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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?

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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 $10^{-3}$.

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  • $\begingroup$ 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. $\endgroup$ – Aza Mar 20 at 9:48

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