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I have some questions related to Chapter 4 of Thomas A Moore's book titled A General Relativity Workbook titled Index Notation. In all the questions $\eta_{\mu\nu}$ refers to the metric tensor and $\Lambda^\mu_{\ \ \ \nu}$ is the Lorentz Transformation matrix. My questions are as follows:

  1. Given below is a derivation given in the textbook for the relation $\eta_{\mu \nu}=\eta_{\alpha \beta}\Lambda^\alpha_{\ \ \ \mu} \Lambda^\beta_{\ \ \ \nu}$.

enter image description here

How has the author rearranged the terms in equation 4.28? As far as I understand the equation can be treated as matrix multiplication which is not supposed to be commutative. What are the rules for rearranging the terms in the equations while using this notation?

  1. How do we prove $(\Lambda^{-1})^\alpha_{\ \ \ \mu}\eta_{\alpha\nu}=\eta_{\mu\beta}\Lambda^\beta_{\ \ \ \nu}$ using the index notation of GR? Are we allowed to write $(\Lambda^{-1})^\alpha_{\ \ \ \mu}\eta_{\alpha\nu}=\eta_{\alpha\nu}(\Lambda^{-1})^\alpha_{\ \ \ \mu}$? If not then how was such a rearrangement done in the derivation above?

  2. One of the practice problems in the book reads,

enter image description here

Here $\delta^\mu_{\ \ \ \mu}$ is the Kronecker Delta which must be equal to 1 since the subscript and superscript are equal. However the hint says otherwise. Can someone explain what the correct answer is?

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  • $\begingroup$ Although quite tedious, it could help if you explicitly write out the sums in $\left(4.28\right)$ to see that this is a legit operation. Also, think about what type of object $\Lambda^\alpha\,_\mu$ is at all. Regarding your third question, think about what it means in the Einstein sum convention when the upper and lower index of an object are equal. As a last point, please type the relevant questions with MathJax and do not upload images of these equations. $\endgroup$
    – Tobias Fünke
    Commented May 27, 2021 at 14:56
  • $\begingroup$ I'd answer P4.3 first. And to answer that, I'd ask myself "what do you do when two of the indices are the same?" Once you've resolved that, the rest of your questions should be easy. $\endgroup$
    – Zo the Relativist
    Commented May 27, 2021 at 15:00
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    $\begingroup$ Hello! It is preferable to type out screenshots; for formulae, one can use MathJax. Thanks! $\endgroup$
    – jng224
    Commented May 27, 2021 at 15:51

2 Answers 2

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How has the author rearranged the terms in equation 4.28? As far as I understand the equation can be treated as matrix multiplication which is not supposed to be commutative. What are the rules for rearranging the terms in the equations while using this notation?

The reason that matrix multiplication is non-commutative (order matters) is that, in matrix notation, we use the order of the matrices to determine how to combine the elements of the matrices. That is, (for square matrices $A$, $B$,) $$(AB)_{ij} = \sum_{k=1}^n A_{ik}B_{kj} \ne \sum_{k=1}^n B_{ik}A_{kj} = (BA)_{ij}.\tag{1}$$ In index notation we avoid this by instead specifying explicitly, with the indices, how to combine the elements of the matrices (or tensors, more generally). In fact, we can see the same idea in the matrix example above by writing, for example $$(AB)_{ij} = \sum_{k=1}^n A_{ik}B_{kj} = \sum_{k=1}^n B_{kj}A_{ik}.\tag{2}$$ This is clearly true because when we write out the indices explicitly in the sum, they make sure that $A$ and $B$ are combined correctly. $A_{ik}$ and $B_{kj}$ are just numbers, and therefore commute.

When using index notation we don't write out the sums explicitly, instead summing over repeated indices, but the principle is the same. The indices take care of combining terms in the correct way, so the order of the terms doesn't matter.

Here $\delta^\mu_{\;\mu}$ is the Kronecker Delta which must be equal to 1 since the subscript and superscript are equal. However the hint says otherwise. Can someone explain what the correct answer is?

Remember that, in index notation, we implicitly sum over repeated indices. Write out the sum explicitly, $$\delta^\mu_{\;\mu} = \sum_\mu \delta^\mu_{\;\mu},\tag{3}$$ and I think you will be able to deduce the value of $\delta^\mu_{\;\mu}$.

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  • $\begingroup$ Thanks for clearing my doubts. I just have one more follow up question. In the derivation how does the author deduce equation 4.31 from equation 4.30? If $dx^\mu=0$ and $dx^\nu=0$ for some $\mu, \nu$ then there is no reason to say $\eta_{\mu \nu}-\eta_{\alpha \beta}\Lambda^{\alpha}_{\ \ \ \mu}\Lambda^{\beta}_{\ \ \ \nu}=0$ for that $\mu, \nu$ right? How then does we get equation 4.31 for all $\mu, \nu$? I did not understand the reasoning provided by the author. $\endgroup$
    – Ethan
    Commented May 27, 2021 at 16:38
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    $\begingroup$ The main point is that (4.30) must be true for every possible $dx^\mu$. So, for example, by setting $dx^\mu=(1,0,0,0)^\mu$ (so that $dx^\mu dx^\nu$ is nonzero only for $\mu=\nu=0$) (4.30) will imply $\eta_{00}-\eta_{\alpha\beta}\Lambda^\alpha_{\;0}\Lambda^\beta_{\;0}=0$. And by setting $dx^\mu=(1,1,0,0)^\mu$ you find that $\eta_{01}-\eta_{\alpha\beta}\Lambda^\alpha_{\;0}\Lambda^\beta_{\;1}=0$, etc. All components can be found this way, and the result is (4.31). $\endgroup$
    – ummg
    Commented May 27, 2021 at 16:52
  • $\begingroup$ A minor correction: setting $dx^\mu=(1,1,0,0)^\mu$ in (4.30) actually gives $(\eta_{00}-\eta_{\alpha\beta}\Lambda^\alpha_0\Lambda^\beta_0) + (\eta_{01}-\eta_{\alpha\beta}\Lambda^\alpha_0\Lambda^\beta_1) + (\eta_{10}-\eta_{\alpha\beta}\Lambda^\alpha_1\Lambda^\beta_0) + (\eta_{11}-\eta_{\alpha\beta}\Lambda^\alpha_1\Lambda^\beta_1) = 0$. But the terms with $\mu=\nu$ are zero by the previous argument, and the remaining two terms are actually identical, due to the symmetry of $\eta_\mu\nu$. ... $\endgroup$
    – ummg
    Commented May 27, 2021 at 17:11
  • $\begingroup$ That is, $\eta_{10}-\eta_{\alpha\beta}\Lambda^\alpha_{\;1}\Lambda^\beta_{\;0}=\eta_{01}-\eta_{\beta\alpha}\Lambda^\alpha_{\;1}\Lambda^\beta_{\;0}=\eta_{01}-\eta_{\beta\alpha}\Lambda^\beta_{\;0}\Lambda^\alpha_{\;1}=\eta_{01}-\eta_{\alpha\beta}\Lambda^\alpha_{\;0}\Lambda^\beta_{\;1}$, where, in the last step, I renamed the dummy indices $\alpha\leftrightarrow\beta$. $\endgroup$
    – ummg
    Commented May 27, 2021 at 17:23
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Matrix multiplication can be written in index notation as : $$(AB)^{\mu}_{~~\nu} = A^{\mu}_{~~\alpha}B^{\alpha}_{~~\nu}$$

However, the nice thing about index notation is that you are dealing with the real numbers numbers which are the component of the geometric objects (vectors, linear operator and higher tensors). Those real numbers do commute. For instance, the equation above can be rewritten as : $$(AB)^{\mu}_{~~\nu} = B^{\alpha}_{~~\nu}A^{\mu}_{~~\alpha}$$ without risk of confusion with the product in the opposite order, which is : $$(BA)^{\mu}_{~~\nu} = B^{\mu}_{~~\alpha}A^{\alpha}_{~~\nu}$$

As for your second question, the keyword is Einstein summation convention.

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