Metric tensor : raising/lowering indices

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SUMMARY

The discussion focuses on the application of tensor notation in quantum electrodynamics (QED), specifically the operation involving the gamma matrices, represented as ##\gamma^\mu g_{\mu\nu} \gamma^\nu = \gamma^\mu \gamma_\mu##. Participants clarify the distinction between raised indices (##\gamma^{\mu\nu}##) and lowered indices (##\gamma_{\mu\nu}##), emphasizing the importance of understanding dual spaces and bilinear forms in tensor calculus. The conversation highlights the inadequacy of learning tensor calculus solely through physics textbooks, recommending more rigorous texts in differential geometry or tensor calculus for a comprehensive understanding.

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tb87
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Hi everyone,

I'm currently studying Griffith's Intro to Elementary Particles and in chapter 7 about QED, there's one part of an operation on tensors I don't follow in applying Feynman's rules to electron-muon scattering :

## \gamma^\mu g_{\mu\nu} \gamma^\nu = \gamma^\mu \gamma_\mu##

My teacher spent very little time on tensors and I'm really not sure 1) what's the difference between ##\gamma^{\mu\nu}## and ##\gamma_{\mu\nu}##, 2) why is the second ##\gamma##'s indice switched from ##\nu## to ##\mu## (and was also lowered).

Besides, I'm still strugling to understand the general difference between lowered and raised indices (e.g. : why I'm never seeing ##\gamma_\mu^\nu##, but other tensors are written that way). In Griffith's electrodynamics, the author says that in ##\Lambda_\mu^\nu##, it is ##\mu## that represents lines and ##\nu## that represents columns. However, this isn't consistent with everything I'm seeing on tensors.

Thanks!
Alex
 
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it is very bad idea to study tensor calculus by physics textbooks. I am sure participants of PF soon prompt you good textbook in differential geometry or in tensor calculus
 
Yes, this notation is not good. As a general guidance, consider the following...

Suppose ##V## an arbitrary finite-dimensional vector space. Then there will always exist a dual space ##V^*:V \to \mathbb{R}## such that for any ##\varphi \in V^*## and any ##v \in V## that ##\varphi(v) = \alpha \in \mathbb{R}##.

And if there exists a bilinear form on ##V##, that is ##b:V \times V \to \mathbb{R},\,\,\,b(v,w) \in \mathbb{R}## ie. an inner product(or norm), then there will always be some particular ##\varphi_v \in V^*## such that ##\varphi_v(w)=b(v,w)## for any ##w \in V##

Now if I fix some ##v \in V## and write, say ##b(v,\,\cdot) \in \mathbb{R}## then I may have the equality ##b(v,\,\cdot)= \varphi_v## That is, ##b(v,\,\cdot)## acts as though it is a dual vector.

Just carry out the obvious notational substitutions, i.e ##b \mapsto g_{\mu\nu}##, and ##v \mapsto\gamma^{\mu},\,\varphi_v \mapsto \gamma_\nu## and you have your answer
 

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