4
$\begingroup$

The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as $$dJ=0$$

Which is really nice because it can all be done without defining a metric tensor!

Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?

$\endgroup$
5
  • 1
    $\begingroup$ Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special. $\endgroup$ Commented Aug 29, 2018 at 2:05
  • $\begingroup$ @AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands. $\endgroup$
    – tparker
    Commented Aug 29, 2018 at 2:09
  • $\begingroup$ @tparker who cares for torsion anyway? :-P $\endgroup$ Commented Aug 29, 2018 at 2:10
  • $\begingroup$ Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round... $\endgroup$
    – Toffomat
    Commented Aug 29, 2018 at 10:56
  • $\begingroup$ @tparker: Thanks for the correction. I deleted my comment. $\endgroup$
    – user4552
    Commented Aug 31, 2018 at 1:48

3 Answers 3

6
$\begingroup$

Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be $$T_{\mu \nu} := -2 \frac{\delta \mathcal{L}}{\delta g^{\mu\nu}} + g_{\mu \nu} \mathcal{L},$$ which suggests to me that it may depend fundamentally on the metric structure of spacetime.

$\endgroup$
4
  • $\begingroup$ Are you sure your second term should be there? The standard definition of $T_{\mu\nu}$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with. $\endgroup$ Commented Aug 29, 2018 at 2:02
  • 1
    $\begingroup$ @AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $\sqrt{g}$ inside the derivative: $-2/\sqrt{g}\ \partial(\sqrt{g} \mathcal{L}) / \partial g^{\mu\nu}$. Doing the product rule eliminates the $\sqrt{g}$'s but leads to the two terms in my answer. $\endgroup$
    – tparker
    Commented Aug 29, 2018 at 2:04
  • 1
    $\begingroup$ Oh, I see. It basically boils down to the redefinition $\mathscr L=\sqrt{g}\mathcal L$. $\endgroup$ Commented Aug 29, 2018 at 2:07
  • $\begingroup$ Yes, exactly... $\endgroup$
    – tparker
    Commented Aug 29, 2018 at 2:07
4
$\begingroup$

If you invite "vector bundle-valued differential forms" then you can define $$ T^\mu=T^\mu_{\ \nu}dx^\nu $$ as a "vector field-valued 1-form", then you have $$ d^\nabla\star T^\mu\sim\nabla_\nu T^{\mu\nu}. $$

However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^\nabla$ and to take the Hodge dual.

Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^\mu=\frac{\delta S_m}{\delta A_\mu}, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the current 3-form $J$, you need to take the Hodge dual $J=\star j$ (needs the metric).

Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.

$\endgroup$
10
  • 2
    $\begingroup$ Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_\mu j^\mu$ in the Lagrangian can be expressed in terms of $J$ as $\star(A \wedge J)$. So the continuity equation really is metric-independent. $\endgroup$
    – tparker
    Commented Aug 31, 2018 at 4:33
  • 1
    $\begingroup$ On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar. $\endgroup$
    – tparker
    Commented Aug 31, 2018 at 4:58
  • 1
    $\begingroup$ @tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $\Omega^\mu_{\ \nu}=\frac{1}{2}R^\mu_{\ \nu\rho\sigma}dx^\rho\wedge dx^\sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.) $\endgroup$ Commented Aug 31, 2018 at 9:33
  • 1
    $\begingroup$ @tparker Okay, then let's try this. From an "elementary" point of view, the charge current 3-vector is $\mathbf{j}=\rho\mathbf{v}$, it is related to the velocity field of the charged continuum - so to turn it into a 3-form, you need a metric. From the point of view of field theory, the current appears as the functional derivative of an interaction Lagrangian. In any sufficiently "fundamental" treatment, the interaction Lagrangian cannot be defined without a metric (it will arise from a charged K-G field or a Dirac field, not a $A\wedge J$) term. $\endgroup$ Commented Aug 31, 2018 at 16:50
  • 1
    $\begingroup$ Fair point. You can formulate a Chern-Simons-type theory on a smooth manifold that really doesn't require a metric at all, but I guess coupling it to any kind of matter field with an effective current density requires introducing a metric. $\endgroup$
    – tparker
    Commented Aug 31, 2018 at 23:14
3
$\begingroup$

With the help of a Killing vector field $\xi$ one can define the current 3-form

$$J_\xi = \star\ \iota_\xi T$$

of which you can then take the exterior derivative to obtain the conservation law

$$\operatorname d J_\xi = 0.$$

Note that the metric is hidden inside both $T$ and $\xi$.

$\endgroup$
5
  • $\begingroup$ Could you explain your notation $\iota_\xi$? $\endgroup$
    – tparker
    Commented Aug 29, 2018 at 1:44
  • 1
    $\begingroup$ @tparker Interior product. $\endgroup$ Commented Aug 29, 2018 at 2:00
  • 1
    $\begingroup$ Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $\iota_\xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $\xi^a T_{ab}$. $\endgroup$
    – Mike
    Commented Aug 29, 2018 at 2:16
  • 2
    $\begingroup$ Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual. $\endgroup$
    – Mike
    Commented Aug 29, 2018 at 2:17
  • 1
    $\begingroup$ ...and definitely in the very notion of a Killing field. $\endgroup$
    – Mike
    Commented Aug 29, 2018 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.