Why do balls follow curved paths in Earth's gravity according to GR?

[my_wan]

No. The magnitude of the integral of ds is zero, not the length of the path. As I mentioned above the term "length" should not be used when discussing relativity because there is a high probability to confuse "length" with Euclidean intuition.

It is the path that I was referring to as defined by an observer. This was a curt response to MeJennifer.

Again I ask - What do you think it means when it is said that two points in spacetime are close to each other? How do you think neighboorhods in spacetime are defined? What do you think the term "open ball" means in gr when referring to a manifold?

Again? Where did you ever ask me or even MeJennifer any of these question?

EDA: You could answer post #23 instead of responding to my responses to MeJennifer.

 

[Ich]

You are the one who is making the extraordinary claim that in GR there is something like a gravitational field without spacetime curvature.

pmb_phy does not make any extraordinary claims, he ist trying to explain the equivalence principle to you. There is indeed something like a gravitational field, and this something makes particles (and even light) move along parabolas instead of straight lines, and you can conjure up or exorcise this something by merely using a different coordinate system. You don't need curvature of spacetime to do that.

 

[pmb_phy]

You are the one who is making the extraordinary claim that in GR there is something like a gravitational field without spacetime curvature. If there is no curvature then there is no gravitation.

This is by far an extraordinary claim. This is basic general relativity. Failure to understand this basic notion is a failure to understand Einstein's Euivalence Principle and hence general relativity. I've given you plenty of references which you can either look up or I can make them availble for you to read. Whether you do so or not is totally up to you of course.

However you have yet to provide one iota of evidence that what I'm explaining to you is wrong. Why is that?

Pete

 

[pmb_phy]

pmb_phy does not make any extraordinary claims, he ist trying to explain the equivalence principle to you. There is indeed something like a gravitational field, and this something makes particles (and even light) move along parabolas instead of straight lines, and you can conjure up or exorcise this something by merely using a different coordinate system. You don't need curvature of spacetime to do that.

Yes. Exactly. Thanks Ich.

Pete

 

[pmb_phy]

In the presence of mass or energy-momentum spacetime is curved.

That is correct. At points in spacetime (event) where the stress-energy-momentum tensor is non-zero the spacetime is curved. However that does not imply that the spacetime outside the matter is curved. E.g. The stress-energy-momentum tensor of a straight cosmic string has a huge mass density and stress. These act to cancel out the gravitational effects. Outside the string the spacetime is totally flat everywhere. Only the global topology is changed (from that of a plane to that of a cone, which has zero curvature off the tip, where the cosmic string is located).

This is an observation all observers would agree on. To claim that any particular slice of spacetime, e.g. space, is flat is strictly an observer and coordinate dependent property. Space and time are interrelated in GR, to understand them separately is doomed to fail.

That is quite wrong. It is impossible to introduce spatial curvature or spatime curvature merely by changing spacetime coordinates. This is because the Reimann tensor is a geometricl object whose vanishing is absolute rather than coordinate dependent. If the tensor does not vanish in one coordinate system then it cannot vanish in any other coordinate system.

Pete

 

[pmb_phy]

So which one is correct?

Both. They are not mutually exclusive. Why would you believe otherwise? I keep asking that question but am getting no response. Why is that?

I believe the confusion here lies in the notion that "gravity is a curvature in spacetime." This is not true, I don't care who says it. It sure wasn't Einstein whho said it. In fact he opposed such an idea. Claiming that gravity is a curvature in spacetime is redefining gravity to something other than how Einstein defined it. It is tidal forces which are a curvature in spacetime. Tidal force is defined quite differently than the gravitational force. The former is a tensor quantity while the later is a force 3-vector.

Consider a straight line through an accelerating object as defined in the frame of the object and perpendicular to the direction of acceleration. Isn't that line a geodesic in another frame of reference?

A geodesic is a geodesic in all coordinate systems. Whether a worldline is a geodesic or not does not depend on the particular coordinate system. I think you may be confusing with spatial lines with worldlines in spacetime. They are different things.

Tidal force tensor is a funny way to put it, ...

Funny? Why? Its not like I'm making this stuff up as I go along. Consider how this is explained in Black Holes & Time Warps, Kip S. Thorne, page 111

Therefore, spacetime curvature and tidal gravity must be precisely the same thing, expressed in different languages.

but no, there is no tidal forces. If you blew up a beach ball to the size and mass of the Earth where it is hollow everywhere but a thin crust then gravity is zero everywhere inside.

That's because the gravitational field inside a spherically symmetric shell is flat and thus there is zero spacetime curvature. In the reference frame of the ball itself the gravitational field inside also zero.

If there are any tidal forces then in what direction? From the center out? This would mean that gravity would accelerate you to the thin crust. Not so. You don't have tidal forces without difference in gravitational forces.

That is correct. However I was responding to the notion that the gravitational field at the center of the Earth is non-zero, not the center of some fictitious hollow sphere. Those are very different objects.

Pete

 

[pmb_phy]

By the way, one can have a curved spacetime and still have a flat space. A perfect example is found in cosmology. When one says that the universe is flat it means that space itself is flat. But spacetime is still curved everywhere.

Again? Where did you ever ask me or even MeJennifer any of these question?

Oops! Sorry. I guess I wanted to ask that but never did. My appologies. But now the question has been asked.

Pete

 

[pmb_phy]

I forgot to mention that the term length is actually used in relativity when speaking about the integral of ds so please don't get me wrong. I merely wanted to indicate that the notion of zero length can be confusing since people often think in Euclidean terms when hearing of length. For example: the "distance" between two events A and B for two distinct events A and B can be zero. This cannot happen in Euclidean geometry where the distance between A and B being zero means that A = B, i.e. the points are not distinct but are the same. This phenomena is called degeneracy and the metric which defines "length/magnitude/distance" is call indefinite.

Pete

 

[yuiop]

Its quite possible to have a gravitational field in a flat spacetime. In fact Einstein's equivalence principle depends on it.

That does not make any sense to me Pete. Again I await an example where an observer could, in principle, measure such a field. For instance it would be interesting to know what actually gravitates in such a field.

It is a null geodesic simply because its length in spacetime is 0.

I think what Pete is getting at the artificial gravity that would be measured in an accelerating rocket. That would be an example of a "gravitational field" in flat spacetime. It can be handled by Rindler Spacetime which is the metric for flat spacetime (I think). I think it is a matter of definitions and terminology. It would be nice if we could clear the matter up so that we are using the same definitions and terminology.

 

[my_wan]

When I first responded to you I wasn't sure how you meant what you said and noted that. The ensuing debate only grew more convoluted so I took the time to consider exactly what the actual issues were.

Both. They are not mutually exclusive. Why would you believe otherwise? I keep asking that question but am getting no response. Why is that?

Again, this is the first I seen this question. I now get where you make the distinction though. It seems our difficulty here is that you take gravity and GR (a gravitational field) as synonymous where I do don't. I as a matter of habit I separate the theoretical description from the thing being defined. When I speak of gravity I am speaking of the acceleration g alone, not its metric description in GR. The acceleration g defined wrt the mass is due to curvature in GR. This in no way implies that the metric of space-time is not influenced by a gravitational field in flat space-time. Flatness is the property of uniformity of the metric.

I believe the confusion here lies in the notion that "gravity is a curvature in spacetime." This is not true, I don't care who says it. It sure wasn't Einstein whho said it. In fact he opposed such an idea. Claiming that gravity is a curvature in spacetime is redefining gravity to something other than how Einstein defined it. It is tidal forces which are a curvature in spacetime. Tidal force is defined quite differently than the gravitational force. The former is a tensor quantity while the later is a force 3-vector.

Yes, you are essentially right about the source of confusion. However, your definition of gravity assumes it refers to it's full description under GR.

The terms gravitation and gravity are mostly interchangeable in everyday use, but in scientific usage a distinction may be made. "Gravitation" is a general term describing the attractive influence that all objects with mass exert on each other, while "gravity" specifically refers to a force that is supposed in some theories (such as Newton's) to be the cause of this attraction.

A geodesic is a geodesic in all coordinate systems. Whether a worldline is a geodesic or not does not depend on the particular coordinate system. I think you may be confusing with spatial lines with worldlines in spacetime. They are different things.

Here I must still disagree. I am not confusing spatial lines with world lines but in some situations a spatial lines can illustrate the world line of a particular observer, i.e., events. Consider, observer A stretches a long thin wire W in space such that it represents a straight line in the word line of A. Observer B passes the a midpoint perpendicular to W with a constant velocity wrt W. The wire is cone shaped in the world line of B. Observer C accelerates through the same path as B. In the world line of B the wire is a geodesic, as per the principle of equivalence.

World lines represent how events are ordered wrt an observer. This ordering defines the geometry of the distribution of things in space as defined by that observer. It would be just as easy to use events, such as flashes of light at various places, to represent this wire stretched in space.

Funny? Why? Its not like I'm making this stuff up as I go along. Consider how this is explained in Black Holes & Time Warps, Kip S. Thorne, page 111

I wasn't implying it was wrong in any way, just never heard it stated that way. LOL

That's because the gravitational field inside a spherically symmetric shell is flat and thus there is zero spacetime curvature. In the reference frame of the ball itself the gravitational field inside also zero.

Here we seem to agree that zero space-time curvature means no acceleration, as per our incongruent definitions of gravity. Only you use the term "gravitational field" is zero whereas I used "gravity" and agree here that zero space-time curvature equals a zero gravitational field. So I can't say you are wrong yet you chose to use my definition of gravity here. So you must have taken issue with what I said purely on the presumption that I mean gravity in the wrong way rather than anything explicitly wrong with my actual statement.

It still leaves the original statement that started this debate to be answered:

This is true if one is approximating the gravitational field near the Earth's surface as a uniform gravitational field. Such a field has zero spacetime curvature. Space is also flat in such a field.

That is correct. However I was responding to the notion that the gravitational field at the center of the Earth is non-zero, not the center of some fictitious hollow sphere. Those are very different objects.

First off hollow spheres are not fictitious objects, that's a red herring. Now you are trying to make a distinction between a hollow sphere and the center of the Earth as if I didn't strictly specify both time I said it, or that it makes any difference whatsoever.

Illustration:
If the Earth was hollow like a large beach ball then anywhere inside this hollow the gravity would be zero, i.e., the space-time is flat, yet the gravitational time dilation would be the same as on the surface with gravity.

If you blew up a beach ball to the size and mass of the Earth where it is hollow everywhere but a thin crust then gravity is zero everywhere inside.

Yet your claim of fictitious doesn't even work because even if the Earth has nothing more than a hollow spot just big enough for you to fit there is still no gravitational forces, tidal or otherwise. In fact the only forces on the mass at the center right now is the pressure from the matter not at the center, no gravitational forces of any kind.


Your original statement that MeJennifer took exception to remains to be explained also, the only argument she made that I concur with:

The trajectory of a particle in the Earth's gravitational field consists of two parts (1) gravitational acceleration and (2) spatial curvature. The space-time associated with the Earth's gravitational field is curved.

You can have gravitational time dilation without curvature but not acceleration. Spatial curvature and acceleration are facets of the same thing.

 

[Antenna Guy]

Again, this is the first I seen this question.

pardon the expresson, but...

It seems our difficulty here is that you take gravity and GR (a gravitational field) as synonymous where I do don't.

If the options are SR and GR, I'm pretty sure that gravity falls under GR.

Regards,

Bill

 

[my_wan]

pardon the expresson, but...

So where was I asked that question before?

If the options are SR and GR, I'm pretty sure that gravity falls under GR.

Regards,

Bill

Gravity is the phenomena, GR is the theory. There is a big difference and as I pointed out in the post even pmb_phy used that distinction in his own response when he said;

That's because the gravitational field inside a spherically symmetric shell is flat and thus there is zero spacetime curvature. In the reference frame of the ball itself the gravitational field inside also zero.

Yet he's been arguing with me for using that same definition, essentially the exact same statement, and claiming I'm somehow conflating gravity and curvature. So the tongue in cheek means little to me.

It's also essentially flat at the center of the Earth as the Earth really is right now.

 

[Dale]

Hi wan and Jennifer, Pete is exactly correct here.

Spacetime curvature is an intrinsic physical characteristic of the spacetime that cannot be transformed away. For example, in a curved spacetime "straight" (i.e. geodesic) worldlines that are initially parallel will not remain parallel. This is an observable fact that can be determined in any reference frame. For example, consider a rock dropped above the north pole simultaneously with a rock dropped above the south pole. They are on geodesics that begin parallel and wind up intersecting. This is a frame-independent observation due to spacetime curvature.

A uniform gravitational field has no such frame-independent curvature. In such a field particles follow parabolic paths. If you drop two rocks in a uniform gravitational field their worldlines remain parallel at each point in time and never intersect. The parabolic path itself can be transformed away by choosing a suitable reference frame. In such a frame the worldlines are seen to be straight.

Remember, one of the key points in doing GR is to take a small volume of space where spacetime can be considered locally flat. Such a volume doesn't get rid of gravity, it gets rid of curvature. In other words, you can make a volume small enough to neglect spacetime curvature on the surface of the earth, but you can never make the volume small enough that an accelerometer will read 0 rather than g.

If you are making the approximation that the field is uniform then you are making the approximation that objects follow parabolic trajectories and you are making the approximation that the spacetime is flat.

 

[Jorrie]

You can have gravitational time dilation without curvature but not acceleration. Spatial curvature and acceleration are facets of the same thing.

How would you then explain the relative acceleration of a particle dropped inside a uniformly accelerating lab? There's surely no curvature inside such a lab?

-J

 

[yuiop]

Hi all,

The attached diagram represents a lab undergoing proper born rigid acceleration in a straight line relative to an inertial observer, drawn on a Minkowski diagram.

Imagine the lab to be multi storey, with observer on each floor with their own clocks. In the diagram two observers on different floors send timing light signals (the blue lines) to each other (Events A and B). They receive these signals simultaneously at events C and D because both these events lie on a line of simultaneity in the lab (The magenta lines). On receiving these signals they echo them back and the signals arrive back at there respective observers simultaneously at events E and F.

Within the accelerating lab, observers spatially separated (horizontally and vertically) observe that once their clocks are synchronised, they remain synchronised so they consider their clocks to run at the same rate. This can not be said for clocks at the top and bottom of a tower on Earth for example. This is because spacetime on Earth is curved while spacetime measured by observers in the accelerating lab is flat. The spacetime within the lab is flat because all spatially separated observers consider their spatial separation to remain constant and their clock rates to remain constant with respect to each other. Despite the apparent flatness of the spacetime they occupy, the observers in the lab observe light moving horizontally to follow a curved path, and any dropped objects to accelerate downwards. Hence you have "gravitation" in flat spacetime.

In an earlier discussion with Jennifer we mentioned that a cylindrical hypersuface can be considered flat in GR while a spherical hypersurface is curved. I made the analogy that the surface of a cylinder is curved in one spatial direction (single curvature) while the surface of a sphere is curved in two spatial directions. (Double curvature). Likewise the accelerating lab can simulate the single curvature of the cylindrical surface but it cannot simulate the double curvature of the spherical surface. Special Relativity, Minkowski diagrams and Rindler spacetime can handle the flat spacetime (single curvature) of an accelerating lab while general Relativity is required to describe motion in the spherical gravitational topology (double curvature) of a typical spherical massive body.

In the Minkowski diagram it is easy to observe that the proper acceleration measured by different observers on different floors of he accelerating lab is proportional to 1/R while the proper acceleration measured by observers in a tower on the Earth varies as 1/R^2.

This is another difference between the flat spacetime of the accelerating lab and the curved spacetime of a spherical massive body.

Free falling observers in the accelerating lab observe light rays in any direction to follow straight paths (straight null geodesics), while the non inertial observers at rest with the lab frame observe light rays to follow curved paths (curved null geodesics).

On the other hand, observers in a lab free falling towards the Earth will not observe light rays to follow exactly straight paths unless the lab is very small and then the paths will approximate straight lines. These free falling observers will not consider their spatial separation to remain constant and they will not consider their clocks to run at the same rate. This is because the curved spacetime with the lab free falling towards the Earth is not the same as the flat spacetime observed by observers freefalling within a lab that is being accelerated by a rocket.

Nearly every reference to a "null geodesic" in a google search suggest a null geodesic is a path and not a point.

Having, thought it through (and rejecting the conclusions I made in post #4 which were only true for an infinitesimal region of space) I tend to think the views and definitions expressed by Pete and Dalespam are the correct ones in this thread.

Admittedly the term "flat spacetime" is a confusing definition for the spacetime found in an accelerating lab, but that seems to be the definition most widely used in texts on the subject. I think that is to differentiate it from the curved spacetime measured in the proximity of a large gravitational body that can not be duplicated by an linearly accelerating rocket except in an infinitesimal region of the gravitational body.

Also as alluded to earlier, it seems that flat spacetime has the property ( possibly a definition?) that all observers at rest with that spacetime and maintaining constant spatial separation measure their clock rates to be the same irrespective of the fact that they may feel or measure different proper accelerations to maintain that constant spatial separation.

 

[Jorrie]

Imagine the lab to be multi storey, with observer on each floor with their own clocks. In the diagram two observers on different floors send timing light signals (the blue lines) to each other (Events A and B). They receive these signals simultaneously at events C and D because both these events lie on a line of simultaneity in the lab (The magenta lines). On receiving these signals they echo them back and the signals arrive back at there respective observers simultaneously at events E and F.
...

Also as alluded to earlier, it seems that flat spacetime has the property ( possibly a definition?) that all observers at rest with that spacetime and maintaining constant spatial separation measure their clock rates to be the same irrespective of the fact that they may feel or measure different proper accelerations to maintain that constant spatial separation.

Hi Kev. I think you have these two parts slightly wrong. The Born-accelerated observers on your 'floors' do not agree that their respective clock rates are the same, because their clocks get more and more out of sync during the acceleration. Think what will happen in your Rindler diagram when the acceleration is stopped - don't you think the clocks will then need to be resynchronized?

Your purple lines of "instantaneous simultaneity" is when the clocks on all floors read the same time, but they realize that the clocks get out of sync. Light signals send forward and backward during the acceleration are red-shifted and blue-shifted respectively, just like in a gravitational field. It's only the tidal effects that differ, AFAIK.

-J

 

[pmb_phy]

I now get where you make the distinction though. It seems our difficulty here is that you take gravity and GR (a gravitational field) as synonymous where I do don't.

That in no way is true. Gravity is the thing that GR describes. I do not use these terms synonymously.

When I speak of gravity I am speaking of the acceleration g alone, not its metric description in GR.

The gravitational acceleration is defined by the metric. Unless the metric is first defined then the affine connections can't be calculated and thus the gravitational acceleration cannot be defined. On the other hand one can state the affine connections but then the metric can be found from those connection coefficients.

The acceleration g defined wrt the mass is due to curvature in GR.

Wrong. That is not true in general. Why do you believe it is?

Yes, you are essentially right about the source of confusion. However, your definition of gravity assumes it refers to it's full description under GR.

My definition? So your saying that although this was defined by Einstein I am to get the credit for it? Gee! Thanks!

Here we seem to agree that zero space-time curvature means no acceleration, as per our incongruent definitions of gravity. Only you use the term "gravitational field" is zero whereas I used "gravity" and agree here that zero space-time curvature equals a zero gravitational field.

You can us the term "gravity" anyway you'd like. I thoiught this was a relativity newsgroup and as such we're talking aboiut how Einstein used these terms.

So I can't say you are wrong yet you chose to use my definition of gravity here. So you must have taken issue with what I said purely on the presumption that I mean gravity in the wrong way rather than anything explicitly wrong with my actual statement.

If we're not talking about Einstein's theory of relativity then I have just lost interest in this discussion. Are we or are we not talking about Einstein's theory of relativity and as such talking about how Einstein defined these terms? Even MTW define the gravitational field as requiring the non-vanishing on the affine connection, which does not require spacetime to be curved.

First off hollow spheres are not fictitious objects, that's a red herring.

Um ... I never said they were!

Now you are trying to make a distinction between a hollow sphere and the center of the Earth as if I didn't strictly specify both time I said it, or that it makes any difference whatsoever.

I said that the actual Earth is not hollow and as such the curvature at the center of the Earth is not zero. However if you hollowed out a cavity centered at the center of the Earth then the gravitational field will vanish throughout that cavity leaving no gravitational field and as such no tidal gradients, hence the spacetime is zero in such a cavity. However, if the center of th hollowed out cavity is not centered at the center of the Earth then the spacetime will still be flat but, in a frame of reference which is at rest with respect to the Earth, the gravitational field will be non-vanishing, in fact it will be a uniform gravitational field. I can show you how to calculate that in the weak field limit if you'd like? Its rather simple in fact.

Yet your claim of fictitious doesn't even work because even if the Earth has nothing more than a hollow spot just big enough for you to fit there is still no gravitational forces, tidal or otherwise.

Excuse me? Who said that I had to fit there? There is a gravitational field inside a matter distribution regardles of whether we measure it or not. Its calculable and therefore exists. Seems to me that you're still confusing gravitational acceleration with spacetime curvature. Why do you keep doing that?

In fact the only forces on the mass at the center right now is the pressure from the matter not at the center, no gravitational forces of any kind.

(sigh) I didn't say that the gravitational field at the center of is non-zero. It is zero. But the fact that its zero does not mean that the curvature is zero. The gravitational field inside the Earth, which is not hollow, is not uniform and therefore the tidal tensor is non-zero. The tidal force tensor can be non-zero and yet leave a zero gravitational field.

You can have gravitational time dilation without curvature but not acceleration.

Correct. I never said otherwise.

Spatial curvature and acceleration are facets of the same thing.

I see that I have no reason to continue responding to your comments. I will therefore be unable to respond unless you explain why you keep saying this. So when you explain or prove that Spatial curvature and acceleration are facets of the same thing. then I see I've stated all the facts and you either accept them, reject them, or have me prove them to you, or let me show you the GR material that explains all of this. Which will it be?

Pete

 

[Dale]

However, if the center of th hollowed out cavity is not centered at the center of the Earth then the spacetime will still be flat but, in a frame of reference which is at rest with respect to the Earth, the gravitational field will be non-vanishing, in fact it will be a uniform gravitational field. I can show you how to calculate that in the weak field limit if you'd like? Its rather simple in fact.

If it is easy enough for you to do I would appreciate that. I assume that the weak field limit is some sort of approximation to GR that is appropriate for "typical" spacetimes instead of black holes.

 

[pmb_phy]

Hi wan and Jennifer, Pete is exactly correct here.

I love to hear those words. :shy:

Spacetime curvature is an intrinsic physical characteristic of the spacetime that cannot be transformed away. For example, in a curved spacetime "straight" (i.e. geodesic) worldlines that are initially parallel will not remain parallel. This is an observable fact that can be determined in any reference frame.

Yes! Quite right!

For example, consider a rock dropped above the north pole simultaneously with a rock dropped above the south pole. They are on geodesics that begin parallel and wind up intersecting.

The geodesics do not start out parallel in this case. In fact its not even very meaningful to compare geodesics which have a finite separation because the parallelness will depend on how a tangent vector is parallel transported in order for them to be compared. A better example is to compare two objects which are very close to each other when they are let go and left to free-fall. The geodesics will start off parallell and then they will diverge.

A uniform gravitational field has no such frame-independent curvature. In such a field particles follow parabolic paths. If you drop two rocks in a uniform gravitational field their worldlines remain parallel at each point in time and never intersect. The parabolic path itself can be transformed away by choosing a suitable reference frame. In such a frame the worldlines are seen to be straight.

I recommend caution be used in such visualizations. Consider the fact that two particles at rest in an inertial frame of reference S will remain at rest in that frame. Consider now a frame S' which has a uniform acceleration with respect to S and directed along a line which is parallel to the separation 3-vector of the two particles. In Newtonian mechanics the distance is invariant, all clocks tick at the same rate and thus the separation remains the same. However in relativity the distance changes with speed and clocks run at different rates. This means the distance between the two particles contracts. This is a result of clocks running at different rates in thefield and thus acclerations of the two particles is different. The particle higher up in the field is larger than that of the particle lower in the field (since clocks higher up run slower when measured locally). However spacetime deviation is an invariant quantity since it is calculated in terms of invariant properties of spacetime.

Remember, one of the key points in doing GR is to take a small volume of space where spacetime can be considered locally flat.

All that means is that the grqavitational field can be transformed away and one can chose the "size" of the volume to be so small that your instruments cannot measure the tidal accelerations. Note that this is more of a limitation of the intruments than the impossibility of detecting the spacetime curvature. Remeber that curvature is a local property, not a global one. Curvature is calculated by using local points in a manifold and taking the limit as the volume approches zero.


Pete

 

[Dale]

The geodesics do not start out parallel in this case.

I thought that parallel was synonymous with "same 4-velocity". So since they both initially start out at rest their geodesics would start out parallel. Am I missing some subtle point here?

 

[yuiop]

Hi Kev. I think you have these two parts slightly wrong. The Born-accelerated observers on your 'floors' do not agree that their respective clock rates are the same, because their clocks get more and more out of sync during the acceleration. Think what will happen in your Rindler diagram when the acceleration is stopped - don't you think the clocks will then need to be resynchronized?

I analysed this in some detail in this old thread (with yet more diagrams) https://www.physicsforums.com/showthread.php?t=216113 and concluded that two clocks would be out of sync after born rigid acceleration when they reach same final constant velocity.

In this thread I am considering the slightly different situation where the two clocks maintain constant born rigid acceleration indefinitely. I would be interested in your opinion of my conclusion in that old thread ;)

Your purple lines of "instantaneous simultaneity" is when the clocks on all floors read the same time, but they realize that the clocks get out of sync. Light signals send forward and backward during the acceleration are red-shifted and blue-shifted respectively, just like in a gravitational field. It's only the tidal effects that differ, AFAIK.

-J

The light signals sent forward and backward should behave as any other light signal in a time space diagram moving at 45 degrees as I have shown on this diagram (the blue lines). If the signals are sent at 0 seconds by the upper and and lower observers and both arrive at 1 seconds as shown by their respective clock and then return at 2 seconds as shown by their respective clocks then presumably the upper and lower clocks must remain in sync (while constant proper born rigid acceleration is maintained) as far as I can tell.

 

[pmb_phy]

I thought that parallel was synonymous with "same 4-velocity". So since they both initially start out at rest their geodesics would start out parallel. Am I missing some subtle point here?

To be called "the same 4-velocity" one must be able to compare the 4-velocities when the particles are located at different points in spacetime. However since there is no way to compare 4-vectors which are located at different points in spacetime the 4-velocities cannot be said to be "the same." This is easily vizualied by using the two-dimensional surface of a sphere such as the Earth. Start with a tangent vector located at the equator which points eastward. First parallel transport this vector to the North Pole by moving along the a great circle which passes through the poles and through the initial position of the tangent vector. Then picture what would happen if you first parallel transported one quarter the way around the sphere and then parallel transport along the great circle which passes through the poles as well as through this new location. You will see that the end results result in tangent vectors which have a 45 degree angle between them.

Pete

 

[Dale]

Yes, that makes sense. I guess to talk about parallel world lines converging or diverging you have to start with two lines close enough together that the spacetime between them at the beginning can be considered flat in order to avoid such problems. Then you have to move them far enough through the curved spacetime that you can see them converge or diverge since it won't be apparent right away.

 

[Jorrie]

I analysed this in some detail in this old thread (with yet more diagrams) https://www.physicsforums.com/showthread.php?t=216113 and concluded that two clocks would be out of sync after born rigid acceleration when they reach same final constant velocity.

As far as I checked, I agree with your conclusions in that thread.

The light signals sent forward and backward should behave as any other light signal in a time space diagram moving at 45 degrees as I have shown on this diagram (the blue lines)

Yep, that's in the inertial reference frame. In an accelerating frame, light does not move isotropically like that, as I'm sure you know.

If the signals are sent at 0 seconds by the upper and and lower observers and both arrive at 1 seconds as shown by their respective clock and then return at 2 seconds as shown by their respective clocks then presumably the upper and lower clocks must remain in sync (while constant proper born rigid acceleration is maintained) as far as I can tell.

I don't think so. Your 'old thread' has shown that when the acceleration stops and both ends of the lab are moving at the same velocity, their clocks will be out of sync. Since the two ends are not experiencing the same proper acceleration, they will reach that 'same velocity' at different times, not only in the reference frame, but also in their own frames. To me that makes their clocks never synchronized during the acceleration (after the start, that is). But I may be wrong...

-J

 

[pmb_phy]

Flat Tilted Plane and a Uniform Gravitaional Field

my_wan = If I've got this right then I imagine that you're visualizing gravity using an embedding diagram. The one for the Earth is like a flat space with a bowling ball placed on it and then depresses into the surface and thus "curving" it. I would imagine that you're thinking of the diagram in terms of a little ball rolling on the surface. Where the surface is level there is no acceleration and where it dips down due to the bowling ball then it accelerates "down" the surface. The meaning of an embedding diagram is to represent the increases in distances as a function of the decrease in the radial coordinate r. If one were to use this vizulization for a uniform gravitational field then you should imagine a tilted plane. In the rest frame of the observer in free-fall all he measures is a flat level plane. Notice that any observer in free-fall can transform the gravitational field away by moving to a free-fall frame. This is to be vizualized by picturing the location a little ball is from the view where the surface normal is directed upward. In a small enough region the the surface looks flat and since the surface is not titled then there is no gravitational field present at that location from that observers point of view.

 

[MeJennifer]

If one were to use this vizulization for a uniform gravitational field then you should imagine a tilted plane. In the rest frame of the observer in free-fall all he measures is a flat level plane.

How could an observer possibly measure a flat plane? If the Riemann curvature tensor vanishes it is flat, and there is nothing to measure. So are you saying there is some kind of force still active that causes gravitation?

Again Peter, what you say about gravitation in flat spacetime does not make any sense to me, I gladly attribute it to my lack of understanding of the matter.

 

[yuiop]

Is the spacetime background experienced by an observer in a rocket experiencing constant proper acceleration flat or curved?

 

[MeJennifer]

Is the spacetime background experienced by an observer in a rocket experiencing constant proper acceleration flat or curved?

There is a difference between something being equivalent and identical.

An accelerating rocket is not identical to gravitation.

 

[yuiop]

...
Yep, that's in the inertial reference frame. In an accelerating frame, light does not move isotropically like that, as I'm sure you know.

I agree, the diagram in post #45 is the point of view of an inertial observer in flat spacetime and to him light follows straight lines in 3 space and in 4D spacetime, as shown in that diagram. Sure, in the reference frame of the observers in the accelerating lab light follows a curved path in 3 space. However, the events are the same from any observers point of view. If A sends a signals at one second intervals as measured by A's clock and they arrive at B at one second intervals as measured by B's clocks then those two clocks are syncronised in the reference frame of A and B. Since the readings are the proper times of those observers then any observer would have to agree that is what they measure.

...

I don't think so. Your 'old thread' has shown that when the acceleration stops and both ends of the lab are moving at the same velocity, their clocks will be out of sync. Since the two ends are not experiencing the same proper acceleration, they will reach that 'same velocity' at different times, not only in the reference frame, but also in their own frames. To me that makes their clocks never synchronized during the acceleration (after the start, that is). But I may be wrong...

-J

I disagree. To the inertial observer in the diagram the observers A and B have different velocities at any given moment. In the reference frame of A and B they consider themselves to have the same velocity at any given instant (which is why they consider their spatial separation to be constant in their reference frame. The line of simultaenity is not only the the line that connects events that simultaneous in the reference frame of the accelerating observers, it is also the line that connects equal velocity as measured by the inertial observer (although of course he does not consider A and B to have equal velocity simultaneously). In the reference frame of the accelerating observers they can consider their velocity to be zero while experiencing different proper accelerations. This is analogous to two observers that are stationary with respect to each but at different altitudes in a tower on the Earth for example.

A subtle difference is that the observers in the Earth tower measure their clock rates to be different while the observers in the accelerating lab do not. This is a tidal effect of the curved spacetime of the Earth that is not present in the lab experiencing artificial constant proper acceleration.

 

[yuiop]

There is a difference between something being equivalent and identical.

An accelerating rocket is not identical to gravitation.

You have avoided the question. Is the spacetime in an accelerating rocket flat or curved?

I did not ask what it is equivalent to.