Relativistic Doppler shift
in [Relativity]
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From: Dono. on 25 Jul 2010 11:03 On Jul 25, 7:25 am, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > 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. you are welcome
From: harald on 25 Jul 2010 16:47 On Jul 24, 6:58 pm, "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. > > 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. > > Thanks again, > > > David A. Smith > > -- > I. Dear I, Indeed there is no Doppler shift in the picture that you sketch, as you take the centre point as being "in rest"; then there is only a redshift due to the motion of the laser: that IS the "time dilation". David already told you that. It appears that your problem is mainly with the term "relativistic Doppler". In fact it consists of the sum (or technically speaking, the product) of the classical Doppler effect and the "time dilation" effect. You seem to have just found this out. Regards, Harald
From: I.N. Galidakis on 25 Jul 2010 17:05 harald wrote: > On Jul 24, 6:58 pm, "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. >> >> 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. >> >> Thanks again, >> >>> David A. Smith >> >> -- >> I. > > Dear I, > > Indeed there is no Doppler shift in the picture that you sketch, as > you take the centre point as being "in rest"; then there is only a > redshift due to the motion of the laser: that IS the "time dilation". > David already told you that. > It appears that your problem is mainly with the term "relativistic > Doppler". In fact it consists of the sum (or technically speaking, the > product) of the classical Doppler effect and the "time dilation" > effect. You seem to have just found this out. The thing that worried me was the "perpendicularity" of the two speeds: linear velocity and that of wave propagation (in this case laser light). After Dono gave me the correct expression, I double-checked it with Wiki and it seems that the perpendicularity resolves into what Dono called "transverse" Doppler effect, given by expression: f_o=f_s/gamma, where gamma is the Lorentz Factor, on Page: http://en.wikipedia.org/wiki/Relativistic_Doppler_effect so all is well. Btw, this expression seems to give a "blue-shift" for a fast-rotating green laser (since I find that the 532nm reduces towards the blue to 520nm for my case), but I understand that the term "redshift" may be used generically for both cases, blue-shift and red-shift. So thanks for all the responses. Back to lurKer mode :-) > Regards, > Harald -- I.
From: harald on 25 Jul 2010 17:35 On Jul 25, 11:05 pm, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > harald wrote: > > On Jul 24, 6:58 pm, "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. > > >> 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. > > >> Thanks again, > > >>> David A. Smith > > >> -- > >> I. > > > Dear I, > > > Indeed there is no Doppler shift in the picture that you sketch, as > > you take the centre point as being "in rest"; then there is only a > > redshift due to the motion of the laser: that IS the "time dilation". > > David already told you that. > > It appears that your problem is mainly with the term "relativistic > > Doppler". In fact it consists of the sum (or technically speaking, the > > product) of the classical Doppler effect and the "time dilation" > > effect. You seem to have just found this out. > > The thing that worried me was the "perpendicularity" of the two speeds: linear > velocity and that of wave propagation (in this case laser light). After Dono > gave me the correct expression, I double-checked it with Wiki and it seems that > the perpendicularity resolves into what Dono called "transverse" Doppler effect, > given by expression: > > f_o=f_s/gamma, where gamma is the Lorentz Factor, on Page: > > http://en.wikipedia.org/wiki/Relativistic_Doppler_effect > > so all is well. Yes, the intro there looks quite OK; "transverse Doppler" is not really a Doppler effect, as you intuitively already understood. > Btw, this expression seems to give a "blue-shift" for a fast-rotating green > laser (since I find that the 532nm reduces towards the blue to 520nm for my > case), but I understand that the term "redshift" may be used generically for > both cases, blue-shift and red-shift. Not exactly: "redshift" means that a "white" spectrum is seen to be shifted towards the red, thus to lower frequencies. > So thanks for all the responses. > > Back to lurKer mode :-) You're welcome ;-)
From: BURT on 25 Jul 2010 17:46
On Jul 25, 2:35 pm, harald <h...(a)swissonline.ch> wrote: > On Jul 25, 11:05 pm, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > > > > > > > harald wrote: > > > On Jul 24, 6:58 pm, "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. > > > >> 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. > > > >> Thanks again, > > > >>> David A. Smith > > > >> -- > > >> I. > > > > Dear I, > > > > Indeed there is no Doppler shift in the picture that you sketch, as > > > you take the centre point as being "in rest"; then there is only a > > > redshift due to the motion of the laser: that IS the "time dilation". > > > David already told you that. > > > It appears that your problem is mainly with the term "relativistic > > > Doppler". In fact it consists of the sum (or technically speaking, the > > > product) of the classical Doppler effect and the "time dilation" > > > effect. You seem to have just found this out. > > > The thing that worried me was the "perpendicularity" of the two speeds: linear > > velocity and that of wave propagation (in this case laser light). After Dono > > gave me the correct expression, I double-checked it with Wiki and it seems that > > the perpendicularity resolves into what Dono called "transverse" Doppler effect, > > given by expression: > > > f_o=f_s/gamma, where gamma is the Lorentz Factor, on Page: > > >http://en.wikipedia.org/wiki/Relativistic_Doppler_effect > > > so all is well. > > Yes, the intro there looks quite OK; "transverse Doppler" is not > really a Doppler effect, as you intuitively already understood. > > > Btw, this expression seems to give a "blue-shift" for a fast-rotating green > > laser (since I find that the 532nm reduces towards the blue to 520nm for my > > case), but I understand that the term "redshift" may be used generically for > > both cases, blue-shift and red-shift. > > Not exactly: "redshift" means that a "white" spectrum is seen to be > shifted towards the red, thus to lower frequencies. > > > So thanks for all the responses. > > > Back to lurKer mode :-) > > You're welcome ;-)- Hide quoted text - > > - Show quoted text - What is the Doppler shift for sideways angles? Is it more or less at 90 degrees absorption? When matter moves sideways to light what is the energy shift? Mitch Raemsch |