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Puk
Puk
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You are correct, if a perfect mirror is moving away from the radiation source, the reflected radiation is red-shifted, and vice versa. The frequency of the reflected waves, as observed by the detector, is $$ f = \frac{1 - v/c}{1 + v/c} f_0 $$ where $f_0$ is the original frequency and $v$ is the ratespeed at which the mirror is receding from the source.

There are various ways this can be derived. This is in fact the classical equation for the Doppler shift when the source is at rest with respect to the "propagation medium." This works because the speed of light is $c$ in the reference frame of the source of the waves, and there is no time dilation since waves are being emitted and received by the source and detector which are at rest with respect to each other. You can also show this result using Maxwell's equations, without using the idea of photons, or even explicitly using special relativity (Maxwell's equations are consistent with special relativity). Finally, as you argue in your question, conservation of energy can also be used to arrive at this result.

You are correct, if a perfect mirror is moving away from the radiation source, the reflected radiation is red-shifted, and vice versa. The frequency of the reflected waves, as observed by the detector, is $$ f = \frac{1 - v/c}{1 + v/c} f_0 $$ where $f_0$ is the original frequency and $v$ is the rate at which the mirror is receding from the source.

There are various ways this can be derived. This is in fact the classical equation for the Doppler shift when the source is at rest with respect to the "propagation medium." This works because the speed of light is $c$ in the reference frame of the source of the waves, and there is no time dilation since waves are being emitted and received by the source and detector which are at rest with respect to each other. You can also show this result using Maxwell's equations, without using the idea of photons, or even explicitly using special relativity (Maxwell's equations are consistent with special relativity). Finally, as you argue in your question, conservation of energy can also be used to arrive at this result.

You are correct, if a perfect mirror is moving away from the radiation source, the reflected radiation is red-shifted, and vice versa. The frequency of the reflected waves, as observed by the detector, is $$ f = \frac{1 - v/c}{1 + v/c} f_0 $$ where $f_0$ is the original frequency and $v$ is the speed at which the mirror is receding from the source.

There are various ways this can be derived. This is in fact the classical equation for the Doppler shift when the source is at rest with respect to the "propagation medium." This works because the speed of light is $c$ in the reference frame of the source of the waves, and there is no time dilation since waves are being emitted and received by the source and detector which are at rest with respect to each other. You can also show this result using Maxwell's equations, without using the idea of photons, or even explicitly using special relativity (Maxwell's equations are consistent with special relativity). Finally, as you argue in your question, conservation of energy can also be used to arrive at this result.

Source Link
Puk
Puk
  • 13.9k
  • 1
  • 23
  • 42

You are correct, if a perfect mirror is moving away from the radiation source, the reflected radiation is red-shifted, and vice versa. The frequency of the reflected waves, as observed by the detector, is $$ f = \frac{1 - v/c}{1 + v/c} f_0 $$ where $f_0$ is the original frequency and $v$ is the rate at which the mirror is receding from the source.

There are various ways this can be derived. This is in fact the classical equation for the Doppler shift when the source is at rest with respect to the "propagation medium." This works because the speed of light is $c$ in the reference frame of the source of the waves, and there is no time dilation since waves are being emitted and received by the source and detector which are at rest with respect to each other. You can also show this result using Maxwell's equations, without using the idea of photons, or even explicitly using special relativity (Maxwell's equations are consistent with special relativity). Finally, as you argue in your question, conservation of energy can also be used to arrive at this result.