Gravity changes produced by a compressed and stretched spring

[Tomas Vencl]

...However, in the above discussion I think that there is one point that we have been glossing over. That is that the energy density and the stress are different components of the stress energy tensor. So it is possible to have two sources with identical energy density but opposite stress. Geometrically that would be akin to having two vectors, one pointing at 89 deg and the other pointing at 91 deg. They wouldn’t be described as “increased” or “decreased”, but they are not “identical” either. The difference is more complicated.

Sorry for my poor English.
This is an interesting topic generally.
Regarding the mentioned above, let's imagine next example:
We have spherically symmetric thin shell of dust. The dust is collapsing by itself gravity (so we do not have static systém of course, this is a difference against the discussion so far). The observer is inside of the shell and is static and not at the centre of the shell. At some moment the shell is just above the observer. Until this moment the observer is in flat space.
After the shell passes by the observer he feels some gravity field. And the question is, how the gravity field at the observer position will change (if ?, it should not, I think) during next collapsing of the shell, and how the gravity field will change (if) after the dust collides and rising pressure will holds the central compact object static ? After the shell pases the observer, there is no energy flow by the observer so I think, that field keeps static.
I always thought, that the pressure in SET is only another form of energy, but from above discussion I feel, that it is not truth and pressure gives extra gravity beside its energy. Maybe I just do not understand the discussion or I am fully wrong. So the example would help me better understand.

 

[Dale]

how the gravity field at the observer position will change (if ?, it should not, I think)

You are correct. The curvature does not change from that point on, unless some mass or radiation escapes back out past the observer.

 

[1977ub]

There is always an offsetting decrease in the gravity elsewhere, right? You compress the spring with your muscles, and now you have less gravitational attraction?

 

[Dale]

There is always an offsetting decrease in the gravity elsewhere, right? You compress the spring with your muscles, and now you have less gravitational attraction?

Because it is a tensor with multiple components the curvature can be different without being either more or less.

 

[timmdeeg]

The curvature could indeed be different, as mentioned above at least inside the springs it would be different.

Is the increase of invariant mass of the spring the same if the work done to compress resp. to stretch the spring is the same? If so why would the curvature different?

 

[PeterDonis]

Is the increase of invariant mass of the spring the same if the work done to compress resp. to stretch the spring is the same?

If we are assuming that the start and end states are both static, then yes. Remember, as has been pointed out several times in this thread, that for a spring that is stretched or compressed to be static, it cannot be in isolation: there must be something else present that is holding it static, and that something else will have a stress that is opposite to the stress in the spring (stretched if spring is compressed, compressed if spring is stressed). The only meaningful invariant mass will be the invariant mass of the system as a whole, and that will be the same if the same amount of work is done on the system, whether the work stretches the spring (and compresses whatever is holding it static) or compresses the spring (and stretches whatever is holding it static).

If so why would the curvature different?

The spacetime curvature in the spring, considered in isolation, will obviously be different, because that's how you specified the problem. But you cannot deduce the invariant mass of the system from the spacetime curvature in the spring considered in isolation.

 

[timmdeeg]

But you cannot deduce the invariant mass of the system from the spacetime curvature in the spring considered in isolation.

Ok, thanks.

 

[Dale]

Is the increase of invariant mass of the spring the same if the work done to compress resp. to stretch the spring is the same?

Yes (with appropriate caveats)

If so why would the curvature different?

Because the curvature (a rank two tensor) depends on the stress energy tensor (a rank two tensor) and not only on the invariant mass (a scalar)

 

[timmdeeg]

Yes (with appropriate caveats)

Because the curvature (a rank two tensor) depends on the stress energy tensor (a rank two tensor) and not only on the invariant mass (a scalar)

Let me please control with your help, if I see it correctly.

1. Work is done to compress or stretch the spring which increases its stress energy.

2. The holding device is fixed at the spring which increases its stress energy too because it is now under pressure or tension.

3. The mass consisting of spring and device as a system has increased.

4. The curvature due to spring and device as a system doesn’t depend on whether the spring was compressed or stretched. Whereas the curvature inside the spring and inside the device does.

 

[Dale]

1. Work is done to compress or stretch the spring which increases its stress energy.

Not really. The stress energy tensor has many components. So you cannot just talk about it increasing or decreasing. It is a more complicated object.

If you put it in compression then the volume will reduce, so the energy density component will increase both due to the smaller volume and the work. There will also be a separate stress component that will increase.

If you put it in tension then the volume will increase, so the energy density component will decrease due to the larger volume and increase due to the work. Which effect is bigger may depend on the properties of the material but the energy density will remain positive. There will also be a separate stress component that will decrease and may become negative.

2. The holding device is fixed at the spring which increases its stress energy too because it is now under pressure or tension.

Similar comments as above.

3. The mass consisting of spring and device as a system has increased.

Yes, but the source of gravity is the stress energy tensor, not the mass. Note, this assumes that the energy came from outside the system which is not always the case.

4. The curvature due to spring and device as a system doesn’t depend on whether the spring was compressed or stretched. Whereas the curvature inside the spring and inside the device does

I don’t know. The math is beyond me and I don’t know of a reference. I already provided a reference that answers this question for spheres, and your statement is true for spheres. I cannot give a definitive answer for springs, but my intuition is that it is not true for all geometries

 

[pervect]

One of the big issues with generalizing the result to non-spherical geometries is that GR is nonlinear. This means that a spring plus a bar holding it is not quite the same solution as a the sum of a spring solution and a bar solution. So in the strong field, every case has to be analyzed separately, and you can't necessarily find the contribution of pieces of the system, because the system is not represented by the sum of its pieces.

If one introduces some additional assumptions of a weak field, one can get around this problem in principle. I have some guesses as to what hapens, but they're only guesses at this point.

So I'd rather stick with the spherical symmetric example. And I think it illustrates the main principles of what's going on quite well, I don't see the huge difference between it and the less-symmetrical and harder to analyze scenario you're asking about.

[add]
I'm not sure what the goal of the question is, if one is trying to come to grips with GR in the strong field realm, the nonlinearity of the theory is a really important feature, which means that one conceptually can't apply the technique of thinking about the spring separately from whatever is holding it stretched. One needs to analyze the entire system as a single, coherent, piece.

This may or may not be a big issue, depending on one's goals.

 

[timmdeeg]

Not really. The stress energy tensor has many components. So you cannot just talk about it increasing or decreasing. It is a more complicated object.

If you put it in compression then the volume will reduce, so the energy density component will increase both due to the smaller volume and the work. There will also be a separate stress component that will increase.

If you put it in tension then the volume will increase, so the energy density component will decrease due to the larger volume and increase due to the work. Which effect is bigger may depend on the properties of the material but the energy density will remain positive. There will also be a separate stress component that will decrease and may become negative.

I was perhaps naively thinking that stretching respectively compressing the spring changes mainly the shear stress components of the SET and the energy density and pressure components only marginally, because I expected the change of volume to be almost negligible.

Yes, but the source of gravity is the stress energy tensor, not the mass. Note, this assumes that the energy came from outside the system which is not always the case.

Given some stress from outside is put on an isolated mass (no bar etc. has to be considered) then as I understood it only the stress energy increases the invariant mass. Is this stress energy change represented by changes of the shear stress components of the SET? And is the change of the curvature due to changes of the shear stress components plus eventual changes of the energy density and pressure components?

 

[Dale]

I expected the change of volume to be almost negligible.

We are talking about relativistic effects on the gravity of a spring. The whole discussion is negligible. I personally would expect that the change in volume would be the largest of these negligible effects.

Given some stress from outside is put on an isolated mass

I don’t understand the question. If there is stress from outside, then how can it be isolated?

And is the change of the curvature due to changes of the shear stress components plus eventual changes of the energy density and pressure components?

Yes. The stress energy tensor includes components for shear stress, pressure, energy density, and momentum density. All contribute to gravity.

 

[timmdeeg]

I don’t understand the question. If there is stress from outside, then how can it be isolated?

You are right, it can't, I've missed this point.

Thanks for your answers.

 

[John K Clark]

The pressure inside a Neutron Star must be enormous and there is no tension factor to counteract it that I can see, so if the stress energy tensor includes components for shear stress, pressure, energy density, and momentum density and if I want to calculate the gravitational field far from the Neutron Star, does that mean knowing just the mass and spin of the star would be insufficient to make that calculation even if I assume the star is spherically symmetrical? To figure out the spacetime curvature would I need to know the details of the stars interior?

John K Clark

 

[Dale]

To figure out the spacetime curvature would I need to know the details of the stars interior?

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

 

[pervect]

The pressure inside a Neutron Star must be enormous and there is no tension factor to counteract it that I can see, so if the stress energy tensor includes components for shear stress, pressure, energy density, and momentum density and if I want to calculate the gravitational field far from the Neutron Star, does that mean knowing just the mass and spin of the star would be insufficient to make that calculation even if I assume the star is spherically symmetrical? To figure out the spacetime curvature would I need to know the details of the stars interior?

John K Clark

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).

 

[PeterDonis]

does that mean knowing just the mass and spin of the star would be insufficient to make that calculation even if I assume the star is spherically symmetrical?

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?

 

[Dale]

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?

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

 

[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

 

[PeterDonis]

Birkhoff's Theorem assumes (I think) that the gravitating object is static

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.

 

[Tomas Vencl]

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?

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 ?

 

[PeterDonis]

from distance observer view, the externally measured mass is constant,

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

this contribution to SET must be counterbalanced by decrease of other part of SET

What process are you talking about?

in my post 36

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).

 

[PeterDonis]

in my post 36

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.

 

[Tomas Vencl]

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.

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.

I was responding to post #50, which talks about a static neutron star).

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.

 

[Tomas Vencl]

... that dust collides (does not matter the details) and creates some static solid object ...

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 beginning 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.

 

[PeterDonis]

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.

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.

And the last part of my question in post 36 is , if after creation such a "star" the observer see any gravity field chances.

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.

 

[PeterDonis]

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.

 

[Tomas Vencl]

...

Thank you very much for detailed answer.

 

[Tomas Vencl]

Once again thanks for reply.
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.

 

[PeterDonis]

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 ?

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.

The pressure we must take into account in this case ?

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.

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 ?

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 canceled 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.

 

[Tomas Vencl]

Thank you. I think I understand.

 

[Dalton Peters]

I don't think so. When you pull on the spring, it stores energy in its bonds, increasing its mass and thus its gravity. If this stored energy is equal to the energy stored in the spring when compressed, then the two should have equal masses. I've never heard of tension causing a repulsion.

When a spring is compressed the stored energy increases its mass? sorry I haven't started under grad physics yet

 

[Drakkith]

When a spring is compressed the stored energy increases its mass? sorry I haven't started under grad physics yet

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.

 

[timmdeeg]

That's right.

With some caveats, see #41.