x,f*y,0) (in the +x,+y direction, with x^2 + y^2 = 1 and the incident angle having tangent y/x) in the stationary reference frame, and consider the mirror to be moving with velocity vector (gamma, 0, gamma*v, 0) (in the +y direction), and *assume* that the mirror is like a rocket that absorbs a photon and reemits it with the same (rocket) observed frequency, with the same (rocket observed) angle, but reflected back (-x direction) we get
f_hat (frequency observed by the mirror as it absorbs the incoming photon) = gamma*f - gamma*v*f*y (which I think agrees with the usual Doppler shift result), and we get a different measured incident angle. But, if we reemit back in the -x, +y direction with the same observed values for the mirror, I think that everything cancels to give the same (reflected) angle and initial frequency in the stationary frame.
So...I think that, although the mirror observes a different frequency and incident angle, it reflects in such a way to cancel out the relativistic effects.
I am not, however, confident that I knocked this off correctly.
Bill
-----Original Message----- From: math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Tuesday, January 09, 2007 8:59 AM To: math-fun@mailman.xmission.com Subject: [math-fun] Reflection from a moving mirror
This is a "test your physics intuition" puzzle. When light reflects from a stationary mirror, we know that (1) the angle of reflection equals the angle of incidence, and (2) the frequency of the reflected light equals the frequency of the incident light, i.e. there is no Doppler shift. Does any of this change if the mirror is moving with a velocity parallel to its surface?
Gene
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-- Thane Plambeck tplambeck@gmail.com http://www.plambeck.org/ehome.htm
Suppose you have two tied parallel mirrors and a double reflection. In the mirrors' rest frame, the outgoing photon comes out with no change in momentum or energy from the incoming photon. So this should be true as well in any frame moving parallel to the mirrors. In such a frame, if a bounce off a mirror were asymmetrical, two bounces would roughly double the asymmetry, and the outgoing photon wouldn't come out with the same momentum and energy. How's that for an intuitive argument? --ms Thane Plambeck wrote:
i think that this is right, what bill wrote below
In the mirror's frame, the photon's frequency is transverse doppler shifted, but i think if you work only in the stationary frame everything cancels on the return under those assumptions
might be fun to work out the angle you should shoot a photon toward a moving perpendicular mirror so that there's no transverse doppler shift observed in the mirror's frame when it arrives
shooting the photon at that angle, it won't come back to you, but when you call the guy riding on top of the mirror on the phone, you'll agree with him on the frequency you sent it
...i think...
corrections welcomed
On 1/9/07, Cordwell, William R <wrcordw@sandia.gov> wrote:
My intuition says that it will. However, suppose that the mirror is sliding in the +y direction, and we shoot a photon at it at an angle from the normal (x direction); if we let f be the frequency of the light, traveling at (f,f*x,f*y,0) (in the +x,+y direction, with x^2 + y^2 = 1 and the incident angle having tangent y/x) in the stationary reference frame, and consider the mirror to be moving with velocity vector (gamma, 0, gamma*v, 0) (in the +y direction), and *assume* that the mirror is like a rocket that absorbs a photon and reemits it with the same (rocket) observed frequency, with the same (rocket observed) angle, but reflected back (-x direction) we get
f_hat (frequency observed by the mirror as it absorbs the incoming photon) = gamma*f - gamma*v*f*y (which I think agrees with the usual Doppler shift result), and we get a different measured incident angle. But, if we reemit back in the -x, +y direction with the same observed values for the mirror, I think that everything cancels to give the same (reflected) angle and initial frequency in the stationary frame.
So...I think that, although the mirror observes a different frequency and incident angle, it reflects in such a way to cancel out the relativistic effects.
I am not, however, confident that I knocked this off correctly.
Bill
-----Original Message----- From: math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Tuesday, January 09, 2007 8:59 AM To: math-fun@mailman.xmission.com Subject: [math-fun] Reflection from a moving mirror
This is a "test your physics intuition" puzzle. When light reflects from a stationary mirror, we know that (1) the angle of reflection equals the angle of incidence, and (2) the frequency of the reflected light equals the frequency of the incident light, i.e. there is no Doppler shift. Does any of this change if the mirror is moving with a velocity parallel to its surface?
Gene
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participants (4)
-
Cordwell, William R -
Eugene Salamin -
Mike Speciner -
Thane Plambeck