Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the frequency of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation."? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true?

  • $\begingroup$ @safesphere Ok, but we don't close questions that are duplicated on other sites, especially non-Stack Exchange sites. ;) However, this question is a spin-off from this SE Astronomy question. $\endgroup$
    – PM 2Ring
    Jan 16, 2019 at 7:57
  • $\begingroup$ @PM2Ring This is not the same question as in the astronomy. Its totally different. In the astronomy case the question is Why light cannot escape from a black hole, my question is photon loses energy in the gravitational redshift. I give an answer to the question but then I had a disagreement with the person who asked the question about the issue that I asked in this question. $\endgroup$
    – seVenVo1d
    Jan 16, 2019 at 8:50
  • $\begingroup$ @Reign I agree it's not the same question, that's why I said it's a spin-off. I've linked these 2 questions together because I think readers of one will be interested in the other, and it will be helpful for them to see the various answers & comments. $\endgroup$
    – PM 2Ring
    Jan 16, 2019 at 9:05
  • $\begingroup$ @safesphere "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation" - Correct, as long as you view from the same reference frame. Can you please elaborate on that a little bit? $\endgroup$ Jan 11, 2020 at 3:01
  • $\begingroup$ @safesphere thank you. What I do not understand is, how can the same observer (same reference frame) measure a photon's frequency in two frames? The observer must stay stationary, and then how can the observer be standing still and observe the photon (traveling at speed c) at two different places? $\endgroup$ Jan 11, 2020 at 15:36

1 Answer 1


The gravitational redshift in Schwarzschild relates the photon energy (frequency) measured by stationary observers at different radial coordinates. It refers to a free falling or free ascending light. The formula in the post assumes the farther observer to be at infinity (far away from the Schwarzschild radius).

If a light ray is free falling from infinity, it is blueshifted as a stationary observer closer to the massive object will measure a higher energy (frequency).

If a light ray is free ascending to infinity, it is redshifted as a stationary observer at infinity will measure a lower energy (frequency).

Informally speaking we may say that the light increases its energy (frequency) as it approaches (free falling) the massive object, while it decreases its energy (frequency) as it moves away (free ascending).

  • $\begingroup$ Your assumption has a strong logic but where does the energy "go"? I am worried about energy conservation rule... $\endgroup$
    – mattia.b89
    Nov 7, 2019 at 17:45
  • 1
    $\begingroup$ In the site there are many questions and related answers to the issue you raised. Basically a photon ascending a Schwarzschild spacetime geometry is similar to a ball thrown up in the gravitational field of the earth. The ball gradually loses its kinetic energy while increasing its gravitational potential energy (i.e. becoming less negative). The total energy is conserved. Consider that the famous equivalence between mass and energy $E = m c^2$ from special relativity means that energy gravitates; hence a photon informally can be thought as having a gravitational potential energy. $\endgroup$ Nov 9, 2019 at 9:01
  • $\begingroup$ With this you reintroduce potential energy. $\endgroup$
    – my2cts
    Jan 11, 2020 at 2:09
  • $\begingroup$ "If a light ray is free falling from infinity, it is blueshifted as a stationary observer closer to the massive object will measure a higher energy (frequency)." higher relative to what? $\endgroup$ Jan 11, 2020 at 2:48
  • 1
    $\begingroup$ @Árpád Szendrei. A higher energy (frequency) relative to the energy (frequency) at infinity. $\endgroup$ Jan 11, 2020 at 9:21

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