I'm reading the Wikipedia page for the Dirac equation:



$J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)$

with the conservation of probability current and density following from the Schrödinger equation:

$\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0.$

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

$\rho = \frac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*).$

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

$J^\mu = \frac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)$

  1. What exactly are $\partial_t$ and $\partial^\mu$?

  2. Are they tensors?

  3. If they are, how are they defined?


$\partial_t\equiv\frac\partial{\partial t}$ and $\partial^\mu\equiv g^{\mu\nu}\frac\partial{\partial x^\nu}=\left(\sum_{\nu=0}^3g^{\mu\nu}\frac\partial{\partial x^\nu}\right)_{\mu=0}^3$ are differential operators. $\partial^\mu$ is formally contravariant (upper index) and obeys the corresponding transformation laws. $\partial_t$ has a lower index and is (up to a constant factor) a component of the formally covariant operator $\partial_\mu$ via $\partial_0=\frac1c\partial_t$, which, in general, is not equal to $\partial^0$, the zeroth component of $\partial^\mu$.

The differential operator $\partial^\mu$ is known as gradient, which derives vector fields from potential functions. The gradient is not a natural operation on arbitrary manifolds and only available if there's a metric. Its dual $\partial_\mu\equiv\frac\partial{\partial x^\mu}$ on the other hand is a natural operation corresponding to the differential $\mathrm d$, taking potentials to 1-forms (covectorfields).

As a side note, $\partial_t$ can also be understood as a local vector field, as one of the intrinsic definitions of vectors on manifolds is via their directional derivatives. In mathematical literature, it is common to write the basis of the tangent space as $\{\frac\partial{\partial x^\mu}\}$ and its dual space as $\{\mathrm dx^\mu\}$.

  • $\begingroup$ Two points: (1) $\partial_{\mu}$ does not correspond to the exterior derivative $\mathrm{d}$, since you need it to be antisymmetric on indices; (2) in general $\partial_{\mu}$ is the basis for a vector field in local coordinates. (I've only seen "local vector field" for "a vector field defined on one coordinate patch and no others.") $\endgroup$ – Alex Nelson Oct 8 '12 at 14:19
  • $\begingroup$ @AlexNelson: re (1), that was indeed a bit sloppy on my part, but it does hold true for the case I explicitly mentioned $\mathrm{d}:\Omega^0(M)=\mathcal{C}^\infty(M)\rightarrow \Omega^1(M)=\Gamma(T^*M),f\mapsto\mathrm{d}f=\partial_\mu f\mathrm{d}x^\mu$; re (2): any element of the local basis is a vectorfield defined on the correcsponding coordinate patch, so I do not see any contradiction $\endgroup$ – Christoph Oct 8 '12 at 14:36
  • $\begingroup$ re (2) my point is a local vector field is not the same as a vector field on the manifold expressed in local coordinates, since the latter is global. $\endgroup$ – Alex Nelson Oct 8 '12 at 20:44
  • $\begingroup$ @AlexNelson: the coordinate expressions for (global) vector fields are local vector fields - the full vector field is given by coordinate expressions for a set of covering patches $\endgroup$ – Christoph Oct 8 '12 at 21:13

Depending on notation used, $\partial^\mu$ could be a tensor, referring to the four-vector $(\partial^0,\partial^1,\partial^2,\partial^3)$. However, in the context of the above equation, you can treat both $\partial_t$ and $\partial^\mu$ as scalars referring to the partial differentiation by $x^t$ and $x_\mu$ correspondingly.

The exact definition of $\partial_t$ is not uniform in literature, sometimes it is set to $\partial_0$, otherwise to $\frac{1}{c}\partial_0$. However, with $c = 1$, this rarely matters.

Note furthermore that time is either assigned the zeroth or fourth component of space-time.

  • 3
    $\begingroup$ Yes, the components individually look like scalars but are not! They transform like tensors. Specifically $\partial^{\mu}$ will transform as a rank-1 contravariant tensor, and $\partial_{t}$ transforms unusually (but not as a scalar!). $\endgroup$ – Alex Nelson Oct 7 '12 at 14:08

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