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Curvature without a mass ?

  1. Oct 17, 2006 #1
    Einsteins field equations are nonlinear. One could interpret this to mean that curvature is itself the source of curvature (thus not only mass). Would it be possible to find a stationary (non-zero) solution of the (non-linearised) field equations without a mass being present - a kind of spacetime deformation which has an existence independent of mass ?
    Has someone proven already that this can not exist ?
  2. jcsd
  3. Oct 17, 2006 #2
    There already is one. Energy as well as mass deform spacetime, not just mass. But no, as far as i know the deformations are not self feeding ie you dont get one spacetime warping causing another. i think.
  4. Oct 17, 2006 #3


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    Is it not the mass of the energy that causes the warping?

    NK, although it has been shown (or so I've read) that gravity itself contributes to its own overall strength (BTW, I find this a trully mind-blowing idea!), I also have never heard of any proposed mechanism by which a gravitational field can be its own cause. I think it sounds rather like being one's own Granpa, or a chicken laying the egg from which that same chicken was hatched. Nothing can be its own cause, all "things" must be caused by some other "thing", and AFAIK, gravity needs mass to cause it.

    However, this might be a way of describing Black Holes. The mass inside the event horizon is completely and forever (?) cut off from this universe, yet the BH itself continues to have an effect on outside objects through the mechanism of gravity. Could this be described as an instance where the mass of the central object of a gravitational field has "dissappeared" (being that it is no longer exactly "in" our universe), yet the mass of the gravitational field left behind is enough to perpetuate itself?

    That doesn't sound right to me, but I'm going to look around a bit and see if I can't find the exact flaw in it.
    Last edited: Oct 17, 2006
  5. Oct 17, 2006 #4
    Hi, Thanks for your comments. I'm not convinced by your chicken and egg argument though. In a propagating wave, one could say that each new cycle is the consequence of the previous one (many chickens and eggs !). In fact, I suspect there exists some kind of soliton like solution to the wave equations , but then of a stationary type. Maybe one could think of it like an eternal vortex in spacetime (no friction or energy dissipation). Maybe that a skilled mathematicion could find such solution. He/she would be famous from day one !!
  6. Oct 17, 2006 #5
    Well, the Schwarzschild solution has no stress-energy term.
  7. Oct 17, 2006 #6


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    The stress-energy tensor (T_uv) is the right hand side of Einstein's field equations, G_uv = 8 pi T_uv, i.e. the stress energy tensor causes gravity, not mass.

    I'm not quite sure what sort of solution you are looking for, but it is worth noting that gravitational radiation is nothing but "ripples" of curvature.

    Thus in an exact solution of Einstein's field equations, gravitational radiation does not appear on the right hand side of Einstein's equations aas a source term.

    However, gravitational radiation is a form of energy, so one would expect it to have some effect on space-time curvature.

    The resolution of this issue is that gravitational radiation does not have an actual stress-energy tensor, but that the non-linear terms give it an "effective" stress-energy tensor, if instead of taking the exact solution of Einstein's equations you ignore the "ripples" and look for the value of overall curvature, ignoring the ripples.
  8. Oct 17, 2006 #7
    Indeed, gravitational energy does not have an actual stress-energy tensor, and the non-linear terms indeed behave as an "effective" stress-energy tensor and the energy in the gravitational field can not be localised (though it is non-zero) but that does not mean that we can say with mathematical certainty that such a massless and non-propagating curvature solution can not exist. As long as I haven't seen the proof, I will not be convinced.
  9. Oct 17, 2006 #8
    Sorry if I introduce a question here. Does it mean that both signs of curvature must exist, and so, both signs of mass as well?
  10. Oct 17, 2006 #9


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    In the standard form of Einstein's field equation:
    (why doesn't my Tex work here?)

    the gravitational 'energy' does not appear, only that of non-gravitational mass, stress, momentum and energy, which appears on the right hand side of the equation.

    All that appears on the left hand side of the equation is the curvature of space-time, which has therefore to include any notion of gravitational energy such as might be transported by gravitational waves. This is one reason why GR is said to be an 'improper energy theorem'.

    This is a non-linear equation in which perturbations in space-time might well be "self feeding".

  11. Oct 17, 2006 #10
    This question might not even make any sense. "Curvature" is just another name for field strength. You're asking if it's possible to have a field without a source. As far as I know that's a matter of definition. You might as well ask for sentences without words.
  12. Oct 17, 2006 #11


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    I'm not saying that "such a solution" doesn't exist either, but mainly because I'm not really clear on exactly what it is you are looking for.

    If you define mass, energy, and momentum as the ADM mass, energy and momentum, we should be able to apply the rules of SR to say that any massless solution must be moving at 'c', for the ADM energy and momentum form a 4-vector with the ADM mass as the invariant of that 4-vector, just as mass of a point particle in SR in the invariant of the energy-momentum 4-vector of that particle.
  13. Oct 17, 2006 #12
    It is not clear to me what you mean with "ADM". However, though the solution I'm looking for has no barionic mass, it does not need to be massless since the gravitational field contains energy (and thus mass). Therefore, the self-feeding-curvature (or "solution") is not massless and can therefore not move at the speed of light.
  14. Oct 17, 2006 #13


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    ADM mass, energy, and momentum are one of the standard ways to measure the mass, energy, and momentum of a composite system in standard General relativity. Certain conditions have to be met, which I'll go into later.

    The name is taken from the intials of Arowitt, Deser and Misner, who first published it as a notion of the total energy and momentum contained in a spatial hyperslice, computed by means of a Hamiltonian formulation. You can find the details in Wald's "General Relativity", and probably a lot of other GR textbooks.

    The concept is applicable to any asymptotically flat space-time. Asymptotic flatness of a space-time is an automatic consequence if you have a source region of space-time with a non-zero stress-energy tensor surrounded by an infinte vacuum region. (Note that our universe doesn't qualify as asymptotically flat, at least in the standard cosmological models such as the FRW (Friedman, Walker, Robertson) models.)

    You can think of the ADM mass, energy, and momentum as arising from Noether's theorem, arising from space and time translations "at infinity", more precisely at "spatial infinity", in the region where space-time is "almost" flat. Space translations give the conserved total momentum via Noether's theorem, time translations give the conserved total energy. The result transforms like a 4-vector. This means that objects with zero ADM mass must move at 'c' as a whole, as measured in the nearly Minkowskian nearly flat region "at infinity".
  15. Oct 18, 2006 #14
    Thanks for this clarification. I found it back now in Gravitation by Misner, Thorne and Wheeler.
    The solution I'm looking for corresponds to a rotating "disk" of spacetime. So, it does not have spherical symmetry, but it has axial symmetry (and a plane of symmetry). A very rough indication of how this can exist is the following: The gravitational field contains energy and should therefore have a tendency to contract itself (or collapse). If it has angular momentum one could have a kind of centrifugal force which balances exactly this contraction tendency such that one obtains a stable solution.
    I know that this very rough and uses an analogon from classical physics which is not valid here. I just mention it to make it a little bit more plausible to the sceptics.
    I am convinced that such a solution exists but I realise that I have not the mathematical skills to find it. It would be wonderfull if someone in this forum could find it. Maybe it requires a genius like Schwarzschild to find it.
  16. Oct 19, 2006 #15


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    The only thing I've seen which might be helpful is a weak-field analysis of a ring laser by Mallett.

    http://www.physics.uconn.edu/~mallett/Mallett2000.pdf [Broken]

    Mallett has a strong field analysis of a cylinder of light, but it has been criticised in the literature and I personally agree with the criticisms.


    I'm not aware of any problems with the Mallett's weak-field analysis, though I have some concerns about the beam profiles

    Finding a beam profile solution that obeys Maxwell's equations is trickier than it seems - see some of the discussion about Gaussian beams on PF, there was another beam profile that looked promising (think it was a Bessell or non-diffractive beam profile, I'm not sure anymore).

    I'm not sure what the equivalent of satisfying Maxwell's equations is for gravity wave beams offhand.

    You can probably sidestep some of the beam profile issues if you imagine that the field is being created by massive particles moving very close to c so that their invariant mass * c^2 is much less than their energy, but this gets you a step away from your original idea.
    Last edited by a moderator: May 2, 2017
  17. Oct 20, 2006 #16
    Thanks Pervect for these interesting links. Still, it would be nice if someone at least could give it a try (original idea) ...
    Last edited by a moderator: May 2, 2017
  18. Oct 20, 2006 #17


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    Okay, try the Einstein-Rosen Bridge, otherwise known as a Lorentzian wormhole.

    Such a wormhole might be formed inside a rotating, or Kerr, Black Hole.

    Imagine a spinning collapsing massive star core after the star itself had gone super-nova. About 3 or more solar masses dissappear into the ring singularity at its centre. You try to follow the mass inside the (now double) event horizon and instead of coming up against the massive singularity you reappear in another part of the universe, or in a different universe! Where has the mass gone?

    Although such a bridge is unstable it may be stabilised by a sphere of large amount on negative pressure exotic matter (DE??). You would then have a black hole in this universe and one in possibly another universe but the mass is nowhere to be seen! :smile:

  19. Oct 20, 2006 #18


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    The picture I get is more of the photon sphere around a black hole. (This isn't a stable orbit, unfortunately).

    Now we get rid of the black hole, and replace it with only a ring of light. Can we use the gravity field of the ring of light itself to eliminate the black hole?

    Light beams going in opposite directions attract. (I've got a reference for this somewhere). So the idea is to use the gravitational interaction of the circulating light itself to keep the light in its circular orbit.

    My intiution says that it wouldn't be stable, and that the light would have to have to be a perfect ring with a zero area delta-function cross section. If you give the light a finite cross section, I don't think it will work.

    Working it out in terms of a metric is beyond me - working it out in the linearized approximation in more detail would be a challenge, but with the Mallett paper as a guide I think it would be possible, but quite time consuming and very involved. (It wouldn't be terribly involved if one trusted Mallett implicitly, but if one really wanted to understand every step and check for errors it would take considerable effort.)

    The next step is to replace the light with a gravity wave.

    A simplification in the opposite direciton would be to replace the light / gravity waves with a beam of particles of non-zero rest mass, but ultra-relativistic in energy. One can then understand the gravitational interaction a bit more intuitively. Gravitomagnetic effects will be quite significant. Gravitomagnetism explains why two parallel light beams attract each other if they are going in opposite directions, and don't attract each other if they are going in the same direction.
    Last edited: Oct 20, 2006
  20. Oct 25, 2006 #19
    But are there any stable orbits in GR?
    I thought that was not the case.
    Last edited: Oct 25, 2006
  21. Oct 25, 2006 #20


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    Sure there are. Take a look at http://www.fourmilab.ch/gravitation/orbits/ for instance, look for the term "stability". It's easiet to talk about this in the context of circular orbits, which don't have to worry about the precession issues that elliptical orbits have.

    Circular orbits are stable when there is a minimum in the effective potential.

    Try for instance, setting L=3.47 in the simulator. You'll see a green line at the minimum of the effective potential. This indicates a stable circular orbit. Click on the effective potential diagram near the green line -you'll launch the simulator generating the stable orbit (it will be exactly circular only if you click right on the green line).

    Now reduce L to 3.46. There will no longer be any minimum of the effective potential, thus there are no stable circular orbits. If you read the detailed derivation you'll see that L=sqrt(12) is the critical point which generated this transistion.
  22. Oct 26, 2006 #21
    I'm pretty sure there is a soln to Einstein's field eqns for a massless universe which contains only energy. In fact, this is what BB cosmology is based upon. Alan Guth, the inventor of the Rapid Inflation Theory, and the Everything From Nothing theory, theorizes that quantum gravity flutuations gave rise to the BB event. That these fluctuations are normal in the preBang realm. If they superimpose just right, a false vacuum forms which becomes unstable and inflates creating space & time, then cooling allows for particle crystalization and matter-based gravity wells.

    The quantum gravity fluctuations in the preBang realm are just that ... gravity-like variations within the medium. So energy exists in the absence of rest mass. However, with no material entity in existence, location in space & time have little meaning (no distict reference) far as how we percieve it in the postBang. Yet, gravity-like fluctuations are CHANGE, and that assumes some form of time and space.

    That said, I have heard that Einstein's GR has a soln for a massless universe, ie energy alone. This is consistent with Guth's false vacuum theory.

  23. Oct 26, 2006 #22
    So when is an orbit stable and when not in GR?
    I thought that an object in orbit is unstable and spirals in because of the loss of energy due to gravitational waves. Are there orbits for which this is not the case?
  24. Oct 26, 2006 #23


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    The defintion of stability that I'm using (and that the webpage I cited is using) is similar to the one used to define the stability (or lack therof) of the Lagrange points, regarding the behavior of the orbits when small pertubations are made in the inital conditions.

    The analysis on the webpage does not include gravitational radiation. However I don't think the situation of a rotating perfect ring has any quadropole moment. And quadropole moment is required to have a source of gravitational radiation. So I think gravitational radiation would be highly surpressed from a rotating ring. It would be a very tiny effect in any event - and it would tend to be supressed even further by the symmetry of the system. Small imperfections in the ring would still radiate, so there would be some gravitational radiation, but it would be very small (ideally zero, one would need information about the imperfections to determine more).

    Circular orbits make the analysis of stability easy, because one can plot the behavior of r vs t in one dimension as the only interesting part of the behavior.

    Unstable orbits act like a pin balanced on its point. They depart very quickly away from the equilibrium point. So the equilibrium point is an exact solution of the orbital equations for an unstable orbit, but if you depart even slightly from the correct initial conditions, your object will quickly spiral outwards or inwards, with some exponential time constant.

    You can see this behavior on the simulator. You can also see this from examining the graph of effective potential. If the object is in a valley, it has the same equations of motion for r vs time that a ball trapped in a valley has. The ball is trapped by the valley walls, and the orbit (and the position of the ball) are said to be stable.

    If the ball is on the top of a hill, though, it is unstable. The slightest push sends the ball off to one side or the other. The same thing is true for the orbit. A small error in initial conditions will send the object (or, in the original case, light beam) either out to infinity, or spiraling into the black hole. It takes some sort of active control system to stabilize an unstable system.

    Formally, stability (in the sense I'm using it) is determined by finding the eignevalues of the linearized system. If those eignevalues all have positve real parts, the exponential response of the system is damped, and the system is stable. If those eignevalues have negative real parts, small errors in initial conditions grow exponentially, leading to active instability. Zero eigenvalues represent marginal stability. In terms of a "ball on a surface" analysis, a zero real part describes a ball on a flat plane. It doesn't "fall off" the top of a hill, but neither is there any restoring force as there would be if the ball was in a valley.
  25. Oct 26, 2006 #24
    So I take it between all the flowery language that the answer is yes orbits in GR are not stable, while the effect is very smalll, orbits spiral towards the center.

    How is it goinig to help anyone understand GR if we cannot get a simple yes/no answer on this.

    I really do not see how flowery language is going to help here.
    So what if you use another definition of stability or that the given website does not handle this, or that the effect is small anyway?

    The fact that orbits are not stable in GR is profound and interesting for anyone to understand GR.
  26. Oct 26, 2006 #25
    Well, what if there are no quadropole moments?
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