5
$\begingroup$

Let's assume two observers $A$ and $B$ hovering in a gravitation field. $A$ sends a radio transmission of frequency $f_1$ to $B$. $B$ receives this transmission and finds it has frequency $f_2$.

As as second experiment $A$ sends an electron beam to $B$. They measure the energies of these electrons on emission and reception.

Particles have de Broglie frequency which is proportional to energy. Will this frequency gets redshifted the same way as photon frequencies in a gravitational field so the rate of the original and redshift frequency/energy will be the same as in the case of photons?

In other words, for example, if a 900 keV photon, fired from $A$, gets red shifted to 850 keV when it arrives at $B$, will any massive 900 keV particle get slowed down to 850 keV after it free falls to $B$? - Given the rest mass is smaller than 850 keV, otherwise it would just fall back and never reach the other observer I guess.

I worked it out in flat spacetime between two accelerating observers that keep fixed distance between them, and $A$ just drops particle and $B$ just catches it. And in that case it seems the rate of the total energy of the received and dropped particle is exactly the rate of acceleration induced time dilation between the two observers. I'm unsure if this only works this way in this specific case or if it works in general in any gravitational or fictitious force fields.

$\endgroup$
  • 2
    $\begingroup$ Effectively a duplicate of Why isn't the De-Broglie wavelength of massive particles redshifted in an expanding universe $\endgroup$ – John Rennie Aug 18 '17 at 14:28
  • $\begingroup$ @JohnRennie But I'm not sure if cosmological redshift is the same kind of redshift as the gravitational one. I think cosmic expansion stretches wavelengths so it lowers momemtum, but the gravitational time dilation stretches frequencies so it lowers energy. Halving energy or momentum results in the same thing for light, but different for massive particles. $\endgroup$ – Calmarius Aug 18 '17 at 14:45
  • 1
    $\begingroup$ In a case where phase and frequency are directly measurable, as for an EM wave, the observable can be expressed as a frequency-wavenumber covector $\boldsymbol{\omega}$. In either a gravitational or a cosmological redshift, we're parallel-transporting $\boldsymbol{\omega}$ from one place to another in spacetime. When an observer then measures a frequency, they are measuring the timelike component of this vector in their own frame of reference. So in this sense, the two effects are exactly the same effect. GR doesn't have a distinction between gravitational redshifts and cosmological ones. $\endgroup$ – Ben Crowell Aug 18 '17 at 17:50
  • 1
    $\begingroup$ @Ben That's most of an answer and the core of a really good one. Why not post it as such. $\endgroup$ – dmckee Aug 18 '17 at 20:22
1
$\begingroup$

Case 1: Convert 1 kg of matter-antimatter to energy

Case 2: Lower 1 kg of matter-antimatter to a gravity well, convert it to energy there, beam the energy up to the original position.

From conservation of energy: Energy produced in case 1 = Energy produced in case 2

Energy in case2 = Energy beamed up + energy generated during lowering

Energy beamed = Energy of 1 kg of matter - energy lost in redshift

It must be so that: Energy lost in redshift = energy generated during lowering

Also: Energy generated during lowering = energy lost, or used, when the lowered thing is lifted back to the original position

So energy lost, or used, when the matter-antimatter is lifted back to the original position = energy lost in redshift

$\endgroup$
  • $\begingroup$ Maybe that's a useful answer, but in a real answer a chunk of matter would be climbing up from a gravity well, using its own energy, instead of being lifted. $\endgroup$ – stuffu Aug 21 '17 at 23:45
-2
$\begingroup$

They simply lose their energy. All of them. If the gravitational field is g, then raising by h in it, a particle of mass m loses

E=mgh

Of course, we are talking about the mass at the moment and changes are differential.

But the difference for photons is that their mass is their energy and vice versa. So, it can lower down to approximately 0. And the mass of an electron, for example, cannot go under 0.5Mev. So, the electron will enlarge its length of wave wore slowly.

After losing all kinetic energy, electron will start to fall back - down. And photon will continue to escape, only there is a height, at which it will lose all its energy. Their behaviour at point of turn won't be even a little bit covered by theory of relativity, but will be described by quantum physics. A standing electron so enlarge its indefinity of position, that looks really large. Almost standing atom can be large as an apple. Or more, of course. Standing still one will be big as the Universe.

$\endgroup$
  • $\begingroup$ The final paragraph is wrong. This is all classical physics. Re the first paragraph, see physics.stackexchange.com/questions/133376/… $\endgroup$ – Ben Crowell Aug 18 '17 at 17:38
  • $\begingroup$ The first paragraph is rather imprecisely stated as well. A free particle will lose that much kinetic energy if it rises ballistically in a region of effectively constant gravitational field (as in a small laboratory near the surface of the Earth). If the particle is not free, then we can't say where the potential energy would come from without further information. If the motion is in a region of non-constant gravitational field the algebraic expression for the energy transfer will be different. $\endgroup$ – dmckee Aug 18 '17 at 20:21
  • $\begingroup$ @dmckee The question IS about the free particles. What are you talking about? $\endgroup$ – Gangnus Aug 19 '17 at 21:31
  • $\begingroup$ @BenCrowell I am talking about the differential changes. The equation then is absolutely true in both GRT and classic physics. $\endgroup$ – Gangnus Aug 19 '17 at 21:34
  • $\begingroup$ @Gangnus I wouldn't have commented if there was only the one ambiguity in your writing here, but you've also silently assumed a constant field model when the most common practical example you can give a layman is GPS which involves long enough radial changes that the constant field approximation is inappropriate. $\endgroup$ – dmckee Aug 19 '17 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.