# Minkowski Inner Product's Shenanigans

In the context of Special Relativity, so in flat spacetime, and with the metric tensor $$g_{\mu \nu}$$ chosen with the signature: $$(+,-,-,-)$$, lets consider the following four vectors: $$x=(x_1,x_2,x_3,x_4)$$ $$y=(y_1,y_2,y_3,y_4)$$ the Minkowski inner product between the two is: $$x^\mu y^\nu g_{\mu \nu}=x_1y_1-x_2y_2-x_3y_3-x_4y_4 \ \ \ \ \ \ \ \ \ \ (1)$$ Wonderful, but we can also interprete the presence of $$g_{\mu \nu}$$ as a "lowering indices agent", such that: $$x^\mu y^\nu g_{\mu \nu}=x^\mu y_\mu$$ this has to be true, but then following the Einstein's summation convention we get: $$x^\mu y^\nu g_{\mu \nu}=x^\mu y_\mu=x_1y_1+x_2y_2+x_3y_3+x_4y_4$$ so a different result than we previously got in $$(1)$$.
This is obviously absurd; where am I wrong?

• Hint: How did you lower the $x$-indices in the last equality? – Qmechanic Jul 4 at 16:33
• I know that probably my mistake is really trivial to spot, but for some reason I can't see it. Maybe I didn't understand the mechanism behind index lowering. Anyway seems to me that I never lowered x's index; I only lowered y index, so your hint seems unhelpful to me. Can you help me out a little bit more? – Noumeno Jul 4 at 16:56
• @Noumeno Your last line should be $x^\mu y_\mu = x^1 y_1 + x^2 y_2 + x^3 y_3 + x^4 y_4$. What is the relation between $x^\mu$ and $x_\mu$? – Prahar Jul 4 at 17:35
• BTW, nobody I know of uses $1,2,3,4$ for $t,x,y,z$. The usual index-numbering conventions are $0,1,2,3$ for $t,x,y,z$ or $1,2,3,4$ for $x,y,z,it$. – G. Smith Jul 4 at 17:42
• Finally, it’s a bad idea to write $x_1$, etc. for the components of $x^\mu$, which is what you appear to be doing. – G. Smith Jul 4 at 17:44

Remember that in general when we lower the index of a vector, the resulting covector isn't just the original vector turned on its side, this is only the case when the metric is the identity matrix (this is why the distinction between row and column vectors is not always strictly necessary, since in this special case they are just the straight transpose of each other).

With the metric and vector you have given, $$\text{diag}(+,-,-,-)$$ and $$(x_1,x_2,x_3,x_4)$$ you will have:

$$\eta:\begin{pmatrix}x_1\\x_2\\x_3\\x_4 \end{pmatrix}\mapsto (x_1,-x_2,-x_3,-x_4) \tag{1}$$

So when you now contract this with $$y^\nu$$ you will have picked up the three minus signs you appear to be missing due to the "lowering" operation. Note also that there are two conventions for the Minkowski metric that differ by a sign, you could have also used $$\text{diag}(-,+,+,+)$$, in which case you would have:

$$\eta:\begin{pmatrix}x_1\\x_2\\x_3\\x_4 \end{pmatrix}\mapsto (-x_1,x_2,x_3,x_4) \tag{2}$$

There is no consensus on which to use, although different fields of physics tend to stick to one convention for various reasons. You do have to make sure however that once you've picked one you stick with it.

In your last equality, you should write $$x^\mu y_\mu=x^1y_1+x^2y_2+x^3y_3+x^4y_4\tag{1}.$$ Note that this is not equal to $$x_1y_1+x_2y_2+x_3y_3+x_4y_4$$. To recover your first equality you should then lower the $$x$$ indices. Remember $$x^\mu=x_\nu g^{\mu\nu}$$, so then $$x^1=x_1g^{11}+x_2g^{12}+x_3g^{13}+x_4g^{14}=x_1+0+0+0$$ But \begin{align} x^2=x_1g^{21}+x_2g^{22}+x_3g^{23}+x_4g^{24}=0-x_2+0+0\\ x^3=x_1g^{31}+x_2g^{32}+x_3g^{33}+x_4g^{34}=0+0-x_3+0\\ x^4=x_1g^{41}+x_2g^{42}+x_3g^{43}+x_4g^{44}=0+0+0-x_4\\ \end{align}

• Isn't your equation 1 wrong? You're missing the negative signs on the last three terms. – Charlie Jul 4 at 23:16
• I don't think so. I'm writing the components of the contravariant vector with an up index and the components of the covariant vector with a down index, which I think is the standard notation. Then I show that the last three components of each differ by a sign, so that if we write the inner product with all indices down we get the minus signs in the last three terms. I think this is just a notation issue; in your answer, you denote the components of the (contravariant) vector $x^\mu$ with subscripts, which can be confusing as someone pointed out in one of the last comments to the question. – Urb Jul 5 at 11:36
• Ah you are right my mistake I see what you've done now, I kept the indices in the same place in my answer to highlight the negative sign picked up from the metric. – Charlie Jul 5 at 15:43