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yuiop

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Given the Doppler shift equation quoted here: http://en.wikipedia.org/wiki/Doppler_effect#General

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) [/tex]

where

c is the velocity of waves in the medium;

v

positive if the receiver is moving towards the source;

v

positive if the source is moving away from the receiver.

Does this equation predict that the redshift seen by the receiver is the same if the observer considers the receiver to be stationary and the source to be moving away at v or if the source is stationary and the receiver is moving away at v?

Special relativity predicts that the redshift seen by the receiver will be the same whatever the speeds of the source and receiver are relative to the observer, as long as the speed of the source relative to the receiver is the same. i.e in SR, if the source is stationary and the reciever is moving away at v, then the redshift will be the same as when the receiver is stationary and the source is moving away at v.

For example (using the classical Doppler shift) let us say the source is moving from the receiver at 0.6c and the receiver (and medium) is stationary with respect to the observer. Using units where the speed of the wave relative to the medium is c=1 we have:

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1 + 0}{1 + 0.6} \right) = 0.625 [/tex]

OK, now we switch to the point of view of an observer that is comoving with the source, so now the receiver appears to be moving at v

If we assume the c in the equation is the speed of the waves relative to the medium (irrespective of the motion of the observer relative to the medium) then the transformed result is:

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1 - 0.6}{1 + 0} \right) = 0.4 [/tex]

which is a different result for the red shift from the one obtained above.

Now if we assume the c in the equation is the speed of the waves relative to the observer (i.e. c = speed of the wave relative to the medium plus the speed of the medium relative to the observer) then the transformed result is:

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1.6 - 0.6}{1.6 + 0} \right) = 0.625 [/tex]

which is the same as the original result.

Which is the correct way to perform the calculation?

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) [/tex]

where

c is the velocity of waves in the medium;

v

_{r}is the velocity of the receiver relative to the medium;positive if the receiver is moving towards the source;

v

_{s}is the velocity of the source relative to the medium;positive if the source is moving away from the receiver.

Does this equation predict that the redshift seen by the receiver is the same if the observer considers the receiver to be stationary and the source to be moving away at v or if the source is stationary and the receiver is moving away at v?

Special relativity predicts that the redshift seen by the receiver will be the same whatever the speeds of the source and receiver are relative to the observer, as long as the speed of the source relative to the receiver is the same. i.e in SR, if the source is stationary and the reciever is moving away at v, then the redshift will be the same as when the receiver is stationary and the source is moving away at v.

For example (using the classical Doppler shift) let us say the source is moving from the receiver at 0.6c and the receiver (and medium) is stationary with respect to the observer. Using units where the speed of the wave relative to the medium is c=1 we have:

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1 + 0}{1 + 0.6} \right) = 0.625 [/tex]

OK, now we switch to the point of view of an observer that is comoving with the source, so now the receiver appears to be moving at v

_{r}= -0.6c.If we assume the c in the equation is the speed of the waves relative to the medium (irrespective of the motion of the observer relative to the medium) then the transformed result is:

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1 - 0.6}{1 + 0} \right) = 0.4 [/tex]

which is a different result for the red shift from the one obtained above.

Now if we assume the c in the equation is the speed of the waves relative to the observer (i.e. c = speed of the wave relative to the medium plus the speed of the medium relative to the observer) then the transformed result is:

[tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1.6 - 0.6}{1.6 + 0} \right) = 0.625 [/tex]

which is the same as the original result.

Which is the correct way to perform the calculation?

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