by snoopies622
Tags: forces, tidal
PF Patron
P: 1,772
 Quote by Altabeh I hope he'd be interested to give us his argument as to how, as you seem to believe in it, in the Rindler spacetime a geodesic is the same as a geodesic in the Minkowski spacetime.
First of all, I wouldn't talk of Rindler spacetime and Minkowski spacetime; I'd talk about Rindler coordinates and Minkowski coordinates as two different coordinate systems within the same spacetime.

The property of "being a geodesic" is a geometrical property that doesn't depend on the choice of coordinates; it is an intrinsic property of a worldline. The 4-acceleration
$$A^\mu = \frac{DU^\mu}{d\tau} = \frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta$$
(where U is 4-velocity) is a 4-vector, i.e. a tensor that transforms between coordinate systems in the correct way. (And the symbol $dU^\mu/d\tau$ does not represent a 4-vector.)

Its magnitude, "proper acceleration", $\sqrt{|A_{\mu} A^{\mu} |}$ is therefore a scalar invariant, the same value in all coordinate systems. The geodesic equation -- in any valid coordinate system you like -- is just the condition that the proper acceleration is zero, or equivalently, that the (spacelike) 4-acceleration is the zero 4-vector.
P: 1,883
 I just said since the components of proper 3-acceleration are position-dependent, as the geodesic equations say, then its magnitude is what an accelerometer being under the control of a comoving observer shows and it is of course frame-dependent.
 But do you disagree with this basic prediction that a constant-velocity path in Minkowski coordinates, when translated into Rindler coordinates using the coordinate transformation, will still be a path that registers zero reading on an ordinary physical accelerometer when we use the Rindler metric to predict the accelerometer's behavior?
He isn't talking about comoving, but free-falling observers. And he isn't talking about coordinate acceleration, but proper acceleration (as measured by an accelerometer).
Your statement that "3-acceleration" is frame dependent, while right, has no relevance whatsoever, and does not back up your disagreement with the statement.
I think there are some misunderstandings that should be clarified.
P: 665
 Quote by Ich JesseM asked: He isn't talking about comoving, but free-falling observers. And he isn't talking about coordinate acceleration, but proper acceleration (as measured by an accelerometer). Your statement that "3-acceleration" is frame dependent, while right, has no relevance whatsoever, and does not back up your disagreement with the statement. I think there are some misunderstandings that should be clarified.
Since the start of this thread, I've been using acceleration as the proper acceleration 3-vector, or similarly, proper 3-acceleration and I clearly made this apparent to everyone when I answered in the affirmative to DrGreg's post:

 Altabeh is thinking of the 3-vector $d^2 x^i / d\tau^2$ (i=1,2,3), or its magnitude, either way, a coordinate-dependent quantity.

 ...that shows DrGreg's guess about my way of looking at the proper acceleration is correct.
I don't think the misunderstanding is from me! But I agree that I was talking about comoving observer rather than a free-falling observer!

 Quote by DrGreg First of all, I wouldn't talk of Rindler spacetime and Minkowski spacetime; I'd talk about Rindler coordinates and Minkowski coordinates as two different coordinate systems within the same spacetime.
Again, the Minkowski spacetime is not the same as the Rindler spacetime. The first one is only an extended version of the latter. See Wald, R. M. General Relativity, page 151. And talking about Rindler spacetime inspires the application of Rindler coordinates immediately so I don't think this was a necessary note to make!

 The property of "being a geodesic" is a geometrical property that doesn't depend on the choice of coordinates; it is an intrinsic property of a worldline. The 4-acceleration $$A^\mu = \frac{DU^\mu}{d\tau} = \frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta$$ (where U is 4-velocity) is a 4-vector, i.e. a tensor that transforms between coordinate systems in the correct way. (And the symbol $dU^\mu/d\tau$ does not represent a 4-vector.)
Edited: Along a geodesic the proper acceleration is zero if such thing is defined to be the magnituse of $DU^\mu/d\tau.$ Otherwise the use of proper acceleration 3-vector combined with a null time-component in the equation $$A=\sqrt{g_{\mu\nu}a^{\mu}a^{\nu}}$$ would not lead to a frame-independent acceleration.

AB
P: 1,883
 I don't think the misunderstanding is from me! But I agree that I was talking about comoving observer rather than a free-falling observer!
I didn't say that the misunderstanding was yours. Your "no" to JesseM's question was clearly wrong, that's why I thought that you are talking about something different, and in that case i think it would have helped if you had read the question carefully.
 with $a^{\mu}=\frac{d^2x^{\mu}}{d\tau^2}[/tex] being the proper four-acceleration That's a coordinate acceleration. It becomes proper acceleration after correcting for coordinate effects (with the help of the Christoffel symbols). That's what DrGreg, JesseM, and I are talking about. P: 665  Quote by Ich That's a coordinate acceleration. It becomes proper acceleration after correcting for coordinate effects (with the help of the Christoffel symbols). That's what DrGreg, JesseM, and I are talking about. This is incorrect! The coordinate acceleration is the derivative of the coordinate velocity with respect to coordinate time, i.e. $$d^2x^{\nu}/dt^2.$$ That is a proper acceleration 3-vector combined with a null time-component. AB Sci Advisor P: 8,430  Quote by DrGreg First of all, I wouldn't talk of Rindler spacetime and Minkowski spacetime; I'd talk about Rindler coordinates and Minkowski coordinates as two different coordinate systems within the same spacetime.  Quote by Altabeh Again, the Minkowski spacetime is not the same as the Rindler spacetime. The first one is only an extended version of the latter. But if the first is an extended version of the latter, you agree they are "the same spacetime" in the region of the Rindler wedge, right? If so, how can you also argue that they disagree about local physical facts on this wedge, like what set of events a particle with no non-gravitational forces acting on it (i.e. a particle following a geodesic path) will cross through? Also, do you disagree that, as DrGreg said, an object on a geodesic path feels zero 4-acceleration everywhere along it rather than zero 3-acceleration? And that the magnitude of the 4-acceleration for a given point on a worldline determines what would be measured by a physical accelerometer moving along that worldline? Finally, do you disagree that if we have a path in Minkowski coordinates on the Rindler wedge that has zero 4-acceleration everywhere as calculated by the Minkowski metric (i.e. a path with constant velocity in Minkowski coordinates), then if we use the coordinate transformation to transform this path into Rindler coordinates, the path will still have zero 4-acceleration everywhere along it as calculated using the Rindler metric? PF Patron Sci Advisor P: 1,772  Quote by Altabeh This is incorrect! The coordinate acceleration is the derivative of the coordinate velocity with respect to coordinate time, i.e. $$d^2x^{\nu}/dt^2.$$ That is a proper acceleration 3-vector combined with a null time-component. AB You are right that [itex]d^2x^i/dt^2$ is coordinate acceleration.

What has caused confusion in this thread is your describing $d^2x^i/d\tau^2$ as "proper acceleration". That is not the terminology that everyone else uses and so everyone has been disagreeing with you.

I'm guessing that maybe you thought of this because some authors describe the 3-vector $dx^i/d\tau$ as "proper velocity". That is not a terminology I like; I prefer to call that by its alternative name "celerity". In relativity, most "proper" things are invariant e.g. proper time, proper length and the correct definition of proper acceleration (& I've seen some people describe (rest) mass as "proper mass"). The 3-vector $dx^i/d\tau$ is, of course, coordinate-dependent. The 4-vector $dx^{\nu}/d\tau$ is described as "4-velocity" (rather than "proper velocity").
P: 665
 Quote by JesseM But if the first is an extended version of the latter, you agree they are "the same spacetime" in the region of the Rindler wedge, right? If so, how can you also argue that they disagree about local physical facts on this wedge, like what set of events a particle with no non-gravitational forces acting on it (i.e. a particle following a geodesic path) will cross through?
I think once I said that these two are the same when it comes to the Rindler wedge. And about the second question, all problems arise from the way I've learned the implication of "proper acceleration"! Since I've been taking the 3-vector $dx^i/d\tau$ as "proper velocity", so automatically I let $d^2x^i/d\tau^2$ be the "proper 3-acceleration" whereas everyone else makes use of some other definition for "proper acceleration", basically the magnitude of the 4-acceleration $D^2x^i/d\tau^2$. This sort of usage of the proper acceleration would mean that such implication is invariant under the change of coordinates and thus is frame-independent. Along a geodesic, as I said earlier, $D^2x^i/d\tau^2$ vanishes whether on the Rindler wedge or in the Minkowski spacetime. But when we use $d^2x^i/d\tau^2$, then we won't end up getting a frame-independent value for its magnitude. This means that if I looked through the angle you see the whole problem, then I would agree with you!

 Also, do you disagree that, as DrGreg said, an object on a geodesic path feels zero 4-acceleration everywhere along it rather than zero 3-acceleration?
The proper 4-acceleration must be zero along any geodesic, but this can't be true for 3-acceleration.

 And that the magnitude of the 4-acceleration for a given point on a worldline determines what would be measured by a physical accelerometer moving along that worldline?
Now I'd like to say "YES"!

 Finally, do you disagree that if we have a path in Minkowski coordinates on the Rindler wedge that has zero 4-acceleration everywhere as calculated by the Minkowski metric (i.e. a path with constant velocity in Minkowski coordinates), then if we use the coordinate transformation to transform this path into Rindler coordinates, the path will still have zero 4-acceleration everywhere along it as calculated using the Rindler metric?
When it comes to motion along a geodesic, then the proper 4-acceleration $D^2x^i/d\tau^2$ automatically vanishes in any spacetime, leading me to not disagree! But if we are talking about a general path, then the proper 4-acceleration $D^2x^i/d\tau^2$ does not neccessarily vanish on the Rindler wedge for a particle that was already supposed to be moving with a constant velocity in the Minkowski spacetime. This is just because the right hand side of geodesic equations is no longer zero for non-geodesic trajectories (i.e. there is no freely moving particle anymore) in the Rindler coordinates.

AB
P: 8,430
 Quote by JesseM Also, do you disagree that, as DrGreg said, an object on a geodesic path feels zero 4-acceleration everywhere along it rather than zero 3-acceleration?
 Quote by Altabeh The proper 4-acceleration must be zero along any geodesic, but this can't be true for 3-acceleration.
 Quote by JesseM And that the magnitude of the 4-acceleration for a given point on a worldline determines what would be measured by a physical accelerometer moving along that worldline?
 Quote by Altabeh Now I'd like to say "YES"!
"YES" you disagree with that last one about the accelerometer? Here you're talking about an accelerometer moving along that worldline at all times and not an accelerometer held by a temporarily co-moving observer, right? If you do disagree that the magnitude of the 4-acceleration determines the reading of such an accelerometer, what do you think determines the reading of the accelerometer?
 Quote by JesseM Finally, do you disagree that if we have a path in Minkowski coordinates on the Rindler wedge that has zero 4-acceleration everywhere as calculated by the Minkowski metric (i.e. a path with constant velocity in Minkowski coordinates), then if we use the coordinate transformation to transform this path into Rindler coordinates, the path will still have zero 4-acceleration everywhere along it as calculated using the Rindler metric?
 Quote by Altabeh When it comes to motion along a geodesic, then the proper 4-acceleration $D^2x^i/d\tau^2$ automatically vanishes in any spacetime, leading me to not disagree! But if we are talking about a general path, then the proper 4-acceleration $D^2x^i/d\tau^2$ does not neccessarily vanish on the Rindler wedge for a particle that was already supposed to be moving with a constant velocity in the Minkowski spacetime.
So if we take a constant-velocity path in Minkowski coordinates (presumably you'd agree this is a geodesic according to the Minkowski metric, and that the Minkowski metric says the 4-acceleration vanishes on this path), then use the coordinate transformation to translate the set of events that constitute the path into Rindler coordinates, you would argue that the 4-acceleration does not vanish along this path according to the Rindler metric?
P: 665
 Quote by JesseM "YES" you disagree with that last one about the accelerometer? Here you're talking about an accelerometer moving along that worldline at all times and not an accelerometer held by a temporarily co-moving observer, right? If you do disagree that the magnitude of the 4-acceleration determines the reading of such an accelerometer, what do you think determines the reading of the accelerometer?
An accelerometer held by a free-falling observer is supposed to be reading the magnitude of the proper 4-acceleration of a particle moving along an arbitrary path. But if the observer is moving parallel to the particle at all times (a co-moving observer) then I think both the magnitudes of proper 3-acceleration and proper 4-acceleration can be read by accelerometer in case the two different scenarios for proper acceleration are considered together.

 So if we take a constant-velocity path in Minkowski coordinates (presumably you'd agree this is a geodesic according to the Minkowski metric, and that the Minkowski metric says the 4-acceleration vanishes on this path), then use the coordinate transformation to translate the set of events that constitute the path into Rindler coordinates, you would argue that the 4-acceleration does not vanish along this path according to the Rindler metric?
If I can get you right, by velocity' you are presumably referring to the coordinate velocity because in case the "proper" is needed, you state that in your sentences! Since the coordinate velocity is frame-dependent, a coordinate transformation must make it change in particular if the secondary metric is position-dependent (e.g. Rindler metric). In Rindler metric with $$c=1$$ the coordinate velocity $$v$$ along geodesics is given by

$$v=(x/2) a$$

where $$a$$ represents the coordinate acceleration. Thus it is clear that if one seeks out a constant $$v$$, it is necessary to let the particle be hovering at a constant $$x.$$ or be moving along paths with constant $$x.$$ Along non-geodesic trajectories, a constant $$v$$ means that the coordinate $$x$$ must remain constant duing the motion and this tells us that the proper 4-acceleration vanishes. But since we are talking about general paths, the constancy of $$v$$ cannot be always guaranteed in the Rindler metric even if the particle has a constant coordinate velocity in Minkowski spacetime.

AB
P: 8,430
 Quote by Altabeh An accelerometer held by a free-falling observer is supposed to be reading the magnitude of the proper 4-acceleration of a particle moving along an arbitrary path. But if the observer is moving parallel to the particle at all times (a co-moving observer) then I think both the magnitudes of proper 3-acceleration and proper 4-acceleration can be read by accelerometer in case the two different scenarios for proper acceleration are considered together.
How can a single accelerometer read both the 3-acceleration and the 4-acceleration? It only shows one reading at any given moment, so surely it must agree with one or the other?
 Quote by Altabeh If I can get you right, by velocity' you are presumably referring to the coordinate velocity because in case the "proper" is needed, you state that in your sentences! Since the coordinate velocity is frame-dependent, a coordinate transformation must make it change in particular if the secondary metric is position-dependent (e.g. Rindler metric). In Rindler metric with $$c=1$$ the coordinate velocity $$v$$ along geodesics is given by $$v=(x/2) a$$ where $$a$$ represents the coordinate acceleration. Thus it is clear that if one seeks out a constant $$v$$
No, I said I wanted to look at the wordline of a particle with constant velocity (coordinate velocity) in Minkowski coordinates, but then I made clear that I wanted to translate the same worldline consisting of the same events into Rindler coordinates using the coordinate transformation. This worldline obviously wouldn't have constant coordinate velocity in Rindler coordinates! Since on the Rindler wedge Minkowski coordinates and Rindler coordinates are just different ways of labeling events in the same physical spacetime, surely you agree that we can talk about the same physical worldline as described in the two different coordinate systems? I don't think there was any ambiguity in my description:
 So if we take a constant-velocity path in Minkowski coordinates (presumably you'd agree this is a geodesic according to the Minkowski metric, and that the Minkowski metric says the 4-acceleration vanishes on this path), then use the coordinate transformation to translate the set of events that constitute the path into Rindler coordinates
Do you understand what I'm saying now? If so, do you agree that if we take a worldline that has constant coordinate velocity in Minkowski coordinates, then use the coordinate transformation to describe the same worldline in Rindler coordinates, and use the Rindler metric to calculate the 4-acceleration on this worldline, then we'll find it has zero 4-acceleration everywhere?
P: 665
 Quote by JesseM How can a single accelerometer read both the 3-acceleration and the 4-acceleration? It only shows one reading at any given moment, so surely it must agree with one or the other?
Our problem is we can't understand each other the way we should! I don't mean an accelerometer can read both of them at the same time! I assume you know that some measurement devices are ad hoc but some aren't, e.g. a clock that shows the current local time in New York and Texas simultananeously which isn't my purpose here! An accelerometer in my opinion can do read both the 3-acceleration and the 4-acceleration but not at the same time and I think I didn't say something like this!

 No, I said I wanted to look at the wordline of a particle with constant velocity (coordinate velocity) in Minkowski coordinates, but then I made clear that I wanted to translate the same worldline consisting of the same events into Rindler coordinates using the coordinate transformation.This worldline obviously wouldn't have constant coordinate velocity in Rindler coordinates!
I think I showed this for a particle moving along a geodesic in Rindler coordinates.

 Since on the Rindler wedge Minkowski coordinates and Rindler coordinates are just different ways of labeling events in the same physical spacetime, surely you agree that we can talk about the same physical worldline as described in the two different coordinate systems? I don't think there was any ambiguity in my description:
Sorry! This time I confused the proper 4-acceleration with coordinate 4-acceleration. I was looking at your 4-acceleration, talking about proper 4-acceleration but trying to make a meaning of coordinate 4-acceleration out of it!

 Do you understand what I'm saying now? If so, do you agree that if we take a worldline that has constant coordinate velocity in Minkowski coordinates, then use the coordinate transformation to describe the same worldline in Rindler coordinates, and use the Rindler metric to calculate the 4-acceleration on this worldline, then we'll find it has zero 4-acceleration everywhere?
Yes!

AB
P: 8,430
 Quote by Altabeh Our problem is we can't understand each other the way we should! I don't mean an accelerometer can read both of them at the same time! I assume you know that some measurement devices are ad hoc but some aren't, e.g. a clock that shows the current local time in New York and Texas simultananeously which isn't my purpose here! An accelerometer in my opinion can do read both the 3-acceleration and the 4-acceleration but not at the same time and I think I didn't say something like this!
I think "accelerometer" has a standard meaning in physics just like "clock". Any of the various accelerometer designs here should measure 4-acceleration, for example (and the 4-acceleration measured in G-forces should also produce the same effects for an observer following this worldline as would be felt if they were at rest in a gravitational field with the same G-forces). Maybe you could produce a different type of device to measure 3-acceleration locally, but I'm not sure how it would work, and I don't think it would qualify as an "accelerometer".
 Quote by JesseM Do you understand what I'm saying now? If so, do you agree that if we take a worldline that has constant coordinate velocity in Minkowski coordinates, then use the coordinate transformation to describe the same worldline in Rindler coordinates, and use the Rindler metric to calculate the 4-acceleration on this worldline, then we'll find it has zero 4-acceleration everywhere?
 Quote by Altabeh Yes!
OK, good. And would you also agree that this worldline is a geodesic according to the Rindler metric? (i.e. it's the worldline that maximizes the proper time between any two events that lie on it)
P: 665
 Quote by JesseM OK, good. And would you also agree that this worldline is a geodesic according to the Rindler metric? (i.e. it's the worldline that maximizes the proper time between any two events that lie on it)
Why not? But let's speak more generally and say that the worldline is not just timelike but arbitrary so the proper "quantity" between any two events lying on it is an extremum.

AB
P: 8,430
 Quote by Altabeh Why not?
Well, because for a long time you were denying my claim that if we looked at a geodesic in Minkowski coordinates, and then used the coordinate transformation to define the same path in Rindler coordinates, then the path would still be a geodesic relative to the Rindler metric. But I guess you were just misunderstanding what I meant when I said things like "same path", not realizing I was talking about using the coordinate transformation to find the new coordinates of the same physical events along a given path which was originally defined in Minkowski coordinates.
 Quote by Altabeh But let's speak more generally and say that the worldline is not just timelike but arbitrary so the proper "quantity" between any two events lying on it is an extremum.
Yeah, spacelike and null geodesics in Minkowski coordinates should also map to spacelike and null geodesics in Rindler coordinates, assuming again that we are looking at the same physical set of events in each coordinate system.

Assuming we've settled this issue, perhaps we can revisit my earlier claim that the equivalence principle would hold in arbitrarily large regions of Rindler coordinates? Hopefully you'd agree that in any flat SR spacetime described in Minkowski coordinates, it should be possible to construct a physical network of rulers and synchronized clocks moving inertially relative to this coordinate system (a network of the type that Einstein used to define the concept of an 'inertial frame'), such that if we use the readings on the rulers and clocks to define a new coordinate system, the laws of physics as seen in this coordinate system (which is just a different Minkowski frame) will be those seen in any SR inertial frame. Now, if you agree that Minkowski coordinates (on the Rindler wedge) and Rindler coordinates are just different descriptions of the same physical spacetime, this same physical network of rulers and clocks could also be described in Rindler coordinates, agreed?

Now suppose we have such a physical network, and we want to describe where events in an experiment happen relative to that network in the context of our Minkowski or Rindler coordinate systems. Suppose a particular event E (a collision, say) happens next to the x=12 light-seconds mark on the physical ruler representing the x-axis in this network, and that the physical clock at that mark reads t=8 seconds when E happens next to it. Then if we are using a separate Minkowski coordinate system with coordinates x',t', to describe events in this spacetime, the coordinates of the physical clock at the x=12 l.s. mark on this ruler reading t=8 s may happen at some completely different coordinates in this system, say x'=23 l.s. and t'=100 s. Likewise, in the Rindler coordinate system, the coordinates x'' and t'' of this event will be different as well. However, both the Minkowski coordinate system and the Rindler coordinate system will agree that whatever the coordinates x' and t' (or x'' and t'') of the event of the physical clock at the x=12 l.s. mark reading t=8 s, the same coordinates x' and t' (or x'' and t'') would be assigned to the event E of the collision--the Minkowski coordinate system and the Rindler coordinate system cannot disagree about local facts like whether two events (in this case the event of that clock at that marking reading 8 s and the event E of the collision) coincide at the same point in spacetime or not!

If both the Minkowski coordinate system and the Rindler coordinate system (along with the metric and laws of physics expressed in those coordinate systems) agree in their predictions about what ruler-markings and clock-readings on the physical network coincide with which events in any physical experiment that takes place within the region covered by that physical network, then both coordinate systems should agree in their predictions about what the equations of the laws of physics will be when expressed in the coordinate system defined by the network (not when expressed in terms of their own coordinates). And of course in flat SR spacetime it should always be possible to construct a physical network of rulers and clocks which define a coordinate system where the laws of physics work the same way as in any other inertial coordinate system, right?

This idea is essentially no different than the idea of the equivalence principle, that even if you have a curved spacetime described in some coordinate system like Schwarzschild coordinates, in any local region you should always be able to construct a network of freefalling rulers and clocks such that that the laws of physics as expressed in the coordinates defined by that network (not when expressed in Schwarzschild coordinates in that region) reduce to the same laws seen in an inertial frame in SR (at least to the first order or something). The only difference is that in this case we need not confine the grid of rulers and clocks to a small local region, they can cover any arbitrarily large region of the Rindler wedge and it'll still be true that the laws of physics as expressed in the coordinates defined by the network (not when expressed in Rindler coordinates in that region) will be exactly those seen in any SR inertial frame.
P: 665
 Quote by JesseM Well, because for a long time you were denying my claim that if we looked at a geodesic in Minkowski coordinates, and then used the coordinate transformation to define the same path in Rindler coordinates, then the path would still be a geodesic relative to the Rindler metric. But I guess you were just misunderstanding what I meant when I said things like "same path", not realizing I was talking about using the coordinate transformation to find the new coordinates of the same physical events along a given path which was originally defined in Minkowski coordinates.
Well, as DrGreg said, the problem lies in the fact that some authors describe the 3-vector $dx^i/d\tau$ as proper velocity and all my focus on the way I'm expected to treat the problems here automatically had been going to this point and don't get me wrong this was the way I learned and tought people until recently that I changed my mind and started using the magnitude of the 4-acceleration instead of $$d^2x^i/d\tau^2$$ as the "proper" acceleration! If you noticed in my early posts here, I doubted that the gravitational field of Rindler spacetime is uniform just because the proper 3-accceleration I mentioned above is dependent on $$x$$, thus it doesn't allow the spacetime to be accompanied by a uniform field. However, the challenge this caused in mind isn't still settled and I hope further studies in the future will help me ponder the problems more than I do now!

I'm in a hurry to go somewhere, so I don't have enough time to read the whole post. I'll be back soon.

AB
P: 665
My own view on the EP not only on Rindler wedge but in any spacetime with a vanishing Riemann tensor but non-vanishing Christoffel symbols is briefly stated in Papaetrou's book "Lectures on General Relativity" at page 56:

 When there are gravitational accelerations present, as for example in the gravitational field of the earth, the space cannot be the flat Minkowski space. Indeed, in the Minkowski space we can have $$\Gamma^{\alpha}_{\mu\nu}=0,$$ everywhere. This should then be interpreted as meaning that the sum of the inertial and the gravitational acceleration could be made equal to zero everywhere. This does, however, not correspond to our experience about gravitational accelerations: When gravitational accelerations exist, it is not possible to make them vanish everywhere. We can only make them vanish at one point, or approximately in a small region, by the use of an appropriate coordinate system. Therefore, when a gravitational field is present, the space will be necessarily a curved Riemannian space.
But since the spacetime is flat and yet there we have non-vanishing Christoffel symbols, there must be a coordinate transformation that does make the these symbols vanish and this is what happens to be true on the Rindler wedge when events can finid co-pairs in the Minkowski spacetime everywhere, leading to the fact that on Rindler wedge the EP is valid overally!

AB

 Related Discussions Classical Physics 1 Special & General Relativity 3 Advanced Physics Homework 1 General Physics 4 Classical Physics 2