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In the presence of a uniform gravitational field two observers at fixed positions obtain different measurements of frequency of the same photon. One observer at the origin of some coordinate system measure $\nu_1$ and one at fixed position $x$ in the same coordinate chart measure $\nu_2$. The relation is

$$\nu_2 = \nu_1 \left( 1 + \frac{\phi}{c^2} \right) \tag1$$

Where the potential difference is $\phi \le 0$, such that $\nu_2 \le \nu_1$. However in Einstein's 1911 paper " On the Influence of Gravitation on the Propagation of Light", there is a formula for the speed of light that each observer obtain. If the observer that measures $\nu_1$ also measures $c_0$ and the observer that measures $\nu_2$ also measures $c$ we have the formula

$$c = c_0 \left( 1 + \frac{\phi}{c^2}\right) $$.

My point is: how the wavelength of the photon should change between these observers? Comparing the two formulas, I'm attempted to think that the observers agree about the measurement of wavelength, but since the frequency changes, the wavelength should also be different for each measurement. Where is my mistake?

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    $\begingroup$ It is good to keep always in mind that light emerges from the quantum mechanical level of photons, but photons are not light. They have energy h*nu, where nu is the frequency of the emergent light from very many photons, but the photons have no wavelength, they are point particles en.wikipedia.org/wiki/Elementary_particle $\endgroup$
    – anna v
    Jul 19, 2021 at 14:28
  • $\begingroup$ @annav thank you so much, this comments helped me a lot. However, if I change "photon" to " a light beam" in my question, would this problem remain? $\endgroup$ Jul 19, 2021 at 15:04
  • $\begingroup$ it would make it consistent with the answer, and with the fact you ask of measuring wavelengths. $\endgroup$
    – anna v
    Jul 19, 2021 at 15:39

2 Answers 2

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Why are you reading and taking seriously a paper from 1911? Einstein had not yet understand the interplay between gravity and time.

The formula you quote for for the speed of light is now known to be the speed in coordinate time. The actual speed of light measured by a clock is $c$ for all observers everywhere. The relation $\lambda \nu=c$ holds for all observers so a reduction in frequency is accompanied by an increase in wavelength.

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  • $\begingroup$ So the "transformation" of the wavelength need to be consistent only with eq. (1) and not with the speed of light equation? $\endgroup$ Jul 19, 2021 at 14:22
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    $\begingroup$ You should ignore the misleading (i.e wrong) "speed of light equation." $\endgroup$
    – mike stone
    Jul 19, 2021 at 15:20
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Consider a different approach, start from the geodesic equation for photon:

$$\frac{dP^{\mu}}{d\tau}+\Gamma^{\mu}_{\rho\nu}P^{\rho}U^{\nu}=0$$, where $P^{\mu}=(E,p,0,0)$ (say) is the 4-momentum of photon with $E=p$. Now use the de-Broglie relation $$p=\frac{h}{\lambda}$$ and substitute this into geodesic equation to obtain the "evolution" of wave-length $\lambda$ between two points on a given geodesic (On solving this equation, you can obtain $\lambda=\lambda(\tau)$, $\tau$ being the affine parameter on the null geodesic.).

For a constant gravitational field, you can construct an approximate metric $g_{ab}$ such that it satisfies (in the non-relativistic limit): $$\ddot{x}^i\approx-\Gamma^{i}_{tt}\left(\frac{dt}{d\tau}\right)^2=g^i$$ i=1,2,3 and $g^i$ are the components of constant gravitational field.

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