In the ADM(Arnowitt – Deser – Misner) formulation, we can foliate a globally hyperbolic spacetime by spacelike hyper-surface(Cauchy surface) $\Sigma_{t}$, which parametrised by global time function $t$. Therefore, in each point of spacelike hyper-surface, we can let $n^{a}$ to be a future -directed timelike unit vector field normal to the hyper surface $\Sigma_{t}$, which satisfies $n^{a}n_{a} = -1 $ and $n_{a} \propto \nabla_{a}t$ ($\nabla_{a}$ is associated with spacetime metric. $g_{ab}$). Therefore, the spacetime metric $g_{ab}$ induces a spatial metric $\gamma_{ab}$ which is defined as

\begin{align} \gamma_{ab} = g_{ab} + n_{a}n_{b} \end{align}

My difficulties are that why the spacetime metric induce such spatial metric $\gamma_{ab}$ and how can I understand the above equation.

  • 2
    $\begingroup$ The spacial metric is just the restriction of $g$ to $\Sigma_t$. $\endgroup$ – Moishe Kohan Apr 25 at 1:37
  • $\begingroup$ Personally i like the exposition in here. $\endgroup$ – Sou Apr 25 at 8:48

Consider a vector of the form,

$$ v^a = \alpha \ \sigma^a + \beta \ n^a,$$

where $\sigma^a$ is a vector orthogonal to $n^a$, i.e, $\sigma_a n^a=0$. Here $\alpha$ and $\beta$ are just some constant scalars.

Now lets contract our vector $v^a$ with the induced metric $\gamma_{ab}$.

$$\gamma_{ab} v^a = \gamma_{ab} ( \alpha \sigma^a + \beta n^a) $$

$$ = (g_{ab} + n_a n_b) ( \alpha \sigma^a + \beta n^a) $$

$$ = \alpha\ g_{ab} \sigma^a + \beta \ g_{ab} n^a + \alpha \ n_b (n_a \sigma^a) + \beta \ n_b (n_a n^a) $$

The ordinary metric just lowers the indices at this step. We also apply $\sigma^a n_a = 0$ and $n_a n^a = -1$.

$$ = \alpha\ \sigma_b + \beta \ n_b + \alpha \ n_b (0) + \beta \ n_b (-1) $$

$$ = \alpha\ \sigma_b + \beta \ n_b + \beta \ n_b (-1) $$

$$ = \alpha\ \sigma_b $$

When $\gamma_{ab}$ is contracted against vectors parallel to the surface $\Sigma$ it lowers their indices as expected (by parallel I mean "orthogonal to $n^a$"). When $\gamma_{ab}$ is contracted against vectors normal to the surface the result is $0$. $\gamma_{ab}$ then acts as a projection onto the hypersurface and as a metric within that surface.

Lets consider a concrete example in Euclidean space. We will be working in $\mathbb{R}^3$ using cartesian coordinates; which means our metric will be $g_{ij} = \delta_{ij}$. We will foliate the space with spheres centered on the origin. Each sphere will have a radius, $r=\sqrt{x^2+y^2+z^2}$.

Our normalized unit vector will be $\hat{n} = \nabla r / \| \nabla r \|$. However this has a norm which is positive $\| \hat{n} \| = 1$, so to obtain $\gamma_{ij}$ we will need to subtract it from $g_{ij}$ rather than add it.

$$\boxed{ \gamma_{ij} = g_{ij} - n_i n_j }$$

Now the following are the normalized basis vectors (vector fields really) for spherical coordinates.

$$ \hat{e}_r = \frac{\nabla r }{\| \nabla r\|} = \begin{bmatrix} x/r \\ y/r \\ z/r \end{bmatrix}$$

$$ \hat{e}_\phi = \frac{\nabla \phi }{\| \nabla \phi\|} = \begin{bmatrix} -y/\sqrt{x^2+y^2} \\ x/\sqrt{x^2+y^2} \\ 0 \end{bmatrix}$$

$$ \hat{e}_\theta = \frac{\nabla \theta}{\| \nabla \theta \|} = \begin{bmatrix} \frac{zx}{r\sqrt{x^2+y^2}} \\ \frac{zy}{r\sqrt{x^2+y^2}} \\ -\frac{x^2+y^2}{r\sqrt{x^2+y^2}}\end{bmatrix}$$

Since these vectors form a basis any vector can be resoled in terms of them.

$$ \vec{v} = \hat{e}_r (\hat{e}_r \cdot \vec{v}) + \hat{e}_\phi (\hat{e}_\phi \cdot \vec{v}) + \hat{e}_\theta (\hat{e}_\theta \cdot \vec{v})$$

$$ \vec{v} = \sum_{\alpha=r,\phi,\theta} \hat{e}_\alpha (\hat{e}_\alpha \cdot \vec{v})$$

$$ \vec{v} = \sum_{\alpha=r,\phi,\theta} \hat{e}_\alpha (\hat{e}_{\alpha})_i v^i$$

$$ v_j = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j (\hat{e}_{\alpha})_i v^i$$

$$ g_{ij} v^i = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j (\hat{e}_{\alpha})_i v^i$$

Since the equality must hold for any vector $\vec{v}$ we can remove it from the equations and we obtain an identity.

$$g_{ij} v^i = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j (\hat{e}_{\alpha})_i v^i$$

$$\boxed{ g_{ij} = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j (\hat{e}_{\alpha})_i}$$

If you have experience in quantum mechanics you should compare this with $\sum |n \rangle \langle n| = 1$ for a complete set of states. Now lets return to our expression for $\gamma_{ij}$ in light of this identity.

$$ \gamma_{ij} = g_{ij} - n_i n_j$$ $$ \gamma_{ij} = g_{ij} - (\hat{e}_r)_i (\hat{e}_r)_j$$ $$ \gamma_{ij} = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j (\hat{e}_{\alpha})_i - (\hat{e}_r)_i (\hat{e}_r)_j$$ $$\boxed{ \gamma_{ij} = \sum_{\alpha=\phi,\theta} (\hat{e}_\alpha)_j (\hat{e}_{\alpha})_i }$$

You can see we are literally subtracting out the part of the metric that has to do with $r$.

You can use the explicit vectors that I wrote down for $\hat{e}_\phi$ and $\hat{e}_\theta$ to construct a matrix for $\gamma_{ij}$ in the Cartesian basis. If you diagonalize this matrix you will find that $\hat{e}_r$, $\hat{e}_\phi$, and $\hat{e}_\theta$ are eigenvectors where $\hat{e}_r$ will have an eigenvalue of $0$. The diagonal form will be the metric for a 2-sphere which you should recognize from your studies, along with a 0 in the diagonal entry corresponding to $r$.

  • $\begingroup$ Thank you for your answer, Spencer. I want to ask further why we can write $\gamma_{ab} = g_{ab} + n_{a}n_{b}$. Is it due to the definition of metric $g_{ab} = \langle \partial_{a}, \partial_{b} \rangle$? Under this definition, we can decompose spacetime metric $g_{ab}$ into following: tangent vectors on $\Sigma_{t}$ form metric $\gamma_{ab}$ and time-like vector form metric $n_{a}n_{b}$ $\endgroup$ – Ricky Pang Apr 25 at 14:47
  • $\begingroup$ I am not familiar with a derivation which follows that approach. The usual conditions are that $\gamma_{ab} n^b = 0$ and $\gamma_{ab}s^b = s_a$ for any $s^a$ tangent to $\Sigma_t$. The idea is we want a symmetric bilinear which is consistent with $g_{ab}$ when acting on vectors tangent to $\Sigma_t$. $\endgroup$ – Spencer Apr 25 at 16:05
  • $\begingroup$ To understand what is going on you might want to try the formula out on explicit metrics from example you know. For instance what will $\gamma_{ab}$ be for the Schwarzchild spacetime? $\endgroup$ – Spencer Apr 25 at 16:06
  • $\begingroup$ Let me know if this new edit clears up your confusion. $\endgroup$ – Spencer Apr 25 at 23:16
  • $\begingroup$ Did I answer your question or is some further explanation needed? $\endgroup$ – Spencer Apr 30 at 20:29

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.