Orbit precessions - General Relativity vs Newton

[Cobalt101]

What is the underlying feature of general relativity that, unlike Newtonian mechanics, results in the correct calculation of orbits i.e. including precession (e.g. Mercury). I not asking for the mathematics (i.e. the additional term in the equation) but rather what underlying "physical" feature causes it to not just be a perfect ellipse ?

 

[A.T.]

http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

 

[Cobalt101]

http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

Thank you

 

[sandy stone]

Those slides certainly help to visualize what is going on, but I have a question. The two slides treat curvature in the time dimension and in spatial dimensions separately. Is this an accurate description of the situation, or is it done to simplify the explanation? For instance, in the depiction of curved space surrounding the sun, is there also curvature in the time dimension as well, that is just too hard to illustrate?

 

[A.T.]

that is just too hard to illustrate?

Yes. You can only show 2 of the 4 space-time dimensions in such diagrams. What you see in part 1 (space-time) happens locally everywhere along the orbit shown in part 2 (space-space).

 

[pervect]

I don't think that precession can be attributed to a single underlying physical feature of general relativity. If onr consider pertubation of a circular orbit - i.e. one takes a particle in a circular orbit and gives it a small perturbing "nudge", the perturbed particle will appear to oscillate relative to the non-preturbed particle. Precession exists when the period of this oscillation is different from an orbital period.

 

[PeterDonis]

what underlying "physical" feature causes it to not just be a perfect ellipse ?

I would say it's the fact that in GR, gravity is not a simple inverse square central force the way it is in Newtonian physics. In general it isn't a force at all; but even in the weak field, slow motion limit, where it can be viewed as a force, it isn't a simple inverse square central force. There is also an extra piece of the interaction which is velocity-dependent (some sources refer to this as the "gravitomagnetic" interaction, and the inverse square Newtonian force as the "gravitoelectric" interaction). But only a pure inverse square force can produce closed elliptical orbits.

 

[haushofer]

Looking at that visualization, you can wonder how scalar theories of gravity account for precession (they do, in fact, but with the wrong amount). So I'm a bit sceptic about such a visualization.

 

[m4r35n357]

What is the underlying feature of general relativity that, unlike Newtonian mechanics, results in the correct calculation of orbits i.e. including precession (e.g. Mercury). I not asking for the mathematics (i.e. the additional term in the equation) but rather what underlying "physical" feature causes it to not just be a perfect ellipse ?

I would say that the precession really is caused by the additional term in the gravitational potential (derived from the Schwarzschild solution). There is simply "more gravity" very close to a black hole than to a sphere (or point) of the same mass in Newtonian gravity. See for example equation 7.48 in Carroll.

As I understand it, it is entirely possible (but masochistic) to use curved spacetime to model Newtonian gravity, so that can't be the explanation.

 

[PAllen]

As I understand it, it is entirely possible (but masochistic) to use curved spacetime to model Newtonian gravity, so that can't be the explanation.

Note, this formulation does not involve a spacetime metric at all, nor spacetime curvature. Only spatial components of the curvature tensor are non-vanishing, same for the connection.

 

[m4r35n357]

Note, this formulation does not involve a spacetime metric at all, nor spacetime curvature. Only spatial components of the curvature tensor are non-vanishing, same for the connection.

OK, thanks for the clarification!

 

[haushofer]

I would say that the precession really is caused by the additional term in the gravitational potential (derived from the Schwarzschild solution). There is simply "more gravity" very close to a black hole than to a sphere (or point) of the same mass in Newtonian gravity. See for example equation 7.48 in Carroll.

As I understand it, it is entirely possible (but masochistic) to use curved spacetime to model Newtonian gravity, so that can't be the explanation.

It isn't. In Newton-Cartan theory there is no precession. I agree with your statement that precession is caused by such an additional term, and I'm not sure if this visualization is really an explanation. As I understand, already for scalar theories in flat space(time) one can find that an orbit does not close after a full rotation, hence giving precession.

 

[haushofer]

Note, this formulation does not involve a spacetime metric at all, nor spacetime curvature. Only spatial components of the curvature tensor are non-vanishing, same for the connection.

Newton-Cartan can be casted in a metrical formulation, but this metric structure is degenerate because there is no spacetime interval being invariant under the local isometry group (rotations+boosts). But this is no problem; the Riemann tensor is then completely given by the degenerate metrics and an additional closed two-form.

 

[Hornbein]

I would say it's the fact that in GR, gravity is not a simple inverse square central force the way it is in Newtonian physics. In general it isn't a force at all; but even in the weak field, slow motion limit, where it can be viewed as a force, it isn't a simple inverse square central force. There is also an extra piece of the interaction which is velocity-dependent (some sources refer to this as the "gravitomagnetic" interaction, and the inverse square Newtonian force as the "gravitoelectric" interaction). But only a pure inverse square force can produce closed elliptical orbits.

Aha. So precession is greatest with Mercury because it has the greatest angular velocity amongst the planets?

 

[PeterDonis]

So it is greatest with Mercury because it has the greatest angular velocity amongst the planets?

It's a combination of angular velocity and closeness to the Sun, because the Sun's "gravitomagnetic field" gets weaker further away, just like the "gravitoelectric" field does.

 

[Hornbein]

It's a combination of angular velocity and closeness to the Sun, because the Sun's "gravitomagnetic field" gets weaker further away, just like the "gravitoelectric" field does.

Right. But these quantities are directly related for an orbiting body.

 

[PeterDonis]

There is simply "more gravity" very close to a black hole than to a sphere (or point) of the same mass in Newtonian gravity.

Note, however, that this "additional gravity" depends on the angular momentum of the test object, so it isn't that the ordinary Newtonian force gets larger; it's that there is a different kind of force present, which is velocity-dependent, and therefore works more like a "magnetic" force than an "electric" force.

 

[PAllen]

Newton-Cartan can be casted in a metrical formulation, but this metric structure is degenerate because there is no spacetime interval being invariant under the local isometry group (rotations+boosts). But this is no problem; the Riemann tensor is then completely given by the degenerate metrics and an additional closed two-form.

What I said is consistent with what you say. There is no spacetime metric. That is equivalent to saying there is no spacetime invariant interval. The artificial combination (which is wholly unnececssary to the formulation of the theory, anyway) is not a spacetime metric because it lacks essential properties. In my view, the logical formulation has connection (and curvature tensor constructed from it), but no metric. Connection is all you need for the geodesic equation and curvature. Trying to make the connection metric compatible adds nothing because the metric is not a true metric.

 

[PeterDonis]

these quantities are directly related for an orbiting body.

Yes, they are, but the relationship depends on the orbital energy and angular momentum, which means it depends on the mean distance from the Sun, and also on the orbital eccentricity. I actually should have mentioned eccentricity as another factor that increases Mercury's precession.

 

[haushofer]

What I said is consistent with what you say. There is no spacetime metric. That is equivalent to saying there is no spacetime invariant interval. The artificial combination (which is wholly unnececssary to the formulation of the theory, anyway) is not a spacetime metric because it lacks essential properties. In my view, the logical formulation has connection (and curvature tensor constructed from it), but no metric. Connection is all you need for the geodesic equation and curvature. Trying to make the connection metric compatible adds nothing because the metric is not a true metric.

If you think in terms of equations of motion. If you e.g. want to derive the geodesic equation from an action principle, you need the metric formulation. The metric formulation also emerges naturally if you consider Newton-Cartan theory as a gauge theory of the underlying Galilean algebra. So it depends on what you call "the formulation of the theory".

 

[PAllen]

If you think in terms of equations of motion. If you e.g. want to derive the geodesic equation from an action principle, you need the metric formulation. The metric formulation also emerges naturally if you consider Newton-Cartan theory as a gauge theory of the underlying Galilean algebra. So it depends on what you call "the formulation of the theory".

Ok, I can see that point of view, up to a point - the degeneracy of the metric flows from the degeneracy of the Galilean algebra (over 4 dimensions). However, I have never seen the value of the action principle derivation of geodesic equation in relativity (as opposed to all other uses of action principles), because making it work for null geodesics is highly artificial. Conceptually (and pedagogically), for relativity, I strongly prefer the 'straightest possible path' formulation of geodesic, made precise via parallel transport. Introduction of metric comes (IMO) logically after discussing geodesics and curvature (and it is important to realize that a metric is wholly unnecessary to define geodesics and curvature).

 

[bcrowell]

But only a pure inverse square force can produce closed elliptical orbits.

Well, $F\propto r^2$ also works, but that's obviously not applicable here.

 

[bcrowell]

The standard way to approach this is to use conserved quantities to reduce it to a one-dimensional problem with an effective potential. That works great, but it's also nice to have a description that works for a less mathematically adept audience.

As Peter Donis notes, any full explanation would need to explain why orbits *are* closed ellipses in the Newtonian limit. However, we can always take that as an established fact and then argue about perturbations on top of Newtonian gravity.

One such approach is to consider the spatial part of the curvature of the spacetime surrounding the sun. This spatial curvature is positive, so a circle's circumference is less than $2\pi$ times its radius. This causes Mercury to get back to a previously visited angular position before it has had time to complete its Newtonian cycle of radial motion. (I've forgotten where I first saw this argument, but I didn't originate it. This web page http://www.relativity.li/en/epstein2/read/i0_en/i1_en/ attributes it to Epstein, Relativity Visualized.)

One way to check that this argument makes some sense is to consider the case where the eccentricity approaches zero. The argument implies that the perihelion advance should approach a nonzero limit in this case, and that is in fact the correct result. (I used to have a different favorite explanation for the effect, but people here on PF convinced me that it was wrong for this reason.)

 

[Jonathan Scott]

Why not be specific? See MTW "Gravitation" section 40.5, which discusses perihelion shift in the PPN approximation, especially equation (40.18):
$$\delta \phi_0 = \frac{2-\beta+2\gamma}{3} \frac{6 \pi M_{\odot}}{a(1-e^2)}$$
In this expression, $\beta$ is the non-linearity correction to the time component and $\gamma$ is the space curvature factor, both equal to 1 in General Relativity. As usual $e$ is the eccentricity and $a$ is approximately the semimajor axis (which is like the average radius of the orbit).
I see that setting both $\beta$ and $\gamma$ to zero does not give the Newtonian result of zero precession; I think that the Newtonian result actually corresponds to $\beta=2$ and $\gamma=0$, although I'm not sure about that.

 

[A.T.]

such approach is to consider the spatial part of the curvature of the spacetime surrounding the sun. This spatial curvature is positive, so a circle's circumference is less than $2\pi$ times its radius. This causes Mercury to get back to a previously visited angular position before it has had time to complete its Newtonian cycle of radial motion. (I've forgotten where I first saw this argument, but I didn't originate it. This web page http://www.relativity.li/en/epstein2/read/i0_en/i1_en/ attributes it to Epstein, Relativity Visualized.)

One way to check that this argument makes some sense is to consider the case where the eccentricity approaches zero. The argument implies that the perihelion advance should approach a nonzero limit in this case, and that is in fact the correct result. (I used to have a different favorite explanation for the effect, but people here on PF convinced me that it was wrong for this reason.)

I think here is the old thread on this:
https://www.physicsforums.com/threa...-Newtons-gravity-and-g-r.626729/#post-4028708

 

[pervect]

I think here is the old thread on this:
https://www.physicsforums.com/threa...-Newtons-gravity-and-g-r.626729/#post-4028708

Yes, this topic has come up before several times. My own argument is a bit technical, that is if we attribute precession entirely to spatial curvature, why does it depend on both the PPN parameters beta and gamma, while spatial curvature only affects $\gamma$?

If we assume the eccentricity e is << 1, the perihelion shift per orbit is proportional to $(2 -\beta + 2 \gamma)/3$. (See for example MTW 1110). So we observe that there is some precession even if $\beta = \gamma = 0$, which would seem to imply that $\beta = \gamma = 0$ is not the Newtonian limit, in which there is no precession

My current thinking on the subject, then, is that precession isn't due to a single effect. Perhaps I'm missing something, but I'd prefer to be cautious and say that it isn't clear what causes precession than to oversimplify the analysis.

 

[bcrowell]

Yes, this topic has come up before several times. My own argument is a bit technical, that is if we attribute precession entirely to spatial curvature, why does it depend on both the PPN parameters beta and gamma, while spatial curvature only affects $\gamma$?

If we assume the eccentricity e is << 1, the perihelion shift per orbit is proportional to $(2 -\beta + 2 \gamma)/3$. (See for example MTW 1110). So we observe that there is some precession even if $\beta = \gamma = 0$, which would seem to imply that $\beta = \gamma = 0$ is not the Newtonian limit, in which there is no precession

My current thinking on the subject, then, is that precession isn't due to a single effect. Perhaps I'm missing something, but I'd prefer to be cautious and say that it isn't clear what causes precession than to oversimplify the analysis.

I'm not quite clear on your reasoning here. Your first paragraph seems persuasive, but your second paragraph muddies the waters. It's not obvious to me that there should be values of $\beta$ and $\gamma$ that correspond to the Newtonian limit. PPN assumes a nondegenerate metric, whereas Newtonian gravity has a degenerate metric. As far as I know (basically just from reading the review article by Will), the standard way of recovering the Newtonian limit from PPN is not by picking values of $\beta$ and $\gamma$ but rather by taking the limit of small velocities and weak fields.

I think it's possible to show explicitly that there are no values of $\beta$ and $\gamma$ that define the Newtonian theory. Perihelion advance is a non-Newtonian effect that is proportional to $2+2\gamma-\beta$, while the Shapiro time delay (or the part of it that Will describes as "non-Newtonian") is proportional to $1+\gamma$. The Nordtvedt effect is proportional to $4\beta-\gamma-3$ (assuming zero values for $\xi$, alphas, and zetas). Setting all three of these to zero gives no solutions for the two parameters.

 

[Jonathan Scott]

...
If we assume the eccentricity e is << 1, the perihelion shift per orbit is proportional to $(2 -\beta + 2 \gamma)/3$. (See for example MTW 1110). So we observe that there is some precession even if $\beta = \gamma = 0$, which would seem to imply that $\beta = \gamma = 0$ is not the Newtonian limit, in which there is no precession

My current thinking on the subject, then, is that precession isn't due to a single effect. Perhaps I'm missing something, but I'd prefer to be cautious and say that it isn't clear what causes precession than to oversimplify the analysis.

I've changed my mind about the values for a Newtonian version. I think the PPN parameter values for a "Special Relativity" Newtonian theory with no precession should be $\beta=0$ and $\gamma=-1$. This is because the space and time factors in the metric need to be equal to preserve $c$ (where in GR they are reciprocals of one another, so $c$ in isotropic coordinates varies as the square of the potential factor). This means that the sign of $\gamma$ needs to be switched compared with GR. That combination gives zero precession in the PPN factor $(2 - \beta + 2\gamma)$.

 

[bcrowell]

I've changed my mind about the values for a Newtonian version. I think the PPN parameter values for a "Special Relativity" Newtonian theory with no precession should be $\beta=0$ and $\gamma=-1$. This is because the space and time factors in the metric need to be equal to preserve $c$ (where in GR they are reciprocals of one another, so $c$ in isotropic coordinates varies as the square of the potential factor). This means that the sign of $\gamma$ needs to be switched compared with GR. That combination gives zero precession in the PPN factor $(2 - \beta + 2\gamma)$.

I'd like to like your argument because it supports my favorite way of explaining the perihelion advance to a nonmathematical audience -- but it leaves me confused. The values of beta and gamma that you suggest would reproduce the Newtonian results for perihelion advance and Shapiro delay, but they would give a nonzero (and therefore non-Newtonian, I think) result for the Nordtvedt effect. It seems strange to me that the parameters should have one Newtonian set of values for one experiment but a different set of Newtonian values for a different experiment. I think all this tells us is that there are no Newtonian values.

I also don't understand what you mean by 'a "Special Relativity" Newtonian theory' -- well, the scare quotes are your own, which I assume means that, like me, you're uncomfortable with trying to reconcile SR with Newtonian gravity. That just doesn't seem possible.

What all of this suggests to me is simply that pervect's argument in #26 is not as solid as it might seem at first glance.

 

[Jonathan Scott]

A simple Newtonian approximation for relativistic gravity would assume a zero value for gamma, so light would still be deflected in the same way as in Newtonian gravity (although not doubled as in GR), and there would still be some perihelion precession. If we want to model gravity to look like a force operating in Newtonian space so as to avoid perihelion precession, we need to be able to describe it relative to a Newtonian Euclidean coordinate space in which light travels in straight lines, with a constant coordinate speed of light, so from the metric point of view any scale factor affecting the time must also be applied to the space, giving $\gamma=-1$ in the PPN formalism. This then means that the tangential acceleration is proportional not to $(1+v^2/c^2)$ as for GR but rather to $(1-v^2/c^2)$, which means that light is not deflected at all.

This is clearly not at all the same thing as a Newtonian approximation to GR. It is just a way of describing the purely Newtonian picture of gravity (in which there is no perihelion precession) within the PPN formalism, for comparison purposes.

 

[Hornbein]

Just think of how disappointed Al would have been if the GR math gave exactly the same result as 1/r^2. Viva la precesion!

 

[pixel]

Actually, Newtonian physics accounted for about 90% of Mercury's precession. General Relativity provided the rest.

 

[bcrowell]

Actually, Newtonian physics accounted for about 90% of Mercury's precession. General Relativity provided the rest.

Yes, what we've been discussing is the non-Newtonian part.

 

[pervect]

I'm not quite clear on your reasoning here. Your first paragraph seems persuasive, but your second paragraph muddies the waters. It's not obvious to me that there should be values of $\beta$ and $\gamma$ that correspond to the Newtonian limit. PPN assumes a nondegenerate metric, whereas Newtonian gravity has a degenerate metric. As far as I know (basically just from reading the review article by Will), the standard way of recovering the Newtonian limit from PPN is not by picking values of $\beta$ and $\gamma$ but rather by taking the limit of small velocities and weak fields.

I think it's possible to show explicitly that there are no values of $\beta$ and $\gamma$ that define the Newtonian theory. Perihelion advance is a non-Newtonian effect that is proportional to $2+2\gamma-\beta$, while the Shapiro time delay (or the part of it that Will describes as "non-Newtonian") is proportional to $1+\gamma$. The Nordtvedt effect is proportional to $4\beta-\gamma-3$ (assuming zero values for $\xi$, alphas, and zetas). Setting all three of these to zero gives no solutions for the two parameters.

My reasoning, in a nutshell, is there are at least two things that affect perihelion shift. The first is spatial curvature, represented by the PPN parameter $\gamma$, the second is the non-linear behavior of the gravitational field, represented by the PPN parameter $\beta$. But these two things alone are not sufficient to explain the exact value of perihelion shift. So I suggest there is probably something else going on as well, a third effect of some sort. At this point I don't have a simple physical explanation for this third effect (I wish I had one). But I feel obligated to mention that the effect is still there, even though I lack a simple "physical" explanation for it.

This leaves me skeptical as to any purported "simple" explanation for a "single physical cause" for perihelion shift, as I see at least two, and those are not sufficient in and of themselves.

I don't claim that my argument is air-tight. That being said, I'm still skeptical of proposed "simple" explanations of perihelion shift.

I have seen analyses based on a formal similarity between some of the differential equations in Schwarzschild coordinates and a classsical analysis, but these in my opinion are a bit of a mathematical trick, the point is they're not physically motivated in a certain sense. What I mean by "physically motivated" is that if you take the best available weak-field approximation of gravity (PPN) and working out the consequences, you get complicated-looking results. The apparently simple looking results only happen if you draw a formal analogy between some equations that fall out of a strong field approach and note the formal similarities in the differential equations and interpret those formal similarities as if they were somehow "physical".

 

[haushofer]

Ok, I can see that point of view, up to a point - the degeneracy of the metric flows from the degeneracy of the Galilean algebra (over 4 dimensions). However, I have never seen the value of the action principle derivation of geodesic equation in relativity (as opposed to all other uses of action principles), because making it work for null geodesics is highly artificial.

You mean by introducing the einbein? This is offtopic here, but I just want to say that in terms of sigma-models this is not so artificial at all; if you want to extend the geodesic equation to strings, branes etc. I think the action principle is quite natural.

But this is maybe for another topic ;)