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I have to do a task for my special relativity class:

Prove that: If $\Gamma_{\mu\nu}a^{\mu}b^{\nu}$ is a scalar for arbitrary four-vectors $a^{\nu}$, $b^{\nu}$, then $\Gamma_{\mu\nu}$ is a four-tensor.

I am totally confused about what to do since the answer seems too simple for me. The only possibility to make this a scalar is to let $\Gamma_{\mu\nu}$ be a 4x4-matrix when $a^{\nu}$ and $b^{\nu}$ are four-vectors, right? I didn't have much linear algebra before, so I'd be thankful for any help on where to start and where to go.

Qmechanic
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RickJames
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1 Answers1

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Well to address your actual question the answer would be ~yes. Since its a scalar and both $a^{\mu}$ and $b^{\mu}$ are 4-vectors $\Gamma_{\mu \nu}$ must be a 2nd rank tensor. In the context of Special Relativity the underlying space is Minkowski space and thats what they mean by 4-tensor (This was already pointed out in the comments by Secavara) and these can be REPRESENTED by a 4x4 matrix.

It might seem a little abstract but that exact configuration is very commonly utilized in SR. For example when you define the inner product which grabs two 4-vectors and gives you a scalar you commonly do this $a \cdot b = a^{\mu} b_{\mu} = a^{\mu} g_{\mu \nu} b^{\nu}$ where $g_{\mu \nu}$ is the Minkowski metric (4x4 representation). You can see that the last term is a special case of the object you have on your question.

As to how you can see that $\Gamma_{\mu \nu}$ is that kind of tensor you can perform a general coordinate transformation. Since a scalar should be invariant under such transformation you can check the only way that it can be satisfied is if Gamma is a 4-tensor.

Fabian Ballar
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