Ciara Rose

Answered

2022-07-17

Suppose $\varphi \left(x\right)$ a scalar field, ${v}^{\mu }$ a $4$-vector. According to my notes a quantity of form ${v}^{\mu }{\mathrm{\partial }}_{\mu }\varphi \left(x\right)$ will not be Lorentz invariant.
But explicitly doing the active transformation the quantity becomes
${\mathrm{\Lambda }}_{\nu }^{\mu }{v}^{\nu }\left({\mathrm{\Lambda }}^{-1}{\right)}_{\mu }^{\rho }{\mathrm{\partial }}_{\rho }\varphi \left(y\right)={v}^{\nu }{\mathrm{\partial }}_{\nu }\varphi \left(y\right)$
where $y={\mathrm{\Lambda }}^{-1}x$ and the partial differentiation is w.r.t. $y$. This seems to suggest that the quantity is a Lorentz scalar, so could be used to construct a Lorentz invariant first order equation of motion.
I'm clearly making a mistake here. But I don't see what I've done wrong. Am I wrong to think that $v$ transforms nontrivially under the active transformation?

Answer & Explanation

Kitamiliseakekw

Expert

2022-07-18Added 23 answers

The function ${v}^{a}{\mathrm{\partial }}_{a}\varphi$ is a scalar field. Nonetheless, an equation like this is ugly because ${v}^{a}$ points in some "preferred" direction.
Here's another point of view, and I think this gets at what you were saying about "no first order equations". Suppose that ${v}^{a}$ is not a vector but is instead just a collection of four fixed real numbers. Suppose we consider the equation $\sum _{\mu }{v}^{\mu }{\mathrm{\partial }}_{\mu }\varphi =0$now. This equation is not Lorentz invariant anymore since the numbers ${v}^{\mu }$ don't change.
Another approach: think of ${v}^{a}$ as a new spacetime-dependent vector field. Then, ${v}^{a}{\mathrm{\partial }}_{a}\varphi =0$ is Lorentz invariant equation, but it involves two fields. This is nicer than choosing a preferred direction.

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