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Precession of Mercury

  1. Mar 17, 2012 #1
    I am having trouble understanding something about the precession of Mercury. The precession takes place in the same direction of motion as the orbit of the planet, so that the ellipse of the orbit becomes angled slightly more forward over time, right? In a gravity simulator, this amounts to an increase of gravity of (G M / r^2) * (1 + 3 *(v/c)^2) to second order in terms of the instantaneous speed of the planet, or likewise with (G M / r^2) (1 + 6 G M / (r c^2)) in terms of the instantaneous distance from the center of the sun. A decrease of the Newtonian gravity gives a precession in the opposite direction.

    Okay, so the local gravity at r matches the Newtonian gravity G M / r^2, correct? A particle travelling radially at r will be gravitationally time dilated and length contracted according to a distant observer, each by sqrt(1 - 2 G M / (r c^2)), so that the distant observer will say that the particle travels (1 - 2 G M / (r c^2)) slower than the local observer measures, right?

    So here's my question. If the local gravity is G M / r^2 and a distant observer uses a time dilation and length contraction of the local measurements for what he measures, then just as the distant observer measures a lesser speed of an infalling particle, so too should he measure a lesser acceleration of the particle, and similarly for orbit, although without the length contraction in the tangent direction, shouldn't he? If so, then since the distant observer measures a lesser acceleration of Mercury, the precession should be in the direction opposite the direction of motion, while precession in the forward direction requires an acceleration measured by the distant observer that is greater, not less, than the Newtonian gravity. So what is going on here? How do we get a precession in the positive direction with a lesser acceleration of gravity?
     
    Last edited: Mar 17, 2012
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  3. Mar 17, 2012 #2

    Bill_K

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    The orbit equation is r(θ), in which the time variable has been eliminated. So I don't think that the relationship between proper time and global time comes into it at all. Mercury's orbit is a circle plus a small sinusoidal variation, so what does come into it is the radial dependence of the gravitational potential. The effective potential will have a minimum at the circular orbit. In particular what determines the frequency of small oscillation is the second derivative of the effective potential at that point.
    If the second derivative is less, the frequency of small oscillation is also less. That implies that if you start counting at one perihelion, after the planet has gone around 2π it has not yet come back to perihelion. The next perihelion will occur slightly later, at 2π + ε, showing that the perihelion is advancing.
     
  4. Mar 17, 2012 #3

    pervect

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    If, by the "local gravity", you mean the acceleration of an accelerometer of a stationary observer at that point, this is not correct.

    Gravity actually increases as measured by local clocks and rulers, to

    (GM/r^2) * (1/sqrt(1-r_s/r))

    But what you really need to do to solve the problem rigorously is to write down the actual differential equations of motion, the geodesic equations. You might (or might not) be able to reasonably interpret them somehow in terms of "forces", but the textbooks do not do describe any such process, though they will point out some of the ambiguitites in even defining the concept of force.

    You can easily measure the proper acceleration of someone who is not orbiting with an accelerometer, but any such reading for someone in orbit will always give "zero". There isn't any local measurement you can do for an orbiting observer to measure the force - so you'd need to compute it. To compute it correctly according to GR you actually need to use GR - you may or may not stumble on something that's "close enough" by trying to appply Newtonian ideas to the problem. But even if you do stumble on something that's close enough, it's likely to be more or less an accident, and very likely to be dependent on your choice of coordinates (which shouldn't matter to any acdtual physics). And it won't have much to say about how GR actually solves the problem - which is really what you're trying to do, at least I hope so.

    There are a variety of ways of varying complexity of deriving these equations - I believe that Kip Thorne's "Exploring Black Holes" has one of the easier ones, but I have nott read it - I"ve read other books by the same author and they've been very good.

    However, I'm not going to attempt to actually derive the equations in this post, I'm simply going to present them:

    http://www.fourmilab.ch/gravitation/orbits/ (you can confirm the equation with MTW's gravitation, or the book I previously mentioned).

    The equations assume you have a geodesic parameterized by tau, so you have

    t(tau), r(tau), phi(tau)

    You have two important constants of motion that make the equations of motion much simpler: L, the angular momentum, and E, the total energy. These are defined differently than they are in classical mechanics, but they're still constant along the orbit - because two of the geodesic equations can be simplified to dL/dtau = 0 and dE/dtau = 0.

    You can then write:

    L = r^2 d (phi/ dtau)
    E = (1-2M/r) (dt / dtau)

    The last equation you need is

    (dr/dtau)^2 + (1-2M/r) (1+L^2/r^2) = E

    You can solve the last equation for r(tau), given initial conditions and E and L. The previous two equations will giv you phi(tau) and t(tau) once you've solved for r(tau).
     
  5. Mar 17, 2012 #4
    Thanks guys. Okay well, in terms of the energy involved, I found (on page 39 of Black Holes, an introduction by Raine and Thomas) that the local acceleration according to a hovering observer at r is indeed (G M / r^2) / sqrt(1 - r_s / r), while that at infinity is measured at just G M / r^2. But then, that would be just the Newtonian acceleration according to what the distant observer sees, giving a regular Newtonian orbit. So it cannot be the proper acceleration of the particle, nor the coordinate acceleration according to an observer travelling inertially at the same instantaneous speed as the particle at r, nor that of a hovering observer at r, nor that of the distant observer. So what is it? Or better yet, how would one derive the actual coordinate acceleration as measured by the distant observer, the difference in the speed of the particle as measured by the distant observer which is measured over infinitesimal time using the distant observer's clock, or half the difference in the square of the speed over infinitesimal distance using the distant observer's ruler?
     
  6. Mar 19, 2012 #5

    Bill_K

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    grav-universe, As pervect has pointed out, the way to handle this type of problem is to solve the geodesic equations. Once you've got the result you're welcome to try to give it some intuitive meaning afterward, but you can't derive the results on the basis of intuition alone! People often do this and then wonder why they came out off by a factor of two or three.

    Some Newtonian concepts are more useful than others. Force and acceleration can be defined in relativity, but in several different ways. You began this post by asking whether "gravity" is stronger or weaker, but I hope we've explained how that concept is questionable.

    The results for this problem are actually pretty simple to state if you don't get bogged down in details. The geodesic condition leads to equations of motion for dr/dτ and dθ/dτ where τ is the proper time. No one wants to see the proper time in there (we don't live on Mercury!) so eliminate τ by dividing one equation by the other, to get an orbit equation for dr/dθ.

    We did exactly the same thing in Newtonian mechanics for the central force problem. We reduced it to an equivalent one-dimensional problem,

    (dr/dθ)2 = (2mr4/L2) (E - Veff)

    where L is the angular momentum (a constant) and Veff is the "equivalent one-dimensional potential."

    Veff = V(r) + L2/2mr2

    is the sum of the original potential plus a "centrifugal barrier". For the Kepler problem,

    Veff = -GMm/r + L2/2mr2

    Well, the orbit equation for a Schwarzschild geodesic has the precisely the same form, only with

    Veff = -GMm/r + L2/2mr2 - GML2/(mc2r3)

    Here r is the usual Schwarzschild coordinate r. This nice simple form is exact - no approximations have been made. You can go on to calculate the circular orbit by setting V'eff = 0. The frequency of small oscillations from the second derivative V''eff will give you the rate of advance of the perihelion.

    Anyway, look at Veff. The first term is the Newtonian 1/r potential, unmodified. The second term is the centrifugal barrier. The third term is the only thing new in relativity. What is it? It's proportional to L2, quite similar to the centrifugal barrier, but with the opposite sign and a different radial dependence. (Also it's smaller, by the ratio between r and the Schwarzschild radius.) In this problem at least gravity is not weakened. The effect of relativity is merely to weaken the centrifugal barrier.
     
  7. Mar 19, 2012 #6
    Thanks, that looks interesting. How do we get the instantaneous coordinate acceleration in the radial and tangent directions according to the distant observer from that? Is V_eff according to the distant observer?
     
  8. Mar 19, 2012 #7

    Bill_K

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    dr/dθ is the radial variation and d2r/dθ2 would be the acceleration. Where r and θ and Veff are locally measurable quantities, not something that we need to have a distant observer measure. The Schwarschild radial coordinate r can be defined everywhere by saying that it's the coordinate r that makes the area of a sphere come out exactly 4πr2. θ of course is defined so as to range between 0 and 2π.
     
  9. Mar 19, 2012 #8
    Okay, here is what I've got so far, at least in terms of the radial coordinate acceleration during freefall according to the distant observer. It might give an idea of what I am looking for overall. Referring back to page 39 of the book I mentioned, it is basically saying that the ratio of initial to final local energies of a particle is the same as that of the redshift of light, the energy there of course being based directly upon the frequency, which in turn depends only upon the gravitational time dilation, sqrt(1 - r_s / r). So for a particle falling from rest at infinity, we have

    E_inf / E_r = sqrt(1 - r_s / r) = [m c^2] / [m c^2 / sqrt(1 - (v/c)^2]

    sqrt(1 - r_s / r) = sqrt(1 - (v/c)^2)

    where v here would be the locally measured speed since the measured energies are local, so we can see immediately that 2 G M / r = v^2, same as for Newtonian gravity. According to the distant observer, however, we have

    v_r = v_loc * time dilation * length contraction = (1 - r_s / r) v_loc = c (1 - r_s / r) sqrt(r_s / r)

    As far as the coordinate radial acceleration according to the distant observer, falling over an infinitesimal distance from r to r - dr, then, we get

    2 a dr = v_{r-dr}^2 - v_r^2

    2 a dr = c^2 (1 - r_s / (r - dr))^2 (r_s / (r - dr)) - c^2 (1 - r_s / r)^2 (1 - r_s / r)

    2 a dr / c^2 = [(r_s / (r - dr)) - 2 (r_s / (r - dr))^2 + (r_s / (r - dr))^3] - [(r_s / r) - 2 (r_s / r)^2 + (r_s / r)^3]

    2 a dr / c^2 = [(r_s / (r - dr)) - (r_s / r)] - 2 [(r_s / (r - dr))^2 - (r_s / r)^2] + [(r_s / (r - dr))^3 - (r_s / r)^3]

    2 a dr / c^2 = [r_s (r - (r - dr)) / (r (r - dr))] - 2 [r_s^2 (r^2 - (r - dr)^2) / (r^2 (r - dr)^2] + [r_s^3 (r^3 - (r - dr)^3) / (r^3 (r - dr)^3)]

    giving to first order of dr,

    2 a dr / c^2 = [r_s dr / r^2] - 2 [r_s^2 (2 r dr) / r^4] + [r_s^3 (3 r^2 dr) / r^6]

    2 a dr / c^2 = r_s dr / r^2 - 4 r_s^2 dr / r^3 + 3 r_s^3 dr / r^4

    a = (r_s c^2 / (2 r^2)) [1 - 4 r_s / r + 3 (r_s / r)^2]

    a = (G M / r^2) (1 - 3 r_s / r) (1 - r_s / r)

    From this we can see that the coordinate acceleration according to the distant observer is positive (attractive) all the way down to 3 r_s, where it becomes zero, then negative (effectively repulsive) thereafter, until the particle reaches r_s where the acceleration becomes zero again. We can see from earlier that the speed of the particle according to the distant observer becomes zero at r_s, so since the acceleration there is also zero, the particle will effectively "freeze" at r_s. The maximum speed will be at 3 r_s where the acceleration turns from positive to negative, with a local speed of

    v_loc = c sqrt(r_s / (3 r_s)) = c / sqrt(3)

    which to the distant observer would become

    v_r = c (1 - r_s / (3 r_s)) sqrt(r_s / (3 r_s))

    v_r = c (2 / 3) / sqrt(3)

    v_r = 2 c / (3 sqrt(3))

    Okay well, that was a good exercise, but we can see that the effective radial coordinate acceleration according to the distant observer to second order is a = (G M / r^2) (1 - 4 r_s / r), while for the precession, it would have to be a_prec = (G M / r^2) (1 + 3 r_s / r). The orbit of a planet would be almost completely tangent rather than radial, but how do we go from 1 - 4 r_s / r to 1 + 3 r_s / r for tangential freefall?
     
    Last edited: Mar 19, 2012
  10. Mar 19, 2012 #9
    Okay, thanks. I will try to apply this to your last post and try to see how it can work out. It might take a minute, though, because I am not used to polar coordinates, preferring cartesian. Could this be translated to cartesian coordinates?
     
  11. Mar 19, 2012 #10
    Let's start with this. Let's say that a planet follows a completely circular orbit around the sun at a distance r according to the distant observer's coordinate system. First of all, is an absolutely circular orbit possible in GR? If so, then what coordinate speed would the planet have according to the distant observer in terms of r? If not, what would be the closest approximation? How is that derived?
     
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