How Can We Understand Geodesics in Curved Spacetime Under Gravity?

[Erland]

If I understand GR correctly, gravity is no real force but only an effect of the curvature of spacetime. Thus, objects subject to no other forces than gravity follow trajectories in spacetime that are geodesics. I find this very hard to understand, because the trajectories of such objects don't look like anything like straight lines.

Let's take a simple example. Consider a 4-dimensional spacetime universe in which we have a constant gravity in the same direction throughout spacetime. This would locally be a good approximation to the situation at a small region on the Earth.
Then, if an object at an event with spacetime coordinates (t0,x0,y0,z0) is thrown upwards, against gravity, it will rise a bit and then fall back and pass the point from which it was thrown, an event with coordinates (t1,x0,y0,z0).
But then, it should be possible to change coordinates, in some way prescribed by GR, such that the trajectory of this thrown object actually becomes a geodesic. How is that possible? How could there be any other geodesic connecting the two events than the 4-dimensional line given by (t,x0,y0,z0) for all t (that is, the object remains at rest at the space coordinates (x0,y0,z0) all the time)?
For although I can understand that such a new coordinate system will distort the space coordinates, I don't understand how the time coordinate can be distorted, except by a constant factor. I can't see how we can get a greater value of

Integral sqrt(dt^2-dx^2-dy^2-dz^2)

along any other spacetime path connecting the two events than along the path where the object is at rest.

Is there any way of explaining what such a curved coordiante system would look like, without using too advanced differential geometry and tensor calculus?

 

[Mentz114]

It all becomes possible if we rewrite the distance formula

Integral sqrt(dt^2-dx^2-dy^2-dz^2)

as ds²=g₀₀dt²-g₁₁dx²-g₂₂dy²-g₃₃dx²

where the g's can vary from place to place. They are functions of position ( and perhaps time also).

 

[Erland]

It all becomes possible if we rewrite the distance formula



as ds²=g₀₀dt²-g₁₁dx²-g₂₂dy²-g₃₃dx²

where the g's can vary from place to place. They are functions of position ( and perhaps time also).

Ok, thanks. But I would like to know what these g-functions will be in this case...

 

[Mentz114]

For the Earths field the g's will be given by the Schwarzschild metric for the space-time outside a spherically symmetric source. I'll give details later ( if no-one has done it for me) because I'm busy right now.

 

[pervect]

For the Earths field the g's will be given by the Schwarzschild metric for the space-time outside a spherically symmetric source. I'll give details later ( if no-one has done it for me) because I'm busy right now.

<br /> ds^2 = (1-2M/r) dt^2 - dr^2 / (1-2M/r) - r^2 d \theta^2 - r^2 sin^2 \theta d \phi^2<br />

 

[HallsofIvy]

But then, it should be possible to change coordinates, in some way prescribed by GR, such that the trajectory of this thrown object actually becomes a geodesic.

You seem to have a misunderstanding of "geodesics" and "coordinates". If a path is a geodesic in one coordinate system, it is a geodesic in any coordinate system. Whether or not a path is a geodesic is independent of the coordinate system.

 

[Erland]

You seem to have a misunderstanding of "geodesics" and "coordinates". If a path is a geodesic in one coordinate system, it is a geodesic in any coordinate system. Whether or not a path is a geodesic is independent of the coordinate system.

But when we describe events on the Earth, we usually use a flat coordinate system, despite the gravity of the Earth. And in a flat coordinate system, the only geodesics are line segments in spacetime, right? But if we throw an object upwards so that it falls back to the starting point, its trajectory is the flat spacetime is not a line segment and thus not a geodesic, despite that it must be a geodesic if we choose a correct curved coordinate system.

What do I misunderstand here?

 

[Mentz114]

But when we describe events on the Earth, we usually use a flat coordinate system, despite the gravity of the Earth. And in a flat coordinate system, the only geodesics are line segments in spacetime, right? But if we throw an object upwards so that it falls back to the starting point, its trajectory is the flat spacetime is not a line segment and thus not a geodesic, despite that it must be a geodesic if we choose a correct curved coordinate system.

What do I misunderstand here?

To get the path of a ball thrown up and returning to the ground using Netwon's laws, we assume a Euclidean 3D space, and an acceleration of about 9.8ms^-2 towards the Earths centre.

To do the GR treatment we can integrate the geodesic equations in Schwarzschild spacetime. ( The differences will be of the order 10^-8.) In this case we are not assuming anything like the spacetime is flat.

 

[Passionflower]

But when we describe events on the Earth, we usually use a flat coordinate system, despite the gravity of the Earth. And in a flat coordinate system, the only geodesics are line segments in spacetime, right? But if we throw an object upwards so that it falls back to the starting point, its trajectory is the flat spacetime is not a line segment and thus not a geodesic, despite that it must be a geodesic if we choose a correct curved coordinate system.

What do I misunderstand here?

Your misunderstanding is that you seem to think that what is a geodesic depends on the coordinate chart, which is not the case.

 

[DrGreg]

But when we describe events on the Earth, we usually use a flat coordinate system, despite the gravity of the Earth. And in a flat coordinate system, the only geodesics are line segments in spacetime, right? But if we throw an object upwards so that it falls back to the starting point, its trajectory is the flat spacetime is not a line segment and thus not a geodesic, despite that it must be a geodesic if we choose a correct curved coordinate system.

What do I misunderstand here?

A geodesic in space-time does not necessarily correspond to a geodesic in space. If you throw a ball in the air its trajectory through space relative to someone on the ground is a parabola (approximately), but its trajectory through space relative to a trampolining observer (whilst in the air between bounces) is a straight line (approximately).

The technical definition of a space-time geodesic does not depend on your choice of coordinates. A geodesic looks locally straight relative to an inertial, free-falling observer, but may look "bent" relative to a non-inertial observer, such as someone on the ground who has an upward proper acceleration of g.

 

[Erland]

OK, thanks all. I'll think of what you said, and do some calculations with the Schwarzschild metric.

 

[pervect]

If you throw a ball up in the air, a short distance, so that if follows a parabolic trajectory and falls back down where it started, it is following a geodesic in space-time. In this one particular special case, one can think of the trajectory as the one that maximizes proper time - however, in general, one must be more careful than this, and say that the trajectory "extremizes" the proper time, rather than maximizes it.

The clock thrown upwards in this example will have a longer proper time than the clock that was "stationary" on the ground the whole time.

The explanation for this is that the metric coefficient g_00 varies with height. This is usually referred to as "gravitational time dilation". Gravitational time dilation is the most important deviation of the space-time of Earth from flatness, and you can see that it involves only time (the g_00 coefficient relates only to time). Under the correct circumstances (gravitational bending of light by the sun, for instance), the other metric coefficients can become important, though.

 

[Erland]

If you throw a ball up in the air, a short distance, so that if follows a parabolic trajectory and falls back down where it started, it is following a geodesic in space-time. In this one particular special case, one can think of the trajectory as the one that maximizes proper time - however, in general, one must be more careful than this, and say that the trajectory "extremizes" the proper time, rather than maximizes it.

The clock thrown upwards in this example will have a longer proper time than the clock that was "stationary" on the ground the whole time.

The explanation for this is that the metric coefficient g_00 varies with height. This is usually referred to as "gravitational time dilation". Gravitational time dilation is the most important deviation of the space-time of Earth from flatness, and you can see that it involves only time (the g_00 coefficient relates only to time). Under the correct circumstances (gravitational bending of light by the sun, for instance), the other metric coefficients can become important, though.

Thanks!

As I understand the formula you gave for the Schwarzschild metric, there will also be a "gravitational length contraction", since g₁₁>1. But I suppose you meant that this will be negligible compared to the gravitational time dilation when we compute the proper time in the case considered.

The space coordinates in general contribute negligibly little to the proper time for times and distances we humans daily experience. The reason is that, with SI-units, there should be a factor c² before the g₀₀-coefficient you gave in the Scharzschild formula, or, alternatively, a factor 1/c² before the other three (spatial) coefficients.

Although it seems that a ball thrown straight up in the air and then falling back (for simplicity, we consider only vertical, no horizontal, motion of the ball, and we disregard the motion of the Earth itself), would follow a trajectory in spacetime that is far from a geodesic, this impression changes if we think of it in terms of units where c=1, which is natural in the theory of relativity.

Suppose, for example, that an observer on the moon watches when I throw the ball on the Earth (a little more than a lightsecond away). Suppose that my throwing takes place at night, on the dark side of the Earth, but still facing the moon, and that the ball contains a very strong light source, so that it can be seen from the moon (we also disregard the motion of the moon).

The observer on the moon will then, unless he/she uses a very big telescope, be unable to see that the ball I throw moves, it is too far away for that. He/she will consider the trajectory of the ball in spacetime as a line segment, with the space coordinates fixed and only the time coordinate changing a few seconds. If it really was a line segment, it would be a geodesic in flat spacetime. The actual changes in the space coordinates, that our observer on the moon only can detect with a big telescope, are so small compared to the change in the time coordinate (if c=1) that it is not hard to believe that the true trajectory is a geodesuc anyway, in a spacetime that is slightly curved due to the gravity of the Earth.

So, the reason that it is hard to see intuitively that the trajectory is actually a geodesic is that with our units, c >> 1.

 

[grav-universe]

Let's take a simple example. Consider a 4-dimensional spacetime universe in which we have a constant gravity in the same direction throughout spacetime. This would locally be a good approximation to the situation at a small region on the Earth.
Then, if an object at an event with spacetime coordinates (t0,x0,y0,z0) is thrown upwards, against gravity, it will rise a bit and then fall back and pass the point from which it was thrown, an event with coordinates (t1,x0,y0,z0).
But then, it should be possible to change coordinates, in some way prescribed by GR, such that the trajectory of this thrown object actually becomes a geodesic.

There is, the equivalence principle. According to GR, if an observer is stationary within a gravitational field and tosses a ball, the ball will follow the geodesics while the observer does not. The ball is in freefall so is considered inertial while the observer is accelerating by remaining stationary within the field.

In terms of the observer's coordinate system, everything that is in freefall is coordinately accelerating, but according to GR, it is really the observer that is accelerating. You can see this from the perspective of considering the ball to travel inertially, but the observer is properly accelerating. The observer, while accelerating, tosses a ball in the direction of acceleration. The ball travels away inertially from the observer at some greater speed than the observer currently travels, but since the observer is accelerating, the observer will continue to speed up until the ball is overtaken. According to the observer's coordinate system, the ball will have initially traveled away, then turned around and traveled back.