Gravitational force of moving objects

[coolmatthew]

Hello guys!

According to relativity, objects with higher kinetic energy have larger mass. Would that affect the gravitational force of the object?

Or in other words, if a neutron moves faster, would it attract other things more strongly?

tyvm

 

[Agerhell]

If you have a box full of a certain number of slowly moving neutrons and another identical box full of the same number of, but faster moving, neutrons, the second box would have a sligthly stronger gravitational attraction on things outside of the box.

 

[WannabeNewton]

You might be interested in the following: http://arxiv.org/pdf/gr-qc/9909014v1.pdf

 

[jtbell]

if a neutron moves faster, would it attract other things more strongly?

If you have a box full of a certain number of slowly moving neutrons and another identical box full of the same number of, but faster moving, neutrons, the second box would have a sligthly stronger gravitational attraction on things outside of the box.

"A box full of neutrons" is different from "a neutron." Gravitational effects are determined by the stress-energy tensor, which I would expect to be different for a box full of randomly-moving neutrons versus the same number of neutrons all moving together in one direction with the same speed.

This question (about the gravitational effect of a moving object) comes up rather often, but I can't find any of the previous threads at the moment. Their titles must not be very obvious, and I can't think of any search keywords that give focused results. We really need an FAQ about it.

Most people's first guess is that an object's gravitational effect depends on the so-called "relativistic mass" $\gamma m_0 = m_0 / \sqrt {1 - v^2 / c^2}$. However, it's not that simple. For one thing, you have to define carefully what you mean by "gravitational effect" in general relativity. I seem to remember a recent thread in which it turned out that (for a certain definition of "gravitational effect") the appropriate quantity is actually $\gamma (1 + \beta)m_0$ where β=v/c.

 

[Bill_K]

I seem to remember a recent thread in which it turned out that (for a certain definition of "gravitational effect") the appropriate quantity is actually $\gamma (1 + \beta)m_0$ where β=v/c.

This may be the thread you're thinking of. And here's the paper that derived the result.

 

[jtbell]

Yes, that's the one, thanks! I saw the title of that thread when I was skimming backwards chronologically, but I didn't look at it because the word "weight" indicated passive gravitational mass to me, that is, the effect of a gravity on a moving object rather than the effect produced by a moving object. Nevertheless, the paper does address active gravitational mass (the "by" case).

 

[Agerhell]

Yes, that's the one, thanks! I saw the title of that thread when I was skimming backwards chronologically, but I didn't look at it because the word "weight" indicated passive gravitational mass to me, that is, the effect of a gravity on a moving object rather than the effect produced by a moving object. Nevertheless, the paper does address active gravitational mass (the "by" case).

How do you know which object it the moving object? The situation with effect of gravity on a moving object from a stationary object is identical to the situation of gravity by a moving object on a stationary object...

 

[Bill_K]

How do you know which object it the moving object? The situation with effect of gravity on a moving object from a stationary object is identical to the situation of gravity by a moving object on a stationary object...

Agerhell, if you take a glance at the paper we're referring to, you'll see that this is exactly what is used to to do the calculation. The deflection of a test particle in the field of a static mass is calculated, and then interpreted in the reverse fashion,