# Taylor expansion of the relativistic Doppler effect?

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1. Aug 24, 2016

### Amara

[Note from mentor: this thread was originally posted in a non-homework forum, therefore it does not use the homework template.]

I have been given an equation for the relativistic doppler effect but I'm struggling to see this as a function and then give a first order Taylor expansion. Any help at all would be appreciated as I'm completely stuck.

fo/fs = √(1 + vrel/c) / √(1-vrel/c)

Where vrel is the relative speed of the source and the observer with respect to each other, c the speed of light, and vrel > 0 is here presumed to mean that the source and
the observer move towards each other. Consider the relativistic Doppler effect in the case of (vrel/c) ≪ 1, but (vrel/c) > 0. Write a first-order Taylor expansion for the relativistic Doppler effect, and show that the result is equivalent to either of the two expressions found for sound.

The expressions for sound are:

i) The source moves with speed vs towards stationary observer

fo/fs = 1 / (1-(vs/v))

ii) The observer moves with speed vo towards the stationary source

fo/fs = 1 + (vo/v)

Last edited by a moderator: Aug 24, 2016
2. Aug 24, 2016

### stevendaryl

Staff Emeritus
This should probably be in the Homework section, right? But just to put this into a more obvious form, let $x = v_{rel}/c$. Then you want to expand the function

$F(x) = \frac{\sqrt{1+x}}{\sqrt{1-x}}$

in powers of $x$. Do you know how to do that?

3. Aug 24, 2016

### Amara

Sorry this is my first question. Because it's university work I thought it would go here.

I started by doing that but I got confused with not having what would be a in a normal Taylor expansion question.

4. Aug 24, 2016

### stevendaryl

Staff Emeritus
I'm not sure what you mean by a "normal Taylor expansion question". The $F(x)$ that I gave is an ordinary function with an ordinary Taylor expansion.

5. Aug 24, 2016

### Amara

Okay I think my Taylor expansion needs some work. But getting confirmation that it's in this form helps. Thanks

6. Aug 25, 2016

### Amara

I still have a problem with this. In a Taylor expansion I have a value to expand about. I am not given a value for this. And the question does not ask for a Maclaurin series.

7. Aug 25, 2016

### stevendaryl

Staff Emeritus
You're expanding $F(x) = \frac{\sqrt{1+x}}{\sqrt{1-x}}$ about $x=0$. Do you know how to do that?

8. Aug 25, 2016

### Amara

Is this a Maclaurin series?

I'm not sure because I don't know if I should use binomial expansion or differentiate withTaylor.

9. Aug 25, 2016

### stevendaryl

Staff Emeritus
Well, they amount to the same thing. You can use the binomial expansion for a fractional exponent:

$(1+x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha -1)}{1 \cdot 2} x^2 + \frac{\alpha (\alpha -1)(\alpha - 2)}{1 \cdot 2\cdot 3} x^3 + ...$

If $\alpha$ is fractional, then the series never terminates, but that doesn't matter.

But that's the same series you would get if you took derivatives:

Letting $G(x) = (1+x)^\alpha$,

$G'(x) = \alpha (1+x)^{\alpha - 1}$
$G''(x) = \alpha (\alpha - 1) (1+x)^{\alpha - 2}$
$G'''(x) = \alpha(\alpha - 1)(\alpha - 2) (1+x)^{\alpha - 3}$
...

So you can write: $G(x) = G(0) + x G'(0) + \frac{x^2}{1 \cdot 2} G''(0) + \frac{x^3}{1 \cdot 2 \cdot 3} G'''(0) + ...$

You get the same series.

For this problem, if you're going to use binomial expansions, you have to compute the case $\alpha = +\frac{1}{2}$ for $\sqrt{1+x}$ and compute the case $\alpha = -\frac{1}{2}$ for $\sqrt{1-x}$ (using $-x$) and then you have to multiply the two series, and group the terms by powers of $x$.

10. Aug 25, 2016

### Amara

Thank you, this is clear now and has helped a lot