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

In summary, 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 this does not imply that the geodesic line of a test particle in space is a straight line. In the presence of a gravitational field, the trajectory of a particle consists of two parts: gravitational acceleration and spatial curvature. The spacetime associated with the Earth's gravitational field is curved, leading to the curved trajectories of objects in its presence. This is due to the observation that spacetime is curved in the presence of mass or energy-momentum, and cannot be considered flat or straight in a gravitational field.
  • #1
aachenmann
11
0
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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
  • #4
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?
 

Attachments

  • geodesic.GIF
    geodesic.GIF
    14.7 KB · Views: 589
  • #5
aachenmann said:
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
 
  • #6
aachenmann said:
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.
 
  • #7
pmb_phy said:
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?
 
  • #8
aachenmann said:
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.
 
  • #9
my_wan said:
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.
 
  • #10
pmb_phy said:
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.
 
  • #11
MeJennifer said:
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.
 
  • #12
MeJennifer said:
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
 
  • #13
my_wan said:
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
 
  • #14
MeJennifer said:
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
 
  • #15
my_wan said:
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.

pmb_phy said:
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.

pmb_phy said:
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?
 
Last edited:
  • #16
my_wan said:
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
 
  • #17
MeJennifer said:
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.
 
  • #18
pmb_phy said:
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.

my_wan said:
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.
 
Last edited:
  • #19
MeJennifer said:
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

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

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
 
  • #20
MeJennifer said:
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
 
  • #21
pmb_phy said:
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

pmb_phy said:
The metric for a uniform gravtiational field is

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

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.
 
Last edited:
  • #22
MeJennifer said:
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
 
  • #23
pmb_phy said:
One cannot introduce spacetime curvature by a mere change in coordinates since spacetime curvature is an intrinsic property of a metric space.

pmb_phy said:
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?

pmb_phy said:
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.

pmb_phy said:
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?
 
  • #24
MeJennifer said:
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.
 
  • #25
my_wan said:
So the length that light travels is zero? Not in my frame of reference.
The length of a null geodesic in spacetime is zero.
 
Last edited:
  • #26
MeJennifer said:
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
 
  • #27
my_wan said:
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
 
  • #28
MeJennifer said:
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.
 
  • #29
pmb_phy said:
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.

pmb_phy said:
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.

my_wan said:
Length wrt space-time doesn't even have meaning.
Sure it does, it is pretty important in the theory of general relativity. :smile:

You are talking about geodesics but you do not want to talk about spacetime distances?
 
Last edited:
  • #30
pmb_phy said:
The magnitude of the integral of ds is zero, not the length of the path.
The arc length of the path is zero.
 
  • #31
pmb_phy said:
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.

pmb_phy said:
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.
 
Last edited:
  • #32
MeJennifer said:
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.
 
  • #33
MeJennifer said:
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
 
  • #34
Ich said:
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
 
  • #35
MeJennifer said:
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
 

Similar threads

  • Special and General Relativity
Replies
8
Views
405
  • Special and General Relativity
Replies
27
Views
4K
  • Special and General Relativity
Replies
8
Views
822
  • Special and General Relativity
Replies
10
Views
1K
  • Special and General Relativity
2
Replies
44
Views
4K
  • Special and General Relativity
3
Replies
95
Views
4K
Replies
21
Views
1K
  • Special and General Relativity
Replies
11
Views
1K
  • Special and General Relativity
Replies
25
Views
1K
  • Special and General Relativity
Replies
5
Views
958
Back
Top