Redshift of spinning star surface as seen on axis

[Jonathan Scott]

I think I heard long ago that the redshift of all parts of the surface of a spinning star as seen from a distant point on the axis is expected to be the same, at least in theory, because of the following argument. Please can anyone confirm or refute this?

A body which is capable of being shaped by its own gravity forms a shape such that locally the surface is "level" or "equipotential" as seen by a local observer. The "potential" for this purpose is a combination of the effect of gravity and the effective potential for the centripetal acceleration caused by the rotation. This means that clocks run at the same rate at different locations on the surface, as seen locally, so clocks over the whole surface can be synchronized. This means that as seen from a distant position on the axis, the clocks rates are also synchronized, and similarly that the redshift of any processes at the surface is also the same.

From the distant point of view, this is because the Special Relativity velocity time dilation for points further from the axis is exactly balanced by being at a higher gravitational potential.

Is this correct? It seems plausible to me, but I'd like a second opinion.

 

[mfb]

I am note sure if that accounts for the transversal Doppler effect as well. If that is negligible, I agree.

 

[Chronos]

No, it is not correct. Doppler shift due to rotation is rather easily measured on the sun, which rotates at a leisurely 2 km/s - http://www.eyes-on-the-skies.org/shs/spec-rot-gb.htm. This is quite difficult to do, however, for other stars.

 

[Jonathan Scott]

No, it is not correct. Doppler shift due to rotation is rather easily measured on the sun, which rotates at a leisurely 2 km/s - http://www.eyes-on-the-skies.org/shs/spec-rot-gb.htm. This is quite difficult to do, however, for other stars.

I'm well aware that one can see Doppler shift from rotation from a point off the axis, thanks.

But I was asking about the redshift as seen from a point on the axis, where all the usual Doppler shifts should cancel out, and the "transverse Doppler shift" is simply the same as the SR velocity dilation. If the argument about effective equipotentials is correct, the poles should be at an extremely slightly lower gravitational potential than the equator, and the resulting difference in time dilation should exactly match the velocity time dilation.

This is of course a much smaller effect than Doppler shift, in that the Doppler shift is related to v/c but the time dilation is roughly 1/2 (v²/c²).

 

[goldsax]

sorry if i have not understood correctly.
if you are looking on the axis of the star. there will be no doppler shift as there is no movement in the radial direction but only in a travserse direction..


waiting to be corrected!

 

[mfb]

@goldsax: There is a transversal doppler effect as result of relativistic time dilation, and gravitational redshift as result of the lower gravitational potential on the surface (compared to free space).

 

[goldsax]

@goldsax: There is a transversal doppler effect as result of relativistic time dilation, and gravitational redshift as result of the lower gravitational potential on the surface (compared to free space).

cheers for that ...
next module... relativistic physics...