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- Thread starter jwes44
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Is this a difficult question? I thought it would be like the corrections for time and mass.

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Jonathan Scott

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For systems dominated by a central local mass, the usual practical convention is to use isotropic coordinates, which have the property that local distances in the x, y and z direction relate to coordinate distances by the same scale factor, and hence that the speed of light relative to the coordinate system is the same in all directions at a given point.

In such coordinates, in the weak approximation (which holds everywhere except in the close vicinity of neutron stars or black holes), the simplest way to look at the effect of gravity is the effect it has on the coordinate momentum

[tex]

\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )

[/tex]

That is, the rate of change of coordinate momentum is (1 + v

This equation applies regardless of whether the object is moving horizontally, vertically or anywhere in between. The total energy E is constant (in Newtonian terms, it is the sum of the potential and kinetic energy) so it can be cancelled in the above equation to show that the field determines the rate of change of

Note that since this force is in the direction of the central object, angular momentum is preserved.

You can work out the acceleration in a given case from the above expression by treating c as a variable function of potential. In the special case where the motion is horizontal, c is constant so the acceleration is simply (1 + v

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There is another kind of acceleration called "proper acceleration". This is the acceleration measured by an on-board accelerometer, and its magnitude is frame invariant (the same in all coordinate systems). For a freely falling object it is 0.

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How do you define the Newtonian gravitational fieldIn such [isotropic] coordinates, in the weak approximation... the simplest way to look at the effect of gravity is the effect it has on the coordinate momentump=Ev/c^{2}where E is the total relativistic energy of the test object,vis its coordinate velocity and c is the coordinate speed of light (not the standard value). This behaves in a very simple way:

[tex]

\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )

[/tex]

That is, the rate of change of coordinate momentum is (1 + v^{2}/c^{2}) times the Newtonian gravitational fieldgtimes the energy in coordinate mass units.

Hmmm... For a pulse of light passing by the Sun, the rate of angular deflection at the point of nearest approach is the same as for Newton's theory for a particle moving at c. (Of course, the total angular deflection is twice the total Newtonian deflection, but this is because of the greater deflection rate at other locations, not at the point of nearest approach.)This depends on the speed, and for objects travelling at or near the speed of light gives twice the Newtonian effect.

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Jonathan Scott

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For weak approximation purposes, it doesn't make much difference which coordinate system you use to define g, as the fractional differences are too small to be significant. It's simply the gradient of the potential in appropriate units.How do you define the Newtonian gravitational fieldgin this context? I don't think there is a unambiguous mapping between any radial coordinate in general relativity and Newton's radial coordinate r. Also, since "g" in Newton's theory is defined in terms of flat coordinates, it's not clear to me what vector in general relativity could be said to correspond to Newton's g.

In this case, the specific definition I'm assuming corresponds to the following expression in isotropic coordinates:

[tex]

\mathbf{g} = - \, \frac{c^2}{\Phi} {\nabla \Phi}

[/tex]

where [itex]\Phi[/itex] is the time dilation factor, approximately given by

[tex]

\Phi = (1 + \phi/c^2)

[/tex]

where [itex]\phi[/itex] is the Newtonian potential, typically approximately equal to [itex]-Gm/r[/itex].

For free fall in a static field, the effective rest energy varies with the potential, and this effect exactly cancels with the change in kinetic energy so that the total relativistic energy is constant.Also, I think the relativistic mass-energy E of the object has a dependence on the object's speed, so it wouldn't correspond to anything in Newton's theory, and this effect would be on top of the last factor in your formula. Can you point me to a reference that gives the details of that formula?

No, that's not right. The rate of deflection (and of acceleration of anything else in free fall) as seen by a local observer at rest is equal to the Newtonian value, but the space of that observer is curved relative to isotropic coordinates (and other spherically symmetrical flat coordinates) so relative to those coordinates the rate of deflection is twice the Newtonian value.Hmmm... For a pulse of light passing by the Sun, the rate of angular deflection at the point of nearest approach is the same as for Newton's theory for a particle moving at c. (Of course, the total angular deflection is twice the total Newtonian deflection, but this is because of the greater deflection rate at other locations, not at the point of nearest approach.)

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I don't think that's true. The choice of coordinate system makes ALL the difference when expressing the acceleration. For example, in the usual Schwarzschild coordinates we have for a radially free-falling particleFor weak approximation purposes, it doesn't make much difference which coordinate system you use to define g, as the fractional differences are too small to be significant. It's simply the gradient of the potential in appropriate units.

d^2 r/dt^2 = -(m/r^2)q[1-3(v/q)^2]

where v = dr/dt and q = 1-2m/r, so one could say the relativististic correction is the factor q[1-3(v/q)^2]. With v = 0 this correction factor is simply q, but if we instead use the particle's proper time as the time coordinate we have

d^2 r/dt^2 = -(m/r^2)

for v = 0, which is FORMALLY identical to Newton's equation, so one could say there is no relativistic correction at all. But that would be misleading, because the quantities r and t do not map unambiguously to the same symbols in Newton's theory. Similarly we could work out the expression for d^r/dt^2 in terms of isotropic coordinates and get a different expression, which would be the product of -m/r^2 times some factor. But again it would be naive and misleading in any of these cases to say that the factor multiplying "-m/r^2" is the relativistic correction, because in none of these cases is there a direct correspondence between the quantities r, t and the Newtonian coordinates.

No, the rate of deflection in Newtonian theory (and also in Einstein's 1911 paper) is ym/r^3 per unit distance for a light ray going roughly parallel to the x axis at roughly constant y, where y=R is the minimum distance from the mass at the origin, and r^2 = x^2 + y^2. Integrating the deflection rate from x = -inf to +inf gives the Newtonian (and 1911) total deflection of 2m/R. This accounted only for the temporal curvature, not the spatial curvature. In 1915 Einstein recognized the spatial curvature effect, and so the rate of deflection is (4x^2 +y^2)ym/r^5, and integrating this from x = -inf to +inf gives the full relativistic deflection 4m/R. But the point is that in both cases the rate of deflection at x = 0 (the point of purely transverse motion) is m/R^2. The spatial curvature does not contribute anything to the deflection rate at the point of tangency. You can read about this in any good book on relativity (or even in Einstein's original papers).The rate of deflection [of light grazing the Sun]... as seen by a local observer at rest is equal to the Newtonian value, but the space of that observer is curved relative to isotropic coordinates (and other spherically symmetrical flat coordinates) so relative to those coordinates the rate of deflection is twice the Newtonian value.

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Jonathan Scott

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The acceleration defined using either set of coordinates differs only by tiny amounts, of order Gm/rc

The curvature of the path is definitely double the Newtonian value for tangential motion when expressed

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I'm still not following you. Let's try to focus on one specific thing, to see if we're in agreement. What is the acceleration of a test particle, moving radially with speed v = dr/dt at a distance r from a spherical gravitating body of mass m? I say that, if r is the Schwarzschild radial parameter, then in terms of the Schwarzschild coordinates we haveI'm talking about the weak approximation usingisotropic coordinates, so the only choice I'm suggesting is between using those or using the local observer's coordinates.

d^2 r/dt^2 = -(m/r^2) q [1 - 3(v/q)^2]

where q = (1-2m/r). Do you agree? Also, what do you think is the Newtonian acceleration in this case? And what is the acceleration in terms of isotropic coordinates?

If by "curvature" you mean the rate of deflection, then you're wrong, as was explained in the previous message. This has nothing to do with isotropic coordinates, because the rate of deflection is the change in angle of the wave normal per angular travel of the light pulse. The angular coordinates are unambiguous.The curvature of the path is definitely double the Newtonian value for tangential motion when expressedrelative to isotropic coordinates.

What part of Einstein's derivation of the deflection do you disagree with?

The spatial curvature gives double the deflection over the entire path, but it does not simply double the rate of deflection at every point on the path. This was explained to you in detail in the previous message. Again, the rate of deflection at the purely transverse point is the same as the Newtonian value, but at other locations along the path the deflection rate is greater due to the spatial curvature.The curvature necessarily has the Newtonian value relative to local space for a static observer, and that local space is curved relative to the isotropic coordinate system, giving double the curvature.

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PAllen

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A well known feature of Schwarzschild coordinates is that they are not isotropic (the local coordinate speed of light is different in different directions). That is why Jonathan Scott put it in bold. I think he assumed you would know. Gullestrand-Panlieve coordinates are isotropic.I'm still not following you. Let's try to focus on one specific thing, to see if we're in agreement. What is the acceleration of a test particle, moving radially with speed v = dr/dt at a distance r from a spherical gravitating body of mass m? I say that, if r is the Schwarzschild radial parameter, then in terms of the Schwarzschild coordinates we have

d^2 r/dt^2 = -(m/r^2) q [1 - 3(v/q)^2]

where q = (1-2m/r). Do you agree? Also, what do you think is the Newtonian acceleration in this case? And what is the acceleration in terms of isotropic coordinates?

If by "curvature" you mean the rate of deflection, then you're wrong, as was explained in the previous message. This has nothing to do with isotropic coordinates, because the rate of deflection is the change in angle of the wave normal per angular travel of the light pulse. The angular coordinates are unambiguous.

What part of Einstein's derivation of the deflection do you disagree with?

The spatial curvature gives double the deflection over the entire path, but it does not simply double the rate of deflection at every point on the path. This was explained to you in detail in the previous message. Again, the rate of deflection at the purely transverse point is the same as the Newtonian value, but at other locations along the path the deflection rate is greater due to the spatial curvature.

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Again, as I said in the message to which you responded (apparently without reading it), this has nothing to do with isotropic coordinates, because the rate of deflection is the change in angle of the wave normal per angular travel of the light pulse. The angular coordinates are unambiguous, whether you are using isotropic coordinates or not. Do you understand this? Do you understand why the rate of deflection of light at the point of closest approach to the gravitating body is the same as the Newtonian rate, even though the overall deflection for the entire path is twice the Newtonian deflection? It was explained in detail in a previous message.A well known feature of Schwarzschild coordinates is that they are not isotropic (the local coordinate speed of light is different in different directions). That is why Jonathan Scott put it in bold. I think he assumed you would know. Gullestrand-Panlieve coordinates are isotropic.

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Jonathan Scott

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[tex]

\frac{d}{dt} \left ( \frac{E\mathbf{v}}{c^2} \right ) = \frac{d\mathbf{p}}{dt} = - \, E \left ( \frac{1}{\Phi_t}{\nabla \Phi_t} - \frac{v^2}{c^2} \frac{1}{\Phi_x} {\nabla \Phi_x} \right )

[/tex]

In this equation, [itex]\Phi_t[/itex] and [itex]\Phi_x[/itex] are the time scale factor and space scale factor in the isotropic coordinate form of the metric for the Schwarzschild solution (see for example MTW exercise 31.7), which are equal to the following (where G, m, r and c are either measured in the isotropic coordinates or in terms of the local observer's view at that point, as the scale factors cancel):

[tex]

\Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}

[/tex]

[tex]

\Phi_x = (1 + Gm/2rc^2)^2

[/tex]

In the weak approximation, we can assume the following:

[tex]

\Phi_t = (1 - Gm/rc^2)

[/tex]

[tex]

\Phi_x = 1/(1 - Gm/rc^2) = 1/\Phi_t

[/tex]

This gives the form of the equation which I gave previously:

[tex]

\frac{d}{dt} \left ( \frac{E\mathbf{v}}{c^2} \right ) = \frac{d\mathbf{p}}{dt} = - \, E \left ( \frac{1}{\Phi_t}{\nabla \Phi_t} \right ) \left ( 1 + \frac{v^2}{c^2} \right ) = \frac{E}{c^2} \mathbf{g} \left (1 + \frac{v^2}{c^2} \right )

[/tex]

where

[tex]

\mathbf{g} = - \, \frac{c^2}{\Phi_t}{\nabla \Phi_t}

[/tex]

It is easy to see from this equation of motion that light moving tangentially (such that c is constant) must accelerate by twice

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I see what you're doing, and I agree that IF we assume (1) that "Newtonian" gravity is represented by just the time-time coefficient of our metric, and (2) that isotropic coordinates represent "Newtonian" coordinates, then the relativistic deflection rate of light at each point on the path is indeed twice the "Newtonian" deflection rate. Assumption (1) is fairly reasonable, but assumption (2) is problematic. Notice that if we accept assumption (1) but we substitute the usual Schwarzschild coordinates in place of isotropic coordinates in assumption (2), then the ratio of the relativistic deflection rate to the "Newtonian" deflection rate varies along the path, and the two rates are actually identical at the point of nearest approach.Just to be really specific and remove any ambiguity about definitions of g, I'll give an exact form (not limited to the weak approximation) of the previous equation of motion for the Schwarzschild solution in isotropic coordinates, as derived from the Euler-Lagrange equations of motion... It is easy to see from this equation of motion that light moving tangentially (such that c is constant) must accelerate by twiceg.

So, depending on whether we identify "Newtonian" gravity with the time-time component of the metric in terms of isotropic coordinates or in terms of Schwarzschild coordinates - or in terms of any other system of coordinates that we might choose, we get very different answers to the question: What is the relativistic correction to gravitational acceleration? That's my point.

It's an interesting question whether isotropic or Schwarzschild coordinates correspond more closely to what we would regard as "Newtonian" coordinates. Each of them is asymptotically equal to flat Newtonian coordinates in the weak field limit. However, I would argue that the Schwarzschild coordinates are closer in spirit to the original Newtonian physics based on Galilean relativity and Euclidean geometry, whereas isotropic coordinates may be closer to special relativistic physics (although with variable c). In any case, it's a historical fact that Einstein's original calculations, with just the time-time component in 1911 and with both the time-time and space-space component in 1915, were based on approximations to what later were called the Schwarzschild coordinates, and hence the rate of deflection at the nearest approach was the same in both calculations, but differed at other points on the path.

So, there's no unambiguous answer to the original poster's question, because there is no unambiguous mapping of the Newtonian concepts into the relativistic context, and the various possible mappings that we might propose give significantly different answers.

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Jonathan Scott

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(For example, E = mc

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It's certainly true that isotropic coordinates have some nice properties (like isotropy!), but they also have some "un-natural" features, not least of which is the fact that the circumference of a circular orbit of radius r is not 2 pi r, as it is for Schwarzschild coordinates.Schwarzschild coordinates are useful in the mathematics of deriving the Schwarzschild solution. However, I think isotropic coordinates provide a much more natural and practical way of describing orbits and similar...

But this is somewhat beside the point. Everyone agrees that various coordinate systems are useful for various purposes when describing the Schwarzschild solution, but the question at hand is not "What are the pros and cons of various coordinate systems?", the question is "What is the relativistic correction for gravitational acceleration?", and to answer this question we need to somehow represent the Newtonian acceleration and the relativistic acceleration on a comparable basis, and the problem is that there is no unambiguous basis for making such a comparison.

A ray that is formally "straight" in terms of the Schwarzschild coordinates is formally curved in terms of isotropic coordinates, and vice versa, so the terms representing the deflection of a ray will obviously be different when expressed in terms of these two systems of coordinates, and moreover the partitioning of the deflection between the time-time and space-space components will be different. If we identify the time-time component as the Newtonian deflection, then our answer is highly dependent on the choice of coordinate system.

What you called the Newtonian

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