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The stress-energy-momentum tensor in General Relativity includes a mass density terms, which is related to energy via $E=mc^2$. How does potential energy figure into this, since potential energy is relative, not absolute.

I was also going to ask if this includes gravitational potential energy, but I found these: Justification for excluding gravitational energy from the stress-energy tensor and Potential Energy in General Relativity

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    $\begingroup$ The stress-energy-momentum tensor … includes a mass density terms. $T^{00}$ is energy density. For a Reissner-Nordstrom black hole, which has electric charge, it is the energy density of the electrostatic field. $\endgroup$
    – Ghoster
    Nov 24, 2023 at 20:05
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    $\begingroup$ Just write down the lagrangian with required potential and compute the stress energy tensor. For gravity, the Newtonian potential actually enters into the free particle lagrangian (in the metric). And by equivalence principle, you can locally move to inertial frame so that stress-energy-momentum for gravity as a tensor field will vanish $\endgroup$
    – paul230_x
    Nov 24, 2023 at 20:28
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    $\begingroup$ There is no “potential energy” in General Relativity. Gravity is not a force field. $\endgroup$
    – safesphere
    Nov 25, 2023 at 9:33
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    $\begingroup$ I feel that people are missing the point. I already answered my own question about gravitational potential energy. But what about other potential energies like electrostatic? The question is where does one assign the arbitrary "zero" because potential energy is relative, not absolute. $\endgroup$ Nov 25, 2023 at 18:46
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    $\begingroup$ I already answered that above : you write down the lagrangian for the system with potential and find stress-energy tensor : it could be the canonical SET, or Einstein-Hilbert SET, or Belinfante-Rosenfeld etc $\endgroup$
    – paul230_x
    Nov 27, 2023 at 6:39

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(References are given at the end)

The short answer is that the usual forms of energy that we consider in Newtonian mechanics, such as kinetic energy, gravitational potential energy, rotational energy, do appear in the energy-momentum(-heat-stress) tensor – depending on the coordinate system. Note that by "gravitational potential energy" I mean the energy that matter has from being in a gravitational field, not the energy of the gravitational field itself, which is non-localizable.

The dependence on the coordinates is a subtle point, which I comment on at the end. But a simple example illustrates the coordinate-dependence of these "kinds" of energies.

Take a spacetime region with a Scharzschild metric, as it could approximately be on and immediately outside the Earth. Using isotropic coordinates $(t,x,y,z)$, and discarding terms of higher order in $1/c$, the metric $\pmb{g}$ has these components (eg Poisson & Will 2014, eqns 8.2): $$ g_{\mu\nu} = \begin{bmatrix} -c^{2}\Bigl(1- \frac{2}{c^2}U\Bigr) & 0 & 0 & 0 \\\\ 0 & 1+ \frac{2}{c^2}U & 0 & 0 \\\\ 0 & 0& 1+ \frac{2}{c^2}U & 0 \\\\ 0 & 0&0& 1+ \frac{2}{c^2}U \end{bmatrix} \ , \qquad U := \frac{G\,M}{r} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}$. We recognize $-U(r)$ as the Newtonian gravitational potential.

Now take a specific spacetime point $P$, with coordinates $(t,x,y,z)$, where there's some baryonic matter moving in the $x$-direction with coordinate velocity $v$. The corresponding 4-velocity $\pmb{V}$ in the present coordinate system has components $$ V^{\mu}=\beta\,(1,v,0,0) \ , \qquad \beta := \frac{c}{\sqrt{-g_{tt}-v^2\,g_{xx}}} \ . $$

Consider an observer having the same 4-velocity, and take a new coordinate system adapted to that observer and in which the observer is in free fall at $P$ (Fermi coordinates). This means that in this new coordinate system, at $P$, the observer's 4-velocity must have new components $(1,0,0,0)$, and the metric has new components with canonical values $\left[\begin{smallmatrix}-c^2&0&0&0\\\\0&1&0&0\\\\0&0&1&0\\\\0&0&0&1\end{smallmatrix}\right]$. I'm emphasizing "at $P$", because the metric in the new coordinate system won't have these components everywhere, as the space is curved.

You can check that, at $P$, the following matrix $$ \varLambda^{\mu'}{}_{\mu} = \begin{bmatrix} -\frac{1}{c^{2}}\beta\,g_{tt} & -\frac{1}{c^{2}}\beta\,v\, g_{xx} & 0 & 0 \\\\ -\frac{1}{c}\beta\,v\,\sqrt{-g_{tt}\,g_{xx}} & \frac{1}{c}\beta\,\sqrt{-g_{tt}\,g_{xx}} & 0 & 0 \\\\ 0 & 0 & \sqrt{g_{yy}} & 0\\\\ 0 & 0 & 0 & \sqrt{g_{zz}} \end{bmatrix} $$ represents the transformation from the coordinates $(t,x,y,z)$ to the local coordinates where the matter is at rest and free-falling. We don't know what the transformation matrix looks like at other points (likely quite complex).

Let's go back to the baryonic matter. In the new coordinates, at $P$, this matter is at rest and free-falling. Its energy-momentum tensor $\pmb{T}$, in its doubly-contravariant form, has components $$ T^{\mu'\nu'} = \begin{bmatrix} \rho + \epsilon/c^{2} & q_{x}/c^{2} & q_{y}/c^{2} & q_{z}/c^{2} \\\\ q_{x}/c^{2} & \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\\\ q_{y}/c^{2} & \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\\\ q_{z}/c^{2} & \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} $$ Here, $\rho$ is (by definition) the rest-mass density at $P$, $\epsilon$ the internal energy density (also by definition), $\pmb{q}$ the heat flux (also by definition), and $\sigma$ the internal stress tensor. If the matter is a fluid, the stress tensor is given by the scalar pressure times the identity matrix; but let's consider a more general situation. The terms $\pmb{q}/c^{2}$ in the first row, negligible in Newtonian approximation, represent the momentum that in relativity is associated with any heat flux (see eg Eckart 1940).

The internal energy $\epsilon$ may include elastic energy, but it does not include kinetic or gravitational-potential terms.

Using the transformation matrix above (or rather its inverse), we can calculate the components at $P$ of the energy-momentum tensor $\pmb{T}$ in the original coordinate system $(t,x,y,z)$. Discarding terms of higher order in $1/c$ (omitted) and some other longer expressions (indicated by "$\dotsb$") we find $$ T^{\mu\nu} = \begin{bmatrix} \rho + \frac{1}{c^{2}}\,(\epsilon + \rho\,v^{2} + 2\rho\,U) & \rho\,v +q_{x}/c^{2}+\dotsb& q_{y}/c^{2}+\dotsb & q_{z}/c^{2}+\dotsb \\\\ \rho\,v +q_{x}/c^{2} +\dotsb & \rho\,v^{2}+\sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\\\ q_{y}/c^{2} +\dotsb& \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\\\ q_{z}/c^{2} +\dotsb& \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} $$

Focusing on the $T^{tt}$ term, we see the appearance of "$\rho\,v^{2}$", suggestive of kinetic energy, and of "$\rho\,U$", suggestive of gravitational potential energy.

This should be an approximate answer to your question: yes, gravitational potential energy does appear, depending on the coordinate system. Let me repeat, though, that we're speaking about the energy that matter has from being in a gravitational field, not the energy of the gravitational field itself.

The fact that a factor $1/2$ seems to be missing, if the terms above are to be interpreted as kinetic and potential energies, may be confusing; see for instance this question. Also, the signs of the kinetic and potential terms should be opposite. These are the additional subtleties I mentioned at the beginning. They boil down to the fact that the energy-momentum tensor is not something that can be integrated, and strictly speaking the term $T^{tt}$ is not quite an energy density, in the sense of being the time component of an energy-flux vector that can be integrated over a 3D hypersurface. I add some remarks about this, just for completeness.

Some extra subtleties

Concretely the difference is this. Take a region of spacetime having some electric charge & current. Now take a 3D spacelike hypersurface there. This surface can have an arbitrary shape, as long as it's spacelike. All observers, all coordinate systems will agree on the "total amount of charge" contained in (or crossing, from a spacetime perspective) that 3D hypersurface. For different observers and in different coordinate systems, this total amount is interpreted differently. For instance, in a coordinate system where the hypersurface is at constant time and there is no electric current, the total charge has the familiar meaning. For other observers or coordinates, the hypersurface may appear as a temporal sequence of spatial 2D surfaces, crossed by charge and current; the "total amount" is the integrated charge & current that flowed through the 2D surfaces. But the final result of the integral is the same for all observers and coordinates.

Not so for energy density. Take a spacetime region having some energy-momentum(-heat-stress). Now take again a 3D spacelike hypersurface there. To define the "total amount of energy" contained there, we need to specify one more thing: a field of timelike vectors defined over that surface. Without this specification, we can't even integrate anything; the very operation of integration is undefined. Once a timelike vector field is specified, then we have an object that can be interpreted as energy density & energy flux, and can be integrated. The situation is then as for charge: for all observers and coordinate systems there's a well-defined "total amount of energy" contained in the 3D hypersurface, and having the same value for all, even if its interpretation may differ.

Mathematically this is reflected in the fact that energy-momentum(-heat-stress) is represented by a 2nd-order tensor, not a 3-form. Only after we contract this tensor with a vector do we obtain a 3-form (or equivalent object), which can be integrated. In passing, if the chosen vector is spacelike then we are defining a momentum density rather than an energy density.

To speak about "energy density" at a spacetime point, we must therefore specify which timelike vector we are referring the energy to, or said otherwise, in terms of which timelike vector we are defining energy. It is this reference vector which determines what "kind" of energy we are considering. Depending on the reference vector, the energy may include kinetic and gravitational potential energy.

As a clear demonstration of this, consider again the point $P$ above and the local coordinate system where the baryonic matter is at rest and in free fall. Its energy-momentum tensor $\pmb{T}$ has components $T^{\mu'\nu'}$ where no kinetic or potential energy terms appear. Contract this tensor with the 4-velocity $\pmb{V}$, which in these coordinates has components $(1,0,0,0)$. We obtain the energy-flux vector $\pmb{J}_{\pmb{V}} := \pmb{T}\cdot\pmb{g}\cdot\pmb{V}$ with components $$ {J_{\pmb{V}}}^{\mu'} = - \begin{bmatrix} \rho\,c^{2}+\epsilon\\\\ q_{x}\\\\q_{y}\\\\q_{z} \end{bmatrix} $$ easily interpreted.

Now take the Killing vector $\pmb{K}$ of the Schwarzschild spacetime. Let's define an energy-flux vector with respect to it: $\pmb{J}_{\pmb{K}} := \pmb{T}\cdot\pmb{g}\cdot\pmb{K}$. In the original coordinates $(t,x,y,z)$, $\pmb{K}$ has components $(1,0,0,0)$; using the transformation matrix we can find its components at $P$ in the new coordinate system, and can calculate the components of $\pmb{J}_{\pmb{K}}$. Discarding higher-order terms in $1/c$ we find: $$ {J_{\pmb{K}}}^{\mu'} = - \begin{bmatrix} \rho\,c^{2}+\epsilon + \frac{1}{2}\rho\,v^{2} - \rho\,U \\\\ q_{x} + v\,\sigma_{xx} \\\\ q_{y} + v\,\sigma_{xy} \\\\ q_{z} + v\,\sigma_{xz} \end{bmatrix} $$ We clearly see the appearance of a kinetic energy density and a gravitational potential energy density, all with the correct signs. We also see "work" terms appearing as energy flux. Yet, these are the components in the coordinate system that's locally at rest and free-falling! Note how neatly the various energy forms appear here: we don't need to worry about taking corrections for volume elements, because the densities $\rho$ and $\epsilon$ are defined in this very coordinate system.

Common choices of the reference vector are these:

  • If at the spacetime point there is baryonic matter, then we can take its 4-velocity as reference timelike vector. One important characteristic of resulting energy-flux vector is that it is only balanced, not conserved.

  • If we have a timelike Killing vector field, this can be used as reference to define the energy. The resulting energy-flux vector has then the property of being not only balanced, but conserved (no energy sources or sink).

  • Another choice of reference timelike vector is the one associated with the timelike coordinate (if any) of the coordinate system. The energy-flux vector so defined has four components $ (T^{\mu}_t)$; note that the "$t$" is in covariant position.

Coordinate systems are often chosen so that $\partial_t$ is a Killing vector field (eg Schwarzschild, isotropic, Kerr, etc). Then the present choice of reference timelike vector coincides with the "Killing choice" described above.

The need of a reference vector to define energy-flux density is treated quite poorly in relativity books, according to my experience. Some exceptions and examples:

  • Choquet-Bruhat & al 2000 (§ II.7.III) are very explicit:

The energy-momentum vector associated to a stress-energy tensor $T$ and a timelike or null vector field $X$ is defined to be $P^{\beta}=X_{\alpha}T^{\alpha\beta}$.

  • Synge 1960 (Chapter VI) is also explicit.

  • Burke 1987 is also somewhat explicit: "stress-momentum-energy tensor [...]. This tensor is a linear map from Killing vectors to conserved currents" (p. 269). See also p. 384ff.

  • Hawking & Ellis 1994 (p. 62, bottom) also refer to this procedure, somewhat implicitly.

  • Gotay & Marsden 1992 can also be read from this perspective. In this and subsequent works (which take a Lagrangean viewpoint) the contraction of the energy-momentum tensor with a vector field is seen as the "response" of matter associated with the tensor, with respect to spacetime deformations generated by the vector field.

  • Ohanian & Ruffini 2013 (§ 2.4, p. 67ff) also show this implicitly.

  • Misner & al 1973 (Box 5.1 A) mention it, but in a somewhat obscure way.

References

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    $\begingroup$ Thank you very much for your detailed answer. I must confess that it is far beyond my capabilities to understand it. If we spent a day or two having coffee together, I could probably understand. Since you are the only person who responded, I'm going to accept your answer anyway. $\endgroup$ Jan 15 at 17:18
  • $\begingroup$ Thank you @StephenMontgomery-Smith – but sad to hear it isn't understandable. In full or in part? Just let me know which parts are too technical and I'll be happy to change it :) And feel free to send a message, of course. $\endgroup$
    – pglpm
    Jan 15 at 18:16

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