# Relativistic Doppler Shift Equation - Derivation

## Homework Statement

The full expression for the relativitic Doppler Shift is

$$v' = v \gamma [1- \beta cos \theta]$$

where v' and v are the frequencies of the light source in its own rest frame and and in the laboraty respectivley. $$\theta$$ is the (laboratory) anglebetween the direction of the photons and the direction in which the source is moving. Show that this is consistent with the equation:

$$v' = v\sqrt{\frac{1 + \beta}{1 - \beta}}$$

for a source moving driectly away from the observer

## Homework Equations

$$v' = v \gamma [1- \beta cos \theta]$$

$$v' = v\sqrt{\frac{1 + \beta}{1 - \beta}}$$

$$\gamma = \frac{1}{\sqrt{1 - \beta}}$$

## The Attempt at a Solution

I have got some way through, but appear to be slightly stuck:

$$v' = v \gamma [1- \beta cos \theta]$$

$$v' = v \frac{[1- \beta cos \theta]}{1 - \beta}$$

for the angle, since moving away, $$\theta = 0$$

thus, $$cos \theta = 1$$

$$v' = v \frac{[1- \beta (1)]}{1 - \beta}$$

But I am not sure how to get the square root over the whole of the equation, or to turn the minus to a plus on the top if the fraction.

Any ideas?

TFM

I might be missing something but i think you forgot the root when you subsituted gamma into the expression. Also, since its moving away you would have theta be pi, i believe, whichw ould give you the plus you're looking for

Yeah it should really be

$$v' = v\frac{1 - \beta}{\sqrt{1 - \beta}}[/$$

Cos of Pi = -1, giving:

$$v' = v \frac{1 + \beta}{\sqrt{1 - \beta}}$$

So all I need to de know is find out how to get the root over the whole of the fraction, not just the bottom.

TFM

Last edited:
Sorry, Latex seems tpo dislike my using ' to represent prime:

Final equation so far:

$$v` = v\frac{1 + \beta}{\sqrt{1 - beta}}$$

Any ideas how to get the square root to cover the whole fraction, noit just the bottom half?

TFM

alphysicist
Homework Helper
Hi TFM,

Your definition of gamma is incorrect. It needs to be:

$$\gamma = \frac{1}{\sqrt{1-\beta^2} }$$

Once you use that, you can factor what's inside the square root on the bottom (it has the form of $a^2-b^2$). After a cancellation, I think you can get the answer. What do you get?

I now get:

$$v^| = v\frac{1 + \beta}{\sqrt{(1 + \beta)(1 - \beta)}}$$

would you now do:

$$v^| = v\frac{(1 + \beta)^{1/2}(1 + \beta)^{1/2}}{(1 + \beta^{1/2})(1 - \beta)^{1/2}}$$

So you would cancel out - does this look correct?

TFM

alphysicist
Homework Helper
You have an exponent in the wrong place in the denominator, but it looks like it's probably just a typo. (The factor $(1+\beta^{1/2})$ needs to be $(1+\beta)^{1/2}$). Other than that, it looks correct, and once you cancel out the identical factors in numerator and denominator you get the desired result.

Yeah, the bracket is in the worng place, should have been:

$$v^| = v\frac{(1 + \beta)^{1/2}(1 + \beta)^{1/2}}{(1 + \beta)^{1/2}(1 - \beta)^{1/2}}$$

So it cancels down to:

$$v^| = v\frac{(1 + \beta)^{1/2}}{(1 - \beta)^{1/2}}$$

Which is the same as:

$$v^| = v \sqrt{\frac{1 + \beta}{1 - \beta}}$$

TFM