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Relativistic Doppler Shift Equation - Derivation

  • Thread starter TFM
  • Start date
  • #1
TFM
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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
 

Answers and Replies

  • #2
208
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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
 
  • #3
TFM
1,026
0
Yeah it should really be

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

Cos of Pi = -1, giving:

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

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:
  • #4
TFM
1,026
0
Sorry, Latex seems tpo dislike my using ' to represent prime:

Final equation so far:

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

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

TFM
 
  • #5
alphysicist
Homework Helper
2,238
1
Hi TFM,

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

[tex]
\gamma = \frac{1}{\sqrt{1-\beta^2} }
[/tex]

Once you use that, you can factor what's inside the square root on the bottom (it has the form of [itex]a^2-b^2[/itex]). After a cancellation, I think you can get the answer. What do you get?
 
  • #6
TFM
1,026
0
I now get:

[tex] v^| = v\frac{1 + \beta}{\sqrt{(1 + \beta)(1 - \beta)}} [/tex]

would you now do:

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

So you would cancel out - does this look correct?

TFM
 
  • #7
alphysicist
Homework Helper
2,238
1
You have an exponent in the wrong place in the denominator, but it looks like it's probably just a typo. (The factor [itex](1+\beta^{1/2})[/itex] needs to be [itex](1+\beta)^{1/2}[/itex]). Other than that, it looks correct, and once you cancel out the identical factors in numerator and denominator you get the desired result.
 
  • #8
TFM
1,026
0
Yeah, the bracket is in the worng place, should have been:

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

So it cancels down to:

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

Which is the same as:

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

TFM
 

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