# ADM Formulation in General Relativity

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.

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

• 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 $n_{a}n_{b}$ – Ricky Pang Apr 25 at 14:47
• 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 $\Sigma_t$. – Spencer Apr 25 at 16:05
• 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? – Spencer Apr 25 at 16:06
• Let me know if this new edit clears up your confusion. – Spencer Apr 25 at 23:16
• Did I answer your question or is some further explanation needed? – Spencer Apr 30 at 20:29