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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 ?

Thanks

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  • $\begingroup$ Duplicate: physicsforums.com/threads/… $\endgroup$ – safesphere Jan 16 at 7:45
  • $\begingroup$ "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. $\endgroup$ – safesphere Jan 16 at 7:54
  • $\begingroup$ @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 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. $\endgroup$ – Reign Jan 16 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 at 9:05
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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).

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