Is the Four-gradient of a Scalar Field a Four-Vector?

Question

Consider a scalar field $\phi$ as a function of spacetime coordinates $x^\mu$. The four-gradient of $\phi$ is given by

\begin{equation} \frac{\partial \phi}{\partial x^\mu} = \left( \frac{\partial \phi}{\partial t}, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) \end{equation}

I'm studying a little bit of classical field theory and in the topic of real scalar fields, it is common to introduce four-gradients of these fields.

My question then is: The components of this four-gradient are not the components of a four-vector. For example, take the first component, and considering $t = t(x',t')$

\begin{align} \frac{\partial \phi}{\partial t} &= \frac{\partial \phi}{\partial t'}\frac{\partial t'}{\partial t} + \frac{\partial \phi}{\partial x'}\frac{\partial x'}{\partial t}\\ &= \gamma \left( \frac{\partial \phi}{\partial t'} - v \frac{\partial \phi}{\partial x'} \right) \end{align}

EDIT: In some books I've read, the four-gradient $\partial _\mu$ is defined as a four-vector, and since the D'alambertian operator $\Box = \partial _\mu \partial ^\mu$ is a lorentz invariant, the four-gradients must be four-vectors. But my computations give me this unsatisfactory result. Am I doing computational mistake or what? Please help me.

Answer 1

The $4$-gradient is a $4$- vector.

Formally, when $x^\mu\to x'^\mu=\Lambda^\mu{}_\nu x^\nu$

$$ \begin{align*} \partial'_\mu &=\frac{\partial}{\partial x'^\mu}\\ &=\frac{\partial}{\partial (\Lambda^\mu {}_\nu x^\nu)}\\ \end{align*} $$ $\therefore$ $$ \begin{align*} \Lambda^\mu {}_\nu\partial'_\mu&=\partial_\nu\\ \end{align*} $$

which makes $\partial_\mu$ a 4 vector and is precisely what you are getting

which is not how the $0^{th}$ component of a four-vector should transform

It is. What you have derived is $\partial_\nu$ in terms of $\partial'_\mu$ which matches above.

This is equivalent to the transform of a 4 covector $$\partial'_\mu=\Lambda_\mu{}^\nu\partial_\nu$$ So, $$ \begin{align*} \frac{\partial}{\partial t'}&=\gamma\left(\frac{\partial}{\partial t}+v\frac{\partial}{\partial x}\right)\\ \frac{\partial}{\partial x'}&=\gamma\left(\frac{\partial}{\partial x}+v\frac{\partial}{\partial t}\right) \end{align*} $$ which leads to $$\frac{\partial}{\partial t}=\gamma\left(\frac{\partial}{\partial t'}-v\frac{\partial}{\partial x'}\right)$$ same as before

Answer 2

The four gradient is a four vector but it transforms covariantly, rather than contravariantly. This makes it a "covector".

It also has a contravariant form, obtained by multiplying it by the metric, which transforms like 4 four position or four momentum.

Answer 3

The "gradient" you wrote is not a four vector (and that's not what should be called a gradient). What you have written is a four co-vector (i.e. a dual vector, which is a linear functional over the space of four-vectors). What you have written is technically the differential of the field $\phi$ which is an exact one-form:

$$d\phi = \frac{\partial\phi}{\partial x^{\mu}} \, dx^{\mu}$$

The gradient depends on an underlying Lorentzian (or Riemannian) metric $g_{\mu \nu}(x)$ tensor, with inverse tensor $g^{\mu \nu}(x)$, that establishes isomorphism between the space of vectors and co-vectors. I.e.

$$\big(\,\nabla \phi \,\big)^{\mu} = g^{\mu \nu}\, \frac{\partial \phi}{\partial x^{\nu}}$$ so when you change the coordinates, $x^{\mu} \mapsto \tilde{x}^{\mu}$, the inverse Lorenzian or Riemannian tensor changes as $$g^{\mu \nu} = \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \tilde{g}^{\,\alpha \beta}\, \frac{\partial {x}^{\nu}}{\partial \tilde{x}^{\beta}}$$ and the partial derivatives transform as $$\frac{\partial\phi}{\partial x^{\nu}} =\frac{\partial \tilde{x}^{\gamma}}{\partial {x}^{\nu}}\, \frac{\partial \phi}{\partial \tilde{x}^{\gamma}}$$ Consequently, the gradient transforms as a four-vector as follows: \begin{align} \big(\,\nabla \phi \,\big)^{\mu} &= g^{\mu \nu}\, \frac{\partial \phi}{\partial x^{\nu}} = \left(\frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \tilde{g}^{\,\alpha \beta}\, \frac{\partial {x}^{\nu}}{\partial \tilde{x}^{\beta}} \right)\, \left(\frac{\partial \tilde{x}^{\gamma}}{\partial {x}^{\nu}}\, \frac{\partial \phi}{\partial \tilde{x}^{\gamma}}\right)\\ &= \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \tilde{g}^{\,\alpha \beta}\, \left(\frac{\partial {x}^{\nu}}{\partial \tilde{x}^{\beta}} \, \frac{\partial \tilde{x}^{\gamma}}{\partial {x}^{\nu}}\right)\, \frac{\partial \phi}{\partial \tilde{x}^{\gamma}}\\ &= \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \tilde{g}^{\,\alpha \beta}\, \left(\frac{\partial \tilde{x}^{\gamma}}{\partial {x}^{\nu}} \, \frac{\partial {x}^{\nu}}{\partial \tilde{x}^{\beta}} \right)\, \frac{\partial \phi}{\partial \tilde{x}^{\gamma}}\\ &= \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \tilde{g}^{\,\alpha \beta}\, \delta^{\gamma}_{\beta} \frac{\partial \phi}{\partial \tilde{x}^{\gamma}}\\ &= \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \tilde{g}^{\,\alpha \beta}\,\frac{\partial \phi}{\partial \tilde{x}^{\beta}}\\ &= \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \left(\tilde{g}^{\,\alpha \beta}\,\frac{\partial \phi}{\partial \tilde{x}^{\beta}}\right)\\ & = \frac{\partial {x}^{\mu}}{\partial \tilde{x}^{\alpha}}\, \big(\,\tilde{\nabla} \phi \,\big)^{\alpha} \end{align} where again, in the current $\tilde{x}^{\alpha}$, the gradient is $$\big(\,\tilde{\nabla} \phi \,\big)^{\alpha} = \tilde{g}^{\,\alpha \beta}\,\frac{\partial \phi}{\partial \tilde{x}^{\beta}}$$

In flat space-time, you have a Minkowskian constant tensor $\eta_{\mu \nu}$.

Answer 4

The language of differential forms can help clarify this. Note in the following what I am calling a k-cofield is what is normally called a diffetential k-form. I prefer using this term as it's shorter and the language of fields is traditional in physics (and not forms).

A scalar function $f$ is a 0-cofield hence we can apply the exterior derivative $d$ to it to obtain $df$, this is a 1-cofield. Assuming that we have a metric available we can raise it a 1-field by the use of the raising operator $\#$. And it turns out that:

$grad(f) = (df)^\#$

Since both $d$ and $\#$ are covariant, then so must $grad(f)$ and as it is a 1-field, aka a tangent field, aka a vector field, then supposing we are working over a 4d manifold, we have a 4-vector (field).

From this, it's worth noting that the exterior derivative is basically a generalisation of the gradient to higher dimensions. And as it maps cofields to cofields, it's probably more suggestively named the cogradient. These higher cogradients when used with a metric yield curl and div (and much else besides).