# Geometric Algebra: Signs of electromagnetic field tensor components?

#### Peeter

[SOLVED] Geometric Algebra: Signs of electromagnetic field tensor components?

Here's a question that may look like an E&M question, but is really just a geometric algebra question. In particular, I've got a sign off by 1 somewhere I think and I wonder if somebody can spot it.

PF isn't accepting what I wrote (my latex appears to trigger an internal database error), so I've converted it to standalone latex and attached my notes and question as a pdf file. I've also attached what I attempted to post to PF, for reply purposes (so it can be cut and pasted from selectively if desired).

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#### Peeter

Here's an abbreviated version of the question I originally attempted to post with full context. Essentially, have a bivector:

$$F = B_1 \gamma_{2} \wedge \gamma_{3} + B_2 \gamma_{3} \wedge \gamma_{1} + B_3 \gamma_{1} \wedge \gamma_{2}$$

Where the $\{\gamma_\mu\}$ vectors are an orthonormal'' basis with respect to a $(+,-,-,-)$ metric dot product.

Calulate the coordinates of the bivector F by taking dot products with the reciprocal frame vectors:

$$F^{\mu\nu} = (\gamma^\nu \wedge \gamma^\mu) \cdot F$$

Where the reciprocal frame vectors $\{\gamma^i\}$ are those vectors defined by:

$$\gamma^\mu \cdot \gamma_\nu = \delta_{\mu\nu}$$

ie:
$\gamma^0 = \gamma_0, and \gamma^i = -\gamma_i$.

If I calculate this for just one pair of index, $12$ say, I get:

\begin{align*} F^{12} &= (\gamma^2 \wedge \gamma^1) \cdot ( \gamma_1 \wedge \gamma_2 ) B_3 \\ &= \gamma^2 \cdot (\gamma^1 \cdot ( \gamma_1 \wedge \gamma_2 )) B_3 \\ &= \gamma^2 \cdot ( \gamma^1 \cdot \gamma_1 \gamma_2 - \gamma^1 \cdot \gamma_2 \gamma_1 ) B_3 \\ &= \gamma^2 \cdot ( -\gamma_1 \cdot \gamma_1 \gamma_2 ) B_3 \\ &= \gamma^2 \cdot \gamma_2 B_3 \\ &= - \gamma_2 \cdot \gamma_2 B_3 \\ &= - (-1) B_3 \\ &= B_3 \\ \end{align*}

I don't see a mistake in my calculation, but the sign is inverted compared to the text (also in the pdf file). Since this isn't listed in the errata even after two editions my assumption was that I had am error in my calculation somewhere.

#### OrderOfThings

Calulate the coordinates of the bivector F by taking dot products with the reciprocal frame vectors:

$$F^{\mu\nu} = (\gamma^\nu \wedge \gamma^\mu) \cdot F$$
This does not seem right. I think that

$$(\gamma_1\wedge\gamma_2)\cdot (\gamma_1\wedge\gamma_2)=1$$

So $\gamma_1\wedge\gamma_2$ is reciprocal to itself.

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#### Peeter

So $\gamma_1\wedge\gamma_2$ is reciprocal to itself.
With the +--- dot product, I don't believe that's the case. The reciprocal may have to include a sign adjustment:

$$(\gamma_1 \wedge \gamma_2) \cdot (\gamma_1 \wedge \gamma_2) = \gamma_1 \cdot (\gamma_2 \cdot (\gamma_1 \wedge \gamma_2)) = \gamma_1 \cdot (-\gamma_2 \cdot \gamma_2 \gamma_1) = -(\gamma_2 \cdot \gamma_2 )(\gamma_1 \cdot \gamma_1) = -(-1)(-1) = -1$$

(for a $\gamma_0 \wedge \gamma_i$ the reciprocal should be itself).

#### anektos

Re-order components

Here's an abbreviated version of the question I originally attempted to post with full context. Essentially, have a bivector:

$$F = B_1 \gamma_{2} \wedge \gamma_{3} + B_2 \gamma_{3} \wedge \gamma_{1} + B_3 \gamma_{1} \wedge \gamma_{2}$$

Where the $\{\gamma_\mu\}$ vectors are an orthonormal'' basis with respect to a $(+,-,-,-)$ metric dot product.

Calulate the coordinates of the bivector F by taking dot products with the reciprocal frame vectors:

$$F^{\mu\nu} = (\gamma^\nu \wedge \gamma^\mu) \cdot F$$

Where the reciprocal frame vectors $\{\gamma^i\}$ are those vectors defined by:

$$\gamma^\mu \cdot \gamma_\nu = \delta_{\mu\nu}$$

ie:
$\gamma^0 = \gamma_0, and \gamma^i = -\gamma_i$.

If I calculate this for just one pair of index, $12$ say, I get:

\begin{align*} F^{12} &= (\gamma^2 \wedge \gamma^1) \cdot ( \gamma_1 \wedge \gamma_2 ) B_3 \\ &= \gamma^2 \cdot (\gamma^1 \cdot ( \gamma_1 \wedge \gamma_2 )) B_3 \\ &= \gamma^2 \cdot ( \gamma^1 \cdot \gamma_1 \gamma_2 - \gamma^1 \cdot \gamma_2 \gamma_1 ) B_3 \\ &= \gamma^2 \cdot ( -\gamma_1 \cdot \gamma_1 \gamma_2 ) B_3 \\ &= \gamma^2 \cdot \gamma_2 B_3 \\ &= - \gamma_2 \cdot \gamma_2 B_3 \\ &= - (-1) B_3 \\ &= B_3 \\ \end{align*}

I don't see a mistake in my calculation, but the sign is inverted compared to the text (also in the pdf file). Since this isn't listed in the errata even after two editions my assumption was that I had am error in my calculation somewhere.
If you want F^{12} you need
\begin{align*} F^{12} &= (\gamma^1 \wedge \gamma^2) \cdot (\gamma_1 \wedge \gamma_2) B_3 \end{align*} [\tex] #### Peeter If you want F^{12} you need [tex] \begin{align*} F^{12} &= (\gamma^1 \wedge \gamma^2) \cdot (\gamma_1 \wedge \gamma_2) B_3 \end{align*}
If that's the typo in their text then the electric field components of E aren't right. With:

$$\mathbf{E} = E_1 \gamma_1 \wedge \gamma_0 + E_2 \gamma_2 \wedge \gamma_0 + E_3 \gamma_3 \wedge \gamma_0$$

I get:

\begin{align*} F^{01} &= (\gamma^1 \wedge \gamma^0) \cdot (\gamma_1 \wedge \gamma_0) E_1 \\ &= -(\gamma^0 \wedge \gamma^1) \cdot (\gamma_1 \wedge \gamma_0) E_1 \\ &= - \gamma^0 \cdot (\gamma^1 \cdot (\gamma_1 \wedge \gamma_0)) E_1 \\ &= - \gamma^0 \cdot \gamma_0 E_1 \\ &= -E_1 \\ \end{align*}

This is consistent with their tensor representation of F:

$$F^{\mu\nu} = \begin{bmatrix} 0 & -E_1 & -E_2 & -E_3 \\ E_1 & 0 & -B_3 & B_2 \\ E_2 & B_3 & 0 & -B_1 \\ E_3 & -B_2 & B_1 & 0 \\ \end{bmatrix}$$

I suppose you could define it either way:

1) the way they've done it:

$$F^{\mu\nu} = (\gamma^{\nu} \wedge \gamma^{\mu}) \cdot F$$

2) your way with inverted sign:

$$F^{\mu\nu} = (\gamma^{\mu} \wedge \gamma^{\nu}) \cdot F$$

Because to reconstruct the bivector the coordinates (ie: the tensor), to reconstruct the bivector from the tensor one needs:

$$F = \pm \frac{1}{2} \sum F^{\mu\nu} \gamma_{\mu} \wedge \gamma_{\nu} = \pm \sum_{\mu < \nu} F^{\mu\nu} \gamma_{\mu} \wedge \gamma_{\nu}$$

with sign depending on which of the 1) or 2) has been used (is there a convention for this?)

Either way there's a typo in the book I think. If 1) is used to define $F^{\mu\nu}$, then the B coordinates of the tensor are off by -1. If it's 2) then the E terms are off by -1.

#### Peeter

Answering my own question about the convention, this is how I'd imagine the tensor components of this bivector ought to be defined implicitly with:

$$F = \sum_{\mu<\nu} F^{\mu\nu} \gamma_{\mu} \wedge \gamma_{\nu}$$

If that's the case, then

\begin{align*} F \cdot (\gamma^{\alpha} \wedge \gamma^{\beta}) &= \sum_{\mu<\nu} F^{\mu\nu} (\gamma_{\mu} \wedge \gamma_{\nu}) \cdot (\gamma^{\alpha} \wedge \gamma^{\beta}) \\ &= \sum_{\mu<\nu} F^{\mu\nu} ((\gamma_{\mu} \wedge \gamma_{\nu}) \cdot (\gamma^{\alpha}) \cdot \gamma^{\beta} \\ &= \sum_{\mu<\nu} F^{\mu\nu} ( \gamma_{\mu} \delta_{\nu}^{\alpha} ) \cdot \gamma^{\beta} - (\delta_{\mu}^{\alpha} \gamma_{\nu} ) \cdot \gamma^{\beta} \\ &= \sum_{\mu<\nu} F^{\mu\nu} \delta_{\mu}^{\beta} \delta_{\nu}^{\alpha} - \delta_{\mu}^{\alpha} \delta_{\nu}^{\beta} \\ &= -2 F^{\alpha\beta} \end{align*}

Therefore I think that what the book should have used was:

$$F^{\mu\nu} = \frac{1}{2} F \cdot (\gamma^{\nu} \wedge \gamma^{\mu})$$

(and there's a typo in their calculation of this for the B components).

#### Peeter

$$F = \sum_{\mu<\nu} F^{\mu\nu} \gamma_{\mu} \wedge \gamma_{\nu}$$

...

Therefore I think that what the book should have used was:

$$F^{\mu\nu} = \frac{1}{2} F \cdot (\gamma^{\nu} \wedge \gamma^{\mu})$$
Actually, I second guess myself about the factor of two. For the sum only one of the pairs of deltas can be non-zero, so you only need the factor of two if one were to define

$$F = \sum F^{\mu\nu} \gamma_{\mu} \wedge \gamma_{\nu}$$

(summing over all indexes instead). Again, a few more details in the book would have been good, unless the intention was for the reader to understand all this a lot better by figuring it out themselves:)

#### Peeter

I found my typo. It was way at the beginning (before what I reposted in the abbreviated version of my original post that the PF software didn't like). They define:

$$F = E + IB = E_1 \gamma_{10} + E_2 \gamma_{20} + E_3 \gamma_{30} + \gamma_{0123}(B_1 \gamma_{10} + B_2 \gamma_{20} + B_3 \gamma_{30})$$

Where I've written the product of the unit vectors:

$$\gamma_{\mu\nu} = \gamma_{\mu} \gamma_{\nu}$$

for short.

Expanding the IB part is where I got my sign mixed up. Example:

$$I B_3 \gamma_3 \gamma_0 = -\gamma_{0120} = \gamma_{21} = \gamma_2 \wedge \gamma_1$$

Note that this is opposite from what I used in all the subsequent calculations, which explains the off by one sign for just the B parts.

So, ... I think I now understand all the notation implied in those few pages, as well as see that there is no typo there (which makes sense given that this is a few times reprinted and its not in the errata).

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#### anektos

Sorry if I misled. I jumped in without reading all the material
and then couldn't find the time to get back and correct.

#### Peeter

Actually, what you said was very helpful. Only when I tried to reply to your post did I realize exactly what was meant by $F^{\mu\nu}$. It really isn't defined without one of the sums above (which isn't in the text). Once you pick one of those as the definition it explains the ordering choice of the reciprocal frame bivector basis to take dot products with.