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

[aachenmann]

The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.

So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks

 

[Ich]

Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

Neither. It's the effect of an accelerated reference frame.
The deviations from a parabola that occur on larger scales are caused mainly by the time component of curvature, so it's curved spacetime but almost flat space, when "space" is defined by observers nearly at rest relative to earth.

 

[A.T.]

What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

Spacetime curvature. Here a nice explanation of space vs. spacetime curvature:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html
More links to curved spacetime visualizations:
https://www.physicsforums.com/showpost.php?p=1557122

 

[yuiop]

Attached is a sketch (to arbitary accuracy) of how I imagine geodesics can interpreted as straight lines. The sketch on the left is the viewpoint of an observer that is stationary with respect to the massive body and the one on the right is the viewpoint of an observer in the free falling elevator. Does anyone think this is a reasonable interpretation? Have I gone too far with the non locality the diagram suggests?

 

[pmb_phy]

The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?).

Yes. That is correct.

Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.

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.

So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

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

Pete

 

[MeJennifer]

The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.

So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks

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

 

[MeJennifer]

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

Hi Pete, unless I misunderstand what you write I do not agree with you on this one. How do you reason the two parts?

 

[my_wan]

The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.

The only geodesic line in space-time that can be defined as a straight line is one defined by the trajectory of a photon, not mass particles.

 

[MeJennifer]

The only geodesic line in space-time that can be defined as a straight line is one defined by the trajectory of a photon, not mass particles.

I am not sure how you conclude this to be true. In curved spacetime there are no straight lines, and the spacetime distance between the emission and absorption of a photon is 0, e.g. a null geodesic.

 

[my_wan]

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.

I'm going to take issue with this statement. Perhaps I am misunderstanding but seems related to the statement MeJennifer rightfully took exception to.

If space-time has no curvature, i.e., flat, then there is no gravity in that space. Just like going to the center of the Earth there is no gravity. You seem to imply here that a uniform gravitational field simply has the same acceleration at different distances but the acceleration remains. Not so.

This seems to be where the notion of separating the acceleration of gravity into the two components that MeJennifer rejected.

1) Without curvature there is no gravity no matter how much mass is around.
2) Ditto for flat space-time.
3) You can still have gravitational time dilation even without an acceleration of gravity.

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.

 

[my_wan]

I am not sure how you conclude this to be true. In curved spacetime there are no straight lines, and the spacetime distance between the emission and absorption of a photon is 0, e.g. a null geodesic.

I didn't say it was a straight line, I said it could be defined as one. A null geodesic is a generalized definition of a straight line in curved space-time in GR.

 

[pmb_phy]

Hi Pete, unless I misunderstand what you write I do not agree with you on this one. How do you reason the two parts?

Consider the deflection of a beam of light in a uniform gravitational field. Such a field has zero spacetime curvatue by definition. Therefore the defelction is due entirely by gravitational acceleration. Now consider Einstein's first derivation of the gravitational deflection of light by the sun. In his first derivation Einstein effectively assumed that space was flat (i.e. globally Euclidean). The amount of deflection was off by a factor of two. Once Einstein took account of the spatial curvature he arrived the correct amount of deflection.

What is it that you object to in my explanation?

Pete

 

[pmb_phy]

The only geodesic line in space-time that can be defined as a straight line is one defined by the trajectory of a photon, not mass particles.

That is incorrect. Timelike geodesics exist for particles with a finite proper mass. E.g. a particle of proper mass m₀ move on timelike geodesics. Luxons, such as a photon, move on null geodesics. Particle's are said to move on the straightest possible lines in a curved spacetime, not a straight line since such a term is undefined for worldlines on curved spacetime manifold.

Pete

 

[pmb_phy]

Sorry my_wan it is not, a null geodesic is not even a line in spacetime, actually it is a point. You only get a "line" out of it by parameterization. Typically the affine parameter is normalized with respect to the momentum four-vector for a null geodesic.

A null geodesic is not a point, it is a worldine in spacetime. This means its a collection of events which are all distinct from each other.

Pete

 

[MeJennifer]

A null geodesic is a generalized definition of a straight line in curved space-time in GR.

Sorry my_wan it is not, a null geodesic is not even a line in spacetime, actually it is a point. You only get a "line" out of it by parameterization. Typically the affine parameter is normalized with respect to the momentum four-vector for a null geodesic.

A null geodesic is not a point, it is a worldine in spacetime. This means its a collection of events which are all distinct from each other.

If you calculate the length of a null geodesic in spacetime you will see it is zero.

Consider the deflection of a beam of light in a uniform gravitational field. Such a field has zero spacetime curvatue by definition. Therefore the deflection is due entirely by gravitational acceleration.

In a uniform gravitational field nothing gets deflected. Perhaps you could provide an example to demonstrate what you mean?

 

[pmb_phy]

If space-time has no curvature, i.e., flat, then there is no gravity in that space.

That is incorrect. That kind of interpretation is based on a misunderstanding of what a gravitational field is. Its quite possible to have a gravitational field in a flat spacetime. In fact Einstein's equivalence principle depends on it. This principle states (weak form)

A uniform gravitational field is equivalent to a uniformly accelerating frame of reference in flat spacetime.

Thus the spacetime of a uniform gravitational field is flat by its very definition. One cannot introduce spacetime curvature by a mere change in coordinates since spacetime curvature is an intrinsic property of a metric space.

Just like going to the center of the Earth there is no gravity.

While the gravitational field is zero at the center of Earth the tidal force tensor is non-zero there. Therefore the spacetime curvature there is non-zero.

You seem to imply here that a uniform gravitational field simply has the same acceleration at different distances but the acceleration remains. Not so.

I didn't say that. I said that a uniform gravitational field has a flat spacetime by its very definition. This is a well-known fact in general relativity.

1) Without curvature there is no gravity no matter how much mass is around.

That is incorrect. A gravitational field can be produced merely by changing spacetime coordinates.

2) Ditto for flat space-time.

Also incorrect.

3) You can still have gravitational time dilation even without an acceleration of gravity.

That too is incorrect. The presence of gravitational time dilation requires that the potential energy be a function of position and as such only the presence of gravitational acceleration is required, not spacetime curvature. Take a look at Einstein's original derivations on gravitational time dilations. He calculated them for flat spacetimes. The Pound-Rebka experiments were compared to the uniform gravitational field approximation and thus the detection of gravtiational time dilation does not mean that the spacetime is curved. A good example is the spacetime associated with a uniformly accelerating frame of reference. E.g. consider a rocket accelerating in a flat spacetime. A photon of frequency f as measured locally and directed towards the tail of the ship will be detected at the tail at which a locally measurement there will yield a different wavelength for the photon, hence gravitational time dilation is detected.

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.

That has no bearing on what I've stated above.

Pete

 

[my_wan]

Sorry my_wan it is not, a null geodesic is not even a line in spacetime, actually it is a point. You only get a "line" out of it by parameterization. Typically the affine parameter is normalized with respect to the momentum four-vector for a null geodesic.

http://en.wikipedia.org/wiki/Null_geodesic

Yes it is parameterized by how a given observer defines the trajectory. It is a null geodesics because the tangent vector norms to zero. A null geodesic is conformally flat, i.e., is a diffeomorphism.

 

[MeJennifer]

Its quite possible to have a gravitational field in a flat spacetime.

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 geodesics because the tangent vector norms to zero. A null geodesic is conformally flat, i.e., is a diffeomorphism.

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

 

[pmb_phy]

If you calculate the length of a null geodesic in spacetime you will see it is zero.

You're misusing the concept of "length" in such an interpretation. The metric for spacetime is non-definite which means that distinct position vectors can give a zero spacetime interval. This means that the spacetime diplacement between two distinct points on a null geodesic have a finite displacement and as such are distinct points, not the same points. As robphy said in another thread, you're using Euclidean intuition to view the non-Euclidean Minkowski spacetime.

In a uniform gravitational field nothing gets deflected.

That is quite incorrect. I fail to see how you came to such a conclusion?

Perhaps you could provide an example to demonstrate what you mean?

The metric for a uniform gravtiational field is

ds^2 = (1+gz/c^2)^2(cdt)^2 - dx^2 - dy^2 - dz^2

This metric is then used to calculat the Christofel symbols which are not all zero. They result in values which yield the delfection of particles, including the deflection of a beam of light.

What I've been explaining is nothing new by the way. It can find many references in the physics literature. Notable examples are

Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173/
Radiation from a Uniformly Accelerated Charge, David G. Boulware, Ann. Phys., 124, (1980), page174.
Relativistic solutions to the falling body in a uniform gravitational field, Carl G. Adler, Robert W. Brehme, Am. J. Phys. 59 (3), March 1991.
Gravitation, Charles. W. Misner, Kip S. Thorne, John Archibald Wheeler, (1973), sect 6.6.
The uniformly accelerated reference frame, J. Dwayne Hamilton, Am. J. Phys., 46(1), Jan. 1978.

If you'd like I can send you the derivations for the gravitational force which results from a uniform gravitational field if you'd like? I can place them in a PM.

Pete

 

[pmb_phy]

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.

Why do say that it makes no sense? Let me guess; you've heard the phrase "Einstein showed that gravity is a curvature in spacetime" right? If so then this assumption is based on false information. Einstein never made such an assertion. In fact when Max Von Laue wrote a general relativity book he interpreted the presence of a gravitational in terms of spacetime curvature because spacetime curvature cannot be transformed away and thus one can always distinguish such a gravitational field from a uniformly accelerating frame of reference. This is how the idea originted, by Max Von Laue. Von Laue sent his book to Einstein for his opinion. Einstein read that part about curved spacetime an objected to it. Einstein told Von Laue that the presence of a gravitational field should be associated, not with the non-vanishing of the Riemann tensor, by from the non-vanishing components of the affine connection (See MTW page 467 on this point).

Are you associating a curved trajectory with spacetime curvature? If so then that is a very big mistake. They have totally different meanings.

Pete

 

[MeJennifer]

You're misusing the concept of "length" in such an interpretation.

It is not an interpretation. Whatever you want to call ds^2 it is zero for a null geodesic. Some references:

Hawking, Ellis - Large Scale Structure of Spacetime. Page 86 - 4.2 Null Curves
Carroll - Spacetime and Geometry. Page 110 - 3.4 Properties of Geodesics

The metric for a uniform gravtiational field is

ds^2 = (1+gz/c^2)^2(cdt)^2 - dx^2 - dy^2 - dz^2

This metric is then used to calculat the Christofel symbols which are not all zero. They result in values which yield the delfection of particles, including the deflection of a beam of light.

Ok, so suppose we have a photon that is emitted and absorped between A and B in this field. We also have an observer, who is also in in this field. How exactly does he observe a deflection? How?

Only observers outside your field would observe a "deflection", but this "deflection" is not due to to the field but due to the fact that the observer is not in the field.

 

[Antenna Guy]

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.

Wouldn't a world line passing between (such that the distance to either is equal at any time) two equally massive bodies constitute a "flat space"?

Regards,

Bill

 

[my_wan]

One cannot introduce spacetime curvature by a mere change in coordinates since spacetime curvature is an intrinsic property of a metric space.

That is incorrect. A gravitational field can be produced merely by changing spacetime coordinates.

So which one is correct?

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?

While the gravitational field is zero at the center of Earth the tidal force tensor is non-zero there.

Tidal force tensor is a funny way to put it, 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. 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 has no bearing on what I've stated above.

Why not. Did the gravity just disappear in the sphere? Did the stress energy just disappear? How do you explain it?

 

[my_wan]

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

So the length that light travels is zero? Not in my frame of reference.

 

[MeJennifer]

So the length that light travels is zero? Not in my frame of reference.

The length of a null geodesic in spacetime is zero.

 

[pmb_phy]

Ok, so suppose we have a photon that is emitted and absorped between A and B in this field. We have an observer, who is also in in this field observer a deflection? How?

Please restate. Your sentance We have an observer, who is also in in this field observer a deflection? How? is meaningless to me.

Only observers outside your field would observer a "deflection", but this "deflection" is not due to to the field but due to the fact that the observer is not in the field.

Huh? What do you mean by "observer is not in the field"? By definition the observer is in the field. That's precisely how one interprets the physical meaning of spacetime coordinates. Let mf give you an example: consider an infinitely large plane of unifrom mass density. This will generate a uniform graviational field. The field is uniform for observers who are at rest with respect to the source, e.g. standing on the plane.

If you'd like me to respond your questions then please respond to mine so that I may better answer yours. Thanks.

By the way, have you ever read any of Einstein's writings such as his book on SR and GR? Think of what an observer in an accelerating elevator would see if a beam of light is initially directed parallel to the floor of the elevator. The elevator would continue to accelerate after the light is emitted and thus the path of the light as observed from the elevators accelerating frame of reference is a curved path. This is identical to what one would view from an observer at rest in a uniform gravitational field.

Pete

 

[pmb_phy]

So the length that light travels is zero? Not in my frame of reference.

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.

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?

Pete

 

[my_wan]

The length of a null geodesic in spacetime is zero.

Length wrt space-time doesn't even have meaning. Length only has meaning wrt an observer like me or a device for measuring. It sounds like you are talking the photon frame of reference which is a pointless and unphysical point of view. It's not something an observer can ever even in principle measure.

 

[MeJennifer]

Let mf give you an example: consider an infinitely large plane of unifrom mass density. This will generate a uniform graviational field. The field is uniform for observers who are at rest with respect to the source, e.g. standing on the plane.

Apart from the issue if the EFE's even allow for such a totally unphysical configuration if observers were to exist above or below such a plane the spacetime would obviously be curved. If only the large plane could exist there would not be nothing to gravitate since nothing above or below could exist.

If you'd like me to respond your questions then please respond to mine so that I may better answer yours. Thanks.

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.

Length wrt space-time doesn't even have meaning.

Sure it does, it is pretty important in the theory of general relativity.

You are talking about geodesics but you do not want to talk about spacetime distances?

 

[MeJennifer]

The magnitude of the integral of ds is zero, not the length of the path.

The arc length of the path is zero.

 

[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 dependant. 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?