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caveman1917
2010-Jul-29, 03:25 AM
I thought the precession of the planets was explained in relativity by the fact that gravity propagates at finite speed (instead of instantaneous), hence the attraction would be towards where the sun 'was' instead of 'is'.

Now i came across this (http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity) wiki article (2 body problem in GR).

Laplace had shown {snip} if gravitational influence does propagate at a finite speed, then at all points in time a planet is attracted to a point where the Sun was some time before, and not towards the instanteneous position of the Sun.

Laplace's estimate for the velocity of gravity is not correct, because in a field theory which respects the principle of relativity, the attraction of a point charge which is moving at a constant velocity is towards the extrapolated instantaneous position, not to the apparent position it seems to occupy when looked at
(my bold)

Could someone explain to me why this is so? This seems to suggest there is a fundamental difference between gravitational attraction and EM radation. In the sense that the photons will point towards the apparent position, but the 'gravitons' (i know they're hypothetical - but i didn't know how else to put the question clearly) will point towards the extrapolated position. At this point in the article we're still in Newtonian + SR, not yet in GR.

Gravity {GR by now} is distinct from the fictitious forces centrifugal force and coriolis force in the sense that the curvature of spacetime is regarded as physically real, whereas the fictitious forces are not regarded as forces.

This would seem to suggest we can consider both perspectives, either consider curvature real and free-fall inertial, or consider curvature not real and free-fall as non-inertial, as equal.

I'm having trouble understanding how this distinction would relate to the bolded part in the quote above this one. It would seem one perspective would make the argument made there to be true (inertial), while the other would make the argument untrue (non-inertial). And it seems to be suggested both perspectives are equivalent, so i'm having a little trouble putting it all together.

Ken G
2010-Jul-29, 04:38 AM
I thought the precession of the planets was explained in relativity by the fact that gravity propagates at finite speed (instead of instantaneous), hence the attraction would be towards where the sun 'was' instead of 'is'.No, precession is a much weaker effect than that. If a planet felt a pull toward a lagged position of the Sun, then there would be component of the pull that was back against the direction of motion of the planet. That doesn't cause precession, that causes orbital decay-- the planets would all have fallen into the Sun by now! Furthermore, the effect you describe, if it happened, would apply to circular orbits too, whereas precession is only something that happens to elliptical orbits.

Could someone explain to me why this is so? This seems to suggest there is a fundamental difference between gravitational attraction and EM radation.There is, but it's not what you might think-- because EM radiation is lagged like you say, but EM forces also point toward the current, not lagged, position of a source moving at constant speed! You know this has to be very subtle if Laplace got it wrong, so I won't pretend to understand it myself, but it turns out that there is an additional term in the force that "corrects" for the direction of motion of the source-- the force ends up being lagged to the prior location of the source, but then "corrected" for the motion of that source at that time. That happens for EM too!

Nevertheless, gravity is different from EM forces, not just from EM radiation, because not only does gravity "correct" for the motion of the source, it also corrects for its acceleration! Thus the gravity will point at the current location of the source even if the source is accelerating. publius has shown the equations for this on other threads, I don't recall the details but it turns out that for two objects in mutual circular orbit, their gravity always points in just the way Newton's instantaneous version would point, not Laplace's lagged version. Indeed, binary star systems wouldn't last long were that not true.

In the sense that the photons will point towards the apparent position, but the 'gravitons' (i know they're hypothetical - but i didn't know how else to put the question clearly) will point towards the extrapolated position. Photons are a bit different from the EM force though, because the EM force is carried by virtual photons. It makes a difference! I cannot begin to say why, but presumably it has to do with the fact that virtual photons are selected from a sea of "possible" photons that can do almost anything, they don't have to propagate from the source to the target like normal radiation does. The constraints on virtual photons are not supplied by classical logic of motion, they are supplied by the vagaries of wave mechanics, to wit, they do whatever does not experience destructive interference. That's why you can get EM attraction, for example-- try getting that with an exchange of real photons!

This would seem to suggest we can consider both perspectives, either consider curvature real and free-fall inertial, or consider curvature not real and free-fall as non-inertial, as equal.I'm not sure if it's necessary to get into GR too deeply, but I don't think one could ever get the correct answer by considering free-fall as non-inertial-- that sounds Newtonian to me. I think the point they are making by contrasting gravity and fictitious forces is that real gravity leaves a tidal imprint that forces you to "pay the piper" at some point when you calculate geometric attributes of inertial (free-fall) motion, whereas fictitious forces are purely coordinate forces that yield trajectories with the same geometric properties as motion in a straight line. In other words, motion in a straight line in Euclidean coordinates maps into curved motion in curvilinear coordinates, but you can tell that the curved motion isn't "really curved", it stems from the curving coordinates. Fictitious forces stem from coordinates that are curvilinear in space and time, not just space alone, but you can still tell that the curved trajectories in spacetime that you get in, say, a rotating reference frame, correspond to straight paths in a nonrotating reference frame. But the curved spacetime paths you get with real gravity are "really curved," it's an invariant type of curvature, not a coordinate time. This all comes from tensor calculus by Riemann and Ricci and that bunch-- I don't know much about it frankly.

caveman1917
2010-Jul-29, 04:50 AM
Thank you for the explanation, Ken. I was indeed forgetting that forces are mediated by virtual particles, not the real versions.

If GR paths are curved in an invariant way, that would indeed suggest we are forced to consider curvature real, and free-fall inertial. I thought that was supposed to be the case, but the statement in the wiki article seemed to suggest otherwise. It is a bit ambiguous as phrased.

publius
2010-Jul-29, 05:31 AM
Thanks Ken, I was going to respond to this, but you beat me to it and saved me a lot of typing.

One small thing though. Circular orbits do radiate. What matters is the "jerk", or change in acceleration during the light travel mean time. Now, elliptical orbits radiate more strongly than circular orbits do, though, there is a (1 - e^2) term in the denominator to a high power, as well as an a^5 term, where a is the semimajor axis. So a tight, highly eccentric orbit radiates like crazy.

To the OP, to see why the "principle of relativity" demands "extrapolation" (to at least velocity), consider this. We have two charges, both at rest in some frame. The forces are obviously equal and opposite, and there's no worry about propagation delay, because the fields are static in that frame. Now, go to a moving frame.

From that frame, the two charges are moving. But yet the forces must remain equal and opposite, otherwise the moving observer would predict something different than in the rest frame! Thus, the forces must somehow extrapolate to at least the velocity of the charges. And, if you solve Maxwell, you'll find just that. In the moving frame, there is a magnetic field, B, due to the motion of the charges, and there is also a dB/dt, as the magnetic field is itself changing. That gives a solenoidal E field, which represents a correction to the "inverse square, Coulomb" force of the rest frame.

It is this latter that is important. Consider a charge moving in some frame acting on a test charge that is *stationary*. Since v = 0 for the test particle, there is no magnetic force. It is entirely the dB/dt correction to the E field that extrapolates and makes the force point to the "instantaneous position" of the source.

That only works for constant velocity. The extrapolation is exact only for constant motion. The force points to where the charge would be in the future if it contiuned to move at constant velocity. If it accelerates in the light travel meantime, the forces will "miss", that is deviate from equal and opposite. This causes mechanical energy and momentum to be lost. And that lost energy and momentum go into the field itself, as radiation.

Now, GR gravity does one better and extrapolates for the acceleration as well as the velocity. Thus it doesn't miss as bad, and radiates less. This is related to the order of radiation. For EM, the first order is dipole, and for gravity, it's quadrapole.

You may wonder why this correction thing and "principle of relativity" wasn't so important pre-Michelson-Morley. Well the answer is that an aether was assumed, and what was important was the velocities relative to the *aether*. Our two charges above would be at rest with respect to the ether, so it was no great surprise that all moving observers would see the same thing and Maxwell's equation would look different in those frames anyway (but note the extrapolation would still happen in the ether frame for a charge moving in the ether).

It was only afterwards when it became that Maxwell was invariant to all observers did all this stuff become more profound.

-Richard

Ken G
2010-Jul-29, 06:34 AM
Thanks for filling in the gaps!

dgavin
2010-Jul-30, 02:21 PM
I've been tryign to understand GR orbital precession for a while now and wanted to ask if i have the understanding of it correct.

Most of the precession of orbits is accounted for in Newton's Gravity, but accounting for gravitation influences of other body's orbiting the sun. But this did not match observered ammount of precession.

GR Presession accounts for the fact that eliptical orbits move away from then back into gravity well's, which adds an addition amount to the precession. Am I uderstanding this part of it correctly?

Can extreme amounts of frame dragging also cause precession? I assume GR would account for that as well, but just wondering.

Hungry4info
2010-Jul-30, 02:54 PM
Can extreme amounts of frame dragging also cause precession?

Not an expert on this, but the frame dragging would rotate the orbit around the rotational axis of the parent body. If the orbit is equatorial to the parent body, then it will seem much like precession. Precession is the advancement of the longitude of perihelion, ω. Frame dragging, by rotating the orbit around the parent body's rotational axis, will advance the ascending node, Ω.

caveman1917
2010-Aug-02, 12:44 PM
not only does gravity "correct" for the motion of the source, it also corrects for its acceleration!

Now, GR gravity does one better and extrapolates for the acceleration as well as the velocity.

Does this have to do with the equivalence of inertial mass and gravitational mass?

Suppose we take the frame of some accelerating (upwards, positive Y-axis) particle P at some point in time. There is a gravitating object O located at (-1,-2) (let's make it 2d). O has a velocity downwards (negative on the Y-axis). We have the real (compensated for both v and a) gravitational vector from P to O as (-1,-2) which we can define as F.

The 'standard' equivalence principle would provide some vector F1, compensated for v but not a (as in the reasoning on EM). So we might say F1 equals (-1,-1).

Now, with inertial mass equal to gravitational mass, and the fact that P 'feels' the acceleration, we can describe this acceleration as another gravitational vector F2, with direction opposite the acceleration (thus downwards on the Y axis in this case) and magnitude related to a.

Will it turn out that this F2 is always such that F1 + F2 = F?
In our example F2 thus being (0,-1); so (-1,-1) + (0,-1) = (-1,-2) (the full 'compensated' gravitational vector).

Or am i totally besides the point here?
If so, i wouldn't mind a pointer to the relevant part of the math.

publius
2010-Aug-03, 01:20 AM
I'm not sure exactly what you're driving at (my system apparently is showing some of the characters, as I just see little boxes after some of your 'F's there).

Here's a paper by Carlip which I always link to when these speed of gravity questions come up:

http://xxx.lanl.gov/abs/gr-qc/9909087

The math there is some pretty high powered tensor stuff for the GR part, but otherwise one can follow it. He starts out by demonstrating the EM velocity "extrapolation" and uses the very elegant 4-vector formulation which makes it look almost trivial. The then does the GR version of that, which is much more complicated. Anyway, the term for the acceleration extrapolation falls right out.

Note one important thing. With EM, we can specify a source charge undergoes any motion we like and plug that right in the EM equations. EM only cares about the charge and the energy of the fields themselves. Energy can enter or leave the system mechanically and the EM equations don't worry about it. That is, we don't have to worry about what's causing the charge to accelerate.

In GR we do. Energy and momentum must be conserved in the source term, the stress-energy tensor, and we must take into account what's causing the acceleration and use a valid stress-energy tensor. Simply specifying an accelerating mass is not valid alone. That is, if we have a rocket in free space, we've got to consider the exhaust energy and momentum as well as that of the rocket payload.

It turns out that there is a well known solution of that, the "photon rocket", where you have a mass which is converting part of itself to radiation "exhaust" and using that to accelerate. In this case the brunt of the gravity will follow the rocket, at least at first, as the exhaust mass, scattered to the winds, is negligible

Carlip groks through that and shows how the gravitational acceleration extrapolates for the acceleration as well as the velocity.

-Richard

publius
2010-Aug-03, 01:33 AM
Now, about the Equivalence Principle. It turns out the EP is what requires a relativistic theory of gravity to be a full rank-2 tensor "field" and not a mere vector field as EM. This refers to the potential. The EM potential is a 4-vector, whose derivatives give the EM field tensor. The equivalent of the potential in GR is the metric itself, which is a rank-2 tensor. Gravity has more "degrees of freedom" than can be expressed in a vector.

Were it not for the EP, you could have a very EM like theory of gravity, where the gravitational mass played the same role as charge and was an invariant. But that would violate the EP, as the inertial mass wouldn't be the same as the gravitational mass.

Now, I'm not sure but I think this may be what requires gravitational radiation to be quadrapole as well. That is, the question is could you have fully relativistic theory of gravity respecting the EP where the first order of radiation was dipole. If not, then you can say the EP is what requires the acceleration extrapolation. I think this is true, but again I'm sure.

-Richard

caveman1917
2010-Aug-03, 02:02 AM
I'm not sure exactly what you're driving at (my system apparently is showing some of the characters, as I just see little boxes after some of your 'F's there).

I edited the post to remove the subscripts.

Basically what i was thinking about was that, starting with a correcting factor for v but not a (as in EM), we would 'miss' the real location of the source by some amount in the direction of acceleration of the rocket. And since we can equate acceleration of the rocket with a gravitational force on the rocket (by the EP) in the other direction, i was wondering wether this sum of gravitational forces (the v-compensated-PoR one and the acceleration-EP one) would then point to the 'right' location of the source. In such manner that this EP-added gravitational force due to acceleration would provide the acceleration compensation needed.

Since you gave the principle of relativity as the factor requiring compensation for v, i wondered wether the equivalence principle would provide somewhere along the same lines the compensation for a.

Thanks for the link, will check out the article in the morning.