Relativistic Doppler Shift Equation - Derivation

[TFM]

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

 

[EngageEngage]

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

 

[TFM]

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

 

[TFM]

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]

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?

 

[TFM]

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]

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.

 

[TFM]

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