Ciara Rose

Answered

2022-07-17

Suppose $\varphi (x)$ a scalar field, ${v}^{\mu}$ a $4$-vector. According to my notes a quantity of form ${v}^{\mu}{\mathrm{\partial}}_{\mu}\varphi (x)$ will not be Lorentz invariant.

But explicitly doing the active transformation the quantity becomes

${\mathrm{\Lambda}}_{\nu}^{\mu}{v}^{\nu}({\mathrm{\Lambda}}^{-1}{)}_{\mu}^{\rho}{\mathrm{\partial}}_{\rho}\varphi (y)={v}^{\nu}{\mathrm{\partial}}_{\nu}\varphi (y)$

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?

But explicitly doing the active transformation the quantity becomes

${\mathrm{\Lambda}}_{\nu}^{\mu}{v}^{\nu}({\mathrm{\Lambda}}^{-1}{)}_{\mu}^{\rho}{\mathrm{\partial}}_{\rho}\varphi (y)={v}^{\nu}{\mathrm{\partial}}_{\nu}\varphi (y)$

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.

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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