"Doppler" Gravity

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  • #1
.Scott
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Main Question or Discussion Point

When a force follows the inverse square law, its effects are stronger as the source approaches than when it recedes.
So light will blue shift (higher energy) and sound is louder and at a higher pitch.

So I would think that gravity would be stronger from an approaching object that from a receding one.
This effect would be important for objects travelling at relativistic velocities and would be in addition to the relativistic increase in mass.

But I haven't seen this mentioned in any articles.

Does this effect exist? If not, why not?
 

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  • #2
Nugatory
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Does this effect exist? If not, why not?
Doppler has nothing to do with whether a phenomenon is governed by an inverse square law or not. Machine gun bullets aren't subject to any inverse square law (in vacuum their momentum and kinetic energy is constant throughout their flight) but do demonstrate the Doppler effect: If you are flying head-on towards the gun you will be hit by more bullets per unit time than if you are flying away from it.
 
  • #3
.Scott
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OK. But my question is whether gravity is more intense from an approaching object than from a receding one.

I put "Doppler" in quotes because Doppler is specific to a frequency shift. In this case, there is no oscillation involved - but there should still be an increase or decrease in intensity.

Actually, the machine gun example further suggests that this effect would apply broadly - perhaps including to gravity.
 
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  • #4
PeterDonis
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So I would think that gravity would be stronger from an approaching object that from a receding one.
Why? Gravity is not EM radiation.

Also, what do you mean by "gravity would be stronger"? I strongly suggest looking at the math rather than trying to use intuition.
 
  • #5
PeterDonis
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This effect would be important for objects travelling at relativistic velocities and would be in addition to the relativistic increase in mass.
The source of gravity is not relativistic mass. It's the stress-energy tensor.
 
  • #6
Grinkle
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I would think that gravity would be stronger from an approaching object that from a receding one.
Waves from an approaching source are not higher in amplitude, they are shifted in frequency.

Gravity waves (which is not what you are talking about, I don't think) don't appear as far as I know just because a massive object is moving through spacetime, they need something exotic like a massive orbiting di-pole. If such a di-pole were approaching a gravity wave detector, I think the waves might have a shifted frequency but not an increased or decreased amplitude vs waves that would be generated from a di-pole not moving with respect to the detector. I am very unqualifed to make that statement - its just supposition on my part.

A massive object moving towards a gravity wave detector won't generate any gravity waves. You aren't talking about gravity waves per se, but your question did make me think some about that aspect.
 
  • #7
PAllen
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Waves from an approaching source are not higher in amplitude, they are shifted in frequency.

Gravity waves (which is not what you are talking about, I don't think) don't appear as far as I know just because a massive object is moving through spacetime, they need something exotic like a massive orbiting di-pole. If such a di-pole were approaching a gravity wave detector, I think the waves might have a shifted frequency but not an increased or decreased amplitude vs waves that would be generated from a di-pole not moving with respect to the detector. I am very unqualifed to make that statement - its just supposition on my part.

A massive object moving towards a gravity wave detector won't generate any gravity waves. You aren't talking about gravity waves per se, but your question did make me think some about that aspect.
Actually, for gravitational waves, dipole oscillation is not enough. You need nonzero second derivative of the quadrupole moment.
 
  • #8
.Scott
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Okay, here's some math:

We start with an observer at location 0 and a massive object (A) travelling along the line from ##-\infty## to ##+\infty## at ##c/2##.

I will assume (please indicate your agreement or objection) that the total change in momentum applied to the observer from the gravitational force of A will be the same as A moves from -4 Km to -3 (##p_{(-4,-3)}##) as when it moves from 3 Km to 4 (##p_{(3,4)}##).
However, from the observer's perspective:
- A reaches -4Km, -3, 3, and 4 at times -8Km/c, -6Km/c, 6Km/c and 8Km/c respectively.
- Gravity from A at -4Km, -3, 3, and 4 reaches the observer at -4Km/c, -3Km/c, 9Km/c and 12Km/c respectively.
So the momentum imparted to the observer from ##p_{(-4,-3)}## occurs over time 1Km/c while that from ##p_{(3,4)}## occurs over 3Km/c.

If my assumption is not correct, then consider the following scenario:
- observers are at -6Km and 6.
- object A materializes at -1Km, travels at ##c/2## and vanishes at +1
- would both observers have the same change in momentum?
 
  • #9
.Scott
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The source of gravity is not relativistic mass. It's the stress-energy tensor.
OK. I will look that up.
 
  • #10
Grinkle
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We start with an observer at location 0 and a massive object (A) travelling along the line from −∞-\infty to +∞+\infty at c/2c/2.
I think this is saying that your observer is falling into the gravity well of the massive object.

- observers are at -6Km and 6.
- object A materializes at -1Km, travels at c/2c/2 and vanishes at +1
- would both observers have the same change in momentum?
I may be missing your point, but when I think about this it just seems like objects moving into or out of gravity wells, not gravity being stronger or weaker?
 
  • #11
jbriggs444
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- object A materializes at -1Km, travels at c/2c/2c/2 and vanishes at +1
Cannot be done. Local energy conservation is a property of general relativity. You cannot materialize or dematerialize objects.
 
  • #12
.Scott
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I think this is saying that your observer is falling into the gravity well of the massive object.

I may be missing your point, but when I think about this it just seems like objects moving into or out of gravity wells, not gravity being stronger or weaker?
My point is that either the effect of gravity (the observers change in momentum) is different based on whether the massive object is approaching or receding. The difference is either:
- the total change in momentum;
- the rate of change (G force); or
- both
 
  • #13
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Cannot be done. Local energy conservation is a property of general relativity. You cannot materialize or dematerialize objects.
That is certainly true. But I can slice up time/space into time-like pieces and consider the changes that occur in each slice.
 
  • #14
jbriggs444
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That is certainly true. But I can slice up time/space into time-like pieces and consider the changes that occur in each slice.
No, you cannot. Because once you've sliced off a chunk of time there can be no change because there is no elapsed time. In any case, that does not let you materialize or dematerialize objects.
 
  • #15
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No, you cannot. Because once you've sliced off a chunk of time there can be no change because there is no elapsed time. In any case, that does not let you materialize or dematerialize objects.
The slice is not zero seconds in length. It is simply from one moment to another moment a few Km/c later.
If you don't like materializing and dematerializing objects, just use those moments at the start and end of the data collection period.
 
  • #16
jbriggs444
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The slice is not zero seconds in length. It is simply from one moment to another moment a few Km/c later.
If you don't like materializing and dematerializing objects, just use those moments at the start and end of the data collection period.
So now that we have permanently gravitating objects, what is the scenario that you are testing?
 
  • #17
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We start with an observer at location 0 and a massive object (A) travelling along the line from ##-\infty## to ##+\infty## at ##c/2##.

I will assume (please indicate your agreement or objection) that the total change in momentum applied to the observer from the gravitational force of A will be the same as A moves from -4 Km to -3 (##p_{(-4,-3)}##) as when it moves from 3 Km to 4 (##p_{(3,4)}##).
However, from the observer's perspective:
- A reaches -4Km, -3, 3, and 4 at times -8Km/c, -6Km/c, 6Km/c and 8Km/c respectively.
- Gravity from A at -4Km, -3, 3, and 4 reaches the observer at -4Km/c, -3Km/c, 9Km/c and 12Km/c respectively.
So the momentum imparted to the observer from ##p_{(-4,-3)}## occurs over time 1Km/c while that from ##p_{(3,4)}## occurs over 3Km/c.
I'm not sure what you mean by "gravity from A .... reaching the observer". Nothing is moving from the gravitational source to the observer, so there's nothing to reach anywhere. However, there's a deeper problem here: How exactly are you defining these points that you're labeling -4km, -3km, 3km, 4km and those times you're labeling -4km/c, -3km/c, 3km/c, 4km/c? Those are coordinates. How are they related to points in spacetime? And how are you comparing the coordinate-dependent three-momenta of the observer at the start and end of the two paths? The distances between 3km and 4km and between -3km and -4km are not 1km, they are "it depends", and not necessarily the same.

I expect that you are implicitly assuming a global inertial frame in which A starts at rest at (0,0), because only then do the answers to these questions appear obvious, and only then would it seem that there is a natural way that any sensible person committed to answering your question instead of quibbling would interpret your description of the thought experiment. Unfortunately, in this thought experiment there can be no such frame - global inertial frames exist only in the absence of gravitational effects, and this problem is specifically about gravitational effects.

To reason about this situation properly, you could either try looking at the Aichelburg-Sexl ultraboost (basically, the Schwarzschild spacetime described using coordinates in which the the gravitating body is moving at relativistic speeds), or work with the equivalent situation: The gravitating body is at rest so you can use Schwarzschold or other more common coordinates, and the observer is approaching it with a non-zero coordinate velocity. The latter approach is better for building an intuitive picture of what's going on, and going through it will demonstrate just how tricky it is to describe the setup accurately.

You should end up with a gravitational slingshot effect like what we see in Newtonian physics when a spacecraft passes by a planet (and identical in the non-relativistic limit).
I withdraw this last sentence. There is a very large can of worms hiding behind it, and getting right would be a huge digression in this thread.
 
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  • #18
.Scott
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So now that we have permanently gravitating objects, what is the scenario that you are testing?
There were a total of three scenarios, four if you count scenario/observer combinations. I described all well.
In all cases, the mass is moving over an interval directly towards or away from an observer - at a relative speed of c/2.

The results are changes in the momentum for the observers. But since the period of time that the observers are affected is different depending on whether the mass is approaching or receding. So either the total momentum is different in those cases or the rate of momentum change is different. I am still not sure which.
 
  • #19
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o either the total momentum is different in those cases or the rate of momentum change is different. I am still not sure which.
Either or both. The three-momentum and its rate of change are both coordinate-dependent, so which it is depends on your choice of coordinates.
 
  • #20
.Scott
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I'm not sure what you mean by "gravity from A .... reaching the observer".
This is the ol' "If the sun disappear now, how long would it take for the Earth to leave orbit?".
Presumable, the effect of any such change will take 8 minutes.

Nothing is moving from the gravitational source to the observer, so there's nothing to reach anywhere.
Would it help if I postulated a mass of photons colliding with another mass and interacting to form object A and the specified velocity and position? Then decaying into two masses of photons and the end position?

However, there's a deeper problem here: How exactly are you defining these points that you're labeling -4km, -3km, 3km, 4km and those times you're labeling -4km/c, -3km/c, 3km/c, 4km/c? Those are coordinates. How are they related to points in spacetime?
There is only one space dimension in these scenarios. Nothing happens in the other two spacial dimensions (except thoses masses of photons I just introduce in this post. So the observer is at the 0Km position and all times are observer times.

And how are you comparing the coordinate-dependent three-momenta of the observer at the start and end of the two paths? The distances between 3km and 4km and between -3km and -4km are not 1km, they are "it depends", and not necessarily the same.
Everything is in the observer's frame. So the question I have about the differences is from the observer's frame.

I expect that you are implicitly assuming a global inertial frame in which A starts at rest at (0,0), because only then do the answers to these questions appear obvious, and only then would it seem that there is a natural way that any sensible person committed to answering your question instead of quibbling would interpret your description of the thought experiment. Unfortunately, in this thought experiment there can be no such frame - global inertial frames exist only in the absence of gravitational effects, and this problem is specifically about gravitational effects.
OK. But my question should still be addressable in terms of what the net effect would be.

To reason about this situation properly, you could either try looking at the Aichelburg-Sexl ultraboost (basically, the Schwarzschild spacetime described using coordinates in which the the gravitating body is moving at relativistic speeds), or work with the equivalent situation: The gravitating body is at rest so you can use Schwarzschold or other more common coordinates, and the observer is approaching it with a non-zero coordinate velocity. The latter approach is better for building an intuitive picture of what's going on, and going through it will demonstrate just how tricky it is to describe the setup accurately.
I will add that to "Tensors" to my homework list. So far, I haven't run into any "meat" in my Gravitational Tensor reading.
 
  • #21
PeterDonis
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This is the ol' "If the sun disappear now, how long would it take for the Earth to leave orbit?".
Which, as has already been pointed out, is impossible because objects like the sun can't disappear; stress-energy is locally conserved.

It turns out to be highly non-trivial to formulate a consistent scenario to test "the speed of gravity". For an introduction to some of the issues involved, I suggest this classic paper by Carlip:

https://arxiv.org/abs/gr-qc/9909087
 
  • #22
jbriggs444
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This is the ol' "If the sun disappear now, how long would it take for the Earth to leave orbit?".
Again, that is not a valid question. The sun cannot disappear.
 
  • #23
.Scott
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Either or both. The three-momentum and its rate of change are both coordinate-dependent, so which it is depends on your choice of coordinates.
OK...
Let me present a somewhat more complicated thought experiment - but one that is easier to describe, although probably much more complicated to address the reasoning:

Let's have a flat universe with many stationary (in their frame) gravitation sources (perhaps galaxies) evenly disbursed.
Now we will introduce another galaxy travelling through this universe at c/2.
Taking this galaxy as our frame, will this galaxy accelerate? If it does, it would be because there is a kind of "Doppler" (for lack of a better term) that is related to gravity's propagation rate. If it does not, then gravity is doing something that I don't understand. Somehow it isn't really propagating at the speed of light - because our moving galaxy isn't "running into" this propagation and causing it to "bunch up".
 
  • #24
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Everything is in the observer's frame. So the question I have about the differences is from the observer's frame.
Well, that makes the problem easy.... The observer's momentum using that frame (which is not global) is initially zero and never changes.
But I expect that what you really meant was "a global inertial frame in which the observer is initially at rest", and there is no such thing. There is a local inertial frame in which the observer is initially at rest, but because it is local it cannot be used to assign times and positions to more distant points.
I will add that to "Tensors" to my homework list. So far, I haven't run into any "meat" in my Gravitational Tensor reading.
The first step (you'll find this towards the start of just about every GR textbook) is to see how ordinary flat-space special relativity works when written in tensor form. This allows you to clearly separate the coordinate artifacts from the actual coordinate-dependent physics, the stuff that we can measure and observe.
 
  • #25
.Scott
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Well, that makes the problem easy.... The observer's momentum using that frame (which is not global) is initially zero and never changes.
But I expect that what you really meant was "a global inertial frame in which the observer is initially at rest", and there is no such thing.
I didn't consider that problem. You're right. I was trying to use a more global frame. But I think the problem with such a global frame is small when compared to the 3:1 ratio in the exposure rate to the receding and approaching masses.
 

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