# Curved space/ spacetime

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.
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?

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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

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.

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?

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.

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.

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.

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.

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

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

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

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?

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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

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.

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.

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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

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

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.

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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

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?

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.

So the length that light travels is zero? Not in my frame of reference.
The length of a null geodesic in spacetime is zero.

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