Spatial gradient of a scalar

Difficulty: ★★★☆☆ undergraduate

Suppose you have a scalar field $\Phi$ , and at a given point in spacetime: a 4-velocity vector interpreted as an “observer”. In which direction does $\Phi$ increase most steeply, when restricted to the observer’s local 3-dimensional space?

Last time I reviewed the gradient 1-form or covector $d\Phi$ , and its associated gradient vector $(d\Phi)^\sharp$ obtained by raising the index as usual. The gradient vector has been described as the direction of greatest increase in $\Phi$ per unit length (Schutz 2009 book §3.3). However this is only guaranteed when the metric is positive definite, meaning a Riemannian manifold, rather than a Lorentzian manifold as used to model spacetime.

The observer’s 4-velocity splits vectors and 1-forms into purely “time” parts parallel to $\mathbf u$ , and purely “space” parts orthogonal to it. (Intuitively, it may help to think of a basis $\{\mathbf e_\alpha\}$ adapted to the observer, meaning $\mathbf e_0 := \mathbf u$ , and the $\mathbf e_i$ vectors are orthogonal to $\mathbf u$ , where $i = 1,2,3$ . Then a purely spatial vector is spanned by the $\mathbf e_i$ . Since vectors and covectors are linear, we need only specify their values on a basis set.)

Consider the tangent space at the specified point. Imagine working within the observer’s local 3-space, by which I mean the 3-dimensional subspace consisting of vectors orthogonal to $\mathbf u$ . Label the gradient as restricted to this subspace by $^{(3)}d\Phi$ . On the subspace the metric has Riemannian signature, hence the corresponding vector $^{(3)}(d\Phi)^\sharp$ is the direction of steepest increase. We can mimic this mathematically by staying in 4 dimensions, but setting the “time” part to zero:

$^{(3)}d\Phi := d\Phi + \langle d\Phi,\mathbf u\rangle \mathbf u^\flat.$

This is a 4-dimensional object, but I reuse the notation “ $^{(3)}$ ” to imply it vanishes in the observer’s time direction. This “3-gradient” is the projection of $d\Phi$ orthogonal to $\mathbf u^\flat$ . The angle brackets signify contraction of the 1-form and vector, and the “flat” symbol denotes the 1-form obtained from $\mathbf u$ by “lowering the index” using the metric. The vector 3-gradient is:

$^{(3)}(d\Phi)^\sharp = (d\Phi)^\sharp + \langle d\Phi,\mathbf u\rangle \mathbf u.$

This follows from “raising the index” using the inverse metric $g^{\mu\nu}$ as usual. Note that on the subspace, the inverse metric coincides with the inverse 3-metric which has components $(g_{ij})^{-1}$ , for $i,j=1,2,3$ . Equivalently, one can apply the spatial projector $g^{\mu\nu}+u^\mu u^\nu$ to either $d\Phi$ or $^{(3)}d\Phi$ , with the same result. This projector agrees with the inverse metric on the 3-space, and is zero on purely timelike covectors. Either way, the essential part of the process is to remove the “time” component of the gradient. I will give examples in the following post.

Spatial gradient of a scalar

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