Gravity changes produced by a compressed and stretched spring

[John K Clark]

Sorry to keep bugging people but I'm still not clear on how pressure and tension fit into General Relativity. It occurs to me that 2 springs with equal energy , a compressed one under pressure and a stretched one under tension (negative pressure) could be both kept static in a arbitrarily light scissor like device and be arbitrarily far apart too, so if the pressure term is positive in one and negative in the other I don't understand how the field equations could say that both springs warp spacetime in identical ways.

John K Clark

 

[PeterDonis]

It occurs to me that 2 springs with equal energy , a compressed one under pressure and a stretched one under tension (negative pressure) could be both kept static in a arbitrarily light scissor like device and be arbitrarily far apart too

No matter how "light" the scissor like device is, or how far apart its arms are, it still has to have stress in it that compensates for the stress in the spring, in order to keep the system as a whole static. There is no way around that; and that is what resolves your question--that you can't just consider the stress in the spring in isolation, you have to consider all of the stresses in the system.

if the pressure term is positive in one and negative in the other I don't understand how the field equations could say that both springs warp spacetime in identical ways

The field equations don't say the springs, in isolation, warp spacetime in identical ways; they say that the static systems, as a whole, warp spacetime in identical ways, because the systems as a whole must include not just the stresses in the spring, but the stresses in whatever it is that is holding the spring static. Those two stresses will be opposite (if the spring is stretched, whatever is holding it will be compressed, and vice versa), so the overall effect is the same either way.

 

[John K Clark]

I understand that far away the effect of tension and pressure would cancel out, but suppose the two springs are far apart and there are 2 observers, one very close to the spring under pressure and the other very close to the spring under tension, would they observe identical spacetime curvature? Are there any circumstances where it would be important to know if the energy in the spring is in the form of pressure or tension?

John K Clark

 

[PeterDonis]

suppose the two springs are far apart

What's holding a particular spring static can't be "far apart" from the spring. So there is no way to just consider a spring under pressure, or a spring under tension, in isolation, if they are static. There has to be something else present that is holding the spring static, that can't be ignored even by an observer close to the spring.

Are there any circumstances where it would be important to know if the energy in the spring is in the form of pressure or tension?

A hypothetical observation made inside one of the springs (i.e., not just close to it but within its internal structure) would obviously see a difference, depending on whether the spring was under pressure or under tension. But under those circumstances you're not talking about the total mass or gravitational field of a static system viewed from the outside.

 

[Dale]

suppose the two springs are far apart and there are 2 observers, one very close to the spring under pressure and the other very close to the spring under tension, would they observe identical spacetime curvature?

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

In the spherically symmetric case you must measure the curvature inside the system to detect any difference whatsoever. Without the spherical symmetry you probably get some differences very nearby the springs. But the math is very complicated so that is just a guess on my part.

Without doing the math I cannot do more than make a guess about the fields. 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.

 

[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.