Relativistic Doppler effect

by fluidistic
Tags: doppler, effect, relativistic
fluidistic is offline
Aug23-10, 05:45 PM
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1. The problem statement, all variables and given/known data
A car is getting closer to a radar as a speed of 135 km/h. If the radar works at a [tex]2 \times 10 ^{9} Hz[/tex], what difference of frequency is observed for the radar?

2. Relevant equations
[tex]\mu=\frac{\mu _0 \sqrt {1- \frac{v^2}{c^2}}}{1-\frac{v}{c}}[/tex]

3. The attempt at a solution
I converted 135 km/h into m/s, which gave me [tex]\frac { 1350 m }{36 s}[/tex].
Then I applied the formula and [tex]\mu - \mu _0[/tex] gave me [tex]1.024 \times 10^{13} Hz[/tex] which seems WAY too big to be realistic. Am I missing something? Is it a wrong formula?!
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kuruman is offline
Aug23-10, 10:13 PM
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The formula is correct and I agree the number is way too large. Can you show how you got it? A word of caution: v/c is too small to plug in a calculator and expect something other than μ0, i.e. no frequency shift. I suggest that you try a Taylor expansion for small values of v/c.

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