# I Gravity changes produced by a compressed and stretched spring

1. Nov 11, 2017

### Staff: Mentor

No. For spherical geometry (outside of the star) see the reference I posted

Last edited: Nov 11, 2017
2. Nov 11, 2017

### pervect

Staff Emeritus
To figure out the space-time curvature in the exterior of the star, Birkhoff's theorem tells you that for a non-rotating star, all you need to know is the mass, the geometry will be the Schwarzschild geometry.

I believe that the space-time geometry of a spinning star is determined only approximately by it's mass and spin. If we imaging a rotating dumbell, it's exterior field will be different than if it had a figure of revolution, for instance. I believe the gravitational wave emission would be different because of the different quadropole moment, (and gravitatoinal radiation is a feature of the space-time geometry). I do have a vague memory that there was some paper that was concerned about the issue, but I'm drawing a blank.

People usually approximate the exterior geometry of a spinning star by the Kerr geometry, I believe. FOr that approximation all you need is mass and spin. Potentially you could add charge to the list, but it's usually assumed there is none (or that if there is, it doesn't significantly effect the gravitational field).

3. Nov 11, 2017

### Staff: Mentor

As @Dale and @pervect have said, the answer to this question is "no". But there is a different question to which the answer is "yes": does the pressure inside a neutron star contribute significantly to its externally measured mass?

4. Nov 11, 2017

### Staff: Mentor

I agree, and if I recall correctly this is also discussed in the paper.

5. Nov 11, 2017

### John K Clark

Thanks to everybody for for taking the time to answer my questions, I think I'm starting to understand what you're trying to say but I have one more dumb question. Birkhoff's Theorem assumes (I think) that the gravitating object is static, but suppose it is not. Would the orbit of a planet around a star that just went supernova change even before the debris from the explosion reached it because the pressure inside the star suddenly increased very dramatically and even a supernova can't propel debris faster than the speed of gravity? I may be wrong but it seems to me the pressure and the resulting increased gravity wouldn't last long but it might be substantial while it did.

John K Clark

6. Nov 11, 2017

### Staff: Mentor

No, it doesn't. It only assumes spherical symmetry and vacuum. It applies to non-static situations like the spherically symmetric collapse of a star, or the spherically symmetric explosion of one--that is, it describes the vacuum region exterior to both of these.

7. Nov 13, 2017

### Tomas Vencl

But from distance observer view, the externally measured mass is constant, so this contribution to SET must be counterbalanced by decrease of other part of SET. For example in my post 36 by vanishing the momentum components (and changing the others, of course). Yes ?

8. Nov 13, 2017

### Staff: Mentor

"Constant" with respect to what? We're talking about a static neutron star. Everything is constant.

What process are you talking about?

Which is talking about a different scenario than the one I was referring to in the post you quoted (I was responding to post #50, which talks about a static neutron star).

9. Nov 13, 2017

### Staff: Mentor

In your post #36, you have a spherically symmetric shell of dust (which has zero pressure) falling past an observer at some fixed altitude. You are correct that spacetime in the observer's immediate area is flat until the shell falls past him; then it becomes curved. Once the shell falls past him, the curvature doesn't change. None of this has anything to do with pressure, since, as I just mentioned, dust has zero pressure.

10. Nov 13, 2017

### Tomas Vencl

At the end of my post 36 I am talking about the scenario that dust collides (does not matter the details) and creates some static solid object (if the mass is big enough it can be neutron star - for this example).which pressure is in counterbalance with gravity forces. And the last part of my question in post 36 is , if after creation such a "star" the observer see any gravity field chances.
OK, sorry, I just thought, that my post 57 is extrapolation of the question to neutron star formation. I should have been more precise, sorry.

11. Nov 13, 2017

### Tomas Vencl

I know that from noninteracting dust at the beginning of experiment I sudennly change the behaviour of the matter to some interacting (this can be a problem, but it can be quite close to reality).
The goal of the experiment was to simplify the conditions of the experiment (compare to experiments with bomb in the shell and so on..). At the begining we have spherical shell of matter (with some rest mass) with momentum and at the end we can have the same matter (rest mass) withn pressure (we can simplify that no changes happen to matter during collision). And finally by Birkhoffs theorem the external gravity field must be the same.

12. Nov 13, 2017

### Staff: Mentor

Ah, ok. Note that this is not going to happen without an additional process happening--a significant amount of energy will have to be radiated away. If that does not happen, the dust will collide in a compact region and just bounce back out again, in the time reverse of the collapse process.

We have to be very precise about the process, since, as I noted just now, you left out a key part of it. Here is a more precise description of the key phases of the process:

(1) The shell is falling inward but is still above the observer. During this phase the observer sees flat spacetime in his vicinity.

(2) The shell has fallen inward past the observer but has not yet gotten compact enough for any collision/radiation to happen. During this phase the observer sees curved spacetime in his vicinity, specifically the Schwarzschild metric with some mass $M_0$.

(3) The shell has fallen into a compact enough region that "collision" is happening--the individual dust particles are now interacting with each other, producing pressure, shock waves, radiation, and all kinds of other fun stuff. However, none of this stuff has come back out to where the observer is. During this phase the spacetime geometry the observer sees is unchanged; it's still the Schwarzschild metric with some mass $M_0$.

(4) A lot of radiation (we're idealizing--probably in a real process there would be outgoing matter--gas jets, gas clouds, etc.--but we're assuming that doesn't happen here, so that all of the matter that falls inward past the observer stays there) flies out past the observer and escapes to infinity. Once it all has passed and the object inside has settled down to its static final state, the observer sees a spacetime geometry in his vicinity which is still the Schwarzschild metric, but now with some smaller mass $M_1$, i.e., $M_1 < M_0$.

Now, having described the process, we can ask a key question: what is this "mass" that appears in the descriptions above? The point being that there are (at least) two ways of defining what this "mass" is:

First, we can define "mass" as the quantity $M$ that appears in the Schwarzschild metric. The observer can measure this quantity by simply measuring the spacetime geometry in his vicinity--for example, by measuring tidal gravity. He doesn't have to know anything at all about the internal structure of the object below him. (Note that in phases 2 and 3, we have very different internal structures, but the mass is the same, and the observer can know this without even knowing anything about how the internal structure is changing from phase 2 to phase 3.) Once this idea is made mathematically rigorous, it is called in GR the "Bondi mass". (A technical point for the cognoscenti: the reason I say "Bondi mass" here instead of "ADM mass", which is the more commonly mentioned version of this concept, is that the Bondi mass decreases when radiation escapes to infinity, whereas the ADM mass does not.)

Second, we can define "mass" as the product of some calculation that add up contributions from all relevant aspects of the object's internal structure to get a total. The way we do this is to integrate the stress-energy tensor over some relevant spacelike hypersurface, paying careful attention to the effect of spacetime curvature on the integration measure. The usual way of doing this, which when made mathematically rigorous is called the "Komar mass", requires the spacetime to be stationary, and our spacetime isn't--or, to put this another way, the Komar mass is only well-defined for a system for which it does not change (for technical reasons that we probably don't need to go into here), and any such integral for the scenario we are discussing would have to change (between phases 3 and 4). But heuristically, we can still think of adding up contributions in each phase, as follows:

For phases 1 and 2, the only contributions are the energy and momentum density of the dust. These change in concert in such a way as to leave the mass integral constant. Tthe only difference between phases 1 and 2 is whether the observer is inside or outside the shell, which affects whether he can measure the mass directly by measuring spacetime curvature in his vicinity. The mass itself, in the sense of the sum of contributions, does not change between phases 1 and 2.

For phase 3, as stated above, the mass is unchanged, but the individual contributions will be changing: basically, the "extra" energy density (i.e., energy density over the rest energy density) and momentum density of the dust is being converted to internal pressure of the final object, plus energy and momentum density in radiation.

For phase 4, the two "pieces" of the mass in phase 3 are now separated, and one (the radiation) is back outside the observer so it doesn't affect the spacetime curvature he sees. That's why he measures a smaller mass. That smaller mass is composed of contributions from the rest energy density and pressure of the matter inside the final object.

I'll follow up with one more comment in a separate post.

13. Nov 13, 2017

### Staff: Mentor

Here is the one more follow-up comment to my previous post.

You might ask, after reading the description in my previous post, what about gravitational potential energy? It is often said that, in addition to pressure, we must also include a (negative) contribution from gravitational potential energy in order to get the right answer for the mass of an object like the static object in phase 4. However, this arises from two things that I would call confusions:

First, note that I said in my previous post that, when integrating all the contributions from stress-energy to get a total mass, one has to pay careful attention to the effect of spacetime curvature on the integration measure. For a static object, it turns out that doing that makes the integral smaller than it would be if we just did a "naive" integral--or, to put it another way, if we just assumed that the "space" we were integrating over was ordinary Euclidean 3-space instead of being a non-Euclidean spatial geometry produced by taking a 3-d spacelike slice out of a curved 4-d spacetime. This reduction can be thought of as taking into account "gravitational potential energy"--or, to put it another way, taking into account the fact that we had to radiate away a bunch of energy in order to form the static object from the collapsing dust cloud. But I think a better way to put it is simply that we are doing the integral correctly in a curved spacetime. (For one thing, like Komar mass, the concept of "gravitational potential energy" is only well-defined in a stationary spacetime, which ours isn't; but the integral I am talking about can be done in a non-stationary spacetime, although there is more arbitrariness in the choice of spacelike hypersurfaces over which to integrate.)

Second, it also turns out that, for a static, spherically symmetric object, if you do the "naive" integral I just described, assuming Euclidean 3-space, but only count the energy density, not the pressure, you get the same answer as the correct integral, which counts the pressure but also accounts for the non-Euclidean spatial geometry. Some people say that this means that pressure "does not contribute" to the mass; others, somewhat more sophisticated, say that it means the contribution from pressure is exactly canceled by the negative contribution from gravitational potential energy. But I think it is better simply to recognize that we are dealing with a regime in which our ordinary intuitions about what questions to ask, and how to describe things, are not necessarily correct.

14. Nov 14, 2017

### Tomas Vencl

Thank you very much for detailed answer.

15. Nov 14, 2017

### Tomas Vencl

Your post 63 is interesting and I have some questions. It is truth that in writing my post 36 was not clear to me, which role plays the gravitational potential energy.
Do I understand it correctly, that if we are integrating in non-Euclidean spatial geometry (whatever it exactly means in sense of choice of spacelike hypersurfaces over which to integrate) , the gravitational potential energy is allready automatically included there ? The pressure we must take into account in this case ?
Or we can simply (as a not correct procedure, but with correct result - which sometimes happen :-)) integrate in flat space but we can not include pressure and gravitational potential energy ?
This is very interesting.

16. Nov 14, 2017

### Staff: Mentor

Not necessarily, because "gravitational potential energy" might not even be well-defined. It is only well-defined in a stationary spacetime. In a stationary spacetime, yes, you can think of the correct integration measure over the non-Euclidean spatial geometry as taking into account the effects of gravitational potential energy.

You must always take the pressure into account. Pressure is part of the stress-energy tensor and the stress-energy tensor is the source of gravity in GR. You can never just leave it out.

You can get away with this in the one particular case I mentioned (a spherically symmetric, static object). It just happens that in this particular case, the effect of including pressure in the integral is exactly cancelled by the effect of using the correct integration measure. But that doesn't mean it's a good idea to just ignore the pressure or the correct integration measure.

17. Nov 14, 2017

### Tomas Vencl

Thank you. I think I understand.

18. Nov 14, 2017

### Dalton Peters

When a spring is compressed the stored energy increases its mass? sorry I havent started under grad physics yet

19. Nov 14, 2017

### Staff: Mentor

That's right. By the $e=mc^2$, if the energy of a system (such as a spring) increases then its mass increases as well. Compressing the spring increases its energy and thus its mass.

20. Nov 15, 2017

### timmdeeg

With some caveats, see #41.

21. Nov 15, 2017

### 1977ub

And moreover, that added mass-energy is removed from the mechanism or agent which did the compressing.