Mass in general relativity Totally Explained

Everything about Mass In General Relativity totally explained

The concept of mass in general relativity (GR) is more complex than the concept of mass in special relativity. In fact, general relativity doesn't offer a single definition for the term mass, but offers several different definitions which are applicable under different circumstances. Under some circumstances, the mass of a system in general relativity may not even be defined.

Review of mass in special relativity

In special relativity, the invariant mass (hereafter simply "mass") of an isolated system, can be defined in terms of the energy and momentum of the system by the relativistic energy-momentum equation:
ť

m = frac dV

When the components of the momentum vector of the system are zero, for example Pⁱ = 0, the approximate mass of the system is just (E+E_binding)/c², E_binding being a negative number representing the Newtonian gravitational self-binding energy.
Hence when one assumes that the system is quasi-static, one assumes that there's no significant energy present in the form of "gravitational waves". When one assumes that the system is in "nearly-flat" space-time, one assumes that the metric coefficients are essentially Minkowskian within acceptable experimental error.

History

In 1918, David Hilbert wrote about the difficulty in assigning an energy to a "field" and "the failure of the energy theorem" in a correspondence with Klein. In this letter, Hilbert conjectured that this failure is a characteristic feature of the general theory, and that instead of "proper energy theorems" one had 'improper energy theorems'. (External Link

).
   This conjecture was soon proved to be correct by one of Hilbert's close associates, Emmy Noether. Noether's theorem applies to any system which can be described by an action principle. Noether's theorem associates conserved energies with time-translation symmetries. When the time-translation symmetry is a finite parameter continuous group, such as the Poincare group, Noether's theorem defines a scalar conserved energy for the system in question. However, when the symmetry is an infinite parameter continuous group, the existence of a conserved energy isn't guaranteed. In a similar manner, Noether's theorem associates conserved momenta with space-translations, when the symmetry group of the translations is finite dimensional. Because General Relativity is a diffeomorphism invariant theory, it has an infinite continuous group of symmetries rather than a finite-parameter group of symmetries, and hence has the wrong group structure to guarantee a conserved energy. Noether's theorem has been extremely influential in inspiring and unifying various ideas of mass, system energy, and system momentum in General Relativity.
   As an example of the application of Noether's theorem is the example of stationary space-times and their associated Komar mass.(Komar 1959). While general space-times lack a finite-parameter time-translation symmetry, stationary space-times have such a symmetry, known as a Killing vector. Noether's theorem proves that such stationary space-times must have an associated conserved energy. This conserved energy defines a conserved mass, the Komar mass.
   ADM mass was introduced (Arnowitt et al, 1960) from an initial-value formulation of general relativity. It was later reformulated in terms of the group of asymptotic symmetries at spatial infinity, the SPI group, by various authors. (Held, 1980). This reformulation did much to clarify the theory, including explaining why ADM momentum and ADM energy transforms as a 4-vector (Held, 1980). Note that the SPI group is actually infinite dimensional. The existence of conserved quantities is due to the fact that the SPI group of "super-translations" has a preferred 4-parameter subgroup of "pure" translations, which, by Noether's theorem, generates a conserved 4-parameter energy-momentum. The norm of this 4-parameter energy-momentum is the ADM mass.
   The Bondi mass was introduced (Bondi, 1962) in a paper that studied the loss of mass of physical systems via gravitational radiation. The Bondi mass is also associated with a group of asymptotic symmetries, the BMS group at null infinity. Like the SPI group at spatial infinity, the BMS group at null infinity is infinite dimensional, and it also has a preferred 4-parameter subgroup of "pure" translations.
   Another approach to the problem of energy in General Relativity is the use of pseudotensors such as the Landau-Lifshitz pseudotensor.(Landau and Lifshitz, 1962). Pseudotensors are not gauge invariant - because of this, they only give consistent gauge-independent answers for the total energy when additional constraints (such as asymptotic flatness) are met. The gauge dependence of pseudotensors also prevents any gauge-independent definition of the local energy density, as every different gauge choice results in a different local energy density.

Questions, answers, and simple examples of mass in general relativity

In special relativity, the invariant mass of a single particle is always Lorentz invariant. Can the same thing be said for the mass of a system of particles in special relativity? ť Surprisingly, the answer is no. A system must either be isolated, or have zero volume, in order for its mass to be Lorentz invariant. While the density of energy momentum, the stress-energy tensor is always Lorentz covariant, the same can't be said for the total energy-momentum. (Nakamura, 2005). Non-covariance of the energy-momentum four-vector implies non-invariance of its length, the invariant mass.

ť What this means in simpler language is that one must use great caution when talking about the mass of a non-isolated system. A non-isolated system is constantly exchanging energy-momentum with its surroundings. Even when the net rate of exchange of energy-momentum with the environment is zero, differences in the definition of simultaneity cause the total amount of energy-momentum contained within the system at a given instant of time to depend on the definition of simultaneity that's adopted by the observer. This causes the invariant mass of a non-isolated system to depend on one's choice of coordinates even in special relativity. Only an isolated system has a coordinate-independent mass.

Can an object move so fast that it turns into a black hole? ť No. An object that isn't a black hole in its rest frame won't be a black hole in any other frame. One of the characteristics of a black hole is that a black hole has an event horizon, which light can't escape. If light can escape from an object to infinity in the object's rest frame, it can also escape to infinity in a frame in which the object is moving. The path that the light takes will be aberrated by the motion of the object, but the light will still escape to infinity(External Link).

If two objects have the same mass, and we heat one of them up from an external source, does the heated object gain mass? If we put both objects on a sensitive enough balance, would the heated object weigh more than the unheated object? Would the heated object have a stronger gravitational field than the unheated object? ť The answer to all of the above questions is yes. The hot object has more energy, so it weighs more and has a higher mass than the cold object. It will also have a higher gravitational field to go along with its higher mass, by the equivalence principle. (Carlip 1999)

Imagine that we've a solid pressure vessel enclosing an ideal gas. We heat the gas up with an external source of energy, adding an amount of energy E to the system. Does the mass of our system increase by E/c²? Does the mass of the gas increase by E/c²? ť The question is somewhat ambiguous as stated. Interpreting the question as a question about the Komar mass, the answers to the questions are yes, and no, respectively. Because the pressure vessel generates a static space-time, the Komar mass exists, and can be found by treating the ideal gas as an ideal fluid. Using the formula for the Komar mass of a small system in a nearly Minkowskian space-time, one finds that the mass of the system in geometrized units is equal to E + ∫ 3 P dV, where E is the total energy of the system, and P is the pressure.

ť The integral ∫ P dV over the entire volume of the system is equal to zero, however. The contribution of the positive pressure in the fluid is exactly canceled out by the contribution of the negative pressure (tension) in the shell. This cancellation isn't accidental, it's a consequence of the relativistic virial theorem (Carlip 1999).

ť If we restrict our region of integration to the fluid itself, however, the integral isn't zero and the pressure contributes to the mass. Because the integral of the pressure is positive, we find that the Komar mass of the fluid increases by more than E/c².

ť The significance of the pressure terms in the Komar formula can best be understood by a thought experiment. If we assume a spherical pressure vessel, the pressure vessel itself won't contribute to the gravitational acceleration measured by an accelerometer inside the shell. The Komar mass formula tells us that the surface acceleration we measure just inside the pressure vessel, at the outer edge of the hot gas will be equal to $Gleft(E + 3 P V
ight) / r^2 c^2$

ť :where E is the total energy (including rest energy) of the hot gas

:G is Newton's Gravitational constant ť :P is the pressure of the hot gas

:V is the volume of the pressure vessel. ť This surface acceleration will be higher than expected because of the pressure terms. In a fully relativistic gas, (this includes a "box of light" as a special case), the contribution of the pressure term 3 P V will be equal to the energy term E, and the acceleration at the surface will be doubled from the value for a non-relativistic gas.

ť One might also ask about the answers to this question if one assumed that one were asking about the mass as it's defined in special relativity rather than the Komar mass. If one assumes that the space-time is nearly Minkowskian, the special relativistic mass exists. In this case, the answer to the first question is still yes, but the second question can't be answered without even more data. Because the system consisting only of the gas isn't an isolated system, its mass isn't invariant, and thus depends on the choice of observational frame. A specific choice of observational frame (such as the rest frame of the system) must be specified in order to answer the second question. If the rest frame of the object is chosen, and special relativistic mass rather than Komar mass is assumed, the answer to the second question becomes yes. This problem illustrates some of the difficulties one faces when talking about the mass of non-isolated systems.

The only difference between the "hot" and "cold" systems in our last question is due to the motion of the particles in the gas inside the pressure vessel. Doesn't this imply that a moving particle has "more gravity" than a stationary particle? ť This remark is probably true in essence, but it's difficult to quantify.

ť Unfortunately, it isn't clear how to measure the "gravitational field" of a single relativistically moving object. It is clear that it's possible to view gravity as a force when one has a stationary metric - but the metric associated with a moving mass isn't stationary.

ť While definitional and measurement issues constrain our ability to quantify the gravitational field of a moving mass, one can measure and quantify the effect of motion on tidal gravitational forces. When one does so, one finds that the tidal gravity of a moving mass isn't spherically symmetrical - it's stronger in some directions than others. One can also say that, averaged over all directions, the tidal gravity increases when an object moves.

ť Some authors have used the total velocity imparted by a "flyby" rather than tidal forces to gain an indirect measure of the increase in gravitational "effective mass" of relativistically moving objects (Olson & Guarino 1985)

ť While there's unfortunately no single definitive way to interpret the space-time curvature caused by a moving mass as a Newtonian force, one can definitely say that the motion of the molecules in a hot object increases the mass of that object.

ť Note that in General Relativity, gravity is caused not by mass, but by the stress-energy tensor. Thus, saying that a moving particle has "more gravity" doesn't imply that the particle has "more mass". It only implies that the moving particle has "more energy".

Suppose the pressure vessel in our previous question fails, and the system explodes - does its mass change? ť The mass of the system doesn't change because the vessel (or the pieces of the vessel after it explodes) form an isolated system. This question does illustrate one of the limitations of the Komar formula - the Komar mass is defined only for stationary systems. If one applies the Komar formula to this non-static non-stationary system, one gets the incorrect result that the mass of the system changes. The pressure and density of the gas remains constant for a short time after the failure, while the tension in the pressure vessel disappears immediately when the pressure vessel fails. One can't correctly apply the Komar formula in this case, however - one needs to apply a different formula, such as the ADM mass formula, the Newtonian limit formula, or the special relativistic formula.

What is the mass of the universe? What is the mass of the observable universe? Does a closed universe have a mass? ť None of the above questions have answers. We know the density of the universe (at least in our local area), but we can only speculate on the extent of the universe, making it impossible for us to give a definitive answer for the mass of the universe. We can't answer the second question, either. Since the observable universe isn't asymptotically flat, nor is it stationary, and since it may not be an isolated system, none of our definitions of mass in General Relativity apply, and there's no way to calculate the mass of the observable universe. The answer to the third question is also no : the following quote from (Misner, et al, pg 457) explains why:

ť :"There is no such thing as the energy (or angular momentum, or charge) of a closed universe, according to general relativity, and this for a simple reason. To weigh something one needs a platform on which to stand to do the weighing ...

ť :"To determine the electric charge of a body, one surrounds it by a large sphere, evaluates the electric field normal to the surface at each point on this sphere, integrates over the sphere, and applies the theorem of Gauss. But within any closed model universe with the topology of a 3-sphere, a Gaussian 2-sphere that's expanded widely enough from one point finds itself collapsing to nothingness at the antipodal point. Also collapsed to nothingness is the attempt to acquire useful information about the "charge of the universe": the charge is trivially zero."

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