# Relativistic Doppler effect

by fluidistic
Tags: doppler, effect, relativistic
 PF Gold P: 3,173 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 $$2 \times 10 ^{9} Hz$$, what difference of frequency is observed for the radar? 2. Relevant equations $$\mu=\frac{\mu _0 \sqrt {1- \frac{v^2}{c^2}}}{1-\frac{v}{c}}$$ 3. The attempt at a solution I converted 135 km/h into m/s, which gave me $$\frac { 1350 m }{36 s}$$. Then I applied the formula and $$\mu - \mu _0$$ gave me $$1.024 \times 10^{13} Hz$$ which seems WAY too big to be realistic. Am I missing something? Is it a wrong formula?!
 HW Helper PF Gold P: 3,444 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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