Space-time curvature as gravity only for speeding particles?

1. Mar 31, 2013

Edi

If gravity rises from the fact that mass bends space-time and stuff falls in because it actually follows a straight line in a curved space as it moves by a gravitating object - doesn't that mean that a relatively stationary particle would not fall in the the claws of gravity as it would NOT be following a line at all?
At least stationary in the plane parallel to the surface because, as I think, moving UP perfectly perpendicular would not encounter curvature in such way to follow it down..

And I mean a stationary fundamental particle, because I know that "stuff" is made of atoms, that move around vibrating and atoms consist of nuclei and they of quarks and gluons and what not speeding inside like crazy (and the speeding quarks and gluons is what makes up most of the mass-energy of matter?).

Faster a particle moves the quicker it follows the line in to the claws of gravity [falls] ?

2. Mar 31, 2013

Jonathan Scott

That would be true if gravity bent space only, not space-time.

The bending with respect to time and space is approximately equal, so you can think of the effect of the time part as being similar to the effect on a particle moving through space at c. When the particle is moving near or at light speed, the effect of the bending of space doubles the effective acceleration to twice the Newtonian acceleration.

3. Mar 31, 2013

Edi

Ok, I did not take moving trough time in account, thank you.

But further thinking.. if it did bed just space, then this: "Faster a particle moves the quicker it follows the line in to the claws of gravity [falls] " would be true.. ?
And if it did bend just space, then moving perfectly up would not feel the bend/ pull of gravity?

4. Mar 31, 2013

A.T.

If stationary in space, it still advances through time.

5. Mar 31, 2013

Jonathan Scott

Yes

It would seem natural to think so, but actually the curvature of space has an additional effect, which is loosely speaking that space varies slightly in "size" because of the curvature, so a ruler which is a fixed size in local space appears slightly smaller closer to a mass and larger further away.

Even if the direction of the particle was straight up or down, this size effect means that the magnitude of its momentum in the downwards direction increases at a rate proportional to its speed, at the same rate as if it were moving sideways at the same speed (in which case the momentum change would be due to change in direction).

(For those who want to be picky about the details, the above description applies to a the path of a test mass described using isotropic coordinates in the weak field approximation for the GR Schwarzschild solution).

6. Mar 31, 2013

1977ub

Half of the light deflection near the Sun is due to distorted space-TIME and half is due to distorted space itself. Theoretically, removing space-TIME distortion, could the space distortion capture a particle into an orbit about the Sun? (i.e. without "gravity")

7. Mar 31, 2013

Jonathan Scott

This is getting very hypothetical, but I guess it's still just about meaningful enough to answer...

No, because for a light beam passing near the surface you've halved the deflection (which is already so small that Eddington had difficulty observing it), and for anything moving more slowly, the effect of the space part is merely that everything still follows the same shape of space, and hence the same path as the light beam.

(Even if the central mass is a black hole, the space component of curvature only closes to form a complete circle at the event horizon).

8. Mar 31, 2013

A.T.

For objects moving fast (lightspeed) the effect of space geometry alone is in the same order as of spacetime geometry. For most objects it is tiny in comparison.
You don't "feel" gravity anyway in free fall. Due to symmetry gravity will not bend a radial path.

9. Mar 31, 2013

Naty1

Code (Text):
...doesn't that mean that a relatively stationary particle would not fall in the the claws of gravity as it would NOT be following a line at all?
'relatively stationary' is upset by Heisenberg uncertainty.....there is always some motion.

10. Mar 31, 2013

1977ub

Interesting. If we take something slower and more massive than light, space-time deflection is even greater than in the case of light, giving rise to orbit or further to collision. So you're saying that the space-deflection is not the same? The deflection due to this for a slow-moving mass would be less rather than more pronounced than in the case of light?

11. Mar 31, 2013

Passionflower

I do not think so.
Why do you think that to be true?

12. Apr 1, 2013

Jonathan Scott

The angle of deflection due to curvature with respect to space is independent of speed. Particles just follow the shape of space.

13. Apr 1, 2013

Jonathan Scott

A fully detailed answer would be out of proportion to the level of this thread, but here's a rough summary.

Describe the space-time in a spherically symmetrical situation around a central mass using isotropic coordinates, where the scale factor relating coordinate space distances to local space distances is the same in all directions.

With all quantities expressed in those coordinates, including letting $c$ represent the coordinate speed of light rather than the standard value, the equation of motion is as follows:

$$\frac{d}{dt}\left ( \frac{v}{c^2} \right ) = \frac{1}{c^2} \, \left ( \mathbf{g}_t + \frac{v^2}{c^2} \mathbf{g}_{xyz} \right )$$

where $\mathbf{g}_t$ is the effective Newtonian gravitational field due to the gradient of the time scale factor and $\mathbf{g}_{xyz}$ is the equivalent field due to the gradient of the space scale factor. For GR (as described by the Schwarzschild solution) these two values are approximately the same, but you can also investigate the equation of motion for other cases, where the two scale factors are specified independently, which is what we have been discussing in this thread.

If you multiply both sides of the above equation by the total energy $E$ of a test particle (which is constant in free fall), the left side becomes the rate of change of coordinate momentum and the right side becomes the coordinate force.

Note that the right hand side is independent of the direction of travel, but depends on the speed. When the direction of travel is horizontal relative to the field, the rate of change of momentum is associated with a curved path. When the direction of travel is vertical, the rate of change of momentum also includes a term relating to the change in the coordinate value of $c$. For a rising or falling photon which is already travelling at $c$ relative to the coordinates, the momentum is changing entirely because $c$ is changing.

14. Apr 1, 2013

A.T.

How about this simple intuitive arguments, as for why a purely spatial distortion doesn't affect stationary objects:

To be affected by the geometry of a manifold you have to move within that manifold. If you are stationary with respect to the spatial dimensions, you are not affected by the spatial geometry.

A purely spatial distortion would mean that there is no gravitational time dilation, no gravitational red shift and therefore no potential difference that would cause a stationary object to start moving.

Are these two valid way to justify it?

15. Apr 1, 2013

Jonathan Scott

Those points sound reasonable to me, but as "intuitive arguments" can lead one astray in relativity, especially if one's intuition isn't particularly experienced in the subject, I feel it is safer to use arguments based on calculations. In this case, the calculated results agree with your statements.

16. Apr 1, 2013

Passionflower

Jonathan, what is the time curvature of a Newton-Cartan spacetime?

17. Apr 1, 2013

Passionflower

Four dimensional spacetime does not evolve in time, spacetime describes everything that exists outside the realm of time. Nothing actually moves in spacetime, the idea of evolution, movement and time is simply a gauge choice.

18. Apr 1, 2013

Jonathan Scott

I'm not sufficiently familiar with that concept to give a specific answer.

However, if it can be expressed as a metric theory with a static solution expressed in isotropic coordinates, then the coordinate acceleration of a particle initially at rest is determined entirely by the time component of the metric. This is what can loosely be described as the "curvature of the path with respect to time" within the coordinate system.

If the time scale factor is $\Phi \approx (1 - Gm/rc^2)$ then the specific expression for $\mathbf{g}_t$ in my previous post is as follows, where all quantities including the speed of light $c$ are as measured in the coordinate system, not local values.

$$\mathbf{g}_t = - c^2 \frac{1}{\Phi} \nabla \Phi$$

19. Apr 1, 2013

1977ub

Okay right. I understand that if a body begins without velocity wrt the Sun, then curvature of space won't move it. If it were on a trajectory toward grazing the edge of the Sun, the space curvature would be getting more pronounced as it got closer to the Sun, and this would cause some curvature of its path. This should be without regard to its mass or its velocity. Okay, thanks.

20. Apr 1, 2013

Passionflower

Jonathan, do you see time on the spacetime manifold as some preferred direction?

Time is simply the evolution of a particular direction on the spacetime manifold, I do not see that if only 3 components of the 4 dimensional are curved things suddenly do not move with respect to each other.

For instance two separated points on a paraboloid. Now evolve the points by increasing h, the points will separate or come together depending on direction of h. No need to curve h.

Last edited: Apr 1, 2013