# Electromagnetic Waves and Relativity

*Most recent answer: 04/13/2011*

Q:

Doppler Effect of light ; and electric , magnetic fields
my query is : The electric and magnetic fields donâ€™t squeeze or stretch then how is Doppler Effect of light possible?
In the phenomenon of Doppler Effect, light emitted from a moving source is detected to have different frequency. If this is taken on terms of detecting the no. of waves passing through the detector in one second it is fine. But according to the principle of relativity, the speed of light is a constant â€¦. It does not change even if its source is movingâ€¦. And that means if still Doppler Effect is observed then the frequency, say, has got increased, then the wavelength should decrease to keep the speed of light constant. And as the wavelength appears to decrease though the original emitted one is largerâ€¦.. it can be visualized as the wave has got squeezed due to its motion. But electric and magnetic fields donâ€™t change their structure even when they are not stationary, i.e., they donâ€™t stretch or squeezeâ€¦.. Then how is the phenomenon of Doppler Effect of light observed?

- apurva (age 18)

rajkot

- apurva (age 18)

rajkot

A:

The key point concerns this line: "electric and magnetic fields don't change their structure even when they are not stationary." I interpret that to mean that you believe that E and B are the same as viewed in references frames in motion with respect to each other. However, that's not true. Neither E nor B is invariant under change of reference frames. As you change reference frames, each becomes a linear combination of the two fields in the initial frame.

The Special Relativistic transformations were originally designed precisely to assure that if Maxwell's equations worked in one inertial frame they would work in all such frames.

Note that the relativistic Doppler effect is not the same as the classical one. The frequency multiplier is sqrt((1+v/c)/(1-v/c)). If you change the sign of the velocity, you get the inverse effect. The classical factor (1+v/c) is not the inverse of (1-v/c), the value for the opposite velocity. The relativistic factor is thus independent of whether you say the receiver or the sender is stationary. The classical factor is not.

Mike W.

The Special Relativistic transformations were originally designed precisely to assure that if Maxwell's equations worked in one inertial frame they would work in all such frames.

Note that the relativistic Doppler effect is not the same as the classical one. The frequency multiplier is sqrt((1+v/c)/(1-v/c)). If you change the sign of the velocity, you get the inverse effect. The classical factor (1+v/c) is not the inverse of (1-v/c), the value for the opposite velocity. The relativistic factor is thus independent of whether you say the receiver or the sender is stationary. The classical factor is not.

Mike W.

*(published on 04/13/2011)*