Listen to Record from Planet w/ Extreme Gravity

[anven]

TL;DR Summary

Listen to a record from a planet with extreme gravity

Hi everyone. I am new here and have a question that has been bothering me for a few days. Maybe you guys can help me out. Suppose a band gives a 60 minute concert on a planet with extremely strong gravity. This concert is recorded on site on a cassette, which has space for 60 minutes. Then a spaceship brings this cassette to me on earth. Would I notice anything unusual when I listen to the cassette? Thank you very much for your help.

 

[Nugatory]

Would I notice anything unusual when I listen to the cassette?

There would be no time dilation effects in the recording.

 

[anven]

There would be no time dilation effects in the recording.

thanks. and if the concert was digitally broadcasted live via glass fibre to earth? still nothing unusual?

 

[Ibix]

thanks. and if the concert was digitally broadcasted live via glass fibre to earth? still nothing unusual?

Then there would be time dilation - the broadcast would be stretched out and frequency reduced.

Not sure that fibres between planets are plausible, but you could use a radio or laser transmission.

 

[anven]

Then there would be time dilation - the broadcast would be stretched out and frequency reduced.

Not sure that fibres between planets are plausible, but you could use a radio or laser transmission.

and what if the concert was broadcasted to Earth and then directly broadcasted back to the other planet. would the band members hear themselves slower or normal again?

 

[PeroK]

Summary:: Listen to a record from a planet with extreme gravity

Hi everyone. I am new here and have a question that has been bothering me for a few days. Maybe you guys can help me out. Suppose a band gives a 60 minute concert on a planet with extremely strong gravity. This concert is recorded on site on a cassette, which has space for 60 minutes. Then a spaceship brings this cassette to me on earth. Would I notice anything unusual when I listen to the cassette? Thank you very much for your help.

Interesting question. It's not clear what effect the extreme gravity would have on the instruments, vocal chords or hearing of the band members. Even if they are playing in a special room that is pressurised to the Earth's atmospheric pressure, there is still the real (extreme) force from the floor of the room. At the very least, they will be moving sluggishly and walking may be difficult. The band members may be flat on the floor and unable to stand. Let alone pick up a guitar!

But, I guess the question is really about the nature of time and gravitational time dilation in particular. So, we need an experiment where the physical effect of the force from the ground is not relevant.

Let's assume that we have a clock (e.g. a cesium clock or perhaps a simple quartz crystal watch would do) that is not mechanically affected by the large local force. Then, we use that to time something and record it on a device that likewise is not affected by the local force. This may be as simple as recording a video on your phone.

When the phone is taken back to Earth and the video replayed, then it will play at normal speed. Perhaps a video of the atomic clock ticking away! In other words, a video that was, say, 1 minute long on the planet and recorded an atomic clock display for a minute will take a minute to play on Earth and the atomic clock in the video will be running normally.

If, however, each tick of the atomic clock is transmitted live from the planet by radio signal, say, then the atomic clock will be measured to be running slow by clocks on Earth.

What this highlights is that gravitational time dilation is not a local phenomenon whereby time "really" runs slower; but a global relativistic phenomenon, where a clock deep in a gravitational potential well will be measured to run slow by a clock not so deep in the well.

In that sense all clocks capable of keeping time locally are running normally. And, it's only the global curvature of spacetime that creates the asymmetry in measurements of time at different locations.

 

[anven]

Interesting question. It's not clear what effect the extreme gravity would have on the instruments, vocal chords or hearing of the band members. Even if they are playing in a special room that is pressurised to the Earth's atmospheric pressure, there is still the real (extreme) force from the floor of the room. At the very least, they will be moving sluggishly and walking may be difficult. The band members may be flat on the floor and unable to stand. Let alone pick up a guitar!

But, I guess the question is really about the nature of time and gravitational time dilation in particular. So, we need an experiment where the physical effect of the force from the ground is not relevant.

Let's assume that we have a clock (e.g. a cesium clock or perhaps a simple quartz crystal watch would do) that is not mechanically affected by the large local force. Then, we use that to time something and record it on a device that likewise is not affected by the local force. This may be as simple as recording a video on your phone.

When the phone is taken back to Earth and the video replayed, then it will play at normal speed. Perhaps a video of the atomic clock ticking away! In other words, a video that was, say, 1 minute long on the planet and recorded an atomic clock display for a minute will take a minute to play on Earth and the atomic clock in the video will be running normally.

If, however, each tick of the atomic clock is transmitted live from the planet by radio signal, say, then the atomic clock will be measured to be running slow by clocks on Earth.

What this highlights is that gravitational time dilation is not a local phenomenon whereby time "really" runs slower; but a global relativistic phenomenon, where a clock deep in a gravitational potential well will be measured to run slow by a clock not so deep in the well.

In that sense all clocks capable of keeping time locally are running normally. And, it's only the global curvature of spacetime that creates the asymmetry in measurements of time at different locations.

thank you very much. i think I'm getting there. so if i would receive the record, make a techno remix of it, and send the new record back to the other planet. this mix would sound perfectly normal on the other planet, right?

 

[PeroK]

thank you very much. i think I'm getting there. so if i would receive the record, make a techno remix of it, and send the new record back to the other planet. this mix would sound perfectly normal on the other planet, right?

Yes. Gravitational time dilation is not a "physical" effect. The whole clock/video thing really acts like a simple clock that is recorded, shipped around and rerun. Wherever it's rerun it will be measured to run at normal speed.

This is where it becomes difficult, perhaps, to find the right words. In general, gravitational time dilation is not physical effect.

There is another concept called differential ageing, which is a physical effect. Take a clock, move it to your other planet, leave it there for a while then bring it home and it will definitely have recorded less time than a clock that stayed on Earth. This is because the traveling clock took a shorter path through spacetime. And, the length of a spacetime path is an invariant quantity, which means it's the same whoever measures it.

Ironically, the traveling clock was never physically running more slowly than the Earth clock. This confounds many people!

The confusion between time dilation and differential ageing is at the heart of the twin paradox and is quite a difficult concept to grasp.

 

[anven]

Yes. Gravitational time dilation is not a "physical" effect. The whole clock/video thing really acts like a simple clock that is recorded, shipped around and rerun. Wherever it's rerun it will be measured to run at normal speed.

This is where it becomes difficult, perhaps, to find the right words. In general, gravitational time dilation is not physical effect.

There is another concept called differential ageing, which is a physical effect. Take a clock, move it to your other planet, leave it there for a while then bring it home and it will definitely have recorded less time than a clock that stayed on Earth. This is because the traveling clock took a shorter path through spacetime. And, the length of a spacetime path is an invariant quantity, which means it's the same whoever measures it.

Ironically, the traveling clock was never physically running more slowly than the Earth clock. This confounds many people!

The confusion between time dilation and differential ageing is at the heart of the twin paradox and is quite a difficult concept to grasp.

This is very helpful, thank you very much!

 

[vanhees71]

Yes. Gravitational time dilation is not a "physical" effect.

I'd not put it like this since the gravitational red or blue shift of light (em. waves) is an observable effect and thus physical. AFAIK it was first realized with $\gamma$ rays in the famous Pound-Rebka experiment:

https://en.wikipedia.org/wiki/Pound–Rebka_experiment

 

[PeroK]

I'd not put it like this since the gravitational red or blue shift of light (em. waves) is an observable effect and thus physical. AFAIK it was first realized with $\gamma$ rays in the famous Pound-Rebka experiment:

https://en.wikipedia.org/wiki/Pound–Rebka_experiment

Yes, this is why I said:

This is where it becomes difficult, perhaps, to find the right words. In general, gravitational time dilation is not physical effect.

The source of the gamma rays was not "physically affected" by gravity. It was the emitted gamma rays traveling through curved spacetime to the receiver that determined the observed wavelength. It's the relationship between the source and receiver that matters.

 

[vanhees71]

Well, for me it's the effect of the gravitational interaction of the em. field. (the gamma rays) due to the gravitational field of the Earth, which within GR of course is described through curved spacetime.

In this point of view the redshift is calculated using the eikonal approximation of the Maxwell equations, which boils down to solve
$$g^{\mu \nu} \partial_{\mu} \Phi \partial_{\nu} \Phi=0$$
for the eikonal (i.e., the phase of the em. wave). The four-vector of the light wave is given by $k_{\mu} = \partial_{\mu} \Phi$. The rays are given by $\Phi=\text{const}$. and thus $k^{\mu}$ are tangential to the rays, i.e., the ray equation can be written as $\dot{x}^{\mu}=k^{\mu}$. From this you get for the covariant derivative
$$\mathrm{D}_{\lambda}^2 x^{\mu}=\dot{x}^{\nu} \nabla_{\nu} k^{\mu}=k^{\nu} \nabla_{\nu} k^{\mu}=0.$$
The latter follows from taking the covariant derivative of
$$g_{\mu \nu} k^{\mu} k^{\mu}=0 \; \Rightarrow \; 2 k^{\nu} \nabla_{\nu} k^{\mu}=0.$$
So the rays are of course given as the null geodesics of spacetime (as usually just stated in the textbooks arguing in a naive photon picture).

Now if you have static metric the eikonal equation can be derived from the Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
Now the Schwarzschild metric is static, i.e., it doesn't depend on $x^{0}$ and thus
$$p_0=\frac{\partial L}{\partial \dot{x}^0}=\text{const}.$$
For the Schwarzschild metric for radial geodesics this means
$$p_0=g_00 \dot{x}^0=g_{00} k^0=\text{const}.$$
Now the measured frequency of an observer "at rest" (i.e., $r,\vartheta,\varphi=\text{const}$) is
$$\omega=\dot{\tau}=\frac{\mathrm{d} \tau}{\mathrm{d} x^0} k^0=\sqrt{g_{00}} k^0$$
and from that
$$\sqrt{g_{00}} \omega = g_{00} k^0=\text{const}.$$
So that (in the notation of and in accordance with the Wikipedia article)
$$\omega_r = \frac{\sqrt{g_{00}(R+h)} \omega_e}{\sqrt{g_{00}(R)}}.$$

 

[Vanadium 50]

he eikonal approximation of the Maxwell equations, which boils down

B-level?

 

[jbriggs444]

and what if the concert was broadcasted to Earth and then directly broadcasted back to the other planet. would the band members hear themselves slower or normal again?

Normal. The round trip signal delay is constant, so there is zero Doppler shift.

 

[PeterDonis]

Normal. The round trip signal delay is constant, so there is zero Doppler shift.

That, plus the fact that the spacetime geometry is stationary, so any gravitational time dilation on the way out is undone on the way back.

 

[alan123hk]

I'm just learning the basics of special relativity, so gravitational time dilation is still a very deep problem for me. But for the Doppler frequency shift of EM Waves, if the relative speed of the distant planet and the Earth is very large, and after the Earth receives the signal, the frequency is not corrected and sent back immediately, I feel like the Doppler effect will be doubled, Am I understanding something wrong ?

http://www.sfu.ca/phys/141/1134/Lectures/SP Lecture 23 - DopplerEffect&EMWaves.pdf
"Police radars get twice the effect since the EM waves make a round trip.."

 

[PeroK]

I'm learning the basics of special relativity, and gravitational time dilation is still a very deep problem for me.

Gravitational time dilation is not part of the theory of SR; it's part of the general theory of relativity and depends on curved spacetime. In particular, it is not the Doppler effect.

 

[Ibix]

I'm learning the basics of special relativity, and gravitational time dilation is still a very deep problem for me. But for the Doppler frequency shift of EM Waves, if the relative speed of the distant planet and the Earth is very large, and after the Earth receives the signal, the frequency is not corrected and sent back immediately, I feel like the Doppler effect will be doubled, Am I understanding something wrong ?

http://www.sfu.ca/phys/141/1134/Lectures/SP Lecture 23 - DopplerEffect&EMWaves.pdf
"Police radars get twice the effect since the EM waves make a round trip.."

Velocity induced Doppler and gravitational time dilation are very different things. In the absence of gravity, yes, if you emit a signal with frequency $\nu$ and I (due to my speed relative to you) receive it with frequency $\nu'=\alpha\nu$ and return the signal to you, it is as if I emitted it at frequency $\nu'$, so you receive it as $\nu''=\alpha\nu'=\alpha^2\nu$. But gravitational time dilation does not work that way. In that case, if I look at your clock and see it running slow, you will see mine running fast. So on a round trip the gravitational redshift cancels out.

 

[alan123hk]

I think I actually understand that gravitational time dilation is part of general relativity.
My English is not good. I have corrected it to be "I'm just learning the basics of special relativity, so gravitational time dilation is still a very deep problem for me.." It seems to be better.

 

[Sagittarius A-Star]

Velocity induced Doppler and gravitational time dilation are very different things.

An exception to this is pseudo-gravity in an accelerated rocket.

... At the front of the rocket is a lamp, which sends a short light pulse. At the rear end of the rocket with length Δh is a sensor, which receives this light pulse. I will show, that it is received blue-shifted. First, I define a “co-moving” inertial reference frame. The accelerated rocket shall have in this frame the velocity Zero at the point in time, when the light-pulse is sent. The light needs approximately Δt ≈ Δh / c until it reaches the sensor. After that time, the sensor has approximately the velocity

v = a * Δt ≈ a * Δh / c.

The sensor moves with that velocity into the light, that was sent out, when the lamp had the velocity Zero in the defined inertial frame. For small “v”, the formula for the classical Doppler effect can be used: $$f(received) / f(sent) \approx 1 + v/c = 1 + \frac {a * \Delta h} {c^2} = 1 + \frac {\Delta\phi} {c^2}$$ In the accelerated rest frame of the sensor, it is not a Doppler effect, but time-dilation between different pseudo-gravitational potentials Φ of lamp and sensor.

So on a round trip the gravitational redshift cancels out.

With an analog argument as shown in the above rocket scenario it can be shown, that if the photon will be reflected and moves up again, it will be red-shifted by the same factor and the blue-shift will be compensated.

Via the equivalence principle, the rocket calculation can be applied to gravity on Earth only locally, but not for a global scenario with another planet, like that in the O.P.

 

[David Lewis]

In general, gravitational time dilation is not physical effect. There is another concept called differential ageing, which is a physical effect.

If you have identical twins, and one is sent to the center of the planet, wouldn't he eventually be younger than his sibling on the surface? The longer he spends at the center of the planet the larger the age difference.

 

[Nugatory]

wouldn't he eventually be younger than his sibling on the surface?

That is only unambiguously true if the two twins eventually reunite so we can directly compare the amount of time (ticks of wristwatch, beats of hearts, …) each has experienced between the separation and reunion events.
As long as they are separated, any comparison of their ages will depend on our choice of simultaneity convention (definition of “at the same time”); this choice is arbitrary and depending on how we make it we can make the comparison of their ages come out either way.

If we do reunite them, we just have a variant of the Twin Paradox demonstrating differential aging, not time dilation. However in this variant the differential aging happens to be consistent with the expectation from time dilation, so instead of a paradox (traveling twin finds Earth clock to be running slow from time dilation yet returns younger) we are tempted to misunderstand the role of time dilation.