# Nambu-Goto action and the World-Sheet Area

I am studying string theory from the book "String theory and M-theory", written by Becker, Becker and Schwartz. My question is:

We are told that the Nambu-Goto action is simply the one that extremizes the area of the world-sheet. The area of the world-sheet is described by the induced metric $$G_{\alpha\beta}=g_{\mu\nu}\frac{\partial X^{\mu}}{\partial \sigma^{\alpha}} \frac{\partial X^{\nu}}{\partial \sigma^{\beta}}$$ Why does the determinant of this so-called induced metric give the area of the world-sheet, or equivalently why does reparametrizing the fields $$X^{\mu}$$ by a set of two coordinates $$\sigma^{\alpha},\ \alpha=0,1$$ yields a metric that describes the geometrical properties of the world-sheet? Can someone give me an example of a simple 2 dimensional object lying in a $$d\ge3$$ dimensional space that is described by an induced metric like that (this is a mathematics question, but since it is related with the Nambu-Goto action, I thought I should ask...)?

I can't comment on your first question, but for the second one you can just imagine the unit sphere in $$\mathbb{R}^3$$. Let's parametrize the sphere via \begin{align*} X^\mu(\theta, \varphi) = \begin{pmatrix}\sin\theta \cos\varphi \\ \sin\theta \sin\varphi \\ \cos\theta \end{pmatrix} \qquad X: [0, \pi] \times [0, 2\pi] \to \mathbb{R}^3 \end{align*} The euclidean metric is just the unit matrix $$g_{\mu\nu} = \delta_{\mu\nu}$$, so the induced metric on the sphere is \begin{align*} G_{ab} = \delta_{\mu\nu} \frac{\partial X^\mu}{\partial \sigma^a} \frac{\partial X^\nu}{\sigma^b} &= \sum_{\mu=1}^3 \frac{\partial X^\mu}{\partial \sigma^a} \frac{\partial X^\mu}{\partial \sigma^b} \\ &= \begin{pmatrix}\cos^2\theta \cos^2\varphi + \cos^2\theta \sin^2\varphi + \sin^2\theta & \cos\theta\cos\varphi\sin\theta (-\sin\varphi) + \cos\theta\sin\varphi \sin\theta\cos\varphi \\ \cos\theta\cos\varphi\sin\theta (-\sin\varphi) + \cos\theta\sin\varphi \sin\theta\cos\varphi & \sin^2\theta\sin^2\varphi + \sin^2\theta\cos^2\varphi \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta \end{pmatrix} \end{align*} Now the determinant is simply $$\det G = \sin^2\theta$$ and so the volume of the sphere is \begin{align*} V(\mathbb{S}^2) = \int\mathrm{d}^2\sigma\ \sqrt{|\det G |} = \int_0^\pi\mathrm{d}\theta \int_0^{2\pi}\mathrm{d}\varphi\ \sin\theta = 2\cdot 2\pi = 4\pi \end{align*} as expected.
Hopefully someone else can flesh the following out a bit more, but here's a rough sketch on why you might expect the square root of the determinant of the metric: the volume-form is completely antisymmetric $$\mathrm{d}\sigma^1 \dots \mathrm{d}\sigma^n = \frac{1}{n!} \varepsilon_{\rho_1 \dots \rho_n} \mathrm{d}\sigma^{\rho_1} \dots \mathrm{d}\sigma^{\rho_n}$$. Under coordinate transformations the Levi-Civita symbol will turn the derivatives from the chain-rule into a Jacobian factor \begin{align*} \frac{1}{n!} \varepsilon_{\rho_1 \dots \rho_n} \mathrm{d}\sigma^{\rho_1} \dots \mathrm{d}\sigma^{\rho_n} = \frac{1}{n!}\det\left(\frac{\partial \sigma}{\partial \tilde{\sigma}}\right)\varepsilon_{\kappa_1 \dots \kappa_n}\mathrm{d}\tilde{\sigma}^{\kappa_1}\dots \mathrm{d}\tilde{\sigma}^{\kappa_n} = J\ \mathrm{d}\tilde{\sigma}^1 \dots \mathrm{d}\tilde{\sigma}^n \end{align*} The metric transforms as \begin{align*} G_{a' b'} = \frac{\partial \sigma^a}{\partial\tilde\sigma^{a'}} \frac{\partial \sigma^b}{\partial \tilde{\sigma}^{b'}} G_{ab} \qquad\Rightarrow\qquad \det G' = \det\left(\frac{\partial \sigma}{\partial \tilde\sigma}\right)^{2}\ \det G = J^2\ \det G \end{align*} By combining the two transformations we find an invariant integrand \begin{align*} \int\mathrm{d}^n\sigma\ \left(\det G\right)^{1/2} = \int J\,\mathrm{d}^n\tilde\sigma\ \left(J^{-2} \det G' \right)^{1/2} = \int\mathrm{d}^n\tilde\sigma\ \left(\det G'\right)^{1/2} \end{align*}
• I'm not sure I quite understand your question. The point is that the underlying object, i.e. the world sheet, is the thing that we care about, but to handle it we need to parametrize it in some way. Since parametrization is somewhat arbitrary we want everything to be independent of the parametrization. A parametrization of $\mathbb{S}^2$ will never include stuff outside of $\mathbb{S}^2$, because otherwise it wouldn't be a parametrization. Mar 21 at 12:08
• Οh okay, so it is because we parametrize the worldsheet (i.e. choosing $\sigma^{\alpha}$ s.t. the new metric describes the surface of the worldsheet) that the area of the worldsheet is given by the new metric (constructed taking into consideration the new parametrization). I think I see it now Mar 21 at 16:10