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General relativity and invariant mass

  1. Nov 17, 2013 #1
    The invariant mass of special relativity:

    [itex]m_0{^2} = E^2 – p^2[/itex]

    There doesn't seem to be any quantity with units of mass that is invariant in general relativity. Invariant mass loses significance, as other than an approximation where space-time is sufficient flat.

    But at the same time, mass is a fundamental unit (or seems to be). As well, it is difficult to see how local propagations at other c can be expressed without something to slow things down. As far as I know, without adding mass, ad hoc, field phase propagates locally at c.

    Is there a quantity that replaces invariant mass in the general theory?
     
    Last edited: Nov 17, 2013
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  3. Nov 17, 2013 #2

    pervect

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    There are at least three different notions of mass in General relativity, but they don't work in a general geometry, they impose requirments on the metric to be defined.

    The three most common are ADM mass, Bondi mass, and Komar mass.
     
  4. Nov 17, 2013 #3

    Bill_K

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    Rena,

    Things like the Bondi mass are only required in the most general case, e.g. with gravitational radiation present. For a test particle in GR, the concept of mass is quite simple, and in fact it's the same as it is in Special Relativity.

    The path of a particle through spacetime is its worldline, a curve whose tangent vector is everywhere timelike and future pointing. The unit tangent vector is its four-velocity, v = (γ, γv), with v·v = -1. The particle has a four-momentum p parallel to v, and p = m v = (γm, γv) = (E, p), where m is the particle's mass, a scalar quantity. The norm of the four-momentum is p·p = - m2, and the relationship m2 = E2 - p2 still holds.
     
  5. Nov 17, 2013 #4

    bcrowell

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    In the OP's terminology, these are really measures of E, energy, not m0, invariant mass of a particle.
     
  6. Nov 17, 2013 #5

    WannabeNewton

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    Where in the world did you get this idea from? Remember that all SR equations hold locally in GR so when we are dealing with a test particle the invariant mass formula ##g(p,p) = -m^2## is definitely valid along the worldline of the particle. You're confusing the mass of an asymptotically flat space-time with the mass of a single test particle.
     
  7. Nov 17, 2013 #6

    pervect

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    I'm not so sure if this is true.

    My interpretation of the fine points of the language usage here is that it's common to say in GR, as in SR, that the energy of a system that is (loosely speaking) "at rest" is equal to its mass multiplied by c^2. And that usually when people are interested in the energy of a system "at rest" they call this energy a mass. Furthermore, if they aren't using geometric units already (in which c=1 and the following point becomes irrelevant), they scale the energy by the appropriate factor of c^2 to turn it into a mass. I.e. when people talk about masses, and aren't using geometric units, they use "mass units" like gm, kg, or solar masses, to describe "masses".

    A study of Wald's language usage (for instance, he refers to ADM Energy, ADM Energy-momentum, and ADM mass around pg 293) seems consistent with my interpretation.

    It's possible the OP really wanted to ask something else. I thought this was at least close to what he was looking for.

    My basic position is that the "invariant mass" of a particle in SR (also, the "total mass of an isolated system" in SR), and the "total mass of a system" of GR are essentially the same concept, which is the "energy at rest".

    The problem in GR comes in is in finding a way to do the totalize operation. The distribution of "mass" - more precisely, the distribution or density of energy and momentum - is well defined by the Stress Energy tensor in GR. Finding a way to reduce this distribution to a single number, to add up the "total rest energy" of a system, turns out to be a big challenge. The details of the notion of "at rest" is also fairly complex and something I've glossed over in the interests of intellgibility and avoiding digression.
     
  8. Nov 17, 2013 #7

    WannabeNewton

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    For a massive test particle, the notion of an instantaneous rest frame always exists and so the above interpretation holds strong. If we are given an asymptotically flat stationary space-time in GR (in essence a stationary isolated system) then a notion of "being at rest with respect to the space-time" exists in a sense and this is the notion of following an orbit of the time-like killing field generated by the time translation symmetry. What if the space-time is not stationary? The energy of the space-time can still be defined (Bondi energy) but what notion is there of "being at rest with respect to the space-time" in the non-stationary case?
     
  9. Nov 17, 2013 #8

    PAllen

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    I agree with the idea that totalling parts is where there get to be complexities in GR compared to SR. To the extent you can say self gravitation is insignificant, and your system is sufficiently isolated from everything else (so you can claim to put your system into empty AF spacetime), you can find that SR computed invariant mass ≈ Bondi mass. However, even for a static mass collection in AF spacetime, when self gravitation is significant, you at least need Komar mass to account (effectively) for the negative contribution of gravitational potential energy. If you give up staticity, then you need Bondi mass.
     
    Last edited: Nov 17, 2013
  10. Nov 17, 2013 #9

    pervect

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    In the ADM case, we can insist that the coordinate components of the total spatial ADM momentum (as defined by Wald, pg 293, 11-2.15) be zero, as our notion of being "at rest".

    I haven't seen an equivalent expression for the Bondi momentum written out, but I rather suspect there should be one.

    See Wald pg 283-285 - the symmetry group at asymptotic infinity is the BMS group of super-translations. According to Wald "There is a unique 4 dimensional subgroup of the supertranslations which is a normal subgroup of the BMS group" and "In the case of Minkowskii spacetime, this four-dimensional subgroup of the BMS group consists precisely of the asymptotic symmetries associated with the exact translational symmetries of Minkowskii spacetime".

    Which leads Wald to define asymptotic translations "at infinity" as the members of this particular subgroup.

    By Noether's theorem, the asymptotic time translation symmetry gives the Bondi energy - while I haven't seen much discussion of the asymptotic space translation symmetries, I think they should define a Bondi momentum. Then "at rest" would be "zero momentum", as in the ADM case.
     
  11. Nov 17, 2013 #10

    PAllen

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    ADM energy includes emitted radiation out to infinity. It is most useful to talk about global conservation in appropriate special cases. Bondi energy allows computation of the mass without radiation, i.e. what's left behind by the radiation.
     
  12. Nov 18, 2013 #11
    From the conserved quantities,

    ##E = m_0(1-2GM/r) (dt/d\tau)##

    and

    ##L = m_0 r^2 (d\phi/d\tau)##

    where E is the total energy and L is the angular momentum of a test particle in the Schwarzschild metric, together with the equations of motion it is possible to work out that:

    ##m_0{^2}= E^2 - m_0{^2} (dr/d\tau)^2 - (1-2GM/r)L^2/r^2 + 2GM m_0{^2}/r##.

    If I define the radial velocity of the in-spiralling and free falling test particle ##dr/d\tau## as v aad the radial momentum as ##p=m_0v##, then the above equation can be written as:

    ##m_0{^2} = E^2 - p^2 - (1-2GM/r)L^2/r^2 + 2GM m_0{^2}/r##.

    It can be seen that the expression for the invariant rest mass in GR is similar to the expression in SR, but with an additional terms for contributions from angular momentum energy and something like gravitational potential energy. (In fact I suspect a full general definition of rest mass in SR would include terms for angular momentum and (non gravitational) potential energy too.) L and E are constants so these quantities remain unchanged as the test particle falls. If the initial velocity and angular momentum of the test mass are zero at infinity, the GR rest mass is equal to the total energy so that ##L=0## and ##m_0 =E## the above equation collapses to:

    ##m_0{^2} - E^2 = 0 = -m_0{^2} v^2 + 2GM m_0{^2}/r##.

    Dividing through by ##2m_0## gives:

    ##\frac{1}{2}m_0v^2 = \frac{GMm_0}{r}##

    which is the suggestive of the relationship between kinetic energy and gravitational potential energy in Newtonian physics, except that the velocity is expressed in terms of proper time and Newtonian physics does not make a distinction between proper time and coordinate time.
     
    Last edited: Nov 18, 2013
  13. Nov 26, 2013 #12
    Thank you all for your responses. And I did misread something.

    However, I think my question, however foolishly initiated, is still of interest.

    I short review, most think in terms of particles, where ipso facto, an invariant mass is impressed on the 4-velocity. From my (rather renegade) point of view classical extended particles should be understood in terms of fields and there are no quantum particles (Consider this attitude an affliction, if you like.) so I look for invariants in space time fields.



    I'm not so sure if what is said of the invariant mass of classical particles, can be said for classical fields.
    Any ideas on invariant mass of fields would be appreciated.

    My apologies. I will certainly read all above posts in detail soon.
     
    Last edited: Nov 26, 2013
  14. Nov 26, 2013 #13

    WannabeNewton

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    Thanks pervect, I took a look at the pages in Wald you referred to. By the way, once Wald defines the Bondi energy as ##E = -\frac{1}{8\pi}\lim_{S_{\alpha}\rightarrow \mathcal{P}} \int _{S_{\alpha}}\epsilon_{abcd}\nabla^{c}\xi^{d}## (eq. 11.2.11) and imposes the gauge condition ##\nabla_a\xi^a## (eq. 11.2.12) on some open subset of ##\mathcal{J}^{+}## he does note that the 4-momentum ##P_a## is subsequently given by acting the linear map associated with the Bondi energy on BMS translations. I will admit however that this notion of 4-momentum is not as physically intuitive to me as the ADM 4-momentum.
     
  15. Nov 26, 2013 #14

    WannabeNewton

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    There is one natural way to view classical extended objects in a space-time ##(M,g_{ab})## and it is microscopic in nature. By considering the constituent particles making up the object and working with a system in which the worldlines of the constituent particles don't intersect, we can note that the worldlines give rise to a vector field ##u^a## that encodes the local kinematical aspects of the extended object and physically represents the 4-velocity field of the constituent particles; for all intents and purposes, the congruence associated with ##u^a## is basically the space-time description of the extended object.

    Letting ##\nabla_a## be the derivative operator associated with ##g_{ab}##, and letting ##h_{ab} = g_{ab}+ u_a u_b## be the spatial 3-metric relative to ##u^a##, we can define the aforementioned kinematical quantities as ##\theta_{ab} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)}## and ##\omega_{ab} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{[c}u_{d]}##.

    ##\theta_{ab}## is called the expansion tensor and it loosely speaking takes into account the relative radial velocities between constituent particles making up the extended object: if ##\xi^a## is an infinitesimal displacement vector between neighboring particles in the congruence then ##u^a \nabla_a (\xi^b\xi_b) = 2\theta_{ab}\xi^a \xi^b##.

    ##\omega_{ab}## on the other hand is called the rotation tensor and it takes into account the relative angular velocities between constituent particles. In fact if we define the twist ##\omega^{a} = \frac{1}{2}\epsilon^{abcd}u_{b}\omega_{cd}## then it is easy to show that ##\omega^{a} = \epsilon^{abcd}u_{b}\nabla_{c}u_{d}## and in the instantaneous rest frame of ##u^a## this is nothing more than ##\vec{\omega} = \vec{\nabla}\times \vec{u}##.

    This is a kinematical decomposition because we can express the covariant derivative of the 4-velocity field as ##\nabla_a u_b = \theta_{ab} + \omega_{ab} + u_a a_b## where ##a^b = u^c\nabla_c u^b## is the 4-acceleration field. Using this essentially hydrodynamical formulation, the kinematics of extended objects in space-times can be worked out in principle.

    Continuing with what was said above, if we can represent an extended object in terms of, for simplicity, a perfect fluid with 4-velocity field ##u^a## then this has associated with it an energy-momentum tensor field ##T_{ab}## and ##\rho = T_{ab}u^a u^b## is the mass density in the instantaneous rest frame of ##u^a##; this is basically the analogue of ##m^2 = -p_a p^a## for a massive test particle.

    Also, say we have a space-time ##(M,g_{ab})## with a time-translation symmetry; this time translation symmetry generates a time-like killing field ##\xi^a##. Note then that ##T_{ab}\xi^b## results in a conserved current because ##\nabla^{a}(T_{ab}\xi^{b}) = T_{(ab)}\nabla^{[a}\xi^{b]} + \xi^{b}\nabla^{a}T_{ab} = 0##.

    If we now assume that ##T_{ab}## has compact support then ##E = \int_{\Sigma} T_{ab}n^{a}\xi^{b}## defines a globally conserved energy associated with ##T_{ab}## in the sense that ##E## is independent of the space-like hypersurface ##\Sigma## as a consequence of the divergence theorem.

    Hope that helps somewhat xP
     
    Last edited: Nov 26, 2013
  16. Nov 26, 2013 #15

    BruceW

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    so for extended objects/fluids, we can get a mass density, but we can't get a mass.

    edit: unless we have a stationary or asymptotically flat spacetime
     
    Last edited: Nov 26, 2013
  17. Nov 26, 2013 #16
    I am wondering if you aware of Frank Wilczek's concept of particles being disturbances of a 'grid'.You might want to look it up, but I won't go on about it here as it is probably 'beyond' the standard model, but he is no crank, as he won a Nobel prize for his ideas.
     
  18. Nov 27, 2013 #17
    Pervect, and all.

    I've been left under the impression tha ADM, Bondi and Komar mass are not general linear transformations of how we would define mass in local Minkowski coordinates, re.: m^2 = E^2-p^2.

    Am I incorrect in this assuption?

    If not, or even if so, is the issue a matter of defining each in terms of a tensor density? I could do the research and find out, but what you all could say on this point could be infinitely better
     
  19. Nov 27, 2013 #18

    PAllen

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    Komar mass is an integral over a body or region of interest (valid only for static (stationary?)) metrics. ADM and Bondi mass are limits of integrals out to spatial or null infinity, respectively. They can technically only answer questions about the "whole universe" with the right asymptotic structure. As a matter of practice, you can ask: suppose I describe a sufficiently isolated system or region as if it were in an empty universe with asymptotically flat structure. For a cosmological spacetime, you argue that for a galaxy over, say, one million years, the expansion and non-zero cosmological constant can be ignored. Such an approach remains adhoc, though.
     
  20. Nov 28, 2013 #19
    Thanks all.

    It seems I'm not reinventing the wheel, after all. Aways a good thing. I've been working on a formulation where a form of generally covariant mass would also drop-out in a natural way. That is, I wasn't looking for it in the first place. I've yet to determine, with rigor, if my formulations are generally covariant or not.
     
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