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I Gravity changes produced by a compressed and stretched spring

  1. Oct 17, 2017 #1
    If I compress a spring beyond its most relaxed point it will have more energy and thus by General Relativity produce a stronger gravitational field than the uncompressed spring will. But suppose instead I pull in the opposite direction and stretch the spring beyond its most relaxed point it is my understanding that it would produce a weaker gravitational field than the compressed spring even if they both had the same amount of energy because tension, unlike pressure, produces a gravitational repulsion that would cancel out some of the attraction made by the mass/energy. Am I correct?

    John K Clark
     
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  3. Oct 17, 2017 #2

    Drakkith

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    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.
     
  4. Oct 17, 2017 #3

    timmdeeg

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    I think if the work done on the spring is the same in both cases then there shouldn't be any difference. The increase of internal energy of the spring is the same.
     
  5. Oct 17, 2017 #4

    PeterDonis

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    No.

    The bit about "tension" producing gravitational repulsion is not a general statement in GR; it's a very specific statement about a very specific type of solution, namely that of an isotropic perfect fluid. A spring is not even close to being an isotropic perfect fluid, and the "tension" you produce in a spring when you stretch it is not the kind of "tension" to which that specific statement applies.

    The correct intuition here is the one that @Drakkith and @timmdeeg have expressed: when you do work on the spring, either by stretching or compressing it, you increase its total stress-energy and therefore increase the spacetime curvature it produces.
     
  6. Oct 17, 2017 #5

    Dale

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    I think that @John K Clark is not too far off. There is a paper I remember reading that discussed a container with a perfect fluid. In such a case the container walls can be in tension in a way that "cancels out" the gravity due to the pressure in the fluid. I will try to find the paper
     
  7. Oct 17, 2017 #6

    PeterDonis

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    Hm, yes, good point; the container walls in this case are not perfect fluids, because the stress in them is not isotropic.

    However, there's still a key difference from the spring case, because in the spring case work is being done on the system to stretch it. That work should still increase the total stress-energy of the spring, and therefore the spacetime curvature it produces. You can't do work on a system and have it produce weaker gravity afterwards: that would violate local conservation of stress-energy.

    I have not tried to model this in more detail mathematically, but I think that might be what would have to be done to resolve the apparent tension (pun intended) between the two statements ("tension cancels out pressure" vs. "doing work adds energy").
     
  8. Oct 17, 2017 #7
    But if the tension wasn't isotropic wouldn't that mean the repulsive gravitational field would also not be isotropic not that the field wouldn't exist? It seems to me that if everything had to be perfect General Relativity would never be applicable in real would situations.

    John K Clark
     
  9. Oct 17, 2017 #8

    Dale

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    Yes, I agree. I think the most you can do is move the gravitational source around a bit. My guess is that if a (static) spring is in tension then something else must be in compression and those two effects cancel.
     
  10. Oct 17, 2017 #9

    Dale

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    Here is the paper I was remembering

    https://arxiv.org/abs/gr-qc/0510041

    Things get complicated...
     
  11. Oct 17, 2017 #10

    PeterDonis

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    No. There might be a small negative contribution from tension to the overall gravitational field, but it won't be anywhere near enough to outweigh the other positive contributions. As has already been said, the overall contribution of doing work on the spring has to be positive as far as the gravitational field is concerned; you can't do work on something and have its gravity become weaker.

    Remember that, for ordinary objects, stresses are something like 20 orders of magnitude smaller than rest energy density, which of course also acts as a source of gravity (and a positive one). So we're talking about miniscule corrections in any case, way below our ability to actually observe.

    Nobody is saying GR is "perfect".
     
  12. Oct 18, 2017 #11
    I know if you put energy into a spring its always going to increase its gravitational field, but I think it matters if that energy is in the form of compression or tension (negative pressure), although the difference would usually be very very small. The reason I got interested in this is I was reading "Black Holes and Time Warps" by Kip Thorne (who won the Nobel Prize for physics just a few days ago) and he indicates that matter/energy density is not the only thing that can make gravity, on page 118 Thorne says to my surprise:

    "Mass and pressure warp spacetime... the curvature near a particle is proportional to the density of the mass in the vicinity of the particle (multiplied by the speed of light squared to convert it into the density of energy) plus 3 times the pressure of matter near the particles vicinity". [...] Under most circumstances the pressure of matter is tiny compared to the mass density times the speed of light squared and therefore the pressure is a unimportant contributor to spacetime curvature. Only deep inside Neutron Stars and a few other exotic places is pressure a significant contributor to the warpage." _Kip Thorne

    So I know it will be small but I just want to be sure I understand what Thorne is saying hence my question: Will 2 identical springs with identical amounts of stored energy in them produce identical gravitational fields if one spring is under pressure and the other in tension? In my interpretation to what Thorne is saying the answer seems to be no. Am I missing something?

    John K Clark
     
  13. Oct 18, 2017 #12

    PeterDonis

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    That's correct. The source of gravity is the stress-energy tensor, which includes contributions from pressure and other stresses as well as energy density.

    Thorne's specific statement, that the curvature is proportional to the density of energy plus 3 times the pressure, assumes a perfect fluid, i.e., the pressure is positive and the same in all directions. Also, when he says "curvature", he means the spacetime curvature inside the object, at some particular point where we are measuring the energy density and the pressure. That is not the same as the external gravitational field produced by the object.

    It depends on what "identical amounts of stored energy" means. Ordinary language is vague. If it means "identical amounts of work done on the spring to get it from its unstressed to its stressed state", then I think the answer has to be yes, their gravitational fields will be identical. But if "identical amounts of stored energy" means something else, the answer might not be yes.
     
  14. Oct 18, 2017 #13
    You say "identical amounts of work done" but work is just a specific sort of energy and as far as I know heat or any other sort of energy can produce a gravitational field so I don't see why there would be any ambiguity in the term "identical amounts of stored energy". Also if pressure can change the spacetime curvature inside an object I don't understand why that change would suddenly stop at the object's boundary unless there were some sort of spacetime discontinuity there and I thought that only happened at the center point of a Black Hole.

    John K Clark
     
  15. Oct 18, 2017 #14

    PeterDonis

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    Because "energy", depending on what you mean by it, can be frame-dependent. But "work done" is an invariant. Mathematically, "work done" must be expressible as a scalar invariant, but "energy" can be just one component of a tensor. They're not necessarily the same thing, because ordinary language is vague. If you told me exactly what you meant by "identical amounts of stored energy" in terms of math, or even in terms of some more precise ordinary language term like "work done" that must be invariant, then there would be no ambiguity. But you didn't do that.

    You're misunderstanding what I was saying. The Einstein Field Equation, which is what Thorne was describing in his statement that you quoted, is local: it tells you the relationship between the stress-energy tensor and spacetime curvature at a particular point. It doesn't directly tell you the relationship between, say, the energy density and pressure inside a neutron star and the gravitational field observed outside the neutron star. To obtain such a relationship, you need to develop a global solution of the Einstein Field Equation, i.e., a description of stress-energy and spacetime curvature throughout some extended region of spacetime, such that, at each point within that region, the relationship between stress-energy and spacetime curvature at that point satisfies the EFE.

    Saying that "pressure can change the spacetime curvature" means that pressure at a particular point in spacetime affects the spacetime curvature at that point in spacetime. That doesn't necessarily mean that, when you develop a global solution, the pressure inside the object will appear in the description of the gravitational field outside the object. As a matter of fact, it can be shown, by an analysis similar to the one given in the paper @Dale linked to, that for the case of a static, spherically symmetric object, the pressure inside the object does not appear in the description of the gravitational field outside the object, i.e., in the externally measured mass of the object. Heuristically, this is because the positive contribution from pressure is exactly cancelled, in equilibrium, by a negative contribution from gravitational potential energy, similar to the way in which the positive contribution from pressure in a gas inside a container is exactly cancelled, in equilibrium, by the negative contribution from tension in the container walls. In both cases, it's not that pressure somehow stops contributing; it's just that its contribution gets cancelled by something else.

    The bottom line is that you can't just say" pressure contributes to gravity" and expect that to give you an answer for a specific case; you have to actually work out, in a specific case, a solution of the EFE that describes that case, and then see what it tells you about the various contributions to spacetime curvature/gravity. And none of the solutions we have discussed so far in this thread apply to the case of a stretched or compressed spring.
     
  16. Oct 18, 2017 #15

    PeterDonis

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    This is quite possible; a common mistake in analyzing situations like this is to leave out a source of gravity, or to put it another way, ignoring the fact that automatic conservation of the source (stress-energy) is built into the Einstein Field Equation. So you can't just say "we'll stretch the spring" in isolation: the energy that produces the work done to stretch the spring has to come from somewhere, and has to be counted in the complete solution of the EFE. So it might, in the end, turn out that it's impossible to model an isolated stretched or compressed spring in GR: you have to model the spring plus whatever it is that stretches or compresses it, and that system as a whole will have an unchanged external gravitational field through the process.
     
  17. Oct 18, 2017 #16

    PAllen

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    An interesting way to motivate the point in #15 is to consider a simple complete system. Imagine a spring connected to the ends of ratchet, with little rocket engines attached to each end of the ratchet. Ignite the engines. The string stretches, the ratchet prevents the string from relaxing after the fuel is consumed. The spring is under tension, while the ratchet is under compression, after equilibrium.
     
  18. Oct 18, 2017 #17

    PeterDonis

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    As an extension to this, one could consider a second complete system in which the two rockets push the spring into compression, with some ratchet-like mechanism to hold it compressed after the rockets' fuel is expended. Then we could impose the condition that the rockets do the same amount of work on the spring in both cases, and compare the two final states.
     
  19. Oct 18, 2017 #18

    Dale

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    As @PeterDonis mentioned above, it definitely does matter locally. You have multiple effects going on locally.

    The biggest effect is that by placing an object in tension or compression you are changing its shape and density. That primarily effects the "energy density" term of the stress energy tensor. The shape must change, even if the material is perfectly incompressible, and real materials are compressible.

    The second, much smaller effect, is the energy density from the external work. This is positive regardless of if the work is done to place the object in compression or tension. It is a small increase to the "energy density" term of the stress energy tensor compared to what you would expect just from the mass density alone.

    The third effect, also much smaller than the first, will be the stress itself. This can be positive (compression) or negative (tension), but it is a completely different part of the stress energy tensor than the energy density term. Essentially, it is a different direction in spacetime. Because it is a different direction it is not generally possible to think of it as just increasing or decreasing gravity. It is a change to a tensor, not a scalar.

    That is precisely the topic studied in the paper I linked to. It is complicated, but if you go through the math that is what you get in certain cases (spherical symmetry and other assumptions described in the paper). Your spring is a more complicated case even than what the paper describes.
     
  20. Oct 18, 2017 #19
    Let's say light bounces between mirrors that hang vertically in a gravity field. Mirrors are charged, attractive Coulomb force keeps them at constant distance from each other.

    How much does the light attract the planet which is the source of the gravity? Well there is some 'extra' attraction between the planet and the light whenever there is some extra bending of light. There is 'extra' bending of light when light can reach its destination faster by bending an extra amount, and that is the situation when the gravity field is non-uniform and the distance between mirrors is so large that the non-uniformity matters.

    I hope I got the 'extra' bending of light correct, I mean the effect where light passes a star and accelerates 200m/s2 where normal, slow speed, free falling matter accelerates 100 m/s2.

    So my point is that there is no 'extra' gravity of pressurized photon gas, when the gas can not detect tidal forces.
     
  21. Oct 18, 2017 #20

    pervect

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    A stumbling block here is that an isolated, static spring in tension or compression can't exist - something needs to hold the ends still. Or the spring needs to be moving. Both present complications.

    In the linearized limit, it can be shown that the net effect of tension and compression on an isolated system balances out. So in the linearized system, there won't be any net gain or loss of mass due to tension or compression, it will just "move it around". And "mass" is a concept that's really only applicable to an isolated system, at least with our current conceptions of mass in General relativity - those concepts don't really (and can't as far as I know) localize "mass" at all, they can only give us the total mass, not assign it a definite location.

    Something similar happens in the simple non-linear case we can analyze, a pressurized sphere. One can argue that a spherical pressure vessel will, by symmetry, not produce any "force" on an object inside the sphere, so the contribution of the pressure vessel to the reading of the proper acceleration of a static observer inside the pressure vessel will be zero.

    Then one changes the pressure inside the pressure vessel by heating up the contents of the sphere. Rather than bring in energy from outside, it's cleaner to have the energy source needed to change the pressure already inside the sphere. We can imagine the pressure containing a bomb. To get detectable effects, I'd suggest a nuclear bomb. Or something better, if available, like a total conversion matter-antimatter bomb. Then the non-pressurized sphere containing the bomb will become a pressurized when we explode the bomb, without importing any external energy. As long as we make it strong enough that the pressure vessel survives the explosion, of course. (That's rather unlikely with any known material, considering the sorts of pressures we need to get measurable effects).

    We can then observe the effects on an accelerometer mounted just inside and just outside the surface of the pressure vessel, one that measures the proper acceleration of a static observer. I haven't seen this analyzed in any text, but working it out for myself. I convinced myself that the reading on the external accelerometer does not change when one explodes the bomb, but the reading on the internal accelerometer (just inside the wall of the pressure vessel) does change. (It goes up, the pressure causes more gravity, meaning the accelerometer has a higher reading). This then is a way of demonstrating via a thought experiment how pressure (and tension) affects gravity. If pressure had no effect on gravity, the reading on the interior accelerometer wouldn't change when one exploded the bomb. But even though pressure affects gravity, conservation laws imply that the reading on the external accelerometer doesn't change.
     
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