# Constant time-like vector fields-Maxwell's equations

1. Nov 17, 2013

### WannabeNewton

Hello there! I have a more or less silly and possibly obvious question that's been bugging me a lot. Here's the premise:

Take a space-time $(M,g_{ab})$ that admits a covariantly constant (unit) time-like vector field $\xi^a$ i.e. $\nabla_a \xi^b = 0$ (such a space-time must necessarily have a degenerate Ricci tensor). Given an electromagnetic field $F_{ab}$, we can decompose it relative to $\xi^a$ into an electric and magnetic field as per $E^a = F^{a}{}{}_{b}\xi^b$ and $B^a = \frac{1}{2}\epsilon^{abcd}F_{cd}\xi_b$.

It can be shown as a consequence of $\nabla_a \xi^b = 0$ that $\nabla^a F_{ab} = J_b$ and $\nabla_{[a}F_{bc]} = 0$ if and only if $D_a B^a =0$, $\epsilon^{abc}D_b E_c = -\xi^b \nabla_b B^a$, $D_a E^a = \rho$, and $\epsilon^{abc}D_b B_c = j^a + \xi^b \nabla_b E^a$ where $D_a$ is the induced derivative operator on the one-parameter family of space-like hypersurfaces orthogonal to $\xi^a$ (such a family exists because $\xi_{[a}\nabla_{b}\xi_{c]} = 0$ trivially), $J^a$ is the 4-current density, $j^a$ is the 3-current density, $\rho$ is the charge density, and $\epsilon_{abc} = \xi^d \epsilon_{dabc}$ is the induced volume element on the space-like hypersurfaces.

If you consider the special case where $g_{ab} = \eta_{ab}$ and $M = \mathbb{R}^{4}$ (Minkowski space-time) then the integral curves of $\xi^a$ are just worldlines of inertial observers and the 4 equations above for the electric and magnetic fields just take on the usual form of Maxwell's equations in global inertial frames: $\vec{\nabla}\cdot \vec{E} = \rho$, $\vec{\nabla}\cdot \vec{B} = 0$, $\vec{\nabla}\times \vec{E} = -\partial_t \vec{B}$, and $\vec{\nabla}\times \vec{B} = \vec{j} + \partial_{t}\vec{E}$.

But what if we aren't in Minkowski space-time? What physical significance/interpretation (if any) does the family of observers following orbits of $\xi^a$ have then? What bothered me is that instead of simply requiring any family of locally inertial observers i.e. simply having $\xi^a \nabla_a \xi^b = 0$, we required a family of observers having $\nabla_a \xi^b = 0$ and such observers are not only locally inertial but also following orbits of a time-like killing field that is also twist-free.

But why physically do we need this (much) stronger restriction? I would imagine that if we wanted the covariant form of Maxwell's equations to decompose into a curved space-time version of the canonical vector calculus form of Maxwell's equations in a global inertial frame in flat space-time then we should be able to get that using any family of locally inertial observers.

2. Nov 17, 2013

### Mentz114

Surely, if the first condition is necessary for current conservation, this gives the strong restriction.

Are you suggesting that a weaker condition on $\xi^a$ could also give the required conservation ?

3. Nov 18, 2013

### WannabeNewton

The necessary condition for current conservation is simply $\nabla^a F_{ab} = J_b$. The condition $\nabla_a \xi^b = 0$ is a condition on the background time-like congruence of observer worldlines relative to whom $F_{ab}$ is decomposed into its electric and magnetic components. It is a necessary and sufficient condition in order for $D_a B^a =0$, $\epsilon^{abc}D_b E_c = -\xi^b \nabla_b B^a$, $D_a E^a = \rho$, and $\epsilon^{abc}D_b B_c = j^a + \xi^b \nabla_b E^a$ to be equivalent to $\nabla^a F_{ab} = J_b$ and $\nabla_{[a}F_{bc]} = 0$.

I'm asking: why physically do we end up requiring more than just an arbitrary family of locally inertial observers to get the aforementioned equivalence? More explicitly: why physically do we need a family of locally inertial observers following orbits of a twist-free time-like killing field $\xi^a$ in order to decompose the covariant form of Maxwell's equations relative to $\xi^a$ into a 3+1 curved space-time version of the usual vector calculus form of Maxwell's equations that inertial observers in flat space-time obtain?

4. Nov 18, 2013

### martinbn

Nice question, I would like to see a good explanation. My very uninformed an uneducated opinion is that the condition is imposed only to make the resulting 3+1 Maxwell equations simple and with a simple derivation. But I may be very wrong and there might be a good reason for the choice. Geroch doesn't make any remarks about it. I am guessing that you are looking at his lecture notes.

5. Nov 18, 2013

### George Jones

Staff Emeritus
What I write below may way off base. I haven't done any calculations, and I haven't even really looked at the math in the above posts, as I am reading this while I am on a short tea break at work.

In Minkowski spacetime, is it possible to have a timelike congruence such that each integral curve has zero 4-acceleration, and such that the congruence has non-zero twist (vorticity)?

If so, do Mawell's equations take their standard form for such a congruence?

6. Nov 20, 2013

### Mentz114

OK, I understand all the words here. But I'm not sure how to distinguish a 'physical reason' from a 'geometric reason'. We are rather hoping the physics is all in the geometry are we not ? The geometry says we need such a family of observers so the physical reason is also that.

I mentioned conservation in my previous ill-worded post and I still think that is the key. Maybe the equations lose their invariance under some transformation in the presence of acceleration. Certainly a worldline that does not allow the possibility of some conserved or invariant quantity cannot be expected to allow such local constructs.

Maybe I'm still missing the point, but that's my best shot right now.

7. Nov 20, 2013

### WannabeNewton

Thanks for the replies guys.

George to answer your question, it would seem the answer is yes you can have a family of inertial observers in flat space-time such that the congruence has non-vanishing twist. For example, if we are in a background global inertial frame $(t,\vec{x})$ whose $x$-axis is oriented in the direction of motion of the family, so that the family is parametrized by $y$, then we can simultaneously boost from each observer in the family to get $\xi^{\mu} = \gamma(y)\partial_{t}^{\mu} + \gamma(y)v(y)\partial_{x}^{\mu}$. Note that $\xi^{\mu}\partial_{\mu}\xi^{\nu} = \xi^{t}\partial_{t}\xi^{\nu} + \xi^{x}\partial_{x}\xi^{\nu} = 0$.

Then $\omega^{\mu} = -\gamma^2 v' \partial_{z}^{\mu}$. Moreover, since $\xi^{\mu}$ is a geodesic congruence, the rotation tensor $\omega_{\mu\nu}$ is simply given by $\omega_{\mu\nu} = \partial_{[\nu}\xi_{\mu]}$ hence since $\omega_{\mu\nu} = 0$ if and only if $\omega^{\mu} = 0$ we have in this case that $\partial_{[\mu}\xi_{\nu]} \neq 0$.

But then $D_{\mu}E^{\mu} = h^{\beta}{}{}_{\alpha}\partial_{\beta}E^{\alpha}\\ = \xi^{\beta}\partial^{\alpha}F_{\alpha\beta} + F^{[\alpha\beta]}\partial_{[\alpha}\xi_{\beta]} + \xi^{\beta}\xi^{[\alpha}\xi^{\gamma]}\partial_{\beta}F_{[\alpha\gamma]}+ F_{\alpha\gamma}\xi^{\alpha}\xi^{\beta}\partial_{\beta}\xi^{\gamma}\\= \rho + F^{[\alpha\beta]}\partial_{[\alpha}\xi_{\beta]}\neq \rho$

The others seem to fail as well. For example, $D_{\mu}B^{\mu} = -\frac{1}{2}\epsilon^{\beta[\alpha\mu\nu]}\xi_{\beta}\partial_{[\alpha}F_{\mu\nu]} + \frac{1}{2}\epsilon^{[\alpha\beta]\mu\nu}F_{\mu\nu}\partial_{[\alpha}\xi_{\beta]}\\ + \frac{1}{2}\epsilon^{[\alpha\beta]\mu\nu}\xi^{\gamma}\xi_{[\alpha}\xi_{\beta]}\partial_{\gamma}F_{\mu\nu} + \frac{1}{2}\epsilon^{\alpha\beta\mu\nu}\xi_{\alpha}F_{\mu\nu}\xi^{ \gamma}\partial_{ \gamma}\xi_{\beta} \\= \frac{1}{2}\epsilon^{[\alpha\beta]\mu\nu}F_{\mu\nu}\partial_{[\alpha}\xi_{\beta]}\neq 0$

So as far as the two Gauss's law equations go, they fail to hold at every point on a time-like congruence even if the congruence is one of inertial observers in flat space-time if the congruence fails to be twist-free i.e. if $\xi^{\mu}$ is not hypersurface orthogonal. I still don't really have much of a physical intuition for why though. Perhaps there isn't any physical intuition for this?

8. Nov 20, 2013

### George Jones

Staff Emeritus
Nice example. I was hoping this was the case, as this means that the Minkowski space case and the cases with non-zero spacetime curvature.

For non-zero twist, even in Minkowski spacetime, I didn't expect (the standard version) of Gauss's law to hold. My guess is that some physical intuition can be built up. As usual I hope to try a very simple-minded calculation to see if any of this can be fleshed out.

My gues is

9. Dec 4, 2013

### George Jones

Staff Emeritus
Unfortunately, this didn't happen, and now I don't think that I am going to get back to this. I was waylaid :uhh: by work, family, and other interesting threads, e.g.,

another thread for which I should find time to make a longer post.

My idea was that the expressions don't take their usual forms even in special relativity when rotating frames are used, so it is not surprising non-zero twist mucks things up.

See, for example, the papers

"Maxwell's Equations in a Rotating Reference Frame", Modesitt, Am. J. Phys, 38, 1487 (1970)

"A Question in General Relativity", Schiff, Pooc. Natl. Acad. Sci., 25, 391 (1939)