Relativistic Doppler shift
in [Relativity]
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From: BURT on 25 Jul 2010 00:00 On Jul 24, 8:56 pm, dlzc <dl...(a)cox.net> wrote: > Dear I.N. Galidakis: > > On Jul 24, 9:58 am, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > > > > > > > dlzc wrote: > > > Dear I.N. Galidakis: > > > [snip] > > > >> In other words, if I rotate a 532 nm laser > > >> beam with a given angular frequency omega, > > >> what would be the Doppler shift for an > > >> observer at radius ~r, where omega*r=c? > > > > You'd have a redshifting of the light source, > > > due to time dilation, then you'd have whatever > > > classical Doppler effect you'd see based on > > > the relative motion. Similar to an SR-only > > > problem, only with the time dilation being > > > more complex due to acceleration. > > > Thanks. The time dilation I was able to > > calculate, but I cannot seem to see why > > I'd get a red/blue-shift (relativistic or not). > > > The way I understand it, the _linear_ velocity > > at the tip of the beam (at distance r) is > > perpendicular to the tip's orbit (circle of > > radius r), so I cannot see how it contributes > > to a Doppler shift for an observer that gets hit > > by the beam head on. > > How do you *not* see a case for classical Doppler shift? It works for > cars at much less than c... even though that is a two-way trip. > > > Any help? I need the actual calcs for the > > red/blue-shift for an observer that gets hit by > > the beam once for every period of the rotation. > > Sorry, I just don't see where you'd have a problem here. Not sure if > you are looking at light as particles, or light as a wave, or how > you'd like to treat it. > > David A. Smith- Hide quoted text - > > - Show quoted text - Red shift is from behind and blueshift is from the front but what about light being absorbed by something moving at a 90 degree angle? It is easy to see that 90 degrees would have no energy shift and that other angles would be partial energy shifts. Mitch Raemsch
From: dlzc on 25 Jul 2010 00:02 Dear I.N. Galidakis: On Jul 24, 9:58 am, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: .... > Any help? I need the actual calcs for the > red/blue-shift for an observer that gets > hit by the beam once for every period of > the rotation. Let's see if this is any help. http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/rigid_disk.html The frequency of the light is indicative not of something unique to the light, but of two components (in flat space): 1) the emitting process, and 2) the relative motion of the emitter and the receiver. David A. Smith
From: I.N. Galidakis on 25 Jul 2010 02:03 dlzc wrote: > Dear I.N. Galidakis: > > On Jul 24, 9:58 am, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: >> dlzc wrote: >>> Dear I.N. Galidakis: >> >> [snip] >> >>>> In other words, if I rotate a 532 nm laser >>>> beam with a given angular frequency omega, >>>> what would be the Doppler shift for an >>>> observer at radius ~r, where omega*r=c? >> >>> You'd have a redshifting of the light source, >>> due to time dilation, then you'd have whatever >>> classical Doppler effect you'd see based on >>> the relative motion. Similar to an SR-only >>> problem, only with the time dilation being >>> more complex due to acceleration. >> >> Thanks. The time dilation I was able to >> calculate, but I cannot seem to see why >> I'd get a red/blue-shift (relativistic or not). >> >> The way I understand it, the _linear_ velocity >> at the tip of the beam (at distance r) is >> perpendicular to the tip's orbit (circle of >> radius r), so I cannot see how it contributes >> to a Doppler shift for an observer that gets hit >> by the beam head on. > > How do you *not* see a case for classical Doppler shift? It works for > cars at much less than c... even though that is a two-way trip. Sorry, my problem is probably something totally trivial and I probably did not express it correctly: I know the Doppler shift expression for relative motion, I just don't know whether to put v=omega*r in this case, or that, multiplied by some cosine or sine, since the velocity of the wave emitter and the linear velocity from rotation are perpendicular when the beam hits the observer. >> Any help? I need the actual calcs for the >> red/blue-shift for an observer that gets hit by >> the beam once for every period of the rotation. > > Sorry, I just don't see where you'd have a problem here. Not sure if > you are looking at light as particles, or light as a wave, or how > you'd like to treat it. Thanks for the link on the other post. Not much, but better than nothing. > David A. Smith -- I.
From: Dono. on 25 Jul 2010 09:48 On Jul 24, 11:03 pm, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > dlzc wrote: > > Dear I.N. Galidakis: > > > On Jul 24, 9:58 am, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > >> dlzc wrote: > >>> Dear I.N. Galidakis: > > >> [snip] > > >>>> In other words, if I rotate a 532 nm laser > >>>> beam with a given angular frequency omega, > >>>> what would be the Doppler shift for an > >>>> observer at radius ~r, where omega*r=c? > > >>> You'd have a redshifting of the light source, > >>> due to time dilation, then you'd have whatever > >>> classical Doppler effect you'd see based on > >>> the relative motion. Similar to an SR-only > >>> problem, only with the time dilation being > >>> more complex due to acceleration. > > >> Thanks. The time dilation I was able to > >> calculate, but I cannot seem to see why > >> I'd get a red/blue-shift (relativistic or not). > > >> The way I understand it, the _linear_ velocity > >> at the tip of the beam (at distance r) is > >> perpendicular to the tip's orbit (circle of > >> radius r), so I cannot see how it contributes > >> to a Doppler shift for an observer that gets hit > >> by the beam head on. > > > How do you *not* see a case for classical Doppler shift? It works for > > cars at much less than c... even though that is a two-way trip. > > Sorry, my problem is probably something totally trivial and I probably did not > express it correctly: > > I know the Doppler shift expression for relative motion, I just don't know > whether to put v=omega*r in this case, or that, multiplied by some cosine or > sine, since the velocity of the wave emitter and the linear velocity from > rotation are perpendicular when the beam hits the observer. > You only get transverse Doppler effect, so you need to put v=omega*r, no cosine, in the formula for transverse Doppler effect. Precisely: f_observed=f_emitted*sqrt(1-(omega*r/c)^2)
From: I.N. Galidakis on 25 Jul 2010 10:25
Dono. wrote: > On Jul 24, 11:03 pm, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: >> dlzc wrote: >>> Dear I.N. Galidakis: >> >>> On Jul 24, 9:58 am, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: >>>> dlzc wrote: >>>>> Dear I.N. Galidakis: >> >>>> [snip] >> >>>>>> In other words, if I rotate a 532 nm laser >>>>>> beam with a given angular frequency omega, >>>>>> what would be the Doppler shift for an >>>>>> observer at radius ~r, where omega*r=c? >> >>>>> You'd have a redshifting of the light source, >>>>> due to time dilation, then you'd have whatever >>>>> classical Doppler effect you'd see based on >>>>> the relative motion. Similar to an SR-only >>>>> problem, only with the time dilation being >>>>> more complex due to acceleration. >> >>>> Thanks. The time dilation I was able to >>>> calculate, but I cannot seem to see why >>>> I'd get a red/blue-shift (relativistic or not). >> >>>> The way I understand it, the _linear_ velocity >>>> at the tip of the beam (at distance r) is >>>> perpendicular to the tip's orbit (circle of >>>> radius r), so I cannot see how it contributes >>>> to a Doppler shift for an observer that gets hit >>>> by the beam head on. >> >>> How do you *not* see a case for classical Doppler shift? It works for >>> cars at much less than c... even though that is a two-way trip. >> >> Sorry, my problem is probably something totally trivial and I probably did >> not express it correctly: >> >> I know the Doppler shift expression for relative motion, I just don't know >> whether to put v=omega*r in this case, or that, multiplied by some cosine or >> sine, since the velocity of the wave emitter and the linear velocity from >> rotation are perpendicular when the beam hits the observer. >> > > > You only get transverse Doppler effect, so you need to put v=omega*r, > no cosine, in the formula for transverse Doppler effect. Precisely: > > f_observed=f_emitted*sqrt(1-(omega*r/c)^2) Thanks. -- I. |