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Formula of doppler effect for sound,relativistic? 
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#1
Nov811, 11:40 PM

P: 764

As you know,the formula for doppler effect is:
[itex]f^{'}=\frac{vv_{o}}{vv_{s}}f[/itex] relativity suggests(at least I think) that [itex]f^{'} [/itex] for [itex] v_{o}=u [/itex] and [itex] v_{s}=0[/itex] should be the same with [itex]f^{'}[/itex] for [itex]v_{o}=0 [/itex] and [itex]v_{s}=u[/itex],otherwise,there will be a way for the listener to know that which one,sound source or listener,is moving with constant velocity. So does relativity suggest a correction or simply I am wrong? thanks 


#2
Nov811, 11:54 PM

P: 1,555

The complete formula is:
[tex]\Large {\frac {\sqrt {1{v}^{2}}}{1+v\cos \left( \theta \right) }}[/tex] which simplifies when theta is 0 to: [tex]\Large {\frac {\sqrt {1{v}^{2}}}{1+v}}[/tex] Note that the Lorentz factor is: [tex]\Large \gamma={\frac {1}{\sqrt {1{v}^{2}}}}[/tex] 


#3
Nov911, 06:09 AM

Sci Advisor
P: 1,256

Sound propagates in a medium, and the usual Doppler formula holds only in the rest system of the medium, so SR does not apply.



#4
Nov1011, 04:35 AM

P: 764

Formula of doppler effect for sound,relativistic?
Good point clem,I should have thought about that.thanks.
But what are the formulas in flower's post? 


#5
Nov1011, 09:08 AM

PF Gold
P: 4,674

The more common formula for when theta=0 is: √[(1β)/(1+β)] But with a little algebra, you can transform the one into the other. These formulas apply when the two observers are moving away from each other, their reciprocal works when the two observers are moving toward each other. You also have to keep in mind that these formulas only apply when the two observers have been in their constant relative motion for a long enough time that the signals have stabilized. Or another way to say that is the formulas tell you the Doppler at an earlier point in time, whatever the signal delay is between the two observers, and, of course, that delay is frame dependent and so in the rest frame of each observer it will be different. But it is this delay that makes Relativistic Doppler an easy illustration of what each observer sees in the Twin Paradox. While they are moving away, they each see the same Doppler, that is they each see the other one's clock running slower than their own. When the traveling twin turns around, he immediately sees the Doppler of the stationary twin go up (it becomes the reciprocal) and so now he sees his twin's clock running faster than his own. Meanwhile, the stationary twin continues to see the traveling twin's clock still going slower than his own. Eventually, the stationary twin will see the Doppler from the traveling twin's clock go up (it becomes the same reciprocal value) and then they reunite. If they both keep track of the cycles of the waveforms they see from the other twin, they can determine the relative age difference between them when they reunite. The traveling twin sees his brother's clock running at the two rates for exactly half the trip while the stationary twin sees his brother's clock run at a slow rate for most of the trip and then the high rate for only the tail end of the trip. 


#6
Nov1011, 11:50 AM

P: 1,555




#7
Nov1011, 12:52 PM

PF Gold
P: 4,674




#8
Nov1011, 12:56 PM

P: 1,555




#9
Nov1011, 01:11 PM

PF Gold
P: 4,674

Maybe my grammar wasn't exactly correct but what I mean is it's the relative velocity (in whatever units you want to use) between the two observers divided by the speed of light, c, (in the same units). Your "v" cannot be a velocity in the normal sense of the word because it is not a unitless value until you divide it by c. The normal nomenclature when doing this is to use β to refer to the ratio of a velocity and the speed of light.
Does that clear it up? 


#10
Nov1011, 01:17 PM

P: 1,555

The ratio of which velocities? You still have not explained by what you mean by the ratio of a velocity.
How is a velocity between A and B relative? There is only one velocity between A and B it is neither a ratio nor is it relative. 


#11
Nov1011, 02:10 PM

PF Gold
P: 4,674

I'm only pointing out that the "v" in your formula needs to be divided by c in order to make it possible to do the computation. 


#12
Nov1011, 02:17 PM

P: 1,555

Are you saying it is not? A relative velocity between two observers does not make any sense to me as there is nothing relative here. 


#13
Nov1011, 02:39 PM

Sci Advisor
PF Gold
P: 1,841

It's fairly standard terminology to say "the relative velocity between A and B" to mean to "the velocity of A relative to B" (in other words, "the velocity of A measured by B"), which is the same as "the velocity of B relative to A" (referring to inertial frames in SR, at least).
All velocities are relative in the sense that it only makes sense to talk "the velocity of A relative to B"; "the velocity of A" is meaningless (unless the "...relative to B" is obvious by context). 


#14
Nov1011, 03:30 PM

PF Gold
P: 4,674




#15
Nov1011, 06:47 PM

P: 1,555




#16
Nov1011, 06:56 PM

PF Gold
P: 4,674

Is it relative?



#17
Nov1011, 10:15 PM

P: 764

Yes PassionFlower,the velocity between two observers is absolute but ghwellsjr is saying that each individual veclocity depends on the frame of reference you choose which is indeed correct. Now I understand that to use the formulas given by PassionFlower,we need to determine the velocity between the two (the procedure differs in different frames)and of course divide it by c.Or simply take the frame of fluid the frame of reference and use the classic formula. And one more point ghwellsjr,you talked about time dilation.The doppler effect is not about the frequency differences because of the different time speeds in different frames.Its sth classical which needs just a little bit modernization.The effect you talked about,is sth that should be calculated in a different context. 


#18
Nov1011, 10:28 PM

PF Gold
P: 4,674




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