# Space-Time Warpage and Kinetic Energy

1. Jan 30, 2014

### sgolson

A mass moves through space-time at constant velocity. Space-time around this mass is warped by the mass and as the mass moves through space-time it causes unwarped space-time to warp and already warped space-time to warp more. In addition, the space-time it has moved through de-warps. Why doesn't this scenario result in kinetic energy being lost by the mass? What don't I understand here?

2. Jan 30, 2014

### Staff: Mentor

If the mass is changing speed (for example, oscillating or moving rapidly in a circular orbit) it will lose energy through gravitational radiation - think about how pushing a floating object up and down causes ripples to radiate out in all directions. But if the object is just moving at a constant speed, coasting, it won't lose kinetic energy. There are several ways of thinking about this; perhaps the easiest to is to consider that if the object is just coasting and subject to no external forces then, as far as the object is concerned, it's at rest so has no kinetic energy to lose. You're thinking in terms of watching an object moving past you, but from the point of view of someone riding on the object they're at rest and you're the one who's moving, in the other direction.

3. Jan 30, 2014

### sgolson

I understand everything you've said, but I don't understand why an action (mass distorts space-time) doesn't result in an equal and opposite reaction which would cause the velocity of the mass to be altered.

4. Jan 30, 2014

### Staff: Mentor

For a hand-wavy explanation (this is a VERY hand-wavy explanation,so don't push it too far - it's not a substitute for working through the math of frame-dependent kinetic energy)...

If you're thinking of the object at rest while you're moving, then it's surrounded by a static and unchanging curvature; nothing is changing so there's no change in kinetic energy (which is, conveniently, zero).

If you're thinking of yourself at rest while the object is moving, then the curvature is increasing in the region in front of the moving object as the distance from a point in that region to the object decreases - but at the same time the region behind the region is flattening out as the object moves away from it. The two effects cancel out so that the total amount of distortion is constant even though it's non-zero; or you could say that there is no net transfer of energy from the object to the gravitational field around it because the field is weakening in one region at the same rate that it's strengthening in another.

Obviously this wouldn't work if space-time were a physical substance that dissipates energy when it's flexed, like a rubber sheet for example. But it's not.

5. Jan 30, 2014

### sgolson

Nugatory, thanks again. I understand now. I didn't imagine that the de-warping space-time in the wake of the moving mass would return any and all energy to the mass that was transfered to the space-time in front.

But one more question:
You write "or you could say that there is no net transfer of energy..."
Do you mean to say that there is, in fact, a transfer of energy; or that there might be but we don't know for sure; or that there really isn't but if it helps me get my head around this issue then I can go ahead and think of it that way but that I shouldn't express that view should I ever go beyond a bachelor's degree?

6. Jan 30, 2014

### Staff: Mentor

Gravitational waves can and do carry away kinetic energy from a system. However, the lowest order of radiation for gravitation is quadrupole, which means that you need something like a spinning barbell to radiate gravitationally. An object in linear motion is not even a dipole source, let alone a quadrupole source.

7. Jan 30, 2014

### Bill_K

An object in linear motion has, relative to a fixed point, a dipole moment that's linearly increasing with time. But of course it doesn't radiate. It's an example of a nonradiative motion.

The moving object also has a quadrupole moment that grows quadratically with time, but this is also nonradiative, since quadrupole radiation is proportional to the third derivative of the moment.

Last edited: Jan 30, 2014
8. Jan 30, 2014

### Staff: Mentor

Yes, that's why I said dipole source rather than dipole moment.

9. Jan 31, 2014

The last