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I believe this is a simple question, but I cannot find the answer anywhere. According to the relativistic Doppler formula, the emitted ($\nu_{\text{em}}$) and observed ($\nu_{\text{obs}}$) frequencies of radiation are related by:

$$\frac{\nu_{\text{obs}}}{\nu_{\text{em}}}=\sqrt{\frac{1+\beta}{1-\beta}} \ \ \ \ where \ \ \ \beta = \frac{v}{c}.$$

But I can only use this formula when $v \ll c$ (and therefore $\beta \ll 1$). If $v\approx c$ (and $\beta \approx 1$), what can I use to determine the emitted or observed frequency?

Brendan Darrer
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Jeff
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2 Answers2

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Would an example help?

Suppose $\nu_{\rm em} = 10^{15}$ Hz and $v = 0.9 c$. Then $\beta = 0.9$ and $(1+\beta)/(1-\beta) = 19$. The square root of 19 is about $4.359$. Hence we obtain $$ \nu_{\rm obs} \simeq 4.359 \times 10^{15}\,{\rm Hz}. $$ This is the case where the source is coming towards you, resulting in a shift to a higher frequency.

Andrew Steane
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This is the relativistic formula for the doppler shift. It is correct for any velocity especially for relativistic cases where $v\approx c$. When dealing with non relativistic velocities $v \ll c$ or $\beta\to 0$ the first order Taylor expansion gives the non-relativistic doppler equation $$\nu_{obs} = (1+\beta)\nu_{em}$$

Níckolas Alves
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Ben
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