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How would you proof that $$ \mathrm {Tr} (\mathbf{S\cdot \bar S })=0$$ where $\mathbf S$ is an element of area delimited for the 4-vectors $\mathbf u$ and $\mathbf v$ given by $$S^{\alpha \beta}\equiv u^\alpha v^\beta-u^\beta v^\alpha$$ and $$\bar S^{\alpha \beta}\equiv \frac{1}{2}\epsilon^{\alpha \beta \gamma \delta} S_{\gamma \delta}$$ is the dual of $\mathbf S$.

I used an analogy with the Maxwell field tensor $\mathbf T$. I know that $\mathrm {Tr} (\mathbf{T\cdot \bar T })=\frac{4}{c}\mathbf E \cdot \mathbf B$. Building the analog vectors $\mathbf E$ and $\mathbf B$ but with $\mathbf S$ I get that $ \mathrm {Tr} (\mathbf{S\cdot \bar S })=0$. But I'm looking for a more ilustrative solution to this problem. Any ideas?

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What is $S \cdot \bar S$ supposed to be? How would you represent it in index notation? –  Muphrid Dec 27 '12 at 3:16
    
Is the matrix multiplication. In index notation you would write something like $S^\alpha _ \beta \bar S^{\beta \delta}$ I think. I'm a beginner with the index notation. –  Anuar Dec 27 '12 at 3:38

2 Answers 2

up vote 4 down vote accepted

We start with the definition

$$\tag{1} S^{\alpha \beta}~:=~u^\alpha v^\beta-u^\beta v^\alpha.$$

Indices are raised and lowered with the metric. Up to an overall factor, one has

$$\tag{2} \bar{S}_{\alpha \beta}~\propto~ \epsilon_{\alpha \beta \gamma \delta} S^{\gamma \delta},$$

so that the matrix trace

$$ \mathrm {Tr} (\mathbf{\bar{S}\cdot S }) ~=~ \bar{S}_{\alpha \beta} S^{\beta\alpha} ~\stackrel{(2)}{\propto}~ \epsilon_{\alpha \beta \gamma \delta} S^{\gamma \delta}S^{\beta\alpha} ~\stackrel{(1)}{\propto}~ \epsilon_{\alpha \beta \gamma \delta} u^{\gamma}v^{\delta}u^{\beta}v^{\alpha}$$ $$\tag{3} ~\stackrel{(4)}\propto~ \det[\mathbf{u,v,u,v}]~=~0$$

is just the determinant of the $4\times 4$ matrix with column vectors $\mathbf{u,v,u,v}$. This is zero, because the determinant is totally antisymmetric in its column vector entries.

--

(4): See e.g. Wikipedia.

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can you tell me how do you know that the last sum is a determinant? –  Anuar Dec 31 '12 at 0:58
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I updated the answer. –  Qmechanic Dec 31 '12 at 1:19
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@Anuar: even if you don't know that it's a determinant, it's a contraction between antisymmetric $\epsilon_{\beta \gamma \dotsm}$ and symmetric $u^\beta u^\gamma$ tensors, so it vanishes automatically. –  Vibert Dec 31 '12 at 11:54

Now that there's an answer in traditional index notation, here's an alternative perspective from geometric algebra.

You have some bivector $S = u \wedge v$ and its dual $\bar S = i (u \wedge v)$, where $i$ is the spacetime pseudoscalar. $\bar S$ is an oriented plane that is entirely orthogonal to $S$. In GA, the quantity you describe would just be denoted $S \cdot \bar S$ (the dot contracts on all indices), but it's really simpler just to look at it as $S \wedge S$, which is obviously zero. An object wedged with itself necessarily results in zero. You can't form a 4-volume from a single planar element, just as you can't form a planar element from a single vector.

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Can you recommend me a book where I can check relativity with geometric algebra? –  Anuar Dec 31 '12 at 0:58
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Give a look to Doran and Lasenby. I'm still learning stuff about the GA approach to things from it, and it devotes a good bit of time to treating gravity as well. –  Muphrid Dec 31 '12 at 1:06

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