1
$\begingroup$

I'm currently reading a book on optics, and have encountered a curious section:

$$\nu = \nu'\sqrt{1-\frac{u^2}{c^2}} = \nu'\left(1-\frac{u^2}{2c^2}+\ldots\right)$$

This is the formula for the transverse Doppler shift, giving the frequency change when the relative motion is at right angles to the direction of observation. The transverse Doppler shift is a second-order effect and is therefore very difficult to measure. It has been verified by using the Mossbauer effect with gamma radiation from radioactive atoms.

What about this specific effect makes it difficult to measure? I understand that the text says it is because it is a second-order effect, but it's not clear to me why that makes a correction term so much more difficult to observe.

Is there a good elucidation on the reasons behind this?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

Because you need to have sufficiently high $\frac{u}{c}$ value in the experiment so that the term $\left(\frac{u}{c}\right)^2$ will be significant and the corresponding change in frequency is detectable.

For eg. If you have $\frac{u}{c} = 0.001$ which is literally having particles moving at a velocity of $300000$ m/s. Then the change you detect is just $10^{-6}$ only so it is difficult to observe when compared to first order $10^{-3}$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Interesting, I was sort of expecting there to be something more to it than that. The same book says the third order correction on the normal Doppler shift was more easily observed. But I'm about to sleep; I'll post the section tomorrow when I get a free moment. $\endgroup$
    – user24081
    Commented Mar 20, 2019 at 9:48
0
$\begingroup$

I think it is a combination of the relative sizes of $v/c$ versus $v^2/c^2$ combined with the requirement to make sure a tangential velocity has no confusing radial component with a high level of precision throughout the measurement.

I suppose ideally you want something moving in a perfect circle and emitting radiation towards the centre. If the tangential velocity is $v_t$, then to isolate the transverse shift, you need to know any radial component to much better than $v_t^2/c^2$.

As an example. If I have an emitter moving at $3\times 10^4$ m/s, then the transverse Doppler effect is of order $10^{-8}$ compared with a "normal" Doppler effect of $10^{-4}$. Thus any radial motion must be eliminated (or characterised) to much better than 1 part in $10^4$ to isolate the transverse shift.

Improve this answer
$\endgroup$
-1
$\begingroup$

The transverse Doppler actually doesn't exist, it is the "relativisic" Doppler Effect that occurs at all angles, which is caused by a slowing down of emission processes by motion in the gravitational field of the earth, which has nothing to do with some kind of time dilatation. There is no need to measure the transverse Doppler Effect at 90 degrees, but if you do it, it is difficult, because the emission angle permanently changes, if measured directly perpendicular to a moving light source.

Improve this answer
$\endgroup$