I was reading the Wikipedia page for the electroweak Lagrangian https://en.wikipedia.org/wiki/Electroweak_interaction

It looks ok to me until you get to the symmetry broken four-point self interaction part $$\mathcal{L}_{WWVV} = -\frac{g^2}{4}\{ [2W_\mu^+W^{-\mu}+(A_\mu \sin \theta_W-Z_\mu \cos \theta_W)^2]^2 \\-[W_\mu^+W_\nu^-+W_\nu^+W_\mu^-+(A_\mu \sin\theta_W-Z_\mu \cos\theta_W)(A_\nu \sin\theta_W-Z_\nu \cos\theta_W)]^2 \}$$

What I do not understand is how this is coordinate invariant. The number of superscript indices is supposed to match the number of subscript indices for it to be invariant. It does not look that way here. Does squaring it make it coordinate invariant? Please convince me that this is coordinate invariant. I expect the Lagrangian to only have quantities like $A_\mu J^\mu$ i.e. where the number of superscript indices matches the number of subscript indices. This is what makes it coordinate invariant.

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    $\begingroup$ Note that: $A^2 := A^\mu A_\mu$ $\endgroup$ Dec 20, 2020 at 15:20
  • $\begingroup$ @AlmostClueless I'm talking about the stuff in the parenthesis being squared such as $(A_\mu sin \theta_W-Z_\mu cos \theta_W)^2$ $\endgroup$ Dec 20, 2020 at 15:23
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    $\begingroup$ Where is the difference? $\endgroup$ Dec 20, 2020 at 15:26
  • $\begingroup$ @AlmostClueless The quantity $A^\mu A_\mu$ is coordinate invariant. The quantity $(A_\mu sin \theta_W-Z_\mu cos \theta_W)$ transforms covariantly. Squaring it in the sum does not make it coordinate invariant. $\endgroup$ Dec 20, 2020 at 15:29
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    $\begingroup$ Every summand of your sum has the same apperent indices, so when you square your sum you will get terms where every term has the pair of indices in the superscript and in the subscript. I think you are just confusing the notation. $\endgroup$ Dec 20, 2020 at 15:34

1 Answer 1


In the interest of this question having an answer:

Define $X_{\mu\nu} \equiv W_\mu^+W_\nu^-$. Since $W_\mu$ transforms as a $(0, 1)$ tensor, the outer product with itself makes $X_{\mu\nu}$ transform as a $(0, 2)$ tensor. As you have mentioned, $Y_\mu \equiv A_\mu \sin\theta_W - Z_\mu \cos\theta_W$ transforms covariantly, and taking the outer product with itself again means that $Y_{\mu\nu}\equiv Y_\mu Y_\nu$ also transforms as an honest-to-goodness tensor.

Rewriting the Lagrangian as $$\mathcal{L}_{WWVV} = -\frac{g^2}{4}\{ [2W_\mu^+W^{-\mu}+Y_\mu Y^\mu]^2 \\-[X_{\mu\nu}+X_{\nu\mu}+Y_{\mu\nu}]^2 \}$$

The first term is obviously invariant, while expanding the second term as $[X_{\mu\nu}+X_{\nu\mu}+Y_{\mu\nu}][X^{\mu\nu}+X^{\nu\mu}+Y^{\mu\nu}]$, you can see that each term is fully contracted. Hence the entire expression is Lorentz invariant, as a Lagrangian should be.


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