Spatial gradient of a scalar

Difficulty:   ★★★☆☆   undergraduate general relativity

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  §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.

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