2
$\begingroup$

If, in a $d$ dimensional space with Euclidean metric, we parametrize a two-dimensional surface with parameters $\xi^1$ and $\xi^2$ then the area can be written as

$$A = \int ~\mathrm d\xi^1 \mathrm d\xi^2 \sqrt{\left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1}\right)^2 \left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^2}\right)^2} \sqrt{1-\dfrac{\left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1} \dot{}{} \dfrac{\mathrm d \vec x}{\mathrm d\xi^2}\right)^2}{\left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1}\right)^2 \left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^2}\right)^2}}$$

Or,

$$A = \int ~\mathrm d\xi^1~\mathrm d\xi^2 \sqrt{\left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1}\right)^2 \left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^2}\right)^2 -\left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1} \dot{}{} \dfrac{\mathrm d \vec x}{d\xi^2}\right)^2}$$

Now, although this expression doesn't explicitly refer to a specific metric (as the inner products involved can be performed using any metric) the definition of elemental area used as well as the use of relation $\vec{A} \dot{}{}\vec{B} = AB \cos\theta$ are specific to Euclidean metric and can't be assumed to hold for a generic metric. Still, Zweibach, in A first course in String Theory, has extended the use of the formula

$$A = \int ~\mathrm d\xi^1 \mathrm d\xi^2 \sqrt{\left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1}\right)^2 \left(\dfrac{\mathrm d \vec x}{\mathrm d\xi^2}\right)^2 -\bigg(\dfrac{\mathrm d \vec x}{\mathrm d\xi^1} \dot{}{} \dfrac{\mathrm d \vec x}{\mathrm d\xi^2}\bigg)^2}$$

(upto a facor of $i$) for a Minkowskian spacetime.

Why? Is it that we can use any scalar quantity of appropriate dimensions to define the area and thus we chose the one we are familiar with in Euclidean geometry or is there any other logical reasoning?

$\endgroup$
1
$\begingroup$

The formula isn't specific to euclidean space. If we are given an abstract vector space $V$ with inner product $\langle,\rangle$ ($V$ is real), we define the angle between two vectors as $$ \cos(\alpha)=\frac{\langle u,v\rangle}{||u||\cdot||v||} .$$ In particular, if $V=T_pM$, and $\langle,\rangle=g_p(,)$, the same still holds.

Given a pseudo-Riemannian manifold $(M,g)$, the volume element is given locally as $$ dM=\sqrt{|\det g|}d^nx, $$ this can be verified by checking that $dM$ is coordinate-invariant (the metric determinant transforms in an inverse way to the coordinate differentials) and in a locally flat, positively oriented coordinate system, it reduces to $d^nx$, which is of course the flat volume element.

Now, if we are given a 2-surface in $M$, and is parametrized by the coordinates $(u,v)$, the induced metric on the 2-surface is given by $$(h_{ij})= \left(\begin{matrix} \frac{\partial \mathbf{x}}{\partial u}^2 & \frac{\partial\mathbf{x}}{\partial u}\cdot\frac{\partial\mathbf{x}}{\partial v} \\ \frac{\partial\mathbf{x}}{\partial v}\cdot\frac{\partial\mathbf{x}}{\partial u} & \frac{\partial \mathbf{x}}{\partial v}^2 \end{matrix}\right) ,$$ where the square and the dots are $g$-inner products. (On manifolds we usually write the coordinate vectors as simply $\partial/\partial u$ since the position vector $\mathbf{x}$ is not well-defined but whatever).

The invariant volume (eg. area) element on the 2-surface is given by $$\sqrt{|\det h|}dudv,$$ which, as you can check, is the same as you wrote.


Edit:

On the angle formula: You are right. I was confused when you said the angle formula holds only in euclidean space, I thought you were doubting if it is also true in curved space, for a moment I forgot about "pseudo-euclidean" spaces.

Of course, in pseudo-euclidean geometry, you can have things, say, imaginary angles, so the whole shiznit is not really well defined. On the other hand, we do say that two vectors are perpendicular/orthogonal to one another if their inner product vanishes in all cases (even for a Hermitean inner product on a complex vector space). In particular, a null vector is orthogonal to itself.

My main point was, which I may have not communicated very clearly is that you don't need the angle formula.

On the uniqueness of the volume element:

First of all, let's check that if $(M,g)$ is an $n$ dimensional pseudo-Riemannian space, then $dM=\sqrt{|\det g|}d^nx$ is invariant!

We know from multilinear calculus that the integration measure $d^nx$ transforms during the coordinate change $x^\mu\mapsto x^{\mu'}$ as $d^nx=\left|\det\frac{\partial x}{\partial x'}\right|d^nx'$. The metric components change as $$g_{\mu\nu}=\frac{\partial x^{\mu'}}{\partial x^\mu}\frac{\partial x^{\nu'}}{\partial x^\nu}g_{\mu'\nu'}.$$ Takint the determinant of both sides gives $$\det g=\left(\det\frac{\partial x'}{\partial x}\right)^2\det g', $$ taking absolute values and a square root gives $$\sqrt{|\det g|}=\left|\det\frac{\partial x'}{\partial x}\right|\sqrt{|\det g'|} .$$ From this we can see that in the product $\sqrt{|\det g|}d^nx$ the two jacobians that occur during coordinate change precisely cancel.

Now on to actual uniqueness. In flat $n$ dimensional spacetime, if $x^{\mu'}$ are pseudo-cartesian coordinates, the volume element is $d^nx'$. If we switch from $x^{\mu'}$ to $x^\mu$, we have $d^nx'=|\det\frac{\partial x'}{\partial x}|d^nx$, but here we have $\sqrt{|\det g|}=|\det\frac{\partial x'}{\partial x}|\sqrt{|\det\eta|}=|\det\frac{\partial x'}{\partial x}|$, so we have $d^nx'=\sqrt{|\det g|}d^nx$ and since $x$ is an arbitrary coordinate system it holds for all. So in flat spacetime, this is the unique form.

Obviously in curved spacetime, we could define different "volume elements", which all reduce to this in flat spacetime. One would be $(1+R)\sqrt{|\det g|}d^nx$, where $R$ is the curvature scalar, but you can take $R$ to be any curvature invariant. We want the volume to be positive, however, and there are spacetimes for which $R=-1$. In that case, the volume element would be identically zero. We clearly do not want that.

There is a stronger requirement than demanding the volume element reduces to the usual form in flat spacetime. As you probably know, on a pseudo-Riemannian space, you can set up so-called "Riemann normal coordinates" about any point $p$. These are also called "locally flat coordinates", since in these coordinates at $p$, $g_{\mu\nu}(p)=\eta_{\mu\nu}$ and $\partial_\sigma g_{\mu\nu}(p)=0$. We can demand that the volume element reduces to the Minkowskian volume element in a locally flat system at $p$. In this case, we can demand that the volume element at $p$ is $dM(p)=d^nx'$, where the $x'$ coordinates are locally flat coordinates. We can then follow the same procedure we did in flat spacetime to obtain that at $p$, in an arbitrary coordinate system $x$, the volume element is $dM(p)=\sqrt{|\det g(p)|}d^nx$, but since $p$ is arbitrary, this holds on the entire manifold.

A more convincing argument can be given in terms of differential forms, but I am not sure if you are familiar with that.

If we consider the above known, then we can see that if we are given a 2-surface $\Sigma$ in an $n$-dimensional pseudo-Riemannian manifold $(M,g)$, and $h$ is the induced metric on $\Sigma$, then the pair $(\Sigma,h)$ is a pseudo-Riemannian manifold (unless $\Sigma$ is a null submanifold, in which case, it is a "degenerate" Riemannian manifold), so the volume element (which in this case is an area element) is given by $\sqrt{|\det h|}d^2\xi$, where $\xi^1 ,\xi^2$ are the surface coordinates. If you calculate $\det h$, this expression will precisely give the formula you started with. We have not used the angle formula at all.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer. I think I understand your answer completely if I get a little resolution on the following two points: 1. In a 2-D Minkowskian spacetime, the dot product of two vectors is zero whenever they are at equal inclinations from the null ray. I don't understand how in such a case, generically calling the angle between them $\dfrac{\pi}{2}$ is justified. 2. Out of all the possible GCT invariant expressions of appropriate dimensions who reduce to the Euclidean area when metric takes Euclidean form, how do we pick one particular expression? Thanks again for your help! $\endgroup$ – Dvij D.C. Dec 23 '16 at 10:17
  • $\begingroup$ @Electrodynamist I'll edit my post when I get home. $\endgroup$ – Bence Racskó Dec 23 '16 at 17:36
  • $\begingroup$ @Electrodynamist Edit added to my answer. Warning: It's a long one. $\endgroup$ – Bence Racskó Dec 23 '16 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.