Gravitational time dilatation and curved spacetime - follow up

In summary: If you do that, you can't use observers at fixed heights, since those worldlines are not geodesics, as I said in my previous post just now. What I described in post #3 in the original thread was an alternative formulation where both of the observers are free-falling--that means each one has to start and end at a fixed height, but in between needs to be in free fall, so they need to be moving upward at the start and downward at the end, i.e., each one follows a parabolic arc that starts and ends at their chosen height, ##z_1## or ##z_2##.But in that case, I did not see any definition of "par
  • #36
vanhees71 said:
You don't need to make a closed loop of geodesics.

Yes, I've already recognized that. Please see post #33.
 
  • Like
Likes vanhees71
Physics news on Phys.org
  • #37
Going back to Fermi-Walker transport, I tend to think of it as the sort of transports that gyroscopes do. But, that view really only works for transporting vectors along the sort of curves that gyroscopes can follow, i.e. along timelike curves.

I suspect the OP want's a more general description of Fermi-Walker transport than that, one that could work along space-like curves. I don't have one to offer at this time, but there were some good posts on the topic by others.

In flat space-time, parallel transport in any inertial frame is just a matter of translating the vector in said frame. The curved space-time case is more interesting, but I don't think I need to rehash what others wrote there.

As far as the difference between Fermi-walker transport and parallel transport goes, I suggest comparing them in flat space time. Then if we restrict the transport to being along timelike curves (which is the way I think of it), Thomas precession of gyroscopes illustrates the difference between the two. In that context, transporting a vector in an inertial frame is just a matter of translating it, and Thomas precision shows that gyroscopes (and Fermi-Walker transport) obey a different transport law.

Wiki has an article on Thomas precession in 3-vector formalism, https://en.wikipedia.org/wiki/Thomas_precession#Statement. In the low velocity limit it's proportional to the cross product of the acceleration times the velocity, there are some additional multiplicative factors in the full relativistic case.
 
  • #38
pervect said:
I suspect the OP want's a more general description of Fermi-Walker transport than that, one that could work along space-like curves.

You can apply the Fermi-Walker transport law along any spacelike curve; you just pick one of the spacelike vectors in the orthonormal tetrad to be the tangent vector to the spacelike curve. This case is not much discussed in the literature, but it works fine mathematically, since the Fermi-Walker transport law does not depend on the curve being timelike.

The case I'm not sure about is applying the Fermi-Walker transport law along a null curve. Obviously you can't choose an orthonormal tetrad in that case if you want to include the curve's tangent vector.
 
  • Like
Likes vanhees71
  • #39
PeterDonis said:
You can apply the Fermi-Walker transport law along any spacelike curve; you just pick one of the spacelike vectors in the orthonormal tetrad to be the tangent vector to the spacelike curve. This case is not much discussed in the literature, but it works fine mathematically, since the Fermi-Walker transport law does not depend on the curve being timelike.

The case I'm not sure about is applying the Fermi-Walker transport law along a null curve. Obviously you can't choose an orthonormal tetrad in that case if you want to include the curve's tangent vector.

I imagine so - the formulas I am used to and looked up are expressed in terms of the acceleration and velocity along a time-like worldline. I would assume that one could use any affinely parameterized curve with suitable adjustments, but I have no intuition for what happens in that case.

The particular formula I use is MTW 6.14

$$\frac{d}{d\tau} v^a = \left( u^a a^b - u^b a^a \right) v_b \quad a^a = \frac{d}{d\tau} u^a$$

Here ##v^a## is the vector being transported, when the equation is satisfied, ##v^a## is fermi-walker transproted. ##u^a## is the unit time-like vector field that does the transporting - physically, it's just the 4-velocity (and 4-acceleration) of the path that the gyroscope takes, while ##v^a## is a vector that represents the spin-axis of the gyroscope.

While I am used to thinking of ##u^a## as a 4-velocity and ##a^a## as a 4-acceleration as above, I imagine it's not essential, ##u^a## would be interpreted as the tangent vector to the generalized curve, and ##a^a## would be it's derivative with respect to the affine parameter along the generalized curve. (I'd use s instead of ##\tau## to paramaterize a non-timelike curve, though of course that's just a matter of convention).

I also am not sure about potential normalization issues in the null case, I imagine that's your concern as well. In the timelike case, it's understood that ##u^a##, the vector field representing the transport, is everywhere of unit length. I think it probably matters that it be of unit length.

Wiki has a more abstract treatment in terms of the Fermi derivative, rather than just writing the components of the transport equation, but that's not the approach I learned the topic from. Also, it's been a while since I've actually done anything like this.
 
  • #40
pervect said:
In flat space-time, parallel transport in any inertial frame is just a matter of translating the vector in said frame. The curved space-time case is more interesting, but I don't think I need to rehash what others wrote there.
In that case (flat space-time in global Lorentzian coordinates) space-time is basically an affine space in affine coordinates thus parallel transporting a tangent vector (by Levi-Civita trivial connection) simply preserves its components.

In the context of this thread, IIUC, Lie transport along the flow of ##\partial_t## (Schwarzschild spacetime in Schwarzschild coordinates KVF) results the same as the Fermi-walker transport along it. Here we're interested at Lie transport because of its feature to preserve the 'shape' of the transported curve (light pulse paths) thus defining the notion "to be congruent"
 
  • #41
pervect said:
I think it probably matters that it be of unit length.

Yes, I think it does, because IIRC the fact that ##u^a## and ##a^a## are orthogonal is used in the derivation of the formula you give. In the null case that orthogonality condition is no longer met.
 
  • #42
But even for a null vector line you have
$$u_{\mu} u^{\mu}=0=\text{const}$$
and thus
$$\dot{u}_{\mu} u^{\mu}=0,$$
where the dot means a derivative wrt. an arbitrary parameter parametrizing the light-like worldline under consideration.

I'm not sure, whether one can define a congruence of "light-cone coordinates" by Fermi-Walker transport though (or a tetrad with two light-like and two real space-like vectros). In special relativity there's no problem to work with a basis consisting of two light-like and two space-like vectors. Of course these are not pseudo-Cartesian coordinates.

The only somewhat related thing I've heard about in GR is the use of Newton-Penrose tetrades, but they use four nullvectors with two of them complex. That's obviously not exactly the same as light-cone coordinates in SR.
 
  • #43
vanhees71 said:
even for a null vector line you have
$$
u_{\mu} u^{\mu}=0=\text{const}
$$
and thus
$$
\dot{u}_{\mu} u^{\mu}=0
$$
where the dot means a derivative wrt. an arbitrary parameter

Yes, you're right. I misstated the concern. The concern is that, for the null case, ##\dot{u}## must be tangent to the worldline. This obviously follows from what you wrote, quoted just above.
 
  • #44
Maybe my notation was not well enough explained; ##u^{\mu}## is the tangent vector already. ##\dot{u}^{\mu}## is then something like the "acceleration", but for light-like curves ##u^{\mu}## is not a four-velocity in the usual sense anymore, and the world-line parameter can of course not be an intrinsic measure of time like proper time for time-like world lines (like the ones describing massive point-like classical particles). So it's not so clear, whether it makes sense to think about ##a^{\mu}=\dot{u}^{\mu}## as an "acceleration" in this case. Anyway, there are no massless classical point particles (even not in SR). So it's anyway an academic question. The one thing null geodesics are good for of course is as a model of "light beams" or rather propagation of em. waves using the eikonal approximation. The usual textbook shortcut is to talk about pointlike photons, which usually gives the right result (e.g., the light bending in gravitational fields) but a wrong qualitative picture, but that's another story.
 
  • #45
vanhees71 said:
Maybe my notation was not well enough explained

Your notation is fine; it's fairly standard to use the dot to denote a derivative with respect to the affine parameter along a curve.

I am simply pointing out that, for the null case, ##\dot{u}## must also be tangent to the null curve, as ##u## itself is; that follows immediately from ##\dot{u}_\mu u^\mu = 0##. To put it another way, null vectors have the counterintuitive property that a vector that is orthogonal to a given null vector is also a multiple of it. Thus, we must have ##\dot{u} = k u##, where ##k## is some function of the curve parameter (I don't think it needs to be constant along the curve). If we plug this into the Fermi derivative formula (using ##\dot{u}## instead of ##a##), it makes the Fermi derivative vanish identically. So I don't think a useful notion of Fermi-Walker transport can be defined for null curves.
 
  • #46
Sure, now I understand what you mean. Of course a vector ##a## that is "orthogonal" to a null vector ##u##, ##a^{\mu} u_{\mu}=g_{\mu \nu} a^{\mu} u^{\nu}## can be either a null vector itself (than it's parallel to ##u##, as you say) or it can be spacelike, depending on the specific light-like curve.

E.g., take the nullvector ##(u^{\mu})=(1,0,0,1)## in SR. Then any null vector ##(a^{\mu})=(a^0,\vec{a})## that is "orthogonal" to ##u## implies ##a^0-a^3=0## and thus ##a^0=a^3## and the from ##(a^0)^2-\vec{a}^2=0## that ##a^1=a^2=0## and thus ##a=a^0 u##. But of course you can have two space-like directions that are "orthogonal" to ##u##, e.g., ##e_1=(0,1,0,0)## and ##e_2=(0,0,1,0)##. Light-cone coordinates are usually chosen by defining the basis consisting of the two light-like vectors ##u_{\pm}=(1,0,0,\pm 1)## and the two space-like vectors ##e_1## and ##e_2##.
 
  • #47
vanhees71 said:
can be spacelike

Ah, that's right, there will always be a spacelike 2-surface orthogonal to the null vector.

vanhees71 said:
depending on the specific light-like curve

I think there is a spacelike 2-surface orthogonal to any null curve; I don't think this depends on the specific curve.
 
  • Like
Likes vanhees71
  • #48
PAllen said:
You might want to study post #177 in that thread for a summary of the accumulated understanding developed int the thread.
Quoting first item in that post:
PAllen said:
1) They posit a theory where the Minkowski metric is the observable metric for distance and time measurements.This is pretty clearly stated."
What do you mean with 'observable' Minkowski metric ? Basically that distance and time measurement are consistent with a Minkowski metric in which the (timelike) paths of observers at fixed height ##z## from ground are assumed to be geodesics of the underlying space-time ?
 
  • #49
cianfa72 said:
Quoting first item in that post:

What do you mean with 'observable' Minkowski metric ? Basically that distance and time measurement are consistent with a Minkowski metric in which the (timelike) paths of observers at fixed height ##z## from ground are assumed to be geodesics of the underlying space-time ?
Yes.
 
  • #50
Quoting third item in that post
PAllen said:
3) They posit it is possible to set up global Lorentz frame physically using a described procedure.
Which is the procedure they posit to be used to setup that global Lorentz frame ?
 
Last edited:
  • #51
cianfa72 said:
Which is the procedure they posit to be used to setup that global Lorentz frame ?

Start with a family of observers at rest at infinity. Then extend to all finite distances ##r## from the origin by adding observers at each ##r## who are at rest relative to the observers at infinity (as they can confirm by exchanging round-trip light signals). The worldlines of all these observers provide the timelike "grid lines" of the global Lorentz frame, and the spacelike "grid lines" are then just mutually orthogonal spatial geodesics that are also orthogonal to those timelike worldlines.
 
  • Like
Likes vanhees71
  • #52
PeterDonis said:
Start with a family of observers at rest at infinity. Then extend to all finite distances ##r## from the origin by adding observers at each ##r## who are at rest relative to the observers at infinity (as they can confirm by exchanging round-trip light signals). The worldlines of all these observers provide the timelike "grid lines" of the global Lorentz frame, and the spacelike "grid lines" are then just mutually orthogonal spatial geodesics that are also orthogonal to those timelike worldlines.
ok, but I'm confused about the following:
The global frame being built this way is a Lorentz one (the observed metric due to space and time measurements results in ##ds^2 = dz^2 - dt^2##) thus coordinate clocks 'wearing' by observers providing timelike 'grid lines' must result 'Einstein synchronized' upon exchanging repetuted round-trip light signals. However that is not the case because to gravitational time dilatation between obsevers at fixed height as discussed in this thread
 
  • #53
cianfa72 said:
The global frame being built this way is a Lorentz one (the observed metric due to space and time measurements results in ##ds^2 = dz^2 - dt^2##)

Not the "observed metric"--the "predicted metric" of the hypothetical theory being described. Remember that this whole "global Lorentz frame" thing is not something we can actually do in the real world. It is something that, at least on one possible reading of Schild's argument, the hypothetical theory of gravity that he is describing (basically, a theory in which gravity is treated as a "force field" like electromagnetism on a flat spacetime background) predicts we should be able to do. But Schild's argument then amounts to the observation that we can't actually do it, because of gravitational time dilation--the actual observed metric doesn't match the predicted metric that we should observe according to the theory.

cianfa72 said:
that is not the case because to gravitational time dilatation between obsevers at fixed height

Yes, which means the hypothetical theory can't be right. See above.

Note also my concern, expressed in the thread, that I'm not sure the hypothetical theory being described is even self-consistent to begin with--in other words, that we don't even need to get to the point where we bring in gravitational time dilation to see that the theory doesn't work.
 
  • #54
PeterDonis said:
the hypothetical theory of gravity that he is describing (basically, a theory in which gravity is treated as a "force field" like electromagnetism on a flat spacetime background) predicts we should be able to do. But Schild's argument then amounts to the observation that we can't actually do it, because of gravitational time dilation--the actual observed metric doesn't match the predicted metric that we should observe according to the theory.
Trying to sum up (sorry but I'm a non-expert :frown:)
In that hypothetical theory of gravity -- assuming a global Lorentz frame built as previously described -- the predicted metric should be Minkowski using that (global) coordinate system. Now about light ray propagation consider the following scenarious:
  1. light ray paths are straights described by the equation ##ds^2=0## (null-like geodesics)
  2. light ray paths are not null-geodesics (or geodesics at all) because of gravity and thus have not equation ##ds^2=0##
In both cases light ray worldlines of Schild's argument at first and second emission are nevertheless 'congruent' -- they are obtained one from the other by a Lie transport along the fixed height observer woldlines (global Lorentz frame timelike 'grid lines') that are actually orbits of ##\partial_t## KVF in those coordinates. Thus we have a quadrilater having two congruent sides (light ray worldlines) and the lower and upper sides (fixed height observer worldlines) parallel -- because based on our 'working hypothesis' they are actually gridlines of our global Lorentz frame.

In both cases gravitational time dilatation for lower and upper observers (as turn out from experiment) gets a contradiction
 
Last edited:
  • #55
cianfa72 said:
the predicted metric should be Minkowski using that (global) coordinate system

Yes, although it's not entirely clear, even leaving out the question of gravitational time dilation, how this predicted metric satisfies the requirements of SR. For example, static observers--observers at rest in the global Minkowski coordinate system--have nonzero proper acceleration, but this is not true in SR of observers at rest in a global inertial frame. This presumably would be attributed to the properties of the "force of gravity" in the hypothetical theory under discussion, but it's not clear to me that such a "force" can be made consistent with SR just on the basis of the proper acceleration, even without taking into account gravitational time dilation. Since nobody has ever actually written down this hypothetical theory, there's no way to know for sure.

cianfa72 said:
In both cases light ray worldlines of Schild's argument at first and second emission are nevertheless 'congruent' -- they are obtained one from the other by a Lie transport along the fixed height observer woldlines (global Lorentz frame timelike 'grid lines') that are actually orbits of ##\partial_t## KVF in those coordinates.

That's correct. Schild's argument as it is presented in MTW makes no assumption at all about whether the light ray worldlines have ##ds^2 = 0## or not in the global Minkowski metric of the hypothetical gravity theory. I don't know whether Schild himself discusses this in the original sources, but I don't see any reason why he would need to; the argument only depends on the two light ray worldlines being congruent, not on their specific value of ##ds^2##.

cianfa72 said:
In both cases gravitational time dilatation for lower and upper observers (as turn out from experiment) gets a contradiction

Yes.
 
  • #56
PeterDonis said:
The worldlines of all these observers provide the timelike "grid lines" of the global Lorentz frame, and the spacelike "grid lines" are then just mutually orthogonal spatial geodesics that are also orthogonal to those timelike worldlines.
Regarding spacelike "grid lines" I'm a bit confused about their actual physical significance. I'm aware of timelike worldlines physical significance as paths taken in spacetime by massive particles, but what about spacelike worldlines ?

ps thanks for your help :wink:
 
  • #57
cianfa72 said:
Regarding spacelike "grid lines" I'm a bit confused about their actual physical significance.

A surface of constant coordinate time picks out events that happen "at the same time" according to the chosen coordinates. In general coordinates don't always have physical significance, but in this particular case the coordinates are matched to a symmetry of the spacetime: the surfaces of constant coordinate time are orthogonal to the orbits of the timelike Killing vector field. So those are the natural surfaces of constant time for static observers.

The spacelike "grid lines" just mark out an ordinary spatial grid in the spacelike surfaces of constant time, and their physical interpretation is similar to that of an ordinary spatial grid. For example, the spatial grid line that connects two events with the same coordinate time on the two static observers' worldlines can be thought of as describing a constant-time "slice" of the "world tube" of a ruler that is placed between them and stays at rest relative to them. The proper distance along that grid line would be the length shown on the ruler.
 
  • #58
cianfa72 said:
ok, but I'm confused about the following:
The global frame being built this way is a Lorentz one (the observed metric due to space and time measurements results in ##ds^2 = dz^2 - dt^2##) thus coordinate clocks 'wearing' by observers providing timelike 'grid lines' must result 'Einstein synchronized' upon exchanging repetuted round-trip light signals. However that is not the case because to gravitational time dilatation between obsevers at fixed height as discussed in this thread

From a metric standpoint ##ds^2 = dz^2 - dt^2## gets replaces with ##ds^2 = dz^2 - z^2 dt^2## - or something equivalent. That's the Rindler metric. I'll leave the details of "something equivalent" as an advanced topic for the reader, for some definitions of equivalent one might want to introduce a few more arbitrary parameters, but the above metric is sufficient for the point I want to make. Note that the path that light follows will obey the differential equation ##dz^2 - z^2 dt^2 = 0##, which implies dz/dt = +z or dz/dt = -z.

From an instantaneous frame point of view, one first have to realize that clocks at different z run at different rates, so one can't synchronize clocks in general unless they are rate-adjusted. After the rate adjustment process, this choice of metric ##dz^2 - z^2 dt^2## is equivalent to using Einstein synchronization over infinitesimal distances. So to synchronize two clocks at a large z difference, you'd need pairwise synchronize a sequence of clocks, and take the limit as you introduce an infinite number of clocks. Again, note that light does not follow a path of constant dz/dt, but rather |dz/dt| = z. So if we take z=2, dz/dt = +2 or dz/dt=-2. In these coordinates, the coordinate speed of light is the same in both directions, but it is not constant, it is a function of z.

Also note that many routine physical formula assume that dz/dt=1, so such formula will need to be modified if one uses Rindler coordinates unless one uses tensor methods. With the particular metric I have given (one of the simplest forms), the origin of the coordinate system of an observer is placed at z=1, not z=0. The behavior at z=0 is a coordinate singularity known as the Rindler horizon.
 
  • #59
pervect said:
From a metric standpoint ##ds^2 = dz^2 - dt^2## gets replaces with ##ds^2 = dz^2 - z^2 dt^2## - or something equivalent. That's the Rindler metric.

Remember that in this discussion, we are not talking about standard SR. We are talking about a hypothetical theory of gravitation as a "force field" on flat spacetime. Since the theory is hypothetical, we don't know exactly how it models the "force field" of gravitation. But we do know that Schild's argument clearly states that static observers--observers at rest in the "gravitational field"--are at rest relative to observers at rest at infinity. But Rindler observers are not at rest relative to observers at rest at infinity. So Schild's static observers cannot be Rindler observers, and the Rindler metric cannot be the metric that static observers will be "at rest" in.
 
  • Like
Likes cianfa72

Similar threads

  • Special and General Relativity
Replies
21
Views
490
  • Special and General Relativity
Replies
27
Views
4K
  • Special and General Relativity
7
Replies
230
Views
17K
  • Special and General Relativity
Replies
30
Views
644
  • Special and General Relativity
Replies
7
Views
1K
  • Special and General Relativity
2
Replies
58
Views
2K
  • Special and General Relativity
2
Replies
57
Views
2K
  • Special and General Relativity
Replies
21
Views
1K
  • Special and General Relativity
2
Replies
37
Views
3K
  • Special and General Relativity
Replies
34
Views
564
Back
Top