- #1

- 1,026

- 0

## Homework Statement

The full expression for the relativitic Doppler Shift is

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

where v' and v are the frequencies of the light source in its own rest frame and and in the laboraty respectivley. [tex] \theta [/tex] 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:

[tex] v' = v\sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

for a source moving driectly away from the observer

## Homework Equations

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

[tex] v' = v\sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

[tex] \gamma = \frac{1}{\sqrt{1 - \beta}} [/tex]

## The Attempt at a Solution

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

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

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

for the angle, since moving away, [tex] \theta = 0 [/tex]

thus, [tex] cos \theta = 1 [/tex]

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

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