In differential geometry, the four-gradient is the four-vector analogue of the gradient from Gibbs-Heaviside vector calculus.

## Definition

The covariant components compactly written in index notation are:[1]

$\dfrac{\partial}{\partial x^\alpha} = \left(\frac{1}{c}\frac{\partial}{\partial t}, \vec{\nabla}\right) = \left(\frac{\partial_t}{c}, \vec{\nabla}\right) = \partial_\alpha = {}_{,\alpha}$

The comma in the last part above ${}_{,\alpha}$ implies the partial differentiation with respect to $x^\alpha$. This is not the same as a semi-colon, used for the covariant derivative.

The contravariant components are:[2]

$\mathbf{\partial} = \partial^\alpha \ = g^{\alpha \beta} \partial_\beta = \left(\frac{1}{c} \frac{\partial}{\partial t}, -\vec{\nabla} \right)= \left(\frac{\partial_t}{c}, -\vec{\nabla}\right) = \left(\frac{\partial_t}{c}, -\partial_x,-\partial_y,-\partial_z\right)$

where gαβ is the metric tensor, which here has been chosen for flat spacetime with the metric signature (+,−,−,−).

Alternative symbols to $\partial_\alpha$ are $\Box$ and D.

## Usage

The 4-Gradient is used in a number of different ways in Special Relativity:

### As a 4-Divergence

The 4-Divergence of the 4-Position $X^\mu$ gives the dimensionality of spacetime:

$\mathbf{\partial} \cdot \mathbf{X} = \partial^\mu \cdot X^\nu = \partial^\mu \eta_{\mu\nu} X^\nu = (\frac{\partial_t}{c},-\vec{\nabla})\cdot (ct,\vec{x}) = \frac{\partial_t}{c}(ct) + \vec{\nabla}\cdot \vec{x} = (\partial_t t) + (\partial_x x+\partial_y y+\partial_z z) = (1) + (3) = 4$

The 4-Divergence of the 4-CurrentDensity $J^\mu = \rho_o U^\mu$ gives a conservation law - the conservation of charge:

$\mathbf{\partial} \cdot \mathbf{J} = \partial^\mu \cdot J^\nu = \partial^\mu \eta_{\mu\nu} J^\nu = \partial_\nu J^\nu = (\frac{\partial_t}{c},-\vec{\nabla})\cdot (\rho c,\vec{j}) = \frac{\partial_t}{c}(\rho c) + \vec{\nabla}\cdot \vec{j} =\partial_t \rho + \vec{\nabla}\cdot \vec{j} = 0$

This means that the time rate of change of the charge density must equal the negative spatial divergence of the current density $\partial_t \rho = -\vec{\nabla}\cdot \vec{j}$. In other words, the charge inside a box cannot just change arbitrarily, it must enter and leave the box via a current. This is a continuity equation.

### As a Jacobian Matrix for the SR Metric Tensor

The 4-Gradient $\partial^\mu$ acting on the 4-Position $X^\nu$ gives the SR Minkowski_space metric $\eta^{\mu\nu}$. :

$\mathbf{\partial} [\mathbf{X}] = \partial^\mu[X^\nu] = X^{\nu_,\mu} = (\frac{\partial_t}{c},-\vec{\nabla})[(ct,\vec{x})] = (\frac{\partial_t}{c},-\partial_x,-\partial_y,-\partial_z)[(ct,x,y,z)],$
$= \begin{bmatrix}\frac{\partial_t}{c} ct & \frac{\partial_t}{c} x & \frac{\partial_t}{c} y & \frac{\partial_t}{c} z \\ -\partial_x ct & -\partial_x x & -\partial_x y & -\partial_x z \\ -\partial_y ct & -\partial_y x & -\partial_y y & -\partial_y z \\ -\partial_z ct & -\partial_z x & -\partial_z y & -\partial_z z\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 & 0 & 0 & -1\end{bmatrix} = Diag[1,-1,-1,-1]$
$\mathbf{\partial} [\mathbf{X}] = \eta^{\mu\nu}$

### As part of the total proper time derivative

The Scalar Product of 4-Velocity $U^\mu$ with the 4-Gradient gives the total derivative with respect to proper time $\frac{d}{d\tau}$:

$\mathbf{U} \cdot \mathbf{\partial} =U^\mu \cdot \partial^\nu = \gamma(c,\vec{u}) \cdot (\frac{\partial_t}{c},-\vec{\nabla}) = \gamma (c \frac{\partial_t}{c} + \vec{u} \cdot \vec{\nabla} )= \gamma (\partial_t + \frac{dx}{dt} \partial_x + \frac{dy}{dt} \partial_y + \frac{dz}{dt} \partial_z) = \gamma \frac{d}{dt} = \frac{d}{d\tau}$

So, for example, the 4-Velocity $U^\mu$ is the proper-time derivative of the 4-Position $X^\mu$:

$\frac{d}{d\tau} \mathbf{X} = (\mathbf{U} \cdot \mathbf{\partial})\mathbf{X} = \mathbf{U} \cdot \mathbf{\partial}[\mathbf{X}] = U^\alpha \cdot \eta^{\mu\nu} = U^\alpha \eta_{\alpha \nu} \eta^{\mu\nu} = U^\alpha \delta_\alpha^\mu = U^\mu = \mathbf{U}$

or

$\frac{d}{d\tau} \mathbf{X} = \gamma\frac{d}{dt} \mathbf{X} = \gamma\frac{d}{dt} (ct,\vec{x}) = \gamma (\frac{d}{dt}ct,\frac{d}{dt}\vec{x}) = \gamma (c,\vec{u}) = \mathbf{U}$

Another example, the 4-Acceleration $A^\mu$ is the proper-time derivative of the 4-Velocity $U^\mu$:

$\frac{d}{d\tau} \mathbf{U} = (\mathbf{U} \cdot \mathbf{\partial})\mathbf{U} = \mathbf{U} \cdot \mathbf{\partial}[\mathbf{U}] = U^\alpha \eta_{\alpha\mu}\partial^\mu[U^\nu]$
$= U^\alpha \eta_{\alpha\mu}\begin{bmatrix} \frac{\partial_t}{c} \gamma c & \frac{\partial_t}{c} \gamma \vec{u} \\ -\vec{\nabla}\gamma c & -\vec{\nabla}\gamma \vec{u} \end{bmatrix} = U^\alpha \begin{bmatrix}\ \frac{\partial_t}{c} \gamma c & 0 \\ 0 & \vec{\nabla}\gamma \vec{u} \end{bmatrix}$
$= \gamma (c \frac{\partial_t}{c} \gamma c , \vec{u} \cdot \nabla\gamma \vec{u} )= \gamma (c \partial_t \gamma, \frac{d}{dt}[\gamma \vec{u}] ) = \gamma (c \dot{\gamma}, \dot{\gamma} \vec{u} + \gamma \dot{\vec{u}} )= \mathbf{A}$

or

$\frac{d}{d\tau} \mathbf{U} =\gamma \frac{d}{dt} (\gamma c,\gamma \vec{u}) =\gamma (\frac{d}{dt}[\gamma c],\frac{d}{dt}[\gamma \vec{u}]) = \gamma (c \dot{\gamma}, \dot{\gamma} \vec{u} + \gamma \dot{\vec{u}} ) = \mathbf{A}$

### As a way to define the 4-WaveVector

The 4-WaveVector is the 4-Gradient of the negative phase, or the negative 4-Gradinet of the phase $\Phi$ of a wave in Minkowski Space

4-WaveVector $\mathbf{K} = \left(\frac{\omega}{c}, \vec{\mathbf{k}}\right) = \mathbf{\partial} [-\Phi]= -\mathbf{\partial} [\Phi]$

This is mathematically equivalent to the definition of the phase of a wave as:

$-\Phi = \mathbf{K} \cdot \mathbf{X} = \omega t - \vec{\mathbf{k}} \cdot \vec{\mathbf{x}}$

where:

4-Position $\mathbf{X} = (ct, \vec{\mathbf{x}})$

### As the d'Alembertian Operator

The square of $\mathbf{\partial}$ is the Four-Laplacian, which is called the d'Alembert operator:

$\mathbf{\partial} \cdot \mathbf{\partial} = \partial^\mu \cdot \partial^\nu = \eta_{\mu\nu} \partial^\mu \partial^\nu = \partial_\nu \partial^\nu = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 = \left(\frac{\partial_t}{c}\right)^2 - \nabla^2$.

As it is the dot product of two four-vectors, the d'Alembertian is a Lorentz invariant scalar.

Occasionally, in analogy with the 3-dimensional notation, the symbols $\Box$ and $\Box^2$ are used for the 4-Gradient and d'Alembertian respectively. More commonly however, the symbol $\Box$ is reserved for the d'Alembertian.

Some examples of the 4-Gradient as used in the d'Alembertian follow:

In the Klein-Gordon relativistic quantum wave equation for spin-0 particles (ex. Higgs_boson):

$[(\mathbf{\partial} \cdot \mathbf{\partial}) + \left(\frac {m_o c}{\hbar}\right)^2]\psi = [\left(\frac{\partial_t^2}{c^2} - \vec{\nabla}^2\right) + \left(\frac {m_o c}{\hbar}\right)^2] \psi = 0$

In the wave equation for the electromagnetic field:

$(\mathbf{\partial} \cdot \mathbf{\partial}) A^{\alpha} = 0$ {in vacuum}
$(\mathbf{\partial} \cdot \mathbf{\partial}) A^\alpha = \mu_0 J^\alpha$ {with a 4-CurrentDensity source}

where:

4-VectorPotential $A^{\alpha} = \left(\frac{\phi}{c},\vec{a}\right)$ is an electromagnetic potential
4-CurrentDensity $J^{\alpha} = (\rho c,\vec{j})$ is an electromagnetic current density

In the 4-dimensional version of Green's_function:

$(\mathbf{\partial} \cdot \mathbf{\partial}) G(x-x') = \delta^4(x-x')$

### As a component of the Schrödinger relations in Quantum Mechanics

The 4-Gradient is connected with Quantum Mechanics. The relation between the 4-Momentum $P$ and the 4-Gradient $\partial$ give the Schrödinger QM relations.

$\mathbf{P} = \left(\frac{E}{c},\vec{p}\right) = i\hbar \mathbf{\partial} = i\hbar \left(\frac{\partial_t}{c},-\vec{\nabla}\right)$

The temporal components give: $E = i\hbar \partial_t$

The spatial components give: $\vec{p} = -i\hbar \vec{\nabla}$

## Derivation

In 3 dimensions, the gradient operator maps a scalar field to a vector field such that the line integral between any two points in the vector field is equal to the difference between the scalar field at these two points. Based on this, it may appear incorrectly that the natural extension of the gradient to four dimensions should be:

 $\partial^\alpha \ = \left( \frac{\partial}{\partial t}, \vec{\nabla} \right)$ incorrect

However, a line integral involves the application of the vector dot product, and when this is extended to four-dimensional space-time, a change of sign is introduced to either the spatial co-ordinates or the time co-ordinate depending on the convention used. This is due to the non-Euclidean nature of space-time. In this article, we place a negative sign on the spatial coordinates (the time-positive Metric convention $\eta^{\mu\nu}=Diag[1,-1,-1,-1]$). The factor of (1/c) is to keep the correct unit dimensionality {1/[length]} for all components of the 4-vector and the (−1) is to keep the 4-gradient Lorentz covariant. Adding these two corrections to the above expression gives the correct definition of four-gradient:

 $\partial^\alpha \ = \left(\frac{1}{c} \frac{\partial}{\partial t}, -\vec{\nabla} \right)$ correct