Parallel plate capacitor in the Rindler metric

[pervect]

Does anyone have a reference or solution for a parallel plate capacitor in the Rindler metric? I'm particularly interested in the case where the capacitor plates are in the xz or yz planes, z being the direction of the acceleration.

The motivation is to get an idea how a transmission line would work in the low frequency limit in the z direction on Einstein's elevator. The inductance of the line wouldn't matter in the low frequency limit, only the capaitance. Thus knowing the solution to the capacitor problem should give information on the signal transmission. In particular, would the electric field between the plates, i.e. in the y direction, be a function of the height, z?

I am unclear if circuit theory is up to this task (I suspect not), but knowing $F_{ab}$, i.e. the field between the plates, should be sufficient.

 

[PeterDonis]

I'm particularly interested in the case where the capacitor plates are in the xz or yz planes, z being the direction of the acceleration.

Meaning that the plates are "vertical" and the direction of the field between them is "horizontal"?

The motivation is to get an idea how a transmission line would work in the low frequency limit in the z direction on Einstein's elevator. The inductance of the line wouldn't matter in the low frequency limit, only the capaitance.

Wouldn't you want plates that were "horizontal" and the field between them "vertical" (instead of the reverse, as above) in order to model the capacitance of a transmission line in the vertical direction?

 

[Paul Colby]

For an x directed E-field to have a z dependence it would have a non-vanishing curl. I see no reason for a time dependent B-field in this static case.

 

[pervect]

I took a stab calculating it myself, not having found a reference that did it. What I'm currently getting is that in a coordinate basis the (t,x) component of the Faraday tensor F is constant, but in an orthonormal basis the (t,x) component of F is proportional to 1/z.

The relevant equation is $\epsilon^{abcd} ( F_{ab;c} + F_{bc;a} + F_{ca;b} ) = 0$

The Levi-Civita tensor, $\epsilon^{abcd}$ isn't strictly necessary, but it's convenient to take the hodges dual of the 3-form with the symbolic algebra program I'm using. Then we set the 4 components of the resulting dual vector to the 3-form equal to zero.

My interpretation of this is that since the plate spacing in the x-direction is uniform, if we use a local voltmeter to measure the voltage between the plates as a function of height, the local voltmeter reading will drop with height, being inversely proportional to z. However, the coordinate value of F(t,x) appears to be constant.

A crude asci diagram

x          x
x          x
x          x
x          x
x          x

z runs vertically, x runs horizontally. The direction of acceleration is the z-direciton, up the page.

 

[Paul Colby]

Isn't it true that,

$F_{ab;c}+F_{bc;a}+F_{ca;b} = F_{ab,c}+F_{bc,a}+F_{ca,b} = 0$

which is independent of the metric?

 

[pervect]

Isn't it true that,

$F_{ab;c}+F_{bc;a}+F_{ca;b} = F_{ab,c}+F_{bc,a}+F_{ca,b} = 0$

which is independent of the metric?

I believe that's true. However, I'm just grinding through the components with a symbolic algebra package. I'd have to deal with setting 64 components to zero if I did it that way. But by taking the dual I can only need to set 4 components to zero. Multiplying the rank-3 zero tensor above by $\epsilon^{abcd}$ still results in a zero tensor, but it reduces the number of components one need to worry about from 64 to 4.

Alternatively I could take the dual earlier, I believe, by taking $\nabla_a M^{ab} = 0$, where M=*F, the Maxwell tensor.

I think we are more or less agreeing that in the coordinate basis, E_x remains constant? If so, that's encouraging - it's easy to make a mistake :(.

 

[Paul Colby]

I think we are more or less agreeing that in the coordinate basis, E_x remains constant?

That would be my point. The equation in #4 less the fully antisymmetric tensor, is just the exterior derivative of the exterior derivative of the potential 1-form. $A$. Thus,

$d^2 A = 0$ (*)

quite independent of geometry or metric. Now the point I raised in #3 is based on this in that, (*) in a 3-space view,

$\nabla\times E= -\frac{\partial B}{\partial t}$ ,

or, I can't have a curl of E without a time dependent B. I can't see being at rest WRT a uniformly accelerating capacitor involves any current flow or any B fields for that matter. This also implies that the voltage drop across the plates must be independent of z since the integral about a closed curve must be zero with no B's in the problem. The voltage across plates is z dependent only if there is also a non-zero tangent field to the plate. This is disallowed for conductors in a static case. Now, I've treated the problem entirely as a classical boundary value problem and have ignored the solid state physics of the electron distribution in the plates. I think this is warranted but don't really know.

 

[pervect]

That would be my point. The equation in #4 less the fully antisymmetric tensor, is just the exterior derivative of the exterior derivative of the potential 1-form. $A$. Thus,

$d^2 A = 0$ (*)

quite independent of geometry or metric. Now the point I raised in #3 is based on this in that, (*) in a 3-space view,

$\nabla\times E= -\frac{\partial B}{\partial t}$ ,

or, I can't have a curl of E without a time dependent B. I can't see being at rest WRT a uniformly accelerating capacitor involves any current flow or any B fields for that matter. This also implies that the voltage drop across the plates must be independent of z since the integral about a closed curve must be zero with no B's in the problem. The voltage across plates is z dependent only if there is also a non-zero tangent field to the plate. This is disallowed for conductors in a static case. Now, I've treated the problem entirely as a classical boundary value problem and have ignored the solid state physics of the electron distribution in the plates. I think this is warranted but don't really know.

We may have a difference in definitions.

If I'm understanding your argument, you are arguing that in any coordinates, regardless of the metric,
F = dt ^ dz is a two form that satisfies Maxwell's equation. I believe this is basically correct as long as t is timelike and z is spacelike, certainly the results I am getting are consistent with this statement.

However, if we consider the metric
$$-z^2 dt^2 + dx^2 + dy^2 + dz^2$$

we see that dt is not in general a unit one-form. The issues is that the coordinate second dt is not the same as the proper second.

I would say that voltage is defined in an orthonormal basis. In the metric above, only at z=1 is dt of unit length, the coordinate where the proper second is the same as a coordinate second. The voltage, as measured in a local orthonormal coordinate system, therefore must vary with height. $z \,dt$ is a unit one-form, $\frac{1}{z} \frac{\partial}{\partial t}$ is a unit vector.

So I would say that this physically significant voltage as measured by a local observer using the local standard proper second at a specific value of z must vary with height.

 

[Paul Colby]

I would say that voltage is defined in an orthonormal basis.

I would say voltage is defined with a volt meter. A voltmeter adds an (idealized) conduction path to the problem (the meter leads) and, oddly enough, the voltage reading depends on this path. For static problems free of B fields one is free to vary the voltage measurement path both in space and time provided the end points remain fixed in time.

Now space and time coordinates are subject to definition here. I'm choosing coordinates in which the capacitor plates, volt meter and observer are at rest and, in a Newtonian sense, all subject to a uniform acceleration. Locally this is identical to the capacitor being on the Earth's surface but with a user selectable value of $g$. In GR, this can be chosen locally as an orthogonal spatial metric with time dependent on z if I recall correctly. In this frame a charged capacitor generates no time dependent magnetic fields using Maxwells equations. (If they did why can't we use tall charged capacitors to generate power )

Which coordinates and volt meter path are you choosing? Just writing a metric down doesn't specify the problem sufficiently.

 

[pervect]

Writing down the metric defines the coordinates, and vica-versa. The metric I'm using is:

$$-z^2 dt^2 + dx^2 + dy^2 + dz^2$$

the specific form of the Rindler metric, which concisely defines the specifics of the coordinates. So that sets the coordinates, I believe.

At any time t, the left plate is defined by x=0, the right plate by x=1, with y and z varying from -infinity to + infinity.

The path is defined to be a path in the z=constant plane, from one plate to the other. Because of the symmetry, I believe it doesn't matter how we get from one plate to the other as long we stay in the z=constant plane. In the region between the plates, E is independent of x and y, and can only depend on z by symmetry.

I've done some more calculations, and pretty much convinced myself that the Maxwell tensor must be

$$M^{ab} = M_{ab} = M^{\hat{a}\hat{b}} = M_{\hat{a}\hat{b}} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & E(z) \\ 0 & 0 & -E(z) & 0 \end{bmatrix}$$

Here E is the E-field in the x direction, the x-direction being $\partial/\partial x$.

It doesn't matter whether we use the coordinate basis $M^{ab}$ the coordinate co-basis $M_{ab}$, the orthonormal basis $M^{\hat{a}\hat{b}}$ or the orthonormal cobasis $M_{\hat{a}\hat{b}}$ because the coordinate basis vectors $\partial / \partial y$ and $\partial / \partial z$ are positive and have unit magnitude. The basis choice would matter for the Faraday tensor, but not the Maxwell.

Next,l we need to set the divergence $\nabla_a M^{ba} = 0$ to find the field between the plates.

The only non-zero term in the divergence is the y-term, index 2. Which is:

$$\frac{\partial}{\partial z} E(z) + \frac{1}{z} E(z) = 0$$

which has the solution E(z) = k/z. To get the voltage we multiply E(z) by the distance between plates, which is 1.

I used a symbolic algebra program to calculate the covariant derivative, but I tried to see if it made sense from looking at a manual calculation, though that's more error prone. I believe the expression for the covariant derivative is:

$$\nabla_a M^{ba} = \partial_a M^{ba} + \sum_d \Gamma^b{}_{ad} M^{da} + \sum_d \Gamma^a{}_{ad} M^{bd}$$

The non-zero Christoffel symbols are $\Gamma^3{}_{00}=z$ , $\Gamma^0{}_{03} = \Gamma^0{}_{30} = 1/z$

The middle term didn't result in any non-zero contributions, the left term gives us $\partial_z M^{23}$ and the right term gives us $\Gamma^0{}_{03} M^{23}$.

 

[Paul Colby]

I would argue that one should be using the Faraday tensor as well which corresponds to

$\nabla \times E +\frac{\partial B}{\partial t} = 0$
$\nabla \cdot B = 0$

which holds independent of metric rather than (in flat space-time leaving off the geometric components you calculate) the Maxwell tensor which corresponds to,

$\nabla\times H - \frac{\partial D}{\partial t} = J$
$\nabla\cdot D = \rho$

Since a time dependent B field doesn't appear in your calculation and is required by the first equation above how is it a solution to the full system?

 

[pervect]

I suspect that the 3d formalism isn't properly handling what happens when $g_{00}$ is a function of position, in this case z.

Using the 4d-formalism, the easiest route I think is from F to *F, ie to M, and then to set the divergence of M to zero, as I did above.

 

[Paul Colby]

I suspect that the 3d formalism isn't properly handling what happens when $g_{00}$ is a function of position, in this case z.

Using the 4d-formalism, the easiest route I think is from F to *F, ie to M, and then to set the divergence of M to zero, as I did above.

your z dependent x directed E field has a nonzero curl. Faraday’s law still holds without modification due to the metric. Your solution violates Faraday’s law so I don’t accept it.

 

[pervect]

your z dependent x directed E field has a nonzero curl. Faraday’s law still holds without modification due to the metric. Your solution violates Faraday’s law so I don’t accept it.

I think we're going around in a loop. I'm not trusting the argument from the 3d formalism in this case, and I've done the problem in enough different ways in the 4d formalism that I'm beginning to believe I haven't made a mistake.

I do get $\nabla[_aF_{bc}] = 0$, as well as M=*F, $\nabla_a M^{ba} = 0$. The second route was much simpler to carry out, as I only needed to compute one divergence rather than 3.

 

[PeterDonis]

in a 3-space view,

$\nabla\times E= -\frac{\partial B}{\partial t}$ ,

or, I can't have a curl of E without a time dependent B.

In the coordinates @pervect is using, the coordinate value of $E$, which is what appears in Maxwell's Equations as you are writing them, is constant. So there is no curl.

 

[Paul Colby]

In the coordinates @pervect is using, the coordinate value of $E$, which is what appears in Maxwell's Equations as you are writing them, is constant. So there is no curl.

Hence the voltage drop (integral of E over the coordinate x) is independent of z since the separation of plates is constant. All is as I have said. Thank you for making my point.

I think the arguments presented till now have blurred the distinction between $E$, $B$ and $D$, $H$. The voltage drop is defined in terms of $E$, not $D$.

 

[PeterDonis]

Hence the voltage drop (integral of E over the coordinate x)

That is not the voltage drop (since you have said that by "voltage drop" you mean what is actually measured by a voltmeter) in Rindler coordinates. If the capacitor were at rest in an inertial frame, it would be, but the capacitor is not at rest in an inertial frame.

 

[Paul Colby]

That is not the voltage drop (since you have said that by "voltage drop" you mean what is actually measured by a voltmeter) in Rindler coordinates. If the capacitor were at rest in an inertial frame, it would be, but the capacitor is not at rest in an inertial frame.

The voltage drop is the line integral of $E$ not $D$ as the OP suggests.

 

[PeterDonis]

The voltage drop is the line integral of $E$ not $D$ as the OP suggests.

As I understand it, @pervect is assuming vacuum between the plates so there is no difference.

 

[Paul Colby]

As I understand it, @pervect is assuming vacuum between the plates so there is no difference.

Apparently not from his results.

 

[Paul Colby]

The tensor relation between the Maxwell tensor and the Faraday tensor is,

$M_{ij} = -\sqrt{\frac{\epsilon_o}{\mu_o}}\left( \frac{1}{2\sqrt{-g}}g_{ia}g_{jb}\epsilon^{abcd}F_{cd} \right)$

This is not a simple relation in general. $D$ and $E$ should not be confused.

 

[Paul Colby]

That is not the voltage drop (since you have said that by "voltage drop" you mean what is actually measured by a voltmeter) in Rindler coordinates. If the capacitor were at rest in an inertial frame, it would be, but the capacitor is not at rest in an inertial frame.

Actually, is this what was said in #9? I did stipulate the end points of the additional conduction path were stationary in the coordinates which means at rest wrt the capacitor in this case. Clearly the voltmeter leads may flop about at will if the B field is zero.

 

[pervect]

In the coordinates @pervect is using, the coordinate value of $E$, which is what appears in Maxwell's Equations as you are writing them, is constant. So there is no curl.

Yes. In particular, I am getting that $F_{ab}$ has constant components in a coordinate basis, though $F^{ab}$ does not. As I write this, though, I realize that $F^{ab}$ doesn't matter, more below.

E is defined in an orthonormal cobasis, in particular the cobasis below. With the metric, we could define the corresponding basis vectors, though as we write this I realize it's not really necessary to do so in this case.

$$\omega^0 = z\,dt \quad \omega^1 = dx \quad \omega^2=dy \quad \omega^3 = dz$$.

I denote the Faraday tensor in this basis as $F_{\hat{a}\hat{b}}$. It has the components

$$F_{\hat{a}\hat{b}} = \begin{bmatrix} 0 & E(z) & 0 & 0 \\ -E(z) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

which we could write concisely as $F = E(z) \, {\omega^1} \wedge {\omega^0}$. (Hopefully I didn't mess up the order or signs, but it doesn't make much difference to the end result if I did, since +0 = -0).

Converting from the orthonormal basis to the coordinate basis $F_{ab}$ we have

$$F_{ab} = \begin{bmatrix} 0 & zE(z) & 0 & 0 \\ -z E(z) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Which we could write concisely as $F = z E(z)\,dx \wedge dt$

With E(z) = k/z, we can re-write this as $F = k \, dx \wedge dt$. So F, by which we mean $F_{ab}$ since we're treating it as a differential form, has constant components when E = k/z.

I'm not super familiar with exterior derivatives, if I were I suppose I'd be more confident in saying that the fact that $F_{ab}$ has constant components implies that the exterior derivative vanishes regardless of the metric, since my references agree with the statement that we don't need a metric to define the exterior derivative. But since I'm not used to using exterior derivatives, and am used to grinding compoonents, I'll settle at this point for grinding through the formula that does need a metric with GRTensor, namely $\nabla[_aF_{bc}]$. Which evaluates to zero. With a note to myself that I'm probably making my life unnecessarily difficult.

As far as $F^{ab}$ goes, a metric would be needed to raise the indices. So it's irrelevant to the metric-independent argument using differential forms.

 

[pervect]

One other thing that's worth mentioning. I was able to confirm, by direct calculation using the methods I'm used to, that $F = dx \wedge dt$ satisfies $\nabla{[_aF_{bc}]}=0$ for a general metric - and the implied Levi-civita connection associated with said general metric - and not just the Rindler metric.

 

[PeterDonis]

One other thing that's worth mentioning. I was able to confirm, by direct calculation using the methods I'm used to, that $F = dx \wedge dt$ satisfies $\nabla{[_aF_{bc}]}=0$ for a general metric - and the implied Levi-civita connection associated with said general metric - and not just the Rindler metric.

Isn't this an identity (the first of the two Maxwell Equations in covariant form), since the commutation of partial derivatives cancels out those terms and the antisymmetry cancels out the connection coefficient terms?

 

[pervect]

Isn't this an identity (the first of the two Maxwell Equations in covariant form), since the commutation of partial derivatives cancels out those terms and the antisymmetry cancels out the connection coefficient terms?

I gather it is an identity yes, but I don't follow the proof of why it's an identity. I see that the partial derivatives vanish, but I don't see how the antisymmetry "cancels out the connection coefficient terms".

 

[PeterDonis]

I don't see how the antisymmetry "cancels out the connection coefficient terms".

All of the connection coefficients are symmetric in the two lower indexes, so any sum of them that is antisymmetric in those indexes must vanish.