Speed of Gravity: Is the Sun 8 Minutes Behind?

In summary: Makes perfect sense. In summary, scientists currently believe that the speed of gravity is between the speed of light (c) and 1.06 times the speed of light.
  • #1
KingBigness
96
0
I understand that Newton thought that the effect of gravity was instantaneous, however Einstein believed it to 'travel' at the speed of light. I also read that in 2002 scientists concluded that gravity traveled at 1.06 times that of the speed of light but with an error of plus or minus 0.21 putting it in the range of the speed of light.

My question that has been keeping me up at night though is, if the light from the sun is 8 minutes old, does that imply that we are in fact not orbiting the sun but orbiting where the sun was 8 minutes ago?

Sorry if this is posted in the wrong section.
 
Physics news on Phys.org
  • #2
Going on the basis that if the sun were to instantaneously disappear and the Earth would keep on following the same orbit for 8m20s before shooting off on a tangent, then yes we are orbiting where the sun was 8m20s ago.
 
  • #3
djsubtronic said:
Going on the basis that if the sun were to instantaneously disappear and the Earth would keep on following the same orbit for 8m20s before shooting off on a tangent, then yes we are orbiting where the sun was 8m20s ago.

This is one way of looking at it, but it leaves out some key points.

(1) Consider light first instead of gravity. The light from the Sun reaching the Earth is not only time delayed; it shows aberration as well. That is, it comes at us from the *direction* the Sun was 500 s ago, not the direction the Sun is now. This effect gives one way of estimating the speed of light.

(2) The apparent "force" of gravity from the Sun on the Earth doesn't come from where the Sun was 500 seconds ago; it comes from where the Sun is now. (Actually that isn't exactly true--see #3 below--but it is to a very good approximation.) That is, the apparent direction of gravity from the Sun, unlike the apparent direction of light from the Sun, does *not* show aberration. So the "apparent" speed of gravity, if you try to deduce it from the "aberration" alone, is *much* faster than c.

(3) However, the deduction of speed of an effect from observed aberration depends on the theory you use. In the case of light, our best theory (at least the one that's good enough for this purpose) is classical electromagnetism, based on Maxwell's Equations, which predict a speed of light equal to c and aberration of light from moving sources. In the case of gravity, however, our best theory is General Relativity, not Newtonian theory; but deducing the "speed of gravity" from the observed (lack of) aberration only works in Newtonian theory. In GR, gravity is not really a "force" at all, it's spacetime curvature; in cases like the Sun and the Earth, gravity in GR can still be viewed as a "force", but it's a force that depends on the Earth's velocity and acceleration as well as its position relative to the Sun, so it's not a Newtonian-type force and you can't deduce its speed from the (lack of) aberration.

When you do the correct GR calculation of the speed of gravity from observations, it comes out to c; as the OP notes, the best current experiments are consistent with that value. So you are right that, if something were to change about the Sun that changed its effect on the Earth, it would take 500 s for the Earth to feel the effect; but that doesn't mean that the apparent "force" from the Sun on the Earth points in the direction the Sun was 500 s ago, because of the velocity and acceleration dependent effects.

Steve Carlip wrote an excellent paper years ago on this subject, which I highly recommend reading:

http://arxiv.org/abs/gr-qc/9909087

This paper gives a more detailed discussion of, and also the math behind, the statements I made above.
 
  • #4
Going on the basis that if the sun were to instantaneously disappear and the Earth would keep on following the same orbit for 8m20s before shooting off on a tangent, then yes we are orbiting where the sun was 8m20s ago.
I disagree. The gravitational attraction of the sun, even in general relativity, does not exhibit retardation. The Earth is orbiting the current location of the sun. This follows immediately from solving for the orbit in the rest frame of the sun and Lorentz transforming it.
 
  • #5
djsubtronic said:
Going on the basis that if the sun were to instantaneously disappear and the Earth would keep on following the same orbit for 8m20s before shooting off on a tangent, then yes we are orbiting where the sun was 8m20s ago.

Wow. So when we are leading the sun, say at midwinter (or whenever) it would follow that we are closer to the sun by something like 8 light-minutes, and when we are trailing the sun, we are further away by a like amount?

So we don't orbit the center of mass, but rather, we orbit the place where the center of mass used to be 8 minutes ago?

Are these correct surmises?

And if the sun is moving very fast relative to some convenient point of reference, then is the Earth's path around it really a huge elipse?
 
  • #6
No, of course not.

Take the solar system, and view it in a rest frame which is traveling "North" at 0.9c. According to this idea, Mercury does not know where the sun is now, it follows it, orbiting the point where the sun used to be 3 minutes ago. For the Earth it's 8 minutes, for Jupiter 40. Each planet lags a bit farther "South". Poor Pluto has to orbit a spot 4 light-hours to the South of the sun's current location. The plane of the ecliptic would not be a plane! This is nonsense.

Newtonian gravitation does not emerge from the sun and travel outward at the speed of light. The planets react to the sun's current gravitational field. The interaction appears instantaneous. Although this seems paradoxical, and even asking for trouble, it is not. Exactly the same thing happens in electromagnetism, where the Coulomb interaction is effectively instantaneous. Nature can get away with this without violating any principles because both the Newtonian field and the Coulomb field are static. Likewise in quantum electrodynamics the Coulomb interaction is instantaneous. Changes, of course, including waves, propagate at c.
 
  • #7
High-amplitude gravitational waves need *not* propagate at c. For example, GR predicts that a gravitational-wave pulse propagating on a background of curved spacetime develops a trailing edge that propagates at less than c.[MTW, p. 957] This effect is weak when the amplitude is small or the wavelength is short compared to the scale of the background curvature.
This is called a wave tail. For example an outward-going wave on a Schwarzschild background will develop two kinds of tail, one an inward-going backscattered wave, the other a monopole field that reduces the Schwarzschild mass to account for the energy that's being carried away.

Discontinuities always travel at c. The Einstein equations are hyperbolic, and their characteristics are null surfaces, bicharacteristic curves are null geodesics.
 
  • #8
Bill_K said:
Exactly the same thing happens in electromagnetism, where the Coulomb interaction is effectively instantaneous.
I like that comparison because it avoids all the curved space time complexity.

When you travel at constant speed relative to a charged object (static E-field), that also emits light (EM-waves), you will see the object at a different direction than you will measure the Coulomb force from:
- The seen position is outdated
- The Coulomb force points to the correct position
And yet when the object is suddenly accelerated, both observed directions will be affected after the same delay by the acceleration.

Bill_K said:
Nature can get away with this without violating any principles because both the Newtonian field and the Coulomb field are static.
I think the key problem people have is accepting that an inertially moving source can produce a static field, that moves along with the source, and doesn't stay behind it like sound waves. A lot of the wrong intuition here comes from the plane noise that comes from a direction where the plane was 15sec ago.
 
  • #9
A.T. said:
.


I think the key problem people have is accepting that an inertially moving source can produce a static field, that moves along with the source, and doesn't stay behind it like sound waves. A lot of the wrong intuition here comes from the plane noise that comes from a direction where the plane was 15sec ago.

My impression (which could be wrong) is that the confusion arises from the rather naive idea of thinking that the force of gravity is carried by little particles, emitted from the sun, and that the "force" occurs when these particles are absorbed.

Unfortunately, this picture is basically wrong :-=(. But it's not clear if it's so wrong if it can't be made to work somehow, if one replaces real particles with virtual particles. However, the details of how to do this aren't particularly clear even in the much simpler electromagnetic case.

It is also clear that conservation laws forbid the sun from disappearing suddenly (or, in the electromagnetic case, forbid charges from disappearing suddenly). So trying to work out what happens mathematically when you assume it happens anyway gives total nonsense, a set of equations that has no solution. But it seems that this observation doesn't seem to be helpful in resolving the dispute - probably because the people having the issues don't actually think about trying to solve the equations mathematically instead relying on rather simple mental pictures (like the particle carrying the force) which they are very, very attached to, while they feel that the math is somehow "beyond their grasp".
 
  • #10
Bill_K said:
No, of course not.

Take the solar system, and view it in a rest frame which is traveling "North" at 0.9c. According to this idea, Mercury does not know where the sun is now, it follows it, orbiting the point where the sun used to be 3 minutes ago. For the Earth it's 8 minutes, for Jupiter 40. Each planet lags a bit farther "South". Poor Pluto has to orbit a spot 4 light-hours to the South of the sun's current location. The plane of the ecliptic would not be a plane! This is nonsense.

Newtonian gravitation does not emerge from the sun and travel outward at the speed of light. The planets react to the sun's current gravitational field. The interaction appears instantaneous. Although this seems paradoxical, and even asking for trouble, it is not. Exactly the same thing happens in electromagnetism, where the Coulomb interaction is effectively instantaneous. Nature can get away with this without violating any principles because both the Newtonian field and the Coulomb field are static. Likewise in quantum electrodynamics the Coulomb interaction is instantaneous. Changes, of course, including waves, propagate at c.
Bill -

Were you addressing me or another poster?

By the sun's "current gravitaional field" do you mean the field which corresponds to the suns present position for a close-by body co-moving with it?

How does anybody which is at a significant distance respond to the current position, when the body "sees" the sun at a position which it occupied in the past (as determined by the close-in, comoving observer)? Why wouldn't a distant body respond to the center of mass as it appears currently to that body? For it to be otherwise, the distant body must anticipate where the sun will be in, say 8 minutes, and not where it appears currently.
 
  • #11
[Note: I have edited this post to correct misstatements about acceleration.]

EskWIRED said:
For it to be otherwise, the distant body must anticipate where the sun will be in, say 8 minutes, and not where it appears currently.

The Carlip paper that I linked to in post #3 explains how this works; it more or less amounts to just what you say here. But as Bill_K notes, there's no paradox or violation of causality involved; what is happening is that the velocity-dependent terms [Edit: removed "acceleration dependent"] in the "force" compensate for the light-speed time delay to make the apparent "force" point in the direction the Sun is "now" instead of where it was 8 minutes ago.

As Carlip puts it, the "force" the Earth feels from the Sun is based, not on the Sun's position 8 minutes ago, but on the Sun's position "quadratically extrapolated" from 8 minutes ago, using the Sun's velocity [Edit: removed "acceleration"] relative to the Earth, 8 minutes ago, to do the extrapolation. (As Carlip points out, in the case of electromagnetism, the extrapolation is only linear; the Coulomb force felt by a charged particle is based on the source "linearly extrapolated" over the light-speed time delay. [Edit: removed "acceleration"])

Of course the extrapolation isn't perfect; the Earth does not "know" exactly where the Sun is "now", but only where it extrapolates the Sun to be "now" based on information from 8 minutes ago. The extra perihelion shift in planetary orbits that is predicted by GR but not by Newtonian theory (so far observed, AFAIK, only for Mercury) is a consequence of the fact that the extrapolation is not perfect (or, to put it another way, a consequence of the fact that the "speed of gravity" is the speed of light).
 
Last edited:
  • #12
the distant body must anticipate where the sun will be in, say 8 minutes, and not where it appears currently.
Yes, that's exactly what does happen! And it's a good thing too. Because here's a further example: consider two equal masses in relativistic motion about each other. Normally the attraction between them is diametrical, towards the center of mass. But if the attraction between them were retarded, each would feel an attraction to a point which lags behind the other, causing the orbital motion to slow down and eventually come to a halt!

But don't take my word for it, here's an article by John Baez, and also a published paper which addresses the question. Both discussions start with the electromagnetic case. I quote:

If a charged particle is moving at a constant velocity, it exerts a force that points toward its present position, not its retarded position, even though electromagnetic interactions certainly move at the speed of light. Here, as in general relativity, subtleties in the nature of the interaction "conspire" to disguise the effect of propagation delay. It should be emphasized that in both electromagnetism and general relativity, this effect is not put in ad hoc but comes out of the equations. Also, the cancellation is nearly exact only for constant velocities. If a charged particle or a gravitating mass suddenly accelerates, the change in the electric or gravitational field propagates outward at the speed of light.

It is well known that if a charged source moves at a constant velocity, the electric field experienced by a test particle points toward the source’s “instantaneous” position rather than its retarded position. Lorentz invariance demands that this be the case, since one may just as well think of the charge as being at rest while the test particle moves. This effect does not mean that the electric field propagates instantaneously; rather, the field of a moving charge has a velocity-dependent component that cancels the effect of propagation delay to first order.
 
  • #13
EskWIRED said:
For it to be otherwise, the distant body must anticipate where the sun will be in, say 8 minutes
This is correct, it extrapolates the movement of the source. If the moving source was suddenly stopped, you would notice this only after a delay, and initially continue to be accelerated towards the wrong extrapolated position.

You can think of this in two ways:

1) We can go to the inertial rest frame of the moving source, where the field is static, and it's obvious that it always pulls the now moving object towards the correct position of the source. Unless the source accelerates.

2) A moving source "emits" a field that points towards the source, but also "inherits" the velocity of the source. So as long the source moves at constant velocity, the distant field-lines continue to point at the current source location. When the source suddenly stops, the information about it travels through the field at c, so initially the distant parts of the field continue to move at the original velocity of the source, and now they point towards a wrong extrapolated position of the source.

This animation shows the opposite scneriow: acceleration of a charge from rest. But stopping it is similar: The moving field-lines continue to move, and point to an extrapolated position, until the disturbance (wave) reaches their position:

field_a.gif


From: http://www.tapir.caltech.edu/~teviet/Waves/empulse.html
 
Last edited by a moderator:
  • #14
This is fascinating stuff.
pervect said:
Unfortunately, this picture is basically wrong :-=(. But it's not clear if it's so wrong if it can't be made to work somehow, if one replaces real particles with virtual particles. However, the details of how to do this aren't particularly clear even in the much simpler electromagnetic case.

So is it currently thought that the graviton doesn't exist?
 
  • #15
Rooted said:
This is fascinating stuff.


So is it currently thought that the graviton doesn't exist?

No, that is not true. A field is mediated by virtual particles, and the field of virtual spin 2, massless bosons matches the predictions of GR to a very high precision.
 
  • #16
PAllen said:
No, that is not true. A field is mediated by virtual particles, and the field of virtual spin 2, massless bosons matches the predictions of GR to a very high precision.

Good, I'm glad about that! Thanks!
 
  • #17
Bill_K said:
Discontinuities always travel at c. The Einstein equations are hyperbolic, and their characteristics are null surfaces, bicharacteristic curves are null geodesics.

I'm inclined to be skeptical of this claim. Logically, there are significant difficulties in defining what would be meant by the speed of propagation of a gravitational wave in the general, strong-field case. They're similar to the difficulties involved in proving geodesic motion from the field equations. You have to somehow distinguish two different metrics, one that would have existed if there had been no wave disturbance, and one in which the wave disturbance exists. Then you have to say what you mean by the speed of propagation of the wave, which presumably you want to measure in one or the other of these two metrics. So it seems likely to me that even if the second quoted sentence holds, it may not be interpretable as claimed in the first sentence, in the general strong-field case.

I had to look up the definitions of the mathematical terms you used:
http://en.wikipedia.org/wiki/Hyperbolic_partial_differential_equation
http://en.wikipedia.org/wiki/Method_of_characteristics
By the definition in the WP article, it seems clear to me that the Einstein field equations do not qualify as hyperbolic in general. This is the reason that we have to define global hyperbolicity (Hawking and Ellis, p. 206). On a spacetime that doesn't have global hyperbolicity in the Hawking and Ellis sense (e.g., a spacetime that has CTCs), we don't generally expect to have uniqueness and existence of solutions to Cauchy problems. H&E refer to a theorem by Geroch, J Math Phys 11 (1970) 437 (I haven't read it), which states that if the spacetime is globally hyperbolic in the H&E sense, then Cauchy surfaces exist. Since the existence of Cauchy surfaces is what WP is giving as the definition of a hyperbolic PDE, I think we can't generally assert that the field equations are hyperbolic -- if we could, then Geroch's 1970 theorem would have been trivially true.

A major application of the method of characteristics seems to be to things like shock waves and fluid dynamics, where we don't have any of the GR issues I worried about above. The second WP article has this, which I think it physically suggestive: "Characteristics may fail to cover part of the domain of the PDE. This is called a rarefaction, and indicates the solution typically exists only in a weak, i.e. integral equation, sense."
 
  • #18
A.T. said:
This is correct, it extrapolates the movement of the source. If the moving source was suddenly stopped, you would notice this only after a delay, and initially continue to be accelerated towards the wrong extrapolated position.

I was reading through, and it stuck to me that if the moving charge is suddenly stopped and we would continue to extrapolate the direction of the force(which is wrong). Does this imply that EM fields must carry momentum in order to preserve conservation of momentum ?

And if EM fields do carry momentum, what does that even mean, when we consider NO net motion of a system carrying static fields?
 
Last edited:
  • #20
Ben,

I'm talking local, not global hyperbolicity, which is a much more difficult subject, hats off to Hawking and Ellis! MTW for example discuss at length the propagation of gravitational waves on a curved background in the short wavelength limit (shock waves are extremely short wavelength) They derive among other things a linear wave equation for the perturbation,

hμν + 2Rαμβν hαβ = 0

which is manifestly hyperbolic. (R is the background Riemann tensor.) See especially their exercise 35.15 on the "geometric optics of gravitational waves of small amplitude propagating in a curved background," which discusses propagation of various quantities along the rays (i.e. null geodesics).

Or, if you're into Italian :biggrin: the original reference,

""Caratteristiche e bicaraterristiche delle equazioni gravitazionali di Einstein", T. Levi-Civita, Rend Acc Lincei Sci Fis Mat Nat 11 (1930) 3, 113.
 
  • #21
PeterDonis said:
This is one way of looking at it, but it leaves out some key points.

(1) Consider light first instead of gravity. The light from the Sun reaching the Earth is not only time delayed; it shows aberration as well. That is, it comes at us from the *direction* the Sun was 500 s ago, not the direction the Sun is now. This effect gives one way of estimating the speed of light.

(2) The apparent "force" of gravity from the Sun on the Earth doesn't come from where the Sun was 500 seconds ago; it comes from where the Sun is now. (Actually that isn't exactly true--see #3 below--but it is to a very good approximation.) That is, the apparent direction of gravity from the Sun, unlike the apparent direction of light from the Sun, does *not* show aberration. So the "apparent" speed of gravity, if you try to deduce it from the "aberration" alone, is *much* faster than c.

(3) However, the deduction of speed of an effect from observed aberration depends on the theory you use. In the case of light, our best theory (at least the one that's good enough for this purpose) is classical electromagnetism, based on Maxwell's Equations, which predict a speed of light equal to c and aberration of light from moving sources. In the case of gravity, however, our best theory is General Relativity, not Newtonian theory; but deducing the "speed of gravity" from the observed (lack of) aberration only works in Newtonian theory. In GR, gravity is not really a "force" at all, it's spacetime curvature; in cases like the Sun and the Earth, gravity in GR can still be viewed as a "force", but it's a force that depends on the Earth's velocity and acceleration as well as its position relative to the Sun, so it's not a Newtonian-type force and you can't deduce its speed from the (lack of) aberration.

When you do the correct GR calculation of the speed of gravity from observations, it comes out to c; as the OP notes, the best current experiments are consistent with that value. So you are right that, if something were to change about the Sun that changed its effect on the Earth, it would take 500 s for the Earth to feel the effect; but that doesn't mean that the apparent "force" from the Sun on the Earth points in the direction the Sun was 500 s ago, because of the velocity and acceleration dependent effects.

Steve Carlip wrote an excellent paper years ago on this subject, which I highly recommend reading:

http://arxiv.org/abs/gr-qc/9909087

This paper gives a more detailed discussion of, and also the math behind, the statements I made above.

Hi
I have been thinking about this phenomenon for awhile and have gone over the linked paper but the math is too advanced to be much help. Could you possibly describe in conseptual terms the principles involved in the explanation. What are the velocity dependent factors that are being exprapolated and how that applies as a hypothetical physical process ??

Having thought about it, it seems to me there is a simple explanation in the case of linear velocity and charged particles.
Regarding a photon , the observed directional vector points to the retarded position due to aberration. But if we consider the wave fronts of the photon as being orthogonal to the emission path then the observed direction is at an angle other than 90deg. to the actual angle of the received wavefront .
If the field and particle accelerations are mediated by virtual photons then it does not seem like too great a leap to assume the resulting field or force vectors would be othogonal to the actual wave orientation rather than the perceived direction of reception. In this case if they are orthogonal to the wavefront ,they would then point at the instantaneous position due simply to aberration.
This would seem to apply equally to closing and reccession .

In the case of gravity in a one body system like the solar system, where the sun can be considered to be in linear motion it might also apply but in a binary system it falls apart completely as far as I can see.

Another mystery of how nature synthesises symmetry and equilibrium out of manifestly asymmetric conditions due to motion and light speed interactions.

SO any help understanding that paper and how a binary system could operate would be appreciated as this question is nagging at me.

Thanks
 
  • #22
Austin0 said:
Could you possibly describe in conceptual terms the principles involved in the explanation. What are the velocity dependent factors that are being extrapolated

First, a general comment: the only reason for even discussing all this stuff about "velocity dependent factors" and so forth is that people insist on some "conceptual" explanation of gravity as a "force". Any such explanation is going to break down at some level, because gravity is not really a "force" in the sense in which the term is being used here. See further comments below.

That said, if we want to view, say, the Sun's gravity acting on the Earth as a "force", the velocity-dependent factors compensate for the light-speed time delay in the propagation of the force from the Sun to the Earth. This compensation makes it appear that the direction of the force is the direction the Sun is in "now", instead of the direction it was in 500 seconds ago (the light-speed time delay). The compensation is not exact; the direction of the force is not the "actual" direction of the Sun "now", but rather the direction the Sun would be in "now" if it moved in an exact straight line from where it was 500 seconds ago, at exactly the relative velocity it had 500 seconds ago. Or, as Carlip puts it, the force points in the direction of the Sun's position "quadratically extrapolated" 500 seconds forward (i.e., to "now") from its position and velocity 500 seconds ago. The "quadratically" refers to how accurately the extrapolation cancels out the effects of the time delay; for electromagnetism, the extrapolation is only linear, which is a less accurate cancellation, so EM effects, such as light, show measurable aberration under circumstances where gravity does not.

Austin0 said:
and how that applies as a hypothetical physical process ??

The best answer to this is that the "extrapolations" and "velocity dependent factors" don't describe any actual physical process; they are just a conceptual crutch to help us understand why gravity, if we insist on viewing it as a "force", doesn't show aberration effects.

I'll defer comment on the rest of your post until I've digested it some more.
 
  • #23
Austin0 said:
In the case of gravity in a one body system like the solar system, where the sun can be considered to be in linear motion it might also apply but in a binary system it falls apart completely as far as I can see.

Not sure it does. When you look at the accelerating charge again, it's obvious that along the line of acceleration the field always points to the correct position. And in an binary system each object is on the the line of acceleration of the other.

But with three objects it indeed seems there must be delayed acceleration.
 
  • #24
Bill_K said:
Yes, that's exactly what does happen! And it's a good thing too. Because here's a further example: consider two equal masses in relativistic motion about each other. Normally the attraction between them is diametrical, towards the center of mass. But if the attraction between them were retarded, each would feel an attraction to a point which lags behind the other, causing the orbital motion to slow down and eventually come to a halt!

:

This is a good example case you are describing here.
In a little more detail:
Two equal masses in identical elliptical orbits (with some eccentricity) around the common barycenter.
During the recessional phase, the retarded postion of the point of emission is a distance Dr ,less than the actual instantaneous position distance Di relative to the point of reception. Dr < Di SO the negative acceleration arising from this is greater than the force should be from the actual position by the inverse square difference in distance

Comparably, during closing the Dr > Di so the positive acceleration is less than the Coulomb factor for the instantaneous distance by the same degree.

Obviously leading to a continual decrease in orbital velocity and catastrophic orbital decay as you described.

I am proposing that the g field associated with the retarded position is anisotropic, polarized, at least partly due to aberration.
So in the first case, of recession, the reduced field flux (varying with angle) due to aberration counters the increase due to less distance.
Likewise in closing. .The increase in field strength due to aberration increases the positive acceleration, countering the decrease due to the greater distance of the retarded position.

I am suggesting that in both cases the flux vector, transported along the null geodesic between the retarded position and the location of the other mass would retain the directional components of the actual (un-aberrated ) direction of emission. Which vector would point to the actual , instantaneous position of the gravitating body.

I don't know the accepted view having never encountered a mention in the context of electrodynamics , but it seems clear to me that the Guass and Coulomb laws lead to the conclusion that the field of a moving charge must be aberrated.
That the animation that A.T. provided is incorrect in it's depiction of the flux lines being isotropically distributed relative to the moving charge.

In the analogous case of light: If we have an isotropic constant emitter in motion. Say an incandescent sphere and we look at the emission from a discrete time interval, we have an expanding shell of photons with a spatial thickness.
That thickness would not be uniform, but would vary, Thickest at the point at the rear (wrt motion) and smoothly decreasing to the point aligned with the direction of travel.
So the photon density would likewise vary, but inversely. Minimal at the rear to maximal at the front.

Then the aberration factor would also result in a further, angle dependent, differential shift of concentration towards the front.

Add the Doppler shift in frequency/energy which also effects a differential energy increase from the back to the front.

So looking at the cloud of photons from constant emission as a photonic energy field surrounding the emitter, it is obvious that it is highly polarized due to three distinct but complementary factors.[ I ignored the dilation factor affecting both frequency and density because it is a scalar with no directional dependence and doesn't affect field distribution, only the overall magnitude of those components.]

SO if EM propagation as photons, EM field propagation as electrostatic interaction between charges and gravity propagation are analogous and mediated by lightspeed interactions through virtual particles, then the question would seem to be; Which, if any, of these three factors would not apply in all three cases?

My hasty inference would be that velocity dependent aberration and anisotropic density would both apply to electrodynamics and gravity but not Doppler.

Any insights regarding the paper posted by PeterDonis or flaws in my reasoning appreciated.
Thanks
 
Last edited:
  • #25
PeterDonis said:
First, a general comment: the only reason for even discussing all this stuff about "velocity dependent factors" and so forth is that people insist on some "conceptual" explanation of gravity as a "force". Any such explanation is going to break down at some level, because gravity is not really a "force" in the sense in which the term is being used here. See further comments below.

That said, if we want to view, say, the Sun's gravity acting on the Earth as a "force", the velocity-dependent factors compensate for the light-speed time delay in the propagation of the force from the Sun to the Earth. This compensation makes it appear that the direction of the force is the direction the Sun is in "now", instead of the direction it was in 500 seconds ago (the light-speed time delay). The compensation is not exact; the direction of the force is not the "actual" direction of the Sun "now", but rather the direction the Sun would be in "now" if it moved in an exact straight line from where it was 500 seconds ago, at exactly the relative velocity it had 500 seconds ago. Or, as Carlip puts it, the force points in the direction of the Sun's position "quadratically extrapolated" 500 seconds forward (i.e., to "now") from its position and velocity 500 seconds ago. The "quadratically" refers to how accurately the extrapolation cancels out the effects of the time delay; for electromagnetism, the extrapolation is only linear, which is a less accurate cancellation, so EM effects, such as light, show measurable aberration under circumstances where gravity does not.

The best answer to this is that the "extrapolations" and "velocity dependent factors" don't describe any actual physical process; they are just a conceptual crutch to help us understand why gravity, if we insist on viewing it as a "force", doesn't show aberration effects..

Well i never mentioned force at all and don't see why it makes a difference. Of course the end result in the world is an acceleration vector whether you attribute it to force or geometry isn't really germane.
You could look at it as information. If displacement of a charge or mass effects a local field fluctuation, then this information has to get from there to all other parts of the field.This seems to imply a mediator wither a wave or a virtual particle. SO if we rule out any non-local quantum instantaneous field responce then the question is specifically how does the motion of the mass/charge affect the wave or particle in a way that is equivalent to instantaneous responce when it arrives at the distant locations in the field. Yes?.
The term extrapolation as I understand it means either projecting into the future based on the past or projecting from a series of values to determine a value outside the range of known values. Both being purely abstract mathematical procedures which I find difficult attaching to any possible physical interaction.
So I still would like to know what the velocity dependent factors could be. Vectors? tensors? They are certainly not any kind of "conceptual crutch" as there is no concept whatsoever to be attached to the term "velocity dependent factors" per se.

i don't mean to give you a hard time as I am appreciative of your efforts at the difficult task of trying to describe the process in simple terms but my curiosity is still vexing me.
Thanks.
PeterDonis said:
I'll defer comment on the rest of your post until I've digested it some more.

Does this mean that I was not sufficiently clear in my description and you are unsure of my meaning or do you understand it but are digesting it to find the flaws in my interpretation or inference?
 
  • #26
I like the animated gif on the previous page... but it is missing something important too.
Remember, for an object to "suddenly accelerate", something else must also do the same in the other direction. And that something else also has its own mass and creates its own gravity, and the combined changes in gravity can actually cancel out in some cases. Remember that the center of mass of a set of objects will never change its motion from forces acting only between those objects.

Especially when that "sudden" acceleration is caused by gravity, I would assume the other objects do take it into account when "predicting" the location of the gravity sources.

(I do not actually know the math for any of that, take me with a grain of salt)
 
  • #27
Austin0 said:
Well i never mentioned force at all and don't see why it makes a difference. Of course the end result in the world is an acceleration vector whether you attribute it to force or geometry isn't really germane.

Yes, it is, because if you attribute it to geometry then there doesn't have to be any "acceleration vector"; the Earth moves on a geodesic, i.e., an unaccelerated world line, and on the geometric view this is perfectly natural since "gravity" is just curvature of the geometry. If you view gravity as a "force" then that at least strongly suggests that there *is* an actual acceleration, i.e., that the object's motion is *not* geodesic; that's the reason why in GR we no longer view things like gravity that don't induce proper acceleration as "forces".

Austin0 said:
You could look at it as information. If displacement of a charge or mass effects a local field fluctuation, then this information has to get from there to all other parts of the field. This seems to imply a mediator wither a wave or a virtual particle.

If by "a wave" you just mean "a disturbance in an underlying medium", then yes, a displacement of a mass produces gravitational waves that carry the information about the change outward.

Austin0 said:
SO if we rule out any non-local quantum instantaneous field responce then the question is specifically how does the motion of the mass/charge affect the wave or particle in a way that is equivalent to instantaneous responce when it arrives at the distant locations in the field. Yes?

It's not exactly equivalent, just approximately equivalent. See below.

Austin0 said:
The term extrapolation as I understand it means either projecting into the future based on the past or projecting from a series of values to determine a value outside the range of known values. Both being purely abstract mathematical procedures which I find difficult attaching to any possible physical interaction.

Don't get hung up on the word "extrapolation". The important point is that the information carried by the field disturbance caused by a displacement of a source (the gravitational waves, in the case of displacement of a mass), as it is modeled in GR, is not just information about the source's position; it's information about the source's position *and* velocity, to a degree of accuracy which is enough to compensate for the light-speed time delay to a certain order of approximation. (For gravity it's second order, i.e., quadratic; for electromagnetism it's first order, i.e., linear.) So the disturbance, when it reaches a distant object (say, the disturbance caused by the Sun reaching the Earth), affects the motion of that object in a way that happens to be the same, to the given order of approximation, as an "instantaneous force" would affect it.

Austin0 said:
So I still would like to know what the velocity dependent factors could be. Vectors? tensors? They are certainly not any kind of "conceptual crutch" as there is no concept whatsoever to be attached to the term "velocity dependent factors" per se.

They're a conceptual crutch because they only appear if you insist on viewing the effect of one body on another (say the Sun on the Earth) as a "force" and expanding it out order by order in the relative velocity. The "velocity dependent factors" are just terms in the expansion. But the fact that we humans need to do the expansion to form a conceptual picture of what's happening does not imply that nature has to "do the expansion".

Austin0 said:
Does this mean that I was not sufficiently clear in my description and you are unsure of my meaning or do you understand it but are digesting it to find the flaws in my interpretation or inference?

It means I didn't have time to read it carefully enough to form a response. I'll respond to it separately.
 
  • #28
Austin0 said:
Regarding a photon , the observed directional vector points to the retarded position due to aberration.

There is a key point here that you may be missing. The observed direction of an incoming photon--say from the Sun, being observed on Earth, to be concrete--does show aberration, but that aberration is *not*, strictly speaking, the "aberration" that Carlip is talking about in his paper. He is talking (in the EM case) about the observed direction of the Coulomb force exerted by a charged source, which also shows "aberration" in the case of an accelerating source. The radiation emitted by the accelerating source is what carries the information about the acceleration, but that information causes a *change* in the observed direction of the Coulomb force. So there's a key distinction between the direction the information about the change in the Coulomb force appears to come from, and the direction that the Coulomb force itself appears to come from.

I won't comment on the rest of what you say about photons and wave fronts because I think you should reconsider it in the light of the above.

I'll post separately about the case of a gravitational binary system since you give more details about that in your response to Bill_K's post.
 
  • #29
Austin0 said:
During the recessional phase, the retarded postion of the point of emission is a distance Dr ,less than the actual instantaneous position distance Di relative to the point of reception. Dr < Di SO the negative acceleration arising from this is greater than the force should be from the actual position by the inverse square difference in distance

No, it isn't, because of the velocity-dependent factors I mentioned. You are assuming that the change in relative distance is the only thing that contributes to a change in "acceleration" (I put that in quotes because, as I mentioned before, both objects are traveling on geodesics, so they have zero proper acceleration; you are talking about coordinate acceleration in the frame in which the common center of mass of the two objects is at rest). It isn't. A given increase in distance during the recessional phase translates into a significantly *larger* change in "acceleration" than your calculation indicates, because of those velocity-dependent factors. Similar remarks apply to the closing phase.

All of the rest of your reasoning appears to depend on the above mistake, so I'm not sure it's worth my making specific comments on it. The only thing I would say is that the above reasoning applies to the EM case as well.

[Edit: Realized that some of the wording in the first paragraph was kind of garbled, so I re-worded it to hopefully make more sense.]
 
Last edited:

1. What is the speed of gravity?

The speed of gravity is approximately 299,792,458 meters per second, which is the same as the speed of light.

2. How is the speed of gravity related to the Sun being 8 minutes behind?

The Sun being 8 minutes behind refers to the time it takes for light from the Sun to reach Earth. This is because the speed of light is the same as the speed of gravity, so it takes approximately 8 minutes for the gravitational effect of the Sun to reach Earth.

3. Is the speed of gravity different for different objects?

No, the speed of gravity is the same for all objects regardless of their mass or size. This is because it is a fundamental constant of the universe.

4. How was the speed of gravity first calculated?

The speed of gravity was first calculated by Sir Isaac Newton in his theory of universal gravitation, which he published in 1687. He used his laws of motion and the law of gravitation to determine the speed of gravity.

5. Can the speed of gravity be measured directly?

No, the speed of gravity cannot be measured directly. It is inferred from observations and calculations based on the effects of gravity on objects. However, scientists are working on developing new techniques to directly measure the speed of gravity in the future.

Similar threads

Replies
9
Views
984
  • Special and General Relativity
Replies
29
Views
1K
  • Special and General Relativity
Replies
10
Views
2K
  • Special and General Relativity
Replies
20
Views
2K
  • Special and General Relativity
2
Replies
39
Views
4K
Replies
3
Views
1K
  • Special and General Relativity
2
Replies
50
Views
3K
  • Special and General Relativity
Replies
28
Views
3K
  • Special and General Relativity
Replies
6
Views
895
  • Special and General Relativity
Replies
27
Views
4K
Back
Top