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Given an embedded surface given by $x_i\rightarrow X^\mu(x)$ where x is 3D and X is 4D. The intrinsic metric is $g_{ab}(x) = \partial_a X^\mu(x) \partial_b X^\mu(x)$.

Apply a small shift at each point given by $\delta X^\mu(x) = N^\mu(x)$ where N is a normal vector at each point on the embedded surface.

Can the change in the metric be written entirely in terms of itself?

I find it changes as $\delta g_{ab}(x) = -N^\mu(x) \partial_a \partial_b X^\mu(x)$ I think. But I don't think the RHS can be written in terms of the metric.

This seems odd to me. Since locally the metric should give all information about the embedded surface.

What is the reason? What information is missing? Would the change in metric be different for a cylinder and a plane, for example, which both have a flat intrinsic metric?

(Also the normal vector can be written as $N^\tau(x) = \varepsilon^{\mu\nu\sigma\tau}\partial_1 X_\mu(x) \partial_2 X_\nu(x) \partial_3 X_\sigma(x)$)

zooby
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1 Answers1

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The change in the metric cannot be expressed only by means of the metric itself. Indeed, it also depends on the second fundamental form. As an example, compare a plane to a cylinder, both embedded in $\mathbb{R}^3$, and both moving in the unit normal direction. Say the plane and the cylinder have the same metric at time $0$. The metric on the plane does not change in time, while the cylinder gets smaller and smaller (or larger and larger, depends on the choice of the normal vector).

Amitai Yuval
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