# Thread: GR coordinate system problem

1. Let's say we have a mass with an object orbitting with constant speed in a circular orbit and a distant observer Bob. According to Bob's coordinate system, the orbit is circular at a speed v and a constant inward coordinate acceleration a. The coordinate acceleration is just what is inferred according to Bob's coordinate system by the usual definitions of acceleration, s = vo t + 1/2 a t^2, 2 a s = vf^2 - vo^2, a = (vf - vo) / t, etc., where s is the distance travelled and vo and vf are the original and final velocities. The relativistic acceleration formulas also reduce to these when considering a body starting at rest and accelerating an infinitesimal distance over infinitesimal time, which we will be doing, so we need not worry with more complicated formulas for this.

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have travelled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass, since the orbit is perfectly circular. The distance being infinitesimal, we can drop higher orders and gain just x = r - r (1 - y^2 / (2 r^2)) = y^2 / (2 r) in the radial direction. We want the radial coordinate acceleration as inferred by Bob, so we can use s = vo t + 1/2 a t^2, where vo = 0 for the original velocity in the radial direction, so x = s = 1/2 a t^2.

So now we have x = 1/2 a t^2 = y^2 / (2 r), and since y = v t, this reduces to just

1/2 a t^2 = y^2 / (2 r)

a t^2 = (v^2 t^2) / r

a = v^2 / r

giving the usual acceleration formula for a circular orbit. Again, this is just the coordinate acceleration inferred by Bob's coordinate system. It is not saying anything about proper acceleration or anything else, just a coordinate effect. Does all of this look okay so far?

2.

3. Originally Posted by grav-universe
Does all of this look okay so far?
No, it doesn't, this is not the way GR deals with equations of motion. Shoehorning basic Newtonian equations is not the way to go.

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have travelled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass
The stuff that you wrote is not correct even in Newtonian mechanics since you are mixing radial (linear) motion with circular motion. So, what you wrote is complete nonsense. If the infinitesimal motion is in the tangential direction, then , in the radial direction you should have , meaning that .

4. Originally Posted by xyzt
Originally Posted by grav-universe
Does all of this look okay so far?
No, it doesn't, this is not the way GR deals with equations of motion. Shoehorning basic Newtonian equations is not the way to go.
All acceleration formulas will reduce to these when starting from rest and measuring over infinitesimal distance and time. If nothing else, we can define the coordinate acceleration in this way. Even the relativistic equations will reduce to the same thing in this case.

s = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1]

where t is infinitesimal, so

s = (c^2 / a) [(1 + (a dt / c)^2 / 2) - 1]

s = (c^2 / a) [a^2 dt^2 / (2 c^2)]

s = a dt^2 / 2

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have travelled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass
The stuff that you wrote is not correct even in Newtonian mechanics since you are mixing radial (linear) motion with circular motion. So, what you wrote is complete nonsense. If the infinitesimal motion is in the tangential direction, then , in the radial direction you should have , meaning that .
For a circular orbit, we would have

X^2 + Y^2 = r^2, so X = sqrt(r^2 - Y^2). For infinitesimal distance y travelled, whereas y = Y, that gives for the distance x travelled in the radial direction

x = r - X = r - sqrt(r^2 - y^2) = y^2 / (2 r)

5. Originally Posted by grav-universe
Originally Posted by xyzt
Originally Posted by grav-universe
Does all of this look okay so far?
No, it doesn't, this is not the way GR deals with equations of motion. Shoehorning basic Newtonian equations is not the way to go.
All acceleration formulas will reduce to these when starting from rest and measuring over infinitesimal distance and time. If nothing else, we can define the coordinate acceleration in this way. Even the relativistic equations will reduce to the same thing in this case.

s = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1]

where t is infinitesimal, so

s = (c^2 / a) [(1 + (a dt / c)^2 / 2) - 1]

s = (c^2 / a) [a^2 dt^2 / (2 c^2)]

s = a dt^2 / 2

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have travelled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass
The stuff that you wrote is not correct even in Newtonian mechanics since you are mixing radial (linear) motion with circular motion. So, what you wrote is complete nonsense. If the infinitesimal motion is in the tangential direction, then , in the radial direction you should have , meaning that .
For a circular orbit, we would have

X^2 + Y^2 = r^2, so X = sqrt(r^2 - Y^2). For infinitesimal y travelled, whereas y = Y, that gives for the distance x travelled in the radial direction

x = r - X = r - sqrt(r^2 - y^2) = y^2 / (2 r)
Why do you insist in posting nonsense? This thread belongs in "Fringe Alternative Theories".

6. Originally Posted by grav-universe
Originally Posted by xyzt
Originally Posted by grav-universe
Does all of this look okay so far?
No, it doesn't, this is not the way GR deals with equations of motion. Shoehorning basic Newtonian equations is not the way to go.
All acceleration formulas will reduce to these when starting from rest and measuring over infinitesimal distance and time. If nothing else, we can define the coordinate acceleration in this way. Even the relativistic equations will reduce to the same thing in this case.

s = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1]

where t is infinitesimal, so

s = (c^2 / a) [(1 + (a dt / c)^2 / 2) - 1]

s = (c^2 / a) [a^2 dt^2 / (2 c^2)]

s = a dt^2 / 2
Doesn't mean that you can do basic Newtonian calculations and pretend that you are doing GR.

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have travelled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass
The stuff that you wrote is not correct even in Newtonian mechanics since you are mixing radial (linear) motion with circular motion. So, what you wrote is complete nonsense. If the infinitesimal motion is in the tangential direction, then , in the radial direction you should have , meaning that .
For a circular orbit, we would have

X^2 + Y^2 = r^2, so X = sqrt(r^2 - Y^2). For infinitesimal distance y travelled, whereas y = Y, that gives for the distance x travelled in the radial direction

x = r - X = r - sqrt(r^2 - y^2) = y^2 / (2 r)
No, you don't. If you move in the tangential direction this means you are moving in the radial direction. Not that it makes any difference since you are writing a bunch of nonsense anyways.

7. I am considering how the object moves along the circular orbit with coordinates X,Y

8. Originally Posted by grav-universe
I am considering how the object moves along the circular orbit with coordinates X,Y
Yes, now that you corrected your elementary derivation, so what? It is still a fringe theory, that has nothing to do with GR.

9. Originally Posted by grav-universe
I am considering how the object moves along the circular orbit with coordinates X,Y
It would be quite troublesome and tedious to consider this problem in Cartesian coordinates; it is far better and far more natural to utilise Schwarzschild coordinates from the outset. The basic equation of motion ( radial projection onto equatorial plane ) then becomes

with the effective potential

As you can see we have two basic constants of motion here, the energy per unit mass E, and the angular momentum per unit mass L. Both of these can be determined from initial conditions. In the simplest case where angular momentum vanishes we get the following orbits :

1. Circular orbit for
2. Precessing orbit if E is little more than
3. Temporary orbiting if E is close to
4. Orbital decay for

You will find a more in-depth treatment here : http://arxiv.org/pdf/1201.5611.pdf
And here is an interactive visualisation of the resulting orbits : Wolfram Demonstrations Project

It should be made explicitly clear that all of this is valid only in Schwarzschild vacuum space-times, i.e. in cases where the central body is spherically symmetric, stationary, uncharged, and has no angular momentum. Other space-times ( e.g. Kerr or Reissner-Nordstrom ) will be more complicated to treat.

10. Thanks Markus. I will cut to the chase. The problem is I am still trying to find relationships and invariants that are independent of the coordinate system used, as we did here. There we found the relations

a' = G M L_t^2 / (z r^2) and a' = L z' c^2 / z

where at r, a' is the locally measured acceleration, z is the time dilation, L is the radial length contraction, and L_t is the length contraction in the tangent direction. From the derivatives of these two relations we can gain the R_00 tensor.

Okay, so now I want to look at what the distant observer infers and find relations and invariants from that. So far we have

a = v_t^2 / r

where v_t is the tangent velocity. From the equivalence principle, if the hovering observer Alice were at r in the same place as the object passes and Alice lets go of a ball at that moment, the ball and object should both accelerate toward the mass at the same rate at that instant. Although the object is travelling tangently, its initial radial velocity is zero, the same as the ball.

Alice measures the local acceleration to be a'. She measures the ball to fall some distance radially along the length of a short rod over some time. The distant observer Bob, however, infers that the time that passes for the ball to travel the length of the rod is 1/z longer and that the length of the rod itself is L shorter, so the coordinate acceleration that Bob measures is z^2 L smaller, whereby a = z^2 L a'. Likewise, the locally measured tangent velocity is v_t', and Bob infers 1/z greater time to travel a distance that is L_t shorter in the tangent direction, so v_t = z L_t v_t'. So now we have

z^2 L a' = (z L_t v_t')^2 / r

L a' = L_t^2 v_t'^2 / r

Here's where the problem lies. Let's look at the invariants. L_t / r is an invariant, having the same value in any coordinate system. a' and v_t' are also invariants since they are what is measured locally, regardless of the distant observer's coordinate system. But that leaves

L / L_t = (L_t / r) v_t'^2 / a'

The right side is all invariant, so the left side should be too. But if the ratio of the radial to tangent length contraction is invariant, then there can be one and only one valid coordinate system. For instance, if one coordinate system gives some ratio L / L_t as an invariant for some spherical shell (with L being found between two very close shells), then the only way to change the coordinate system such that this ratio remains the same for that shell is to change the radius of all shells by the same ratio. But if we make them all .99 the original radiuses, then the distant observer's distance from the mass also changes by .99, whereas it should verge upon 1, so there can be only one coordinate system where this is also true.

11. Originally Posted by grav-universe
Thanks Markus. I will cut to the chase. The problem is I am still trying to find relationships and invariants that are independent of the coordinate system used, as we did here.
But this is NOT what you are doing, what you are doing is a series of (incorrect) basic hacks due to your inability/unwillingness to study GR. You are trying to force your Newtonian hacks into GR. This is not how things work, Markus gave you the general equation of motion, derived from the Euler-Lagrange equations, you should start studying in earnest.

12. Let's go ahead and try this for the Schwarzschild coordinate system and see if we can pinpoint where the problem actually lies. What is the locally measured orbittal speed as measured by a hovering observer at r for an object in a circular orbit at r?

13. Originally Posted by grav-universe
see if we can pinpoint where the problem actually lies.
There's a fair chance it's between your ears...

14. Okay well, looks like the orbittal speed the distant observer infers is just v_t^2 = G M / r and the local hovering observer measures v_t' = (G M / r) / (1 - 2 G M / (r c^2)). So starting with the local acceleration and backtracking, we can get

a' = G M L_t^2 / (z r^2)

a' = L_t^2 v_t^2 / (z r)

a' = L_t^2 (z L_t v_t')^2 / (z r)

and picking out the invariants, that gives

a' / [(v_t'^2) (L_t / r)] = L_t^3 z

which says that L_t^3 z is invariant. That doesn't look right either.

15. Oh, z is also an invariant, so we are left with L_t^3. It will work out, however, if v_t^2 = L_t^3 G M / r, although I don't immediately see why that should be the case. Plugging that into the coordinate acceleration formula for the distant observer, we get

a = v_t^2 / r

a = G M L_t^3 / r^2

z^2 L a' = G M L_t^3 / r^2

and separating the invariants,

L / L_t = G M (L_t / r)^2 / z^2

so we are still left with L / L_t as an invariant. It would be an easy fix if a = z^2 L_t a' instead of a = z^2 L a', but the acceleration is in the radial direction, not the tangent direction.

16. As far as v_t^2 = L_t^3 G M / r, I can almost see it now with

v_t^2 = L_t^3 G M / r

z^2 L_t^2 v_t'^2 = L_t^3 G M / r

z^2 v_t'^2 = G M (L_t / r)

which carries only invariants with nothing left over as it should be. But I still don't yet see the reason for the L_t^3 there.

18. Here is an awesome post where much of what I am trying to do has been worked out.
Okay, so far from pervect's post #12 in the other thread, converting to the units I am using, for time dilation, for radial length contraction, and for the tangent length contraction, changes the metric from

to

and then we have

and first derivatives with

For the proper acceleration of a hovering observer, or measured acceleration by the same observer for a ball dropped from rest, we have

The coordinate acceleration for a distant observer becomes

whereby , so everything agrees so far with what I've got for this much.

19. Originally Posted by grav-universe
Let's go ahead and try this for the Schwarzschild coordinate system and see if we can pinpoint where the problem actually lies.
There is no "problem", other than you are just a pretender. That's all.

20. Originally Posted by grav-universe
The problem is I am still trying to find relationships and invariants that are independent of the coordinate system used
The only invariants here that are coordinate independent are the Riemann curvature tensor and its contractions. So far as the orbits are concerned, they are solutions to the geodesic equations - the equations themselves are again covariant, but their solutions are explicitly dependent on a coordinate system chosen, so I am not sure what you are hoping to achieve here. The best you can do is consider what all orbits have in common - namely that the bodies are in free fall, and hence

This is independent of any coordinate system, so the covariant vanishing of acceleration is an invariant of motion.

21. I think this guy suffers from a rare form of graphomania: writing total BS and trying to pass it as physics. He managed to totally bamboozle some pretty good guys, see here. I wonder how long will take them to catch on the fact that his posts are total nonsense.

22. Originally Posted by Markus Hanke
Originally Posted by grav-universe
The problem is I am still trying to find relationships and invariants that are independent of the coordinate system used
The only invariants here that are coordinate independent are the Riemann curvature tensor and its contractions. So far as the orbits are concerned, they are solutions to the geodesic equations - the equations themselves are again covariant, but their solutions are explicitly dependent on a coordinate system chosen, so I am not sure what you are hoping to achieve here. The best you can do is consider what all orbits have in common - namely that the bodies are in free fall, and hence

This is independent of any coordinate system, so the covariant vanishing of acceleration is an invariant of motion.
Thanks, Markus. What does that tensor say in algebraic form?

As far as what I am attempting to achieve, this thread is a continuation of another thread here (post #4). In that thread, we could find through somewhat straight-forward and intuitive methods a couple of relationships for local acceleration.

and

Combining those gives

and finding the derivatives of that gives the R_00 tensor.

Basically, what I am trying to do is bypass the Einstein field equations and find other relations that will solve for the unknowns, z, L, and L_t, through natural relations that can be found here and there and/or working through the invariants. With what we have so far, we can make a coordinate choice for one and find the relation between the other two, but there is not enough information to solve for all three. I couldn't determine anything more about the local acceleration, so instead I am now attempting it by looking at what is inferred by a distant observer.

23. Originally Posted by grav-universe

Basically, what I am trying to do is bypass the Einstein field equations
Basically, you are just putting together some crank hacks . I wonder how long it will take the others to figure your BS.

24. Okay, so the last equation in Pervect's post gives

which becomes

So now we can gain

and bringing all of the obvious invariants to the left side, we are left with

This means that is the invariant associated here while I had found just . I'm not sure where the additional term of comes from yet but I will continue to look into it.

25. Okay, so to verify that really quick, in Schwarzschild coordinates, we would have

so for , the invariant

would give

In Eddington's isotropic coordinates, we would have

giving

So it does indeed work out the same, as an invariant, in both coordinate systems.

26. Oh wait. What I had originally would give

while pervect's post gives

so that's quite different actually. Backtracking from what pervect has gives

That is very nonintuitive, so I'm not sure I can use it. I'll need something much more straight-forward.

27.

28. Originally Posted by grav-universe
Oh wait. What I had originally would give

...

so that's quite different actually. Backtracking from what pervect has gives
...

That is very nonintuitive, so I'm not sure I can use it. I'll need something much more straight-forward.
I am unsure of what the point is of all of these posts by you. They read much like an undergraduate's homework notes. Perhaps you are doing UG homework? If so, maybe asking for homework help would be more directly useful.

Or is there some other purpose for presenting all of these tedious stream-of-consciousness algebraic manipulations on a public forum??

29. Originally Posted by tk421
Originally Posted by grav-universe
Oh wait. What I had originally would give

...

so that's quite different actually. Backtracking from what pervect has gives
...

That is very nonintuitive, so I'm not sure I can use it. I'll need something much more straight-forward.
I am unsure of what the point is of all of these posts by you. They read much like an undergraduate's homework notes. Perhaps you are doing UG homework? If so, maybe asking for homework help would be more directly useful.

Or is there some other purpose for presenting all of these tedious stream-of-consciousness algebraic manipulations on a public forum??
Well, I'm looking for a somewhat intuitive way to solve for the unknowns z, L, and L_t while bypassing the Einstein field equations. On another forum, it has been shown that the invariant I just found is just z. If that can be demonstrated, then we will have a way to solve for all three unknowns. For instance, let's make the original coordinate choice, L_t = 1, whereby L_t' = 0. Then for the two relations

and

we can reduce to just

and

and putting those two together we get

and then solving for z using Wolfram, we find

where c_1 is a numerical constant. So since z must verge upon 1 for large r, c_1 must be 1, and since L = z, we therefore have

30. Originally Posted by grav-universe
Thanks, Markus. What does that tensor say in algebraic form?
I am not certain what you mean by that. a is the acceleration 4-vector, and the double pipe denotes covariant differentiation, in this case with respect to the proper time.

As far as what I am attempting to achieve, this thread is a continuation of another thread here (post #4).
Ok, I remember that one. I calculated the Ricci tensor for you in cases where the system isn't necessarily spherically symmetric.

Basically, what I am trying to do is bypass the Einstein field equations and find other relations that will solve for the unknowns, z, L, and L_t, through natural relations that can be found here and there and/or working through the invariants.
I'm afraid that doesn't make much sense. All the maths you have been typing is ( so far as I can see ) based on Schwarzschild geometry, which is only one particular solution to the field equations. What the EFEs do is allow you to calculate the components of the metric tensor in the general case, i.e. given an energy-momentum tensor and some initial conditions. You can't replace that by looking at invariants, because these invariants are themselves dependent on a particular metric tensor.

31. Originally Posted by Markus Hanke
I'm afraid that doesn't make much sense. All the maths you have been typing is ( so far as I can see ) based on Schwarzschild geometry, which is only one particular solution to the field equations. What the EFEs do is allow you to calculate the components of the metric tensor in the general case, i.e. given an energy-momentum tensor and some initial conditions. You can't replace that by looking at invariants, because these invariants are themselves dependent on a particular metric tensor.
I am looking at what is invariant regardless of the geometry with the metric in the general form

Some obvious invariants are z, dr / L, r / L_t, and locally measured values such as a' and v_t'.

I am trying to find another relation like the two we already have for the local acceleration that will solve for the unknowns z, L, and L_t once a coordinate choice has been made for one of them as in post #28 where the solution for Schwarzschild was found, although we could just as easily have made a different coordinate choice and found a different solution.

32. Originally Posted by grav-universe
I am looking at what is invariant regardless of the geometry with the metric in the general form

This metric is only three dimensional...?

In general, if you want to know the invariants for a given geometry, the best way to do that is to examine the Killing vectors of the space-time in question. Are you familiar with that concept ?

I am trying to find another relation like the two we already have for the local acceleration that will solve for the unknowns z, L, and L_t once a coordinate choice has been made for one of them as in post #28 where the solution for Schwarzschild was found, although we could just as easily have made a different coordinate choice and found a different solution.
See post #19 - the vanishing of the covariant derivative of the acceleration 4-vector wrt proper time does just that.

33. Originally Posted by grav-universe
Originally Posted by tk421
Originally Posted by grav-universe
Oh wait. What I had originally would give

...

so that's quite different actually. Backtracking from what pervect has gives
...

That is very nonintuitive, so I'm not sure I can use it. I'll need something much more straight-forward.
I am unsure of what the point is of all of these posts by you. They read much like an undergraduate's homework notes. Perhaps you are doing UG homework? If so, maybe asking for homework help would be more directly useful.

Or is there some other purpose for presenting all of these tedious stream-of-consciousness algebraic manipulations on a public forum??
Well, I'm looking for a somewhat intuitive way to solve for the unknowns z, L, and L_t while bypassing the Einstein field equations.
This is what happens when you don't understand the basics. The coefficients of the line element (the stuff that you call " the unknowns z, L, and L_t") are components of the metric tensor. The metric tensor is integral part of the EFE's, so you can find its components ONLY by solving the EFE's. So, your attempt at "solving for the unknowns z, L, and L_t while bypassing the Einstein field equations", is pure crackpottery. How long will you keep this up is a function of the patience of the moderators.

34. I can now visualize why there can be only one coordinate system where the coordinate acceleration is measured the same. Let's say we have such a coordinate system for the distant observer. Locally at r, the radial acceleration of a particle falling from rest is the same as that of a particle travelling tangently so that the equivalence principle will hold. As with the falling elevator example, the elevator falling from rest and a particle travelling tangently must accelerate radially together so that according to the elevator observer, the particle will travel straight across inertially, ignoring the gravitational gradient. The distant observer agrees that both the particle travelling tangently at r and another particle that falls from rest at the moment the first particle passes will both accelerate radially together, so having the same coordinate radial acceleration.

Now let's change the coordinate system of the distant observer. Think of the space around the mass as composed of spherical shells at all r, each with their own values for z, L, and L_t, although L is measured between shells that are an infinitesimal distance apart. We will change the coordinate system with the simple coordinate transform r1 = r - 3 r_s. This simply pushes all spherical shells inward by the same amount, 3 r_s. The distances between shells, however, remains the same, since all of the shells move inward radially by the same amount, so the value of L also remains the same, since that is the length contraction of local radial rulers, which the distant observer measures the same as before. Of course the time dilation z is an invariant for each particular shell, so that remains the same as well. Since the shells are now smaller, though, the value for the length contraction in the tangent direction L_t will be smaller, so the distant observer will say that a ruler that is pointing tangently at each shell is now smaller by a factor of r1 / r = 1 - 3 r_s / r.

Okay, so the particle falling from rest will now fall over the same distance between shells in the same time, so that coordinate acceleration will remain the same as before. The orbitting particle however, travelling tangently at the same place as the particle dropped from rest, will now still travel the same angle in the same time around the orbit as before, but since the shell is now smaller, that means it will have travelled a lesser distance along the x and y directions than before, so falling inward at a lesser rate than the particle falling from rest by the same factor as that between the old and new shell radii. The radial coordinate acceleration of both particles will no longer be measured the same.

In order for both particles to still be measured to have the same coordinate radial acceleration according to the distant observer, L and L_t must change by the same amount, so the entire space around the mass can only be scaled up or down by some numerical ratio only, all shells at once. But any scaling of this nature, say by a factor of .99 for example, will also leave measurements at large radii working toward .99 rather than 1 as they should. So there is one and only one coordinate system where the coordinate acceleration is measured the same.

All observers agree upon events that coincide in the same place, such as the readings upon two clocks as they pass. It seems to me that as the two particles pass, coinciding in the same place, that if the local observer says that each accelerates radially at the same rate, then all observers should agree. Since acceleration is measured over a distance and time that the particles travel, however, there is some difference between that and the readings upon two clocks at the exact moment they pass. The distances and times the accelerations are measured over, however, are infinitesimal, so it would still seem that all observers should agree, but it turns out there would be only one coordinate system where that would be possible, an isotropic coordinate system it appears from what I have further worked out if I have done it correctly, where L_t = L.

35. Originally Posted by grav-universe
All observers agree upon events that coincide in the same place, such as the readings upon two clocks as they pass. It seems to me that as the two particles pass, coinciding in the same place, that if the local observer says that each accelerates radially at the same rate, then all observers should agree. Since acceleration is measured over a distance and time that the particles travel, however, there is some difference between that and the readings upon two clocks at the exact moment they pass. The distances and times the accelerations are measured over, however, are infinitesimal, so it would still seem that all observers should agree, but it turns out there would be only one coordinate system where that would be possible, an isotropic coordinate system it appears from what I have further worked out if I have done it correctly, where L_t = L.
In GR, you need to distinguish carefully between coordinate and proper quantities; the former are dependent on the observer ( and hence on the chosen coordinate system ), whereas the latter are not. If you define radial acceleration via distances and times, and consider only its magnitude, you will run the danger of arriving at a coordinate-dependent quantity; on the other hand, you can avoid the issue altogether by formulating it using a 4-vector quantity. All observers in free fall will then satisfy the very simple equation

where the double pipes denote covariant differentiation. This holds true for all observers in all coordinate systems, and is thus a fully covariant expression. If you expand this using the definition of the covariant derivative, you get the geodesic equation - which makes perfect sense, since test particles in free fall trace out geodesics in space-time.

EDIT : Embarrassing mistake in the indices within the equation above - surprised that no one has pointed that out ! Fixed now.

36. Thanks Markus. Would you happen to have a good link handy that expands that in algebraic form?

37. Originally Posted by grav-universe
Thanks Markus. Would you happen to have a good link handy that expands that in algebraic form?
I'll just do it for you right here. Recall the definition of the covariant derivative ( of a 4-vector in this instance ) :

wherein the single pipe denotes the "normal" partial derivative. Also recall that

as usual. Now expand the expression I gave in post #34 using these definitions :

which is precisely the geodesic equation. In fact, this is really the definition of a geodesic :

It's quite simply a curve that parallel-transports its own tangent vector; it doesn't really get any more intuitive than this

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement