Does Gravity affect the frequency of an electromagnetic wave? If so, does it increases or decreases it? Please explain me. Thank you!
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2Read gravitational redshift here https://en.wikipedia.org/wiki/Gravitational_redshift – Paul Mar 10 '21 at 16:28
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Pehaps this is interesting for you https://physics.stackexchange.com/questions/276196/does-the-speed-of-light-change-in-a-gravitational-field/276233#276233 – HolgerFiedler Mar 11 '21 at 05:06
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If we describe a plane-wave (or a single photon) by its wave vector:
$$ k^{\mu} = (\omega/c, \vec k) $$
what pops out is that the frequency is the time-like part of a 4-vector. We can look at energy and momentum:
$$ p^{\mu} = \hbar k^{\mu} = (E/c, \vec p) $$
to try to find a natural rest frame:
$$ p^{\mu}p_{\mu} = \frac 1 {c^2}[(E)^2-||p||^2c^2] = \frac {\hbar^2} {c^2}[\omega^2 - (ck)^2] = 0 $$
so that there is no rest frame.
Energy is a property of the observer's frame, not the wave. The wave just goes at $c$, locally.
JEB
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@KarimChahine Frequency isn't a property of the wave, it depends on the reference frame. – JEB Mar 11 '21 at 02:40
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@JEB - But doesn't a gravitational field blue-shift (or is it red-shift, I always mix them up) EM radiation entering from infinity? – honeste_vivere Mar 12 '21 at 15:43
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1@honeste_vivere yes, you need 2 reference frames to compare frequencies. You didn't mention infinity in the question. – JEB Mar 12 '21 at 19:07
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