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Sorry this question is really vague- as a result I'm very confused. Perhaps you can help me make sense of it.

"Radar measurements: Construct a space-time diagram of the radar tracking of an asteroid moving at speed $v$. By using the postulates of special relativity verify the expressions on (a book I do not have). Set $v \ll c$ and obtain the non-relativistic expressions for the relative velocity of asteroid, the one-way and two-way Doppler measurements."

So postulates of SR: 1) laws of physics are the same in all frames of reference. 2) $c$ is always constant.

To me Doppler shifts are to do with $f'$, $f$ and angular velocities...

When it says set $v \ll c$ it makes me think that as a result $\gamma$ ($v/c$) is meant to tend to $0$ therefore it comes into the question somewhere I'm just not sure where. Perhaps

$$f' ~=~( 1-\beta/1+\gamma) ^{1/2}.$$

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user, I've adjusted your math markup a little bit, but I am almost certain that you meant the term in the parenthesis to be either a fraction (markup with \frac{1 - \beta}{1 + \gamma} to get $\frac{1 - \beta}{1 + \gamma}$) or one minus a fraction. In either case you'll want scaled parenthesis (markup with \left( \dots \right) to get $\left( \dots \right)$). –  dmckee Feb 15 '13 at 1:13
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1 Answer 1

$v\ll c$ doesn't reaaallly imply $v/c=0$. It just implies that $v/c \ll 1$. I don't know if that notation is expressible in pure math, so there's a bit of handwaving, but usually what it means to do is to use calculus to find the series approximation of a function, and drop higher order terms of $v/c$ (so, ignore $(v/c)^2$, $(v/c)^3$ etc. in the series expansion).

The equation is: $f'=\left( \frac{1-\beta}{1+\beta} \right)^{\frac{1}{2}}f$, where $f'$ is the frequency in a frame $S'$, moving with velocity $V$ as observed in $S$, and $f$ is the frequency as observed in $S$.

So anyways, the expansion of $(1-\beta)^{1/2}$ around $\beta=0$ is $1-\beta+O(\beta^2)$. Of $(1+\beta)^{-1/2}$ is $1+\beta/2+O(\beta^2)$. So:

$$f'=f \left( \frac{1-\beta}{1+\beta} \right)^{\frac{1}{2}}\approx f\left( (1-\beta)(1+\beta/2)+O(\beta^2)\right)=f \left( 1-\beta+O(\beta^2)\right)$$

Or, in other words, $f'=f(1-\beta)$ as $\beta\ll 1$.

So, it's just a matter of using approximations. If you're not familiar with this level calculus mention it in the comments. You can figure these formulas from calculus/taylor series, or from the generalized binomial theorem. http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem

(And, if you use mathematica, here'scode I used to check to make sure my coefficients were right: Series[((1 - \[Beta])/(1 + \[Beta]))^(1/2), {\[Beta], 0, 1}], outputs 1-\[Beta]+O[\[Beta]]^2. [imagine "\[Beta]" being replaced with a beautiful rendering of the greek character])

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