# Is a volume element in relativity represented by a vector?

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## Main Question or Discussion Point

Can we say that a volume element can be represented by a vector, or is there some hidden complication that makes this inadvisable?

For some background, the stress-energy tensor has been described as the density of energy and momentum, in for instance MTW. So if one says that the represention of a volume element is a vector, one can basically say that a rank 2 tensor (the stress energy tensor) maps the volume element (a rank 1 tensor) into another rank 1 tensor (the energy-momentum 4-vector), where the energy-momentum 4-vector representing the total energy and momentum contained in the specified volume.

But MTW, for instance, does not, as far as I've seen, ever explicitly say that a volume element can be represented by a vector. Instead, they say that one multiples the stress-energy tensor by the 4-velocity of an observer.

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Orodruin
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The volume element is a 4-form (or, in an N-dimensional manifold, an N-form) that maps N tangent vectors to the volume spanned by them. When you write down your integration in terms of coordinates, the vectors you supply to it are the tangent vectors of the coordinate lines.

The energy-momentum tensor is more akin to a current (or a collection of currents), as should be clear from the divergence relation $\nabla_\mu T^{\mu\nu} = 0$, which is just a current conservation law. The divergence theorem relates the volume integral to a surface integral. In terms of relativity, this relates an integral over a 4-dimensional space-time integral to an integral over its boundary. If the energy-momentum is spatially located (i.e., vanishes sufficiently fast as you go to spatial infinity) and you have a foliation of spacetime into space-like hypersurfaces (that you call "simultaneities"), then
$$\int_\Omega \nabla_\mu(T^{\mu\nu}K_\nu) \sqrt{|g|}d^4x = \int_{\Sigma_2} T^{\mu\nu} K_\nu dS_\mu - \int_{\Sigma_1} T^{\mu\nu} K_\nu dS_\mu,$$
where $dS_\mu$ is the 3-hypersurface element of the simultaneities $\Sigma_2$ and $\Sigma_1$, respectively.

Note that if $K$ is a Killing field, then $\nabla_\mu K_\nu = - \nabla_\nu K_\mu$ and so
$$\nabla_\mu(T^{\mu\nu}K_\nu) = \{T^{\mu\nu}\mbox{ conserved} \leftrightarrow \nabla_\mu T^{\mu\nu} = 0\} = T^{\mu\nu} \nabla_\mu K_\nu = 0,$$
since $T^{\mu\nu}$ is symmetric and $\nabla_\mu K_\nu$ is anti-symmetric, and you end up with a conservation law.

The energy-momentum tensor has some components that are the spatial density of energy and momentum (namely the components $T^{0\nu}$ in a local Minkowski frame). They are not the spacetime densities and therefore are not integrated with the spacetime volume element. Just like the surface element of a 2-surface in regular vector calculus is a vector, the hypersurface element of a 3-hypersurface in a 4-dimensional manifold is a vector (or, more precisely, what you integrate over the 3-surface needs to be a 3-form and that 3-form is given by $i_X\eta$, where $\eta$ is the spacetime volume element and $X$ is the current you are integrating over it).

Edit: Sorry if that was a bit garbled up. I started typing thinking you were referring to the spacetime volume element (i.e., the 4-form) but as I progressed through it I started doubting this and thinking you might be meaning the 3-surface element, which for a space-like hypersurface would act as the volume element ... I could not really tell from your question which you were intending.

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Orodruin
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Instead, they say that one multiples the stress-energy tensor by the 4-velocity of an observer.
Just to comment on this also. In the standard simultaneity convention in Minkowski space, the 4-velocity of an observer is orthogonal to the simultaneity surface and of length one. Correspondingly, the 3-hypersurface element (i.e., spatial volume element) is then exactly the 4-velocity of the observer multiplied by the spatial 3-volume.

Edit: Also - best question this month!

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Edit: Sorry if that was a bit garbled up. I started typing thinking you were referring to the spacetime volume element (i.e., the 4-form) but as I progressed through it I started doubting this and thinking you might be meaning the 3-surface element, which for a space-like hypersurface would act as the volume element ... I could not really tell from your question which you were intending.
Yes, I was indeed thinking of the spatial volume element. Does it make sense to say that a spatial volume element is the wedge product of three spatial vectors - and then to go further and say that that the hodges dual of this 3-form is a vector, so that we then wind up saying that a spatial volume element can be represented by a vector?

Sorry that the question wasn't too clear, trying to get the semantics down is probably a large part of my issue.

Orodruin
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Before I shoot a mosquito with a cannon: Are you asking in the context of SR or GR? (Minkowski space vs general Lorentzian manifold)

Edit: Also, what level of differential geometry are you comfortable with? Are you comfortable with the difference between tangent vectors and dual vectors? Are you familiar with the language of differential forms?

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I'm not terribly familiar with exterior derivatives, though I'm comfortable with differential forms as totally anti-symmetric tensors. Hopefully that gives you some idea of where I'm coming from.

I'm comfortable with dual vectors.

I'm interested in both SR and GR. If one is simpler, it might be better to start with the simpler one first.

Orodruin
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I am at work now and what I would like to write would take some time to formulate. I will take try to write something more extensive tonight (European time).

In the meantime, let me just say that this is not an issue that is restricted to relativity. You will find the same kind of construction for integration on any manifold (sublime commercial: I discuss integration on general manifolds in chapter 9 of my book if anyone reading this has access to it and wants more detail). Also, due to the ambiguity of simultaneity, different observers will disagree on whether the 3-volume is actually a 3-volume or a general 3-dimensional slice of spacetime.

vanhees71
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The three-volume is a hypersurface in Minkwski space (in special relativity; I assume we discuss only this for now). Then the three-volume element is a four-vector (as in usual 3D vector analysis the surface element is a vector):
$$\mathrm{d}^3 \sigma_{\rho}=\mathrm{d}^3 q \epsilon_{\rho \mu \nu \sigma} \frac{\partial x^{\mu}}{\partial q_1} \frac{\partial x^{\rho}}{\partial q_2} \frac{\partial x^{\sigma}}{\partial q_3},$$
where
$$\epsilon^{\mu \nu \rho \sigma}=-\epsilon_{\mu \nu \rho \sigma}$$
is the Levi-Civita symbol with $\epsilon^{0123}=1$ and totally antisymmetric under permutations of the indices.

Orodruin
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So here is my general take, stop me and ask if you have any questions or something flies above your head.

Let's start by going back to the volume of a parallelepiped in $\mathbb R^3$. As you are certainly familiar with, the volume of the parallelepiped spanned by three vectors $\vec u$, $\vec v$, and $\vec w$ is given by the triple product
$$V(\vec u,\vec v,\vec w) = \vec u \cdot (\vec v \times \vec w)$$
and is defined as positive whenever the vectors form a right-handed system and negative if they form a left-handed system. In component form in Cartesian coordinates, this translates into
$$V = \epsilon_{ijk} u^i v^j w^k.$$
Going to the three-dimensional volume element that we are to integrate in volume integrals in $\mathbb R^3$, we look for the volume that is within $\chi_0^a < \chi^a < \chi_0^a + d\chi^a$, where $\chi^a$ are general coordinates (e.g., cylinder or spherical or any other coordinate system). That volume is spanned by the vectors $(\partial \vec x/\partial \chi^a) d\chi^a$ (no sum over $a$) and therefore the volume element is
$$dV = V(\vec E_1, \vec E_2, \vec E_3) d\chi^1 d\chi^2 d\chi^3 \equiv \eta_{123} d\chi^1 d\chi^2 d\chi^3,$$
where $\vec E_a$ are the tangent basis vectors $\vec E_a = \partial \vec x/\partial \chi^a$ and we defined $\eta_{abc} = V(\vec E_a,\vec E_b,\vec E_c)$. Note that $\eta_{abc}$ is completely asymmetric, but not equal to $\epsilon_{abc}$. In fact, it is quite straight-forward to show that $\eta_{abc} = \sqrt{g} \epsilon_{abc}$, where $g$ is the metric determinant.

Now comes the question: What are the properties of the triple product $V(\vec u, \vec v, \vec w)$? Well, it maps three (tangent) vectors to a number and so it must be a (0,3) tensor. It is also completely anti-symmetric, so it is actually a 3-form (which is just a fancy name for a completely anti-symmetric (0,3) tensor). This is the property that we will use to define integrations on a manifold or on a submanifold of a manifold (which, of course, is a manifold in itself). The integral of a $k$-form $\omega$ over a $k$-dimensional submanifold $S$ can be defined according to
$$\int_S \omega = \int_{S^*} \omega(\dot\gamma_1, \ldots, \dot\gamma_k) dt^1 \ldots dt^k,$$
where the $t^i$ ($i = 1, \ldots, k$) parametrise $S$, $S^*$ is the domain of those parameters, and $\dot\gamma_i$ is the tangent vector of the curve of constant $t^i$. Again, it is rather straight-forward to show that the $N$-volume element in an $N$-dimensional manifold with a metric must be of the form $\sqrt{g} \epsilon_{a_1\ldots a_N}$ in order for a set of orthogonal vectors to span the appropriate volume (this is where the $\sqrt{g}$ that you see in all GR action integrals comes from!).

So what happens when you have a $N-1$-dimensional submanifold that you would like to integrate over? Well, to have the integral properly defined, you need an $N-1$-form, but what you have access to is the $N$-volume element so you are essentially missing one of its argument. You can define an $N-1$-form by already plugging in a vector field into one (typically the first) of its arguments (for an $N$-form $\omega$ and a vector field $J$, this would typically be denoted $i_J\omega$). In a slightly different point of view, if you want to integrate over a $N-1$-dimensional hypersurface, you plug the $N-1$ tangent vectors into the volume form and you are left with a 1-form so the $N-1$-hypersurface element is a dual vector - it effectively maps a tangent vector to a number which is the volume spanned by the tangents in the hypersurface and the vector field you are integrating. This is just the flux of the field through the hypersurface as can be glanced in the $\mathbb R^3$ case, where $d\vec S = [(\partial \vec x/\partial s) \times (\partial\vec x/\partial t)] ds\, dt$ and therefore
$$\vec J \cdot d\vec S = V(\vec J, \partial \vec x/\partial s,\partial \vec x/\partial t) ds\, dt.$$

With those general preliminaries out of the way, what does this do for you in terms of integrals in spacetime? Well, everything carries over essentially unchanged (although the metric determinant is negative so you replace $g \to |g|$). The only thing is how you interpret the things physically. The 4-volume element is a 4-form that can be integrated over an entire region of space-time. Depending on the orientation of a 3-hypersurface, you may interpret it differently. If your 3-hypersurface is a simultaneity (however you have defined that), then the surface element is your spatial volume element with a unit normal (this is where the "observer 4-velocity" comes from). This can be seen if you take a simultaneity from a standard coordinate frame in Minkowski space, i.e., $t = t_0$, then the vectors tangent to your coordinate lines are $\partial_1$, $\partial_2$, and $\partial_3$ and therefore
$$dS_\mu = \epsilon_{\mu 123} dx^1 dx^2 dx^3 = \epsilon_{\mu 123} dV.$$
Thus, $J^\mu dS_\mu = J^0 dV = J^\mu U_\mu dV$, where $U_\mu$ is the dual vector related (by the metric) to the 4-velocity $U^\mu$ of an observer at rest in the frame we took the simultaneity surface from. However, this is not the only type of 3-hypersurface you can have. Leaving out other spacelike surfaces such as simultaneities of other frames for the moment, you can have 3-hypersurfaces that have time as a tangent direction, for example the surfaces of constant $x^3 = x^3_0$. These surfaces would have a 3-hypersurface element given by (order of indices depending on the direction of the surface)
$$dS_\mu = \epsilon_{\mu 012} dt\, dx^1dx^2 = B_\mu dt\, dA_{12},$$
where $dA_{12}$ is the area element of the $x^1$-$x^2$-region integrated over and $B_\mu$ is a dual vector corresponding to picking out the third element of the tangent vector you are contracting with the 3-hypersurface element. In effect
$$J^\mu dS_\mu = J^3 dt\, dA_{12} = \vec J \cdot d\vec A_{12} dt.$$
The integral of $\vec J \cdot d\vec A_{12}$ is just the flux through the surface $x^3 = x^3_0$ and integrating it over time gives you the total amount of whatever the current $J^\mu$ is a current of that was carried through the surface over the time interval you integrated over.

This discussion also underlines the role of the 0-component of any current $J^\mu$ as the spatial density of whatever $J^\mu$ is a current of. The zero component is what you integrate over a simultaneity to obtain how much of that thing is carried over that simultaneity into the future.

Now, in GR, your hypersurfaces will typically be more involved, but as long as they are space- or time-like, they can be locally interpreted in a similar manner. If you go back to #2, you can see one application of this type of integrals, the conservation laws. For example, go back to SR again and consider an energy-momentum tensor that is located so that $T^{\mu\nu} = 0$ for sufficiently large spatial coordinates. Then you can integrate the divergence of the energy-momentum tensor over the spacetime volume $\Omega$ between two simultaneities $\Sigma_1$ and $\Sigma_2$ to obtain
$$\int_\Omega T^{\mu\nu} (\nabla_\mu K_\nu) \sqrt{|g|} d^4 x = \int_{\Sigma_2} T^{\mu\nu} K_\nu dS_\mu - \int_{\Sigma_1} T^{\mu\nu} K_\nu dS_\mu.$$
As we discussed, the 3-hypersurface elements for these simultaneities are just $\epsilon_{\mu 123} dV$ and so
$$\int_{\Sigma_2} T^{0\nu} K_\nu dV - \int_{\Sigma_1} T^{0\nu} K_\nu dV = \int_\Omega T^{\mu\nu} (\nabla_\mu K_\nu) \sqrt{|g|} d^4 x.$$
Letting $K$ be any Killing field means (as discussed in #2) that the right-hand side here vanishes due to the symmetry of $T^{\mu\nu}$. In particular, letting $K = \partial_0$ would imply that $K_0 = 1$ with all other components vanishing and therefore
$$\int_{\Sigma_2} T^{00} dV = \int_{\Sigma_1} T^{00} dV,$$
which is just the conservation of energy. The corresponding argument for the Killing fields $\partial_i$ instead give you the conservation of the momenta.

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Thank you. Sorry for the delayed response, real life (tm) distracted me for a bit. I can see now where I was taking some shortcuts. Nothing really seriously wrong (as in getting the wrong answer) but it made the way I was thinking about the problem and talking about it a bit non-standard. Seeing the standardized notational description was quite helpful.

Orodruin
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Thank you for asking interesting questions. I enjoyed writing that post rather than the usual "you-need-to-take-relativity-of-simultaneity-into-account" post. For once an A-level thread that is actually on A-level ...

Yes, I was indeed thinking of the spatial volume element. Does it make sense to say that a spatial volume element is the wedge product of three spatial vectors - and then to go further and say that that the hodges dual of this 3-form is a vector, so that we then wind up saying that a spatial volume element can be represented by a vector?

Sorry that the question wasn't too clear, trying to get the semantics down is probably a large part of my issue.
You're asking if it makes sense to say that $dV= dx \wedge dy \wedge dz$ which is true, however, i get a scalar if I'm ONLY working in 3 space. If I'm working in 4 space, then I will get a vector, but it will be in another direction.

So to make this a bit more clear, I will show the work for 3 space, then 4 space. We can even add a time element if you want for 4 space!

3 space: Let my basis be $\omega = \{dx, dy, dz\}=dx \wedge dy \wedge dz$ the hodge dual has the property of $a \wedge \star a = g(a,a) \omega$ where $g(a,a)$ is the inner product. Thus, if I let $dV= dx \wedge dy \wedge dz$ then $dV \wedge \star dV = g(dV,dV) dx \wedge dy \wedge dz$ It is clear that the inner product will return one. So, $\star dV = 1$ aka, the hodge dual will give us a scalar, not a vector.

4 space: Let my basis be $\omega =\{dt, dx, dy, dz \} = dt \wedge dx \wedge dy \wedge dz$ We ask ourselves what will the hodge dual return if we do the same exercise as above? Well, let us see. $dV \wedge \star dV = g(v,v) dt \wedge dx \wedge dy \wedge dz$ once again, the inner product will return one. So what we have is $dx \wedge dy \wedge dz \wedge \star dV = dt \wedge dx \wedge dy \wedge dz$ so, we know that $\star dV = A dt$ because we have a dt on the right side, but nothing on the left side. The only think left is to get the $\star dV$ to the front. Wedge products have the property that $dx \wedge dy = -dy \wedge dx$ which i think is the anti-symmetric property. Using this on our left side, we will go step by step so we are very clear on what happens. $dx \wedge dy \wedge dz \wedge \star dV = -dx \wedge dy \wedge \star dV \wedge dz = dx \wedge \star dV \wedge dy \wedge dz = -\star dV \wedge dx \wedge dy \wedge dz$. So, to make this equal to the right side, $\star dV = -dt$ which would be a vector.

I think for your question on moving observers, this page will be of more use than I: http://physics.oregonstate.edu/coursewikis/GGR/book/ggr/dust