Extreme confusion with the Lorentz transformation law for vector fields

Question

Let $\Lambda$ be a Lorentz transformation represented as $4 \times 4$ matrix.

Then, following What does it mean to transform as a scalar or vector? , it seems that a vector field $f : \mathbb{R}^4 \to \mathbb{R}^4$ should satisfy the transformation property \begin{equation} [\Lambda^\mu_\nu f^\nu](x) = f^\mu ( \Lambda x) \end{equation}

However, this leads to some annoying confusion for me. For example, let $g : \mathbb{R}^4 \to \mathbb{R}$ be a scalar. Then $\nabla g = (\partial^0 g, \partial^1 g, \partial^2 g,\partial^3 g)$ is supposed to be a vector field.

However, it seems to me that \begin{equation} (\Lambda^\mu_\nu \partial^\nu g)(x) \neq (\partial^\mu g)(\Lambda x) \end{equation}

I do not see how to express my difficulty clearly, but the problem is that in the expression $(\partial^\mu g)(\Lambda x)$, the gradient $\partial^\nu$ acts "before" $\Lambda x$ and something like chain rule does not apply to bring $\Lambda$ "outside".

It is getting more confusing and frustrating now...Could anyone please clarify for me please?

Answer 1

Taking the active point of view, that is rotating or boosting the field while leaving the coordinates the same, as opposed to the passive view where one leaves the field alone and rotates or boosts the coordinates.

An ordinary 4-vector transforms as $x^\mu\rightarrow x^{\prime\mu}=\Lambda^\mu_{\;\nu}x^\nu$, the 4-vector has been boosted as it were through some angle. Now it is worth discussing how scalar fields transform so that we can consider the transformation of its gradient, before finally discussing the notion of transforming a vector field, i.e. something like $V^\nu(x)$, where $x$ is a point in Minkowski space.

So given what we know about transforming 4-vectors, how does a scalar field transform under the active view? Obviously we have something like: $$\phi(x)\rightarrow\phi^{\prime}(x),$$ where $\phi^{\prime}(x)$ is the scalar field which has been rotated or boosted, note that we did not alter the variable $x$; this is because in the active view the point $x$ was left alone and the entire field was transformed instead. However, we could have wrote: $$\phi(x)\rightarrow\phi^{\prime}(x)=\phi(\Lambda^{-1}x).$$ Thus, the transformed field $\phi^{\prime}(x)$ is evaluated in terms of the "old" untransformed field, however, at the expense of inverse transforming the coordinate. Thus, actively transforming the scalar field can be understood as leaving the field alone; the coordinates, rather than the field, have been actively transformed. So what about the gradient?

Since actively transforming both 4-vectors and scalar fields can be understood as above, it follows that: $$\Lambda(\partial_\mu\phi(x))=\Lambda_\mu^{\;\nu}\partial_\nu\phi(\Lambda^{-1}x).$$ Similarly one can write: $$\Lambda(\partial^\mu\phi(x))=\Lambda^\mu_{\;\nu}\partial^\nu\phi(\Lambda^{-1}x).$$ The result stems from combining what we know about active transformations of both 4-vectors and scalar fields. There is no need to invoke the chain rule in order to get the external factor of $\Lambda$. Transform the gradient operator $\partial^\mu$ as one would a vector while transforming the scalar field according to the rule for scalar fields respectively.

Answer 2

For a spacetime $1+1$: $$\frac{\partial g}{\partial x^{0}} = \frac{\partial g}{\partial y^0}\frac{\partial y^0}{\partial x^{0}} + \frac{\partial g}{\partial y^1}\frac{\partial y^1}{\partial x^{0}}$$

$$\frac{\partial g}{\partial x^{1}} = \frac{\partial g}{\partial y^0}\frac{\partial y^0}{\partial x^{1}} + \frac{\partial g}{\partial y^1}\frac{\partial y^1}{\partial x^{1}}$$

Supposing that the transformation is $x \to y$, we see that, in matrix notation: $\nabla g = \Lambda \nabla g'$, or $\nabla g' = \Lambda^{-1} \nabla g$

Expressing in tensor notation in the old coordinates:

\begin{equation} (\partial^\mu g(x)) = (\Lambda^\mu_\nu \partial^\nu g(\Lambda x)) \end{equation}

Or expressing in the transformed coordinates:

\begin{equation} (\partial^\mu g(x)) = (\Lambda^{-1})^{\mu}_{\nu} (\partial^\nu g(\Lambda^{-1} x)) \end{equation}

It is the transformation of a covariant vector (the case of the gradient). The rule for a contravariant vector is different.