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From Schwartz "QFT And The Standard Model", section 8.2.2. $$ \ L=-\frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu+\frac{1}{2} m^2 A_\mu^2.\tag{8.17} $$

The author says that this is not a Lagrangian for a spin-1 particle, but for four scalar particles. Whereas the Lagrangian $$ \ L=\frac{a}{2} A_\mu \square A_\mu+\frac{b}{2} A_\mu \partial_\mu \partial_\nu A_\nu+\frac{1}{2} m^2 A_\mu^2.\tag{8.20} $$

gives a Lagrangian for spin-1 particles, because "the $\partial_\mu A_\mu $ contraction forces $A_\mu $ to transform as a 4-vector".

I need some (more explicit) explanation of that. Why can't I contract in the first Lagrangian and force $ \ A_\mu $ to transform as a 4-vector? What contraction takes place in the second Lagrangian?

Qmechanic
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Fortinbras
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  • and if someone knows a resource with explicit examples of tensor "gymnastics" for physicists that would be great – Fortinbras Apr 10 '23 at 10:18
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    The second example is clearer if we shift by a total derivative to make the second term $-\frac12b(\partial_\mu A_\mu)^2$, not that I'm happy all the Greek indices are downstairs. The two examples differ in what's contracted with what. Hint: if you write the first in the form $\sum_\mu L_\mu$, how does each $L_\mu$ behave? – J.G. Apr 10 '23 at 10:26
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    Closely related: https://physics.stackexchange.com/q/149018/226902 ...see also, from Schwartz's book, related: https://physics.stackexchange.com/q/548314/226902 https://physics.stackexchange.com/q/639982/226902 https://physics.stackexchange.com/q/697430/226902 – Quillo Apr 10 '23 at 10:34

1 Answers1

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Let's write the first lagrangian differently:

$$ L=-\frac{1}{2} \partial_\nu A_i \partial^\nu A^i +\frac{1}{2}m^2 A_iA^i. $$

Notice that we can let $i$ run from $1$ to $n$ for any value of $n$. It doesn't have to be a spacetime-index. The only restriction on $A_i$ is that it has to be part of a vector space with some inner-product.

This is not the case if we have terms like $\partial_\mu A^\mu$. Now $\mu$ has to have $D$ values (where $D$ is the number of spacetime dimensions). Also, for this term to be a scalar, $A^\mu$ has to transform like a contravariant vector under Lorentz transformations (to cancel the transformation acting on the covariant $\partial_\mu$).

In other words, in the second case, the indicis of $A$ are contracted with the spacetime indices, and thus symmetry forces them to be spacetime indices, thus making $A$ a spacetime-vector field, or a spin-1 field.

Rd Basha
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