3.40. Differential Geometry

3.40.1. Manifolds

Scalars, Vectors, Tensors

Differentiable manifold U is a space covered by an atlas of maps, each map covers part of the manifold and is a one to one mapping to an euclidean space \R^n:

\phi:U\to \R^n

Let’s have a one-to-one transformation between x^\mu and x'^\mu coordinates (we simply write x\equiv x^\mu, etc.):

x' = x'(x)

x = x(x')

Scalar \phi(x) is such a field that transforms as (\phi'(x') is it’s value in x' coordinates):

\phi'(x')=\phi(x)

One form p_\alpha(x) is such a field that transforms the same as the gradient \partial\phi(x)\over\partial x^\mu of a scalar, that transforms as (\partial\phi'(x')\over\partial x'^\mu is it’s value in x' coordinates):

{\partial\phi'(x')\over\partial x'^\mu} ={\partial x^\nu\over\partial x'^\mu} {\partial\phi'(x')\over\partial x^\nu} ={\partial x^\nu\over\partial x'^\mu} {\partial\phi(x)\over\partial x^\nu}

so

p'_\mu(x') ={\partial x^\nu\over\partial x'^\mu} p_\nu(x)

Vector V^\alpha is such a field that produces a scalar \phi=V^\alpha p_\alpha when contracted with a one form and this fact is used to deduce how it transforms:

\phi' = V'^\alpha p'_\alpha = V'^\alpha {\partial x^\beta\over\partial x'^\alpha} p_\beta =\phi = V^\beta p_\beta

so we have

V'^\alpha {\partial x^\beta\over\partial x'^\alpha} = V^\beta

multiplying by {\partial x'^\mu\over\partial x^\beta} and using the fact that {\partial x^\beta\over\partial x'^\alpha} {\partial x'^\mu\over\partial x^\beta} = {\partial x'^\mu\over\partial x'^\alpha} =\delta^\mu_\alpha we get

V'^\mu = {\partial x'^\mu\over\partial x^\beta} V^\beta

Higher tensors are build up and their transformation properties derived from the fact, that by contracting with either a vector or a form we get a lower rank tensor that we already know how it transforms.

Having now defined scalar, vector and tensor fields, one may then choose a basis at each point for each field, the only requirement being that the basis is not singular. For example for vectors, each point in U has a basis \vec e_\alpha, so a vector (field) \vec V has components V^\alpha with respect to this basis:

\vec V = V^\alpha\vec e_\alpha

Covariant differentiation

The derivative of the basis vector {\partial \vec e_\alpha\over\partial x^\beta} is a vector, thus it can be written as a linear combination of the basis vectors:

{\partial \vec e_\alpha\over\partial x^\beta}=\Gamma^\mu_{\alpha\beta} \vec e_\mu

Differentiating a vector is then easy:

{\partial\vec V\over\partial x^\beta}\equiv\nabla_\beta \vec V ={\partial V^\alpha\over\partial x^\beta}\vec e_\alpha+ V^\alpha {\partial \vec e_\alpha\over\partial x^\beta} ={\partial V^\alpha\over\partial x^\beta}\vec e_\alpha+ V^\alpha \Gamma^\mu_{\alpha\beta} \vec e_\mu =\left({\partial V^\alpha\over\partial x^\beta}+ \Gamma^\alpha_{\mu\beta}V^\mu \right) \vec e_\alpha

So we define a covariant derivative:

\nabla_\beta V^\alpha = {\partial V^\alpha\over\partial x^\beta}+ \Gamma^\alpha_{\mu\beta}V^\mu

and write

{\partial\vec V\over\partial x^\beta}=\nabla_\beta \vec V =\left(\nabla_\beta \vec V\right)^\alpha\vec e_\alpha =\left(\nabla_\beta V^\alpha\right)\vec e_\alpha

I.e. we have:

\nabla_\beta \vec V = \nabla_\beta(V^\alpha\vec e_\alpha) = (\nabla_\beta V^\alpha)\vec e_\alpha

We also define:

\nabla_{\vec X} \vec V = \nabla_{X^\beta \vec e_\beta} \vec V \equiv X^\beta\nabla_\beta\vec V = X^\beta(\nabla_\beta V^\alpha)\vec e_\alpha

A scalar doesn’t depend on basis vectors, so its covariant derivative is just its partial derivative

\nabla_\alpha \phi={\partial \phi\over\partial x^\alpha}

Differentiating a one form p_\alpha is done using the fact, that \phi=p_\alpha V^\alpha is a scalar, thus

\nabla_\beta \phi={\partial p_\alpha V^\alpha\over\partial x^\beta} ={\partial p_\alpha \over\partial x^\beta}V^\alpha+ p_\alpha{\partial V^\alpha\over\partial x^\beta} ={\partial p_\alpha \over\partial x^\beta}V^\alpha+ p_\alpha\left(\nabla_\beta V^\alpha- \Gamma^\alpha_{\mu\beta}V^\mu\right)=

=V^\alpha\left({\partial p_\alpha \over\partial x^\beta}-\Gamma^\mu_{\alpha\beta}p_\mu\right)+ p_\alpha\nabla_\beta V^\alpha =V^\alpha\nabla_\beta p_\alpha+ p_\alpha\nabla_\beta V^\alpha

where we have defined

\nabla_\beta p_\alpha = {\partial p_\alpha \over\partial x^\beta}-\Gamma^\mu_{\alpha\beta}p_\mu

This is obviously a tensor, because the above equation has a tensor on the left hand side (\nabla_\beta \phi) and tensors on the right hand side (p_\alpha\nabla_\beta V^\alpha and V^\alpha). Similarly for the derivative of the tensor A^{\mu\nu} we use the fact that V^\mu=A^{\mu\nu}p_\nu is a vector:

\nabla_\beta V^\mu=\nabla_\beta (A^{\mu\nu}p_\nu)=\partial_\beta (A^{\mu\nu}p_\nu)+\Gamma^\mu_{\alpha\beta}A^{\alpha\nu}p_\nu =p_\nu\partial_\beta A^{\mu\nu}+ A^{\mu\nu}\partial_\beta p_\nu+\Gamma^\mu_{\alpha\beta}A^{\alpha\nu}p_\nu=

=p_\nu\partial_\beta A^{\mu\nu}+ A^{\mu\nu}\left(\nabla_\beta p_\nu+\Gamma^\mu_{\nu\beta}p_\mu\right) +\Gamma^\mu_{\alpha\beta}A^{\alpha\nu}p_\nu =p_\nu\nabla_\beta A^{\mu\nu}+ A^{\mu\nu}\nabla_\beta p_\nu

where we define

\nabla_\beta A^{\mu\nu}=\partial_\beta A^{\mu\nu} +\Gamma^\mu_{\alpha\beta}A^{\alpha\nu} +\Gamma^\nu_{\alpha\beta}A^{\mu\alpha}

and so on for other tensors, for example:

\nabla_\beta A^\mu{}_\nu=\partial_\beta A^\mu{}_\nu +\Gamma^\mu_{\alpha\beta}A^\alpha{}_\nu -\Gamma^\alpha_{\nu\beta}A^\mu{}_\alpha

\nabla_\beta A_{\mu\nu}=\partial_\beta A_{\mu\nu} -\Gamma^\alpha_{\mu\beta}A_{\alpha\nu} -\Gamma^\alpha_{\nu\beta}A_{\mu\alpha}

One can now easily proof some common relations simply by rewriting it to components and back:

\nabla_{\vec X}(f\vec Y) = (\nabla_{\vec X}f)\vec Y + f\nabla_{\vec X}\vec Y

\nabla_{\vec X}(\vec Y+\vec Z) = \nabla_{\vec X}\vec Y + \nabla_{\vec X}\vec Z

\nabla_{f\vec X}\vec Y = f\nabla_{\vec X}\vec Y

Change of variable:

\Gamma'^\alpha{}_{\beta\gamma}= {\partial x^\mu\over\partial x'^\beta} {\partial x^\nu\over\partial x'^\gamma} \Gamma^\sigma{}_{\mu\nu} {\partial x'^\alpha\over\partial x^\sigma} + {\partial x'^\alpha\over\partial x^\sigma} {\partial^2 x^\sigma\over\partial x'^\beta\partial x'^\gamma}

Parallel transport

If the vectors \vec V at infinitesimally close points of the curve x^\mu(\lambda) are parallel and of equal length, then \vec V is said to be parallel transported along the curve, i.e.:

{\d \vec V\over\d\lambda} = 0

So

{\d \vec V\over\d\lambda} = {\d (V^\alpha\vec e_\alpha)\over\d\lambda} = {\d x^\beta\over\d\lambda}\partial_\beta (V^\alpha\vec e_\alpha) = {\d x^\beta\over\d\lambda}(\nabla_\beta V^\alpha) \vec e_\alpha = 0

In components (using the tangent vector U^\beta = {\d x^\beta\over\d\lambda}):

{\d V^\alpha\over\d\lambda} = U^\beta\nabla_\beta V^\alpha = 0

Fermi-Walker transport

In local inertial frame:

U^\lambda_0 = (1, 0, 0, 0)

{\d S^i\over\d t} = 0

We require orthogonality S_\mu U^\mu = 0, in a general frame:

{\d S^\alpha\over\d \tau} = \lambda U^\alpha = S_\mu {\d U^\mu\over\d \tau} U^\alpha

where \lambda was calculated by differentiating the orthogonality condition. This is called a Thomas precession.

For any vector, we define: the vector X^\mu is Fermi-Walker tranported along the curve if:

{\d X^\mu\over\d\lambda} = X_\alpha{\d U^\alpha\over\d\lambda}U^\mu -X_\alpha U^\alpha{\d U^\mu\over\d\lambda}

If X^\mu is perpendicular to U^\mu, the second term is zero and the result is called a Fermi transport.

Why: the U^\mu is transported by Fermi-Walker and also this is the equation for gyroscopes, so the natural, nonrotating tetrade is the one with \vec e^\mu_0 \equiv U^\mu, which is then correctly transported along any curve (not just geodesics).

Geodesics

Geodesics is a curve x^\alpha(\lambda) that locally looks like a line, i.e. it parallel transports its own tangent vector:

U^\beta\nabla_\beta U^\alpha = 0

so

U^\beta\partial_\beta U^\alpha + \Gamma^\alpha_{\beta\gamma}U^\beta U^\gamma = 0

or equivalently (using the fact U^\beta\partial_\beta U^\alpha= {\d x^\beta\over\d\lambda}{\partial\over\partial x^\beta} {\d x^\alpha\over\d\lambda} = {\d^2 x^\alpha\over\d\lambda^2}):

(3.40.1.1){\d^2 x^\alpha\over\d\lambda^2} + \Gamma^\alpha_{\beta\gamma} {\d x^\beta\over\d\lambda}{\d x^\gamma\over\d\lambda} = 0

Let’s determine all possible reparametrizations that leave the geodesic equation invariant:

\lambda' = f(\lambda) {\d x^\alpha\over\d\lambda} ={\d \lambda'\over\d\lambda} {\d x^\alpha\over\d\lambda'} = f'(\lambda){\d x^\alpha\over\d\lambda'} {\d^2 x^\alpha\over\d\lambda^2} ={\d \over\d\lambda} \left( f'(\lambda){\d x^\alpha\over\d\lambda'} \right) = f''(\lambda) {\d x^\alpha\over\d\lambda'} + f'(\lambda){\d \over\d\lambda}{\d x^\alpha\over\d\lambda'} = = f''(\lambda) {\d x^\alpha\over\d\lambda'} + f'^2(\lambda){\d \over\d\lambda'}{\d x^\alpha\over\d\lambda'} = = f''(\lambda) {\d x^\alpha\over\d\lambda'} + f'^2(\lambda){\d^2 x^\alpha\over\d\lambda'^2}

Substituting into the geodesic equation, we get:

f''(\lambda) {\d x^\alpha\over\d\lambda'} + f'^2(\lambda)\left( {\d^2 x^\alpha\over\d\lambda'^2} + \Gamma^\alpha_{\beta\gamma} {\d x^\beta\over\d\lambda'} {\d x^\gamma\over\d\lambda'} \right) = 0

So we can see that the equation is invariant as long as f''(\lambda) = 0, which gives:

f(\lambda) = a \lambda + b

This is called an affine reparametrization.

Another way to derive the geodesic equation is by finding a curve that extremizes the proper time:

\tau = \int \d \tau = \int \sqrt{-{1\over c^2} \d s^2} = \int \sqrt{-{1\over c^2} g_{\mu\nu} \d x^\mu \d x^\nu} = \int \sqrt{-{1\over c^2} g_{\mu\nu} {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda}} \d\lambda = \int K \d\lambda

Here \lambda can be any parametrization. We have introduced K to make the formulas shorter:

K = \sqrt{-{1\over c^2} g_{\mu\nu} {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda}}

We vary this action with respect to x^\mu:

\delta \tau = \delta \int K \d\lambda = = \delta \int \sqrt{-{1\over c^2} g_{\mu\nu} {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda}} \d\lambda = = \int {-{1\over c^2}\left( (\delta g_{\mu\nu}) {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} +g_{\mu\nu} \left(\delta {\d x^\mu\over\d \lambda}\right) {\d x^\nu\over\d\lambda} +g_{\mu\nu} {\d x^\mu\over\d \lambda}\left(\delta {\d x^\nu\over\d\lambda} \right) \right) \over 2 K } \d\lambda = = {1\over c^2}\int {\left(-\half (\delta g_{\mu\nu}) {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} - g_{\mu\nu} \left(\delta {\d x^\mu\over\d \lambda}\right) {\d x^\nu\over\d\lambda} \right) \over K } \d\lambda = = {1\over c^2}\int {\left(-\half (\delta x^\alpha) \partial_\alpha g_{\mu\nu} {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} - g_{\alpha\nu} {\d (\delta x^\alpha)\over\d \lambda} {\d x^\nu\over\d\lambda} \right) \over K} \d\lambda = = {1\over c^2}\int \left({-\half (\delta x^\alpha) \partial_\alpha g_{\mu\nu} {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} \over K } + {\d\over\d\lambda}\left({g_{\alpha\nu} {\d x^\nu\over\d\lambda} \over K}\right)(\delta x^\alpha)\right) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left({g_{\alpha\nu} {\d x^\nu\over\d\lambda} \over K}\right) -\half \partial_\alpha g_{\mu\nu} {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} \right) (\delta x^\alpha) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left(1\over K\right)+ {\d g_{\alpha\nu} \over\d\lambda} {\d x^\nu\over\d\lambda} + g_{\alpha\nu} {\d^2 x^\nu\over\d\lambda^2} -\half \partial_\alpha g_{\mu\nu} {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} \right) (\delta x^\alpha) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left(1\over K\right)+ g_{\alpha\nu} {\d^2 x^\nu\over\d\lambda^2} +\partial_\mu g_{\alpha\nu} {\d x^\mu\over\d\lambda} {\d x^\nu\over\d\lambda} -\half \partial_\alpha g_{\mu\nu} {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} \right) (\delta x^\alpha) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left(1\over K\right)+ g_{\alpha\nu} {\d^2 x^\nu\over\d\lambda^2} +\half(\partial_\mu g_{\alpha\nu}+\partial_\nu g_{\alpha\mu} - \partial_\alpha g_{\mu\nu}) {\d x^\mu\over\d \lambda}{\d x^\nu\over\d\lambda} \right) (\delta x^\alpha) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left(1\over K\right)+ g_{\alpha\nu} {\d^2 x^\nu\over\d\lambda^2} +\half( \partial_\mu g_{\alpha\nu} + \partial_\nu g_{\alpha\mu} - \partial_\alpha g_{\mu\nu} ) {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda} \right)g^{\alpha\rho}(\delta x_\rho) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left(1\over K\right)g^{\alpha\rho}+ \delta_\nu{}^\rho {\d^2 x^\nu\over\d\lambda^2} +\half g^{\alpha\rho} ( \partial_\mu g_{\alpha\nu} + \partial_\nu g_{\alpha\mu} - \partial_\alpha g_{\mu\nu} ) {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda} \right)(\delta x_\rho) \d\lambda = = {1\over c^2}\int {1\over K} \left( K{\d\over\d\lambda}\left(1\over K\right)g^{\alpha\rho}+ {\d^2 x^\rho\over\d\lambda^2} +\Gamma^\rho_{\mu\nu} {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda} \right)(\delta x_\rho) \d\lambda

By setting the variation \delta\tau=0 we obtain the geodesic equation:

(3.40.1.2)K{\d\over\d\lambda}\left(1\over K\right)g^{\alpha\rho}+ {\d^2 x^\rho\over\d\lambda^2} +\Gamma^\rho_{\mu\nu} {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda} = 0

We have a freedom of choosing \lambda, so we choose such parametrization so that \d\lambda = \d\tau, which makes K=1 and we recover (3.40.1.1):

{\d^2 x^\rho\over\d\lambda^2} +\Gamma^\rho_{\mu\nu} {\d x^\mu\over\d \lambda} {\d x^\nu\over\d\lambda} = 0

Note that the equation (3.40.1.2) is parametrization invariant, but (3.40.1.1) is not (only affine reparametrization leaves (3.40.1.1) invariant).

Riemann Curvature Tensor

Curvature means that we take a vector V^\mu, parallel transport it around a closed loop (which is just applying a commutator of the covariant derivatives [\nabla_\alpha, \nabla_\beta]V^\mu) and see how it changes. We express the result in terms of the vector V^\mu:

[\nabla_\alpha, \nabla_\beta]V^\mu\equiv R^\mu{}_{\nu\alpha\beta}V^\nu

The coefficients R^\mu{}_{\nu\alpha\beta} form a tensor called Riemann curvature tensor. Expanding the left hand side:

[\nabla_\alpha, \nabla_\beta]V^\mu =\nabla_\alpha \nabla_\beta V^\mu - \nabla_\alpha \nabla_\beta V^\mu = =\nabla_\alpha \nabla_\beta V^\mu - (\alpha \leftrightarrow \beta) = =\partial_\alpha \nabla_\beta V^\mu - \Gamma^\sigma_{\alpha \beta} \nabla_\sigma V^\mu + \Gamma^\mu_{\alpha \sigma} \nabla_\beta V^\sigma - (\alpha \leftrightarrow \beta) = =\partial_\alpha \left(\partial_\beta V^\mu +\Gamma^\mu_{\beta\sigma} V^\sigma\right) - \Gamma^\sigma_{\alpha \beta} \left(\partial_\sigma V^\mu +\Gamma^\mu_{\sigma\nu} V^\nu\right) + \Gamma^\mu_{\alpha \sigma} \left(\partial_\beta V^\sigma +\Gamma^\sigma_{\beta\nu} V^\nu\right) - (\alpha \leftrightarrow \beta) = =\partial_\alpha \left(\partial_\beta V^\mu +\Gamma^\mu_{\beta\sigma} V^\sigma\right) + \Gamma^\mu_{\alpha \sigma} \left(\partial_\beta V^\sigma +\Gamma^\sigma_{\beta\nu} V^\nu\right) - (\alpha \leftrightarrow \beta) = = \partial_\alpha \partial_\beta V^\mu + \partial_\alpha\Gamma^\mu_{\beta\sigma} V^\sigma + \Gamma^\mu_{\beta\sigma} \partial_\alpha V^\sigma + \Gamma^\mu_{\alpha \sigma} \partial_\beta V^\sigma +\Gamma^\mu_{\alpha \sigma}\Gamma^\sigma_{\beta\nu} V^\nu - (\alpha \leftrightarrow \beta) = = \partial_\alpha\Gamma^\mu_{\beta\sigma} V^\sigma +\Gamma^\mu_{\alpha \sigma}\Gamma^\sigma_{\beta\nu} V^\nu - (\alpha \leftrightarrow \beta) = = \left(\partial_\alpha\Gamma^\mu_{\beta\nu} +\Gamma^\mu_{\alpha \sigma}\Gamma^\sigma_{\beta\nu} - (\alpha \leftrightarrow \beta) \right) V^\nu

Where we have used the fact that all terms symmetric in \alpha\beta (in particular \Gamma^\sigma_{\alpha\beta} and \partial_\alpha \partial_\beta V^\mu and \Gamma^\mu_{\beta\sigma} \partial_\alpha V^\sigma + \Gamma^\mu_{\alpha \sigma} \partial_\beta V^\sigma) get canceled by the same term in the (\alpha \leftrightarrow \beta). We get

R^\mu{}_{\nu\alpha\beta} = \partial_\alpha\Gamma^\mu_{\beta \nu} + \Gamma^\mu_{\alpha\sigma}\Gamma^\sigma_{\beta \nu} - (\alpha \leftrightarrow \beta) = \partial_\alpha\Gamma^\mu_{\beta \nu} - \partial_\beta \Gamma^\mu_{\alpha\nu} + \Gamma^\mu_{\alpha\sigma}\Gamma^\sigma_{\beta \nu} - \Gamma^\mu_{\beta \sigma}\Gamma^\sigma_{\alpha\nu}

In order to see all the symmetries, that the Riemann tensor has, we lower the first index

R_{\mu\nu\alpha\beta} = g_{\mu\lambda} R^\lambda{}_{\nu\alpha\beta} = g_{\mu\lambda} \left( \partial_\alpha\Gamma^\lambda_{\beta \nu} + \Gamma^\lambda_{\alpha\sigma}\Gamma^\sigma_{\beta \nu} - (\alpha \leftrightarrow \beta) \right)

and use local inertial frame coordinates, where all Christoffel symbols vanish (not their derivatives though):

R_{\mu\nu\alpha\beta} = g_{\mu\lambda} \left( \partial_\alpha\Gamma^\lambda_{\beta \nu} - (\alpha \leftrightarrow \beta) \right) = g_{\mu\lambda}\left(\partial_\alpha\left( \half g^{\lambda\sigma}( \partial_\beta g_{\sigma\nu} +\partial_\nu g_{\sigma\beta} -\partial_\sigma g_{\beta\nu} )\right) - (\alpha \leftrightarrow \beta) \right) = = \half g_{\mu\lambda}g^{\lambda\sigma} \left(\partial_\alpha\left( \partial_\beta g_{\sigma\nu} +\partial_\nu g_{\sigma\beta} -\partial_\sigma g_{\beta\nu} \right) - (\alpha \leftrightarrow \beta) \right) = = \half \delta_\mu{}^\sigma \left(\partial_\alpha\left( \partial_\beta g_{\sigma\nu} +\partial_\nu g_{\sigma\beta} -\partial_\sigma g_{\beta\nu} \right) - (\alpha \leftrightarrow \beta) \right) = = \half \left( \partial_\alpha\partial_\beta g_{\mu\nu} +\partial_\alpha\partial_\nu g_{\mu\beta} -\partial_\alpha\partial_\mu g_{\beta\nu} - (\alpha \leftrightarrow \beta) \right) = = \half \left( \partial_\alpha\partial_\nu g_{\mu\beta} -\partial_\alpha\partial_\mu g_{\beta\nu} - (\alpha \leftrightarrow \beta) \right) = = \half \left( \partial_\alpha\partial_\nu g_{\mu\beta} -\partial_\alpha\partial_\mu g_{\beta\nu} -\partial_\beta\partial_\nu g_{\mu\alpha} +\partial_\beta\partial_\mu g_{\alpha\nu} \right)

We will also need:

\nabla_\lambda R_{\mu\nu\alpha\beta} = \half \partial_\lambda \left( \partial_\alpha\partial_\nu g_{\mu\beta} -\partial_\alpha\partial_\mu g_{\beta\nu} - (\alpha \leftrightarrow \beta) \right) = = \half \left( \partial_\lambda\partial_\alpha\partial_\nu g_{\mu\beta} -\partial_\lambda\partial_\alpha\partial_\mu g_{\beta\nu} -\partial_\lambda\partial_\beta\partial_\nu g_{\mu\alpha} +\partial_\lambda\partial_\beta\partial_\mu g_{\alpha\nu} \right)

Using these expressions for the curvature tensor in a local inertial frame, we derive the following 5 symmetries of the curvature tensor by simply substituting for the left hand side and verify that it is equal to the right hand side:

R_{\mu\nu\alpha\beta} &= -R_{\mu\nu\beta\alpha} \\ R_{\mu\nu\alpha\beta} &= -R_{\nu\mu\alpha\beta} \\ R_{\mu\nu\alpha\beta} &= R_{\alpha\beta\mu\nu} \\ R_{\mu\nu\alpha\beta} + R_{\mu\alpha\beta\nu} + R_{\mu\beta\nu\alpha} &= 0\\ \nabla_\lambda R_{\mu\nu\alpha\beta} +\nabla_\mu R_{\nu\lambda\alpha\beta} +\nabla_\nu R_{\lambda\mu\alpha\beta} &= 0

These are tensor expressions and so even though we derived them in a local inertial frame, they hold in all coordinates. The last identity is called a Bianchi identity.

The Ricci tensor is defined as:

R_{\mu\nu} = R^\lambda{}_{\mu\lambda\nu} = g^{\lambda\sigma} R_{\sigma\mu\lambda\nu}

From the last equality we can see that it is symmetric in \mu\nu. A Ricci scalar is defined as:

R = R^\mu{}_\mu = g^{\mu\nu} R_{\mu\nu}

The Einstein tensor is defined as:

G_{\mu\nu} = R_{\mu\nu} - \half R g_{\mu\nu}

It is symmetric in \mu\nu due to the symmetry of the metric and Ricci tensors. By contracting the Bianchi identity twice, we can show that Einstein tensor has zero divergence:

g^{\lambda\alpha} g^{\nu\beta}\left( \nabla_\lambda R_{\mu\nu\alpha\beta} +\nabla_\mu R_{\nu\lambda\alpha\beta} +\nabla_\nu R_{\lambda\mu\alpha\beta}\right) = = \nabla^\alpha g^{\nu\beta} R_{\mu\nu\alpha\beta} +\nabla_\mu g^{\lambda\alpha} R^\beta{}_{\lambda\alpha\beta} +\nabla^\beta g^{\lambda\alpha} R_{\lambda\mu\alpha\beta} = = \nabla^\alpha g^{\nu\beta} R_{\nu\mu\beta\alpha} -\nabla_\mu g^{\lambda\alpha} R^\beta{}_{\lambda\beta\alpha} +\nabla^\beta R^\alpha{}_{\mu\alpha\beta} = = \nabla^\alpha R^\beta{}_{\mu\beta\alpha} -\nabla_\mu g^{\lambda\alpha} R_{\lambda\alpha} +\nabla^\beta R_{\mu\beta} = = \nabla^\alpha R_{\mu\alpha} -\nabla_\mu R^\alpha{}_\alpha +\nabla^\beta R_{\mu\beta} = = 2\nabla^\alpha R_{\mu\alpha} -\nabla_\mu R = = 2\nabla^\alpha\left(R_{\mu\alpha} -\half g_{\mu\alpha} R \right) = = 2\nabla^\alpha G_{\mu\alpha} = 0

Lie derivative

Definition of the Lie derivative of any tensor T is:

\L_{\vec U} T=\lim_{t\to0}{\phi_{t*}T(\phi_t(p))-T(p)\over t}

it can be shown directly from this definition, that the Lie derivative of a vector is the same as a Lie bracket:

\L_{\vec U}\vec V \equiv [\vec U, \vec V]

and in components

\L_{\vec U} V^\alpha = [\vec U, \vec V]^\alpha\equiv U^\beta\nabla_\beta V^\alpha- V^\beta\nabla_\beta U^\alpha = U^\beta\partial_\beta V^\alpha- V^\beta\partial_\beta U^\alpha

Lie derivative of a scalar is

\L_{\vec V} f = V^\mu\partial_\mu f

and of a one form p_\mu is derived using the observation that f=p_\mu V^\mu is a scalar:

\L_{\vec V} p_\mu = V^\nu\nabla_\nu p_\mu+p_\nu\nabla_\mu V^\nu = V^\nu\partial_\nu p_\mu+p_\nu\partial_\mu V^\nu

and so on for other tensors, for example:

\L_{\vec V} g_{\mu\nu} = V^\alpha\nabla_\alpha g_{\mu\nu} +g_{\alpha\nu}\nabla_\mu V^\alpha +g_{\mu\alpha}\nabla_\nu V^\alpha = V^\alpha\partial_\alpha g_{\mu\nu} +g_{\alpha\nu}\partial_\mu V^\alpha +g_{\mu\alpha}\partial_\nu V^\alpha

Metric

In general, the Christoffel symbols are not symmetric and there is no metric that generates them. However, if the manifold is equipped with metrics, then the fundamental theorem of Riemannian geometry states that there is a unique Levi-Civita connection, for which the metric tensor is preserved by parallel transport:

\nabla_\mu g_{\alpha\beta}=0

We define the commutation coefficients of the basis c^\alpha{}_{\mu\nu} by

c^\alpha{}_{\mu\nu}\vec e_\alpha = \nabla_{\vec e_\mu}\vec e_\nu- \nabla_{\vec e_\nu}\vec e_\mu

In general these coefficients are not zero (as an example, take the units vectors in spherical or cylindrical coordinates), but for coordinate bases they are. It can be proven, that

\Gamma^\mu_{\alpha\beta}=\half g^{\mu\sigma} \left(\partial_\beta g_{\sigma\alpha}+\partial_\alpha g_{\sigma\beta}- \partial_\sigma g_{\alpha\beta}+c_{\alpha\sigma\beta}+c_{\beta\sigma\alpha} -c_{\sigma\alpha\beta}\right)

and for coordinate bases c^\alpha{}_{\mu\nu}=0, so

\Gamma^\mu_{\alpha\beta}=\Gamma^\mu_{\beta\alpha} \Gamma^\mu_{\alpha\beta}=\half g^{\mu\sigma} \left(\partial_\beta g_{\sigma\alpha}+\partial_\alpha g_{\sigma\beta}- \partial_\sigma g_{\alpha\beta}\right)

As a special case:

\Gamma^\mu_{\mu\beta}=\half g^{\mu\sigma} \left(\partial_\beta g_{\sigma\mu}+\partial_\mu g_{\sigma\beta}- \partial_\sigma g_{\mu\beta}\right)=\half g^{\mu\sigma}\partial_\beta g_{\sigma\mu}= =\half \Tr g^{-1}\partial_\beta g =\half \Tr\partial_\beta\log g =\half \partial_\beta\Tr\log g =\half \partial_\beta\log|\det g| =\partial_\beta\log\sqrt{|\det g|} = ={1\over2\det g}\partial_\beta\det g ={1\over\sqrt{|\det g|}}\partial_\beta\sqrt{|\det g|}

All last 3 expressions are used (but the last one is probably the most common). g is the matrix of coefficients g_{\mu\nu}. At the beginning we used the usual trick that g^{\mu\sigma} is symmetric but \partial_\mu g_{\sigma\beta}- \partial_\sigma g_{\mu\beta} is antisymmetric. Later we used the identity \Tr\log g = \log|\det g|, which follows from the well-known identity \det\exp A = \exp\Tr A by substituting A=\log g and taking the logarithm of both sides.

Diagonal Metric

Many times the metric is diagonal, e.g. in 3D:

g_{ij} = \mat{h_1^2 & 0 & 0\cr 0 & h_2^2 & 0\cr 0 & 0 & h_3^2\cr}

(in general g_{ij} = \diag(h_1^2, h_2^2, \dots)), then the Christoffel symbols \Gamma^k_{ij} can be calculated very easily (below we do not sum over i, j and k):

\Gamma^k_{ij} =\half g^{kl} \left(\partial_j g_{li}+\partial_i g_{lj}- \partial_l g_{ij}\right) =\half g^{kk} \left(\partial_j g_{ki}+\partial_i g_{kj}- \partial_k g_{ij}\right)

If k=i or k=j then

(3.40.1.3)\Gamma^k_{ij} = \Gamma^i_{ij} = \Gamma^i_{ji} =\half g^{ii} \left(\partial_j g_{ii}+\partial_i g_{ij}- \partial_i g_{ij}\right) =\half g^{ii} \partial_j g_{ii} =\half {1\over h_i^2} \partial_j h_i^2 ={1\over h_i} \partial_j h_i

otherwise (i.e. k\neq i and k\neq j) then either i=j:

(3.40.1.4)\Gamma^k_{ij} = \Gamma^k_{ii} =\half g^{kk} \left(\partial_i g_{ki}+\partial_i g_{ki}- \partial_k g_{ii}\right) =-\half g^{kk} \partial_k g_{ii} =-\half {1\over h_k^2} \partial_k h_i^2 =-{h_i\over h_k^2} \partial_k h_i

or i\neq j (i.e. i\neq j\neq k):

\Gamma^k_{ij} =\half g^{kk} \left(\partial_j g_{ki}+\partial_i g_{kj}- \partial_k g_{ij}\right) =0

In other words, the symbols can only be nonzero if at least two of i, j or k are the same and one can use the two formulas (3.40.1.3) and (3.40.1.4) to quickly evaluate them. A systematic way to do it is to write (3.40.1.3) and (3.40.1.4) in the following form:

(3.40.1.5)\Gamma^i_{ij} = \Gamma^i_{ji} ={1\over h_i} \partial_j h_i \quad\quad \mbox{$i$, $j$ arbitrary} \Gamma^j_{ii} =-{h_i\over h_j^2} \partial_j h_i \quad\quad \mbox{$i\neq j$}

Then find all i and j for which \partial_j h_i is nonzero and then immediately write all nonzero Christoffel symbols using the equations (3.40.1.5).

For example for cylindrical coordinates we have h_\rho=h_z =1 and h_\phi = \rho, so \partial_j h_i is only nonzero for i=\phi and j=\rho and we get:

\Gamma^\phi_{\phi\rho} = \Gamma^\phi_{\rho\phi} = {1\over h_\phi} \partial_\rho h_\phi = {1\over\rho}\partial_\rho \rho = {1\over\rho} \Gamma^\rho_{\phi\phi} = -{h_\phi\over h_\rho^2} \partial_\rho h_\phi = -{\rho\over 1^2}\partial_\rho \rho = -\rho

all other Christoffel symbols are zero. For spherical coordinates we have h_\rho=1, h_\theta=\rho and h_\phi=\rho\sin\theta, so \partial_j h_i is only nonzero for i=\theta, j=\rho or i=\phi, j=\rho or i=\phi, j=\theta and we get:

\Gamma^\theta_{\theta\rho} = \Gamma^\theta_{\rho\theta} = {1\over h_\theta}\partial_\rho h_\theta = {1\over \rho}\partial_\rho \rho = {1\over\rho} \Gamma^\rho_{\theta\theta} = -{h_\theta\over h_\rho^2}\partial_\rho h_\theta = -{\rho \over 1^2}\partial_\rho \rho = -\rho \Gamma^\phi_{\phi\rho} = \Gamma^\phi_{\rho\phi} = {1\over h_\phi}\partial_\rho h_\phi = {1\over \rho\sin\theta}\partial_\rho (\rho\sin\theta) = {1\over\rho} \Gamma^\rho_{\phi\phi} = -{h_\phi\over h_\rho^2}\partial_\rho h_\phi = -{\rho\sin\theta \over 1^2}\partial_\rho (\rho\sin\theta) = -\rho\sin^2\theta \Gamma^\phi_{\phi\theta} = \Gamma^\phi_{\theta\phi} = {1\over h_\phi}\partial_\theta h_\phi = {1\over \rho\sin\theta}\partial_\theta (\rho\sin\theta) = {\cos\theta\over\sin\theta} \Gamma^\theta_{\phi\phi} = -{h_\phi\over h_\theta^2}\partial_\theta h_\phi = -{\rho\sin\theta \over \rho^2}\partial_\theta (\rho\sin\theta) = -\sin\theta\cos\theta

All other symbols are zero.

Symmetries, Killing vectors

We say that a diffeomorphism \phi is a symmetry of some tensor T if the tensor is invariant after being pulled back under \phi:

\phi_*T = T

Let the one-parameter family of symmetries \phi_t be generated by a vector field V^\mu(x), then the above equation is equivalent to:

\L_{\vec V} T = 0

If T is the metric g_{\mu\nu} then the symmetry is called isometry and V^\mu is called a Killing vector field and can be calculated from:

\L_{\vec V} g_{\mu\nu} = V^\alpha\nabla_\alpha g_{\mu\nu} +g_{\alpha\nu}\nabla_\mu V^\alpha +g_{\mu\alpha}\nabla_\nu V^\alpha = \nabla_\mu V_\nu +\nabla_\nu V_\mu = 0

The last equality is Killing’s equation. If x^\mu is a geodesics with a tangent vector U^\mu and V^\mu is a Killing vector, then the quantity V_\mu U^\mu is conserved along the geodesics, because:

{\d (V_\mu U^\mu)\over\d\lambda} = U^\nu\nabla_\nu(V_\mu U^\mu)=U^\nu U^\mu\nabla_\nu V_\mu +V_\mu U^\nu\nabla_\nu U^\mu = 0

where the first term is both symmetric and antisymmetric in (\mu, \nu), thus zero, and the second term is the geodesics equation, thus also zero.

Symmetry and Antisymmetry

Every tensor can be decomposed into symmetric and antisymmetric parts:

T_{\alpha\beta} = \half(T_{\alpha\beta} + T_{\beta\alpha}) + \half(T_{\alpha\beta} - T_{\beta\alpha})

In particular, for a symmetric tensor S_{\alpha\beta}=S_{\beta\alpha} we get:

S_{\alpha\beta} = \half(S_{\alpha\beta} + S_{\beta\alpha})

and for antisymmetric tensor A_{\alpha\beta}=-A_{\beta\alpha} we get:

A_{\alpha\beta} = \half(A_{\alpha\beta} - A_{\beta\alpha})

When contracting a symmetric tensor with an antisymmetric tensor we get zero:

S_{\alpha\beta} A^{\alpha\beta} = = \half S_{\alpha\beta} (A^{\alpha\beta} - A^{\beta\alpha}) = = \half (S_{\alpha\beta} A^{\alpha\beta} - S_{\alpha\beta} A^{\beta\alpha}) = = \half (S_{\alpha\beta} A^{\alpha\beta} - S_{\beta\alpha} A^{\beta\alpha}) = = \half (S_{\alpha\beta} A^{\alpha\beta} - S_{\alpha\beta} A^{\alpha\beta}) = = 0

When contracting a general tensor T with a symmetric tensor S, only the symmetric part of T contributes:

T_{\alpha\beta} S^{\alpha\beta} = = \half(T_{\alpha\beta} + T_{\beta\alpha})S^{\alpha\beta} + \half(T_{\alpha\beta} - T_{\beta\alpha})S^{\alpha\beta} = = \half(T_{\alpha\beta} + T_{\beta\alpha})S^{\alpha\beta}

When contracting a general tensor T with an antisymmetric tensor A, only the antisymmetric part of T contributes:

T_{\alpha\beta} A^{\alpha\beta} = = \half(T_{\alpha\beta} + T_{\beta\alpha})A^{\alpha\beta} + \half(T_{\alpha\beta} - T_{\beta\alpha})A^{\alpha\beta} = = \half(T_{\alpha\beta} - T_{\beta\alpha})A^{\alpha\beta}

Example I

We want to rewrite:

2\partial_\nu g_{\mu\lambda} {\d x^\mu\over\d s} {\d x^\nu\over\d s}

So we write the left part as a sum of symmetric and antisymmetric parts:

(\partial_\nu g_{\mu\lambda} + \partial_\mu g_{\nu\lambda} +\partial_\nu g_{\mu\lambda} - \partial_\mu g_{\nu\lambda}) {\d x^\mu\over\d s} {\d x^\nu\over\d s}

Here \partial_\nu g_{\mu\lambda} - \partial_\mu g_{\nu\lambda} is antisymmetric and {\d x^\mu\over\d s} {\d x^\nu\over\d s} is symmetric in \mu,\nu, so the contraction is zero. The final result is:

(\partial_\nu g_{\mu\lambda} + \partial_\mu g_{\nu\lambda}) {\d x^\mu\over\d s} {\d x^\nu\over\d s}

Example II

Let F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. Then we can simplify:

F^{\mu\nu} F_{\mu\nu} = F^{\mu\nu} (\partial_\mu A_\nu - \partial_\nu A_\mu) = =2 F^{\mu\nu} \half (\partial_\mu A_\nu - \partial_\nu A_\mu) =2 F^{\mu\nu} \partial_\mu A_\nu

Here \half (\partial_\mu A_\nu - \partial_\nu A_\mu) is the antisymmetric part (the only one that contributes, because F^{\mu\nu} is antisymmetric) of \partial_\mu A_\nu.

Example III

Let F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. Then we get

F_{\mu\nu} v^\mu v^\nu = 0\,,

because F_{\mu\nu} is an antisymmetric tensor, while v^\mu v^\nu is a symmetric tensor.

Divergence Operator

\nabla_\mu A^\mu =\partial_\mu A^\mu+\Gamma^\mu_{\mu\sigma}A^\sigma = =\partial_\mu A^\mu+{1\over\sqrt{|\det g|}}\left(\partial_\sigma\sqrt{|\det g|}\right) A^\sigma = ={1\over\sqrt{|\det g|}} \partial_\mu\left(\sqrt{|\det g|}\, A^\mu\right)

If the metric is diagonal (let’s show this in 3D):

g_{ij} = \mat{h_1^2 & 0 & 0\cr 0 & h_2^2 & 0\cr 0 & 0 & h_3^2\cr}

then

\sqrt{|\det g_{ij}|}=h_1 h_2 h_3 g^{ij} = \mat{{1\over h_1^2} & 0 & 0\cr 0 & {1\over h_2^2} & 0\cr 0 & 0 & {1\over h_3^2}\cr}

and

\nabla\cdot{\bf A}= \nabla_i A^i ={1\over h_1 h_2 h_3}\partial_i \left(h_1 h_2 h_3 A^i \right)

Laplace Operator

\nabla^2\varphi =\nabla_\mu\nabla^\mu\varphi =\partial_\mu\nabla^\mu\varphi+\Gamma^\mu_{\mu\sigma}\nabla^\sigma\varphi =\partial_\mu\partial^\mu\varphi+\Gamma^\mu_{\mu\sigma}\partial^\sigma\varphi = =\partial_\mu\partial^\mu\varphi+{1\over\sqrt{|\det g|}}\left(\partial_\sigma\sqrt{|\det g|}\right) \partial^\sigma\varphi = ={1\over\sqrt{|\det g|}} \partial_\mu\left(\sqrt{|\det g|}\, \partial^\mu\varphi\right) ={1\over\sqrt{|\det g|}} \partial_\mu\left(\sqrt{|\det g|}\, g^{\mu\sigma}\partial_\sigma\varphi\right)

If the metric is diagonal (let’s show this in 3D):

g_{ij} = \mat{h_1^2 & 0 & 0\cr 0 & h_2^2 & 0\cr 0 & 0 & h_3^2\cr}

then

\sqrt{|\det g_{ij}|}=h_1 h_2 h_3 g^{ij} = \mat{{1\over h_1^2} & 0 & 0\cr 0 & {1\over h_2^2} & 0\cr 0 & 0 & {1\over h_3^2}\cr}

and

\nabla^2\varphi =\sum_i{1\over h_1 h_2 h_3}\partial_i \left({h_1 h_2 h_3\over h_i^2}\partial_i\varphi \right)

Covariant integration

If f(x) is a scalar, then the integral \int f(x) \d^4 x depends on coordinates. The correct way to integrate f(x) in any coordinates is:

\int f(x) \sqrt{|g|}\d^4 x

where g\equiv\det g_{\mu\nu}. The Gauss theorem in curvilinear coordinates is:

\int_\Omega \nabla_\mu u^\mu \sqrt{|g|}\d^4 x =\int_\Omega {1\over\sqrt{|g|}}\partial_\mu\left(\sqrt{|g|} u^\mu\right) \sqrt{|g|}\d^4 x =\int_\Omega \partial_\mu\left(\sqrt{|g|} u^\mu\right)\d^4 x = =\int_{\partial\Omega} \sqrt{|g|} u^\mu n_\mu\d^3 x =\int_{\partial\Omega} u^\mu n_\mu \sqrt{|g|}\d^3 x

where \partial\Omega is the boundary (surface) of \Omega and n_\nu is the normal vector to this surface.

3.40.2. Examples

Weak Formulation of Laplace Equation

As an example, we write the weak formulation of the Laplace equation in arbitrary coordintes:

\nabla^2\varphi - f = 0 \int \left(\nabla^2\varphi v - f v\right) \sqrt{|g|}d^3 x = 0 \int \left( {1\over\sqrt{|g|}}\partial_i\left(\sqrt{|g|}g^{ij}\partial_j\varphi\right) v - f v\right) \sqrt{|g|}d^3 x = 0 \int \left( \partial_i\left(\sqrt{|g|}g^{ij}\partial_j\varphi\right) v - f v\sqrt{|g|}\right) d^3 x = 0

Now we apply per-partes (assuming the boundary integral vanishes):

\int \left( -\sqrt{|g|}g^{ij}\partial_j\varphi \partial_i v - f v\sqrt{|g|}\right) d^3 x = 0 \int \left( -g^{ij}\partial_j\varphi \partial_i v - f v\right) \sqrt{|g|} d^3 x = 0

For diagonal metric this evaluates to:

\int \left( -\sum_i {1\over h_i^2}\partial_i\varphi \partial_i v - f v\right) h_1 h_2 h_3 d^3 x = 0

Cylindrical Coordinates

x = \rho\cos\phi

y = \rho\sin\phi

z = z

The transformation matrix is

{\partial (x, y, z)\over\partial(\rho, \phi, z)} =\mat{\cos\phi & -\rho\sin\phi & 0 \cr \sin\phi & \rho\cos\phi & 0 \cr 0 & 0 & 1 \cr}

The metric tensor of the cartesian coordinate system \hat x^a=(x, y, z) is \hat g_{ab}={\rm diag}(1, 1, 1), so by transformation we get the metric tensor g_{ij} in the cylindrical coordinates x^i=(\rho, \phi, z):

g_{ij} = {\partial \hat x^a\over\partial x^i} {\partial \hat x^b\over\partial x^j} \hat g_{ab} = \left({\partial \hat x\over\partial x}\right)^T \hat g {\partial \hat x\over\partial x} =

= \left({\partial (x, y, z)\over\partial(\rho, \phi, z)}\right)^T \mat{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr} {\partial (x, y, z)\over\partial(\rho, \phi, z)}=

= \mat{\cos\phi &\sin\phi & 0 \cr -\rho\sin\phi & \rho\cos\phi & 0 \cr 0 & 0 & 1 \cr} \mat{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr} \mat{\cos\phi & -\rho\sin\phi & 0 \cr \sin\phi & \rho\cos\phi & 0 \cr 0 & 0 & 1 \cr}= \mat{1 & 0 & 0\cr 0 & \rho^2 & 0\cr 0 & 0 & 1\cr}

g^{ij} = \mat{1 & 0 & 0\cr 0 & 1\over\rho^2 & 0\cr 0 & 0 & 1\cr} \det g = \det g_{ij} = \rho^2 h_\rho=h_z =1 h_\phi = \rho \Gamma^\phi_{\phi\rho} = \Gamma^\phi_{\rho\phi} = {1\over h_\phi} \partial_\rho h_\phi = {1\over\rho}\partial_\rho \rho = {1\over\rho} \Gamma^\rho_{\phi\phi} = -{h_\phi\over h_\rho^2} \partial_\rho h_\phi = -{\rho\over 1^2}\partial_\rho \rho = -\rho \nabla\cdot{\bf A}= \nabla_i A^i ={1\over h_1 h_2 h_3}\partial_i \left(h_1 h_2 h_3 A^i \right) ={1\over \rho}\partial_i \left(\rho A^i \right) = ={1\over\rho}\partial_\rho(\rho A^\rho) + \partial_\theta A^\theta + \partial_z A^z =\partial_\rho A^\rho + {1\over\rho}A^\rho + \partial_\theta A^\theta + \partial_z A^z

\nabla^2 \varphi = \nabla^i\nabla_i\varphi ={1\over\sqrt{|\det g|}} \partial_i\left(\sqrt{|\det g|}\, g^{ij}\partial_j\varphi\right) =

={1\over\rho}\partial_i\left(\rho g^{ij}\partial_j\varphi\right) ={1\over\rho}\partial_\rho\left(\rho \partial_\rho\varphi\right) +{1\over\rho}\partial_\phi\left(\rho {1\over\rho^2}\partial_\phi\varphi\right) +{1\over\rho}\partial_z\left(\rho \partial_z\varphi\right)= ={1\over\rho}\partial_\rho\left(\rho \partial_\rho\varphi\right) +{1\over\rho^2}\partial_\phi\partial_\phi\varphi +\partial_z\partial_z\varphi=

= \partial_\rho\partial_\rho\varphi +{1\over\rho}\partial_\rho\varphi +{1\over\rho^2}\partial_\phi\partial_\phi\varphi +\partial_z\partial_z\varphi

As a particular example, let’s write the Laplace equation with nonconstant conductivity for axially symmetric field. The Laplace equation is:

\nabla\cdot\sigma\nabla\varphi=0

so we use the formulas above to get:

0=\nabla\cdot\sigma\nabla\varphi=\nabla^i\sigma\nabla_i\varphi = {\partial\over\partial\rho}\sigma{\partial\varphi\over\partial\rho} + {1\over\rho^2} {\partial\over\partial\phi}\sigma{\partial\varphi\over\partial\phi} + {\partial\over\partial z}\sigma{\partial\varphi\over\partial z} + {\sigma\over\rho}{\partial\varphi\over\partial\rho}

but we know that \varphi=\varphi(\rho, z), so {\partial\varphi\over\partial\phi}=0 and the final equation is:

{\partial\over\partial\rho}\sigma{\partial\varphi\over\partial\rho} + {\partial\over\partial z}\sigma{\partial\varphi\over\partial z} + {\sigma\over\rho}{\partial\varphi\over\partial\rho} =0

To write the weak formulation for it, we need to integrate covariantly (e.g. \rho\,\d\rho\d\phi\d z in our case) and rewrite it using per partes. We did exactly this in the previous example in a coordinate free maner, so we just use the final formula we got there for a diagonal metric:

\int\left( -\partial_\rho\varphi\partial_\rho v -{1\over\rho^2}\partial_\phi\varphi\partial_\phi v -\partial_z\varphi\partial_z v \right)\sigma\rho\,\d\rho\d\phi\d z=0

and for \partial_\phi\varphi=0, we get:

-2\pi\int\left( \partial_\rho\varphi\partial_\rho v +\partial_z\varphi\partial_z v \right)\sigma\rho\,\d\rho\d z=0

Spherical Coordinates

The relation between cartesian coordinates \hat x^a=(x, y, z) and spherical coordinates x^i=(\rho, \theta, \phi) is:

(3.40.2.1)x &= \rho\sin\theta\cos\phi \\ y &= \rho\sin\theta\sin\phi \\ z &= \rho\cos\theta \\

The transformation matrix (Jacobian) is calculated by differentiating (3.40.2.1):

(3.40.2.2){\partial \hat x\over\partial x} = {\partial (x, y, z)\over\partial(\rho, \theta, \phi)} = \mat{ \sin\theta\cos\phi & \rho\cos\theta\cos\phi & -\rho\sin\theta\sin\phi \cr \sin\theta\sin\phi & \rho\cos\theta\sin\phi & \rho\sin\theta\cos\phi \cr \cos\theta & -\rho\sin\theta & 0 \cr}

The inverse Jacobian is calculated by inverting the matrix (3.40.2.2):

{\partial x\over\partial \hat x} = {\partial(\rho, \theta, \phi)\over\partial(x, y, z)} = \mat{ \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \cr {\cos\theta\cos\phi\over\rho} & {\cos\theta\sin\phi\over\rho} & -{\sin\theta\over\rho} \cr -{\sin\phi\over\rho\sin\theta} & {\cos\phi\over\rho\sin\theta} & 0 \cr }

We expressed the above Jacobians using \rho, \theta, \phi and we can use (3.40.2.1) to express them using x, y, z. Code:

from sympy import var, sin, cos, zeros, Matrix, simplify, latex
var("rho theta phi")
x_hat = Matrix([
    rho * sin(theta) * cos(phi),
    rho * sin(theta) * sin(phi),
    rho * cos(theta)])
x = Matrix([rho, theta, phi])

M = zeros(3, 3)
for i in range(3):
    for j in range(3):
        M[i, j] = x_hat[i].diff(x[j])

N = M.inv(method="ADJ")
one = sin(phi)**2*sin(theta)**2 - cos(phi)**2*cos(theta)**2 + \
        cos(phi)**2 + cos(theta)**2
one_simple = one.subs(sin(phi)**2, 1-cos(phi)**2).expand().simplify()
N.simplify()
# one_simple is equal to 1, but simplify() can't do this automatically yet:
N = N.subs(one, one_simple)

print "J =", latex(M)
print
print "J^{-1} =", latex(N)

Output:

J = \left[\begin{smallmatrix}\sin{\left (\theta \right )} \cos{\left (\phi \right )} & \rho \cos{\left (\phi \right )} \cos{\left (\theta \right )} & - \rho \sin{\left (\phi \right )} \sin{\left (\theta \right )}\\\sin{\left (\phi \right )} \sin{\left (\theta \right )} & \rho \sin{\left (\phi \right )} \cos{\left (\theta \right )} & \rho \sin{\left (\theta \right )} \cos{\left (\phi \right )}\\\cos{\left (\theta \right )} & - \rho \sin{\left (\theta \right )} & 0\end{smallmatrix}\right] J^{-1} = \left[\begin{smallmatrix}\sin{\left (\theta \right )} \cos{\left (\phi \right )} & \sin{\left (\phi \right )} \sin{\left (\theta \right )} & \cos{\left (\theta \right )}\\\frac{\cos{\left (\phi \right )} \cos{\left (\theta \right )}}{\rho} & \frac{\sin{\left (\phi \right )} \cos{\left (\theta \right )}}{\rho} & - \frac{\sin{\left (\theta \right )}}{\rho}\\- \frac{\sin{\left (\phi \right )}}{\rho \sin{\left (\theta \right )}} & \frac{\cos{\left (\phi \right )}}{\rho \sin{\left (\theta \right )}} & 0\end{smallmatrix}\right]

The transformation matrices (Jacobians) are then used to convert vectors

V_i = {\partial \hat x^a\over\partial x^i} \hat V_a

and tensors

T_{ij} = {\partial \hat x^a\over\partial x^i} {\partial \hat x^b\over\partial x^j} \hat T_{ab}

between spherical and cartesian coordinates. For example the partial derivatives from cartesian to spherical coordinates transform as:

\partial_i = {\partial \hat x^a\over\partial x^i} \hat \partial_a \mat{ \partial_\rho & \partial_\theta & \partial_\phi \cr } = \mat{ \partial_x & \partial_y & \partial_z \cr } \mat{ \sin\theta\cos\phi & \rho\cos\theta\cos\phi & -\rho\sin\theta\sin\phi \cr \sin\theta\sin\phi & \rho\cos\theta\sin\phi & \rho\sin\theta\cos\phi \cr \cos\theta & -\rho\sin\theta & 0 \cr}

and from spherical to cartesian as:

\hat \partial_a = {\partial x^i\over\partial \hat x^a} \partial_i \mat{ \partial_x & \partial_y & \partial_z \cr } = \mat{ \partial_\rho & \partial_\theta & \partial_\phi \cr } \mat{ \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \cr {\cos\theta\cos\phi\over\rho} & {\cos\theta\sin\phi\over\rho} & -{\sin\theta\over\rho} \cr -{\sin\phi\over\rho\sin\theta} & {\cos\phi\over\rho\sin\theta} & 0 \cr }

Care must be taken when rewriting the index expression into matrices – the top index of the Jacobian is the row index, the bottom index is the column index.

The metric tensor of the cartesian coordinate system \hat x^a is \hat g_{ab}={\rm diag}(1, 1, 1), so by transformation we get the metric tensor g_{ij} in the spherical coordinates x^i:

g_{ij} = {\partial \hat x^a\over\partial x^i} {\partial \hat x^b\over\partial x^j} \hat g_{ab} = \left({\partial \hat x\over\partial x}\right)^T \hat g {\partial \hat x\over\partial x} = = \left({\partial (x, y, z)\over\partial(\rho, \theta, \phi)}\right)^T \mat{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr} {\partial (x, y, z)\over\partial(\rho, \theta, \phi)}= = \mat{ \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \cr \rho\cos\theta\cos\phi & \rho\cos\theta\sin\phi & -\rho\sin\theta \cr -\rho\sin\theta\sin\phi & \rho\sin\theta\cos\phi & 0 \cr } \mat{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr} \mat{\sin\theta\cos\phi & \rho\cos\theta\cos\phi & -\rho\sin\theta\sin\phi \cr \sin\theta\sin\phi & \rho\cos\theta\sin\phi & \rho\sin\theta\cos\phi \cr \cos\theta & -\rho\sin\theta & 0 \cr} = = \mat{1 & 0 & 0\cr 0 & \rho^2 & 0\cr 0 & 0 & \rho^2\sin^2\theta\cr}

Once we have the metric tensor expressed in spherical coordinates, we don’t need the cartesian coordinates anymore. All formulas only contain the spherical coordinates and the metric tensor.

g^{ij} = \mat{1 & 0 & 0\cr 0 & 1\over\rho^2 & 0\cr 0 & 0 & 1\over\rho^2\sin^2\theta\cr}

\det g = \det g_{ij} = \rho^4\sin^2\theta

\nabla^i\nabla_i\varphi = \partial^i\partial_i\varphi+{1\over{2\det g}}\partial_j(\det g)\,\, g^{jk}\partial_k\varphi=

= g^{ij}\partial_i\partial_j\varphi+{1\over{2\rho^4\sin^2\theta }}\left( \partial_\rho(\rho^4\sin^2\theta)\,\,g^{\rho \rho}\partial_\rho\varphi + \partial_\theta(\rho^4\sin^2\theta)\,\,g^{\theta \theta}\partial_\theta\varphi \right)

= g^{ij}\partial_i\partial_j\varphi+{2\over\rho}\partial_\rho\varphi +{\cos\theta\over\rho^2\sin\theta}\partial_\theta\varphi =

= \partial_\rho\partial_\rho\varphi +{1\over\rho^2}\partial_\theta\partial_\theta\varphi +{1\over\rho^2\sin^2\theta}\partial_\phi\partial_\phi\varphi +{2\over\rho}\partial_\rho\varphi +{\cos\theta\over\rho^2\sin\theta}\partial_\theta\varphi

Rotating Disk

Let’s have a laboratory Euclidean system x^\mu = (t, x, y, z) and a rotating disk system x'^\mu = (t', x', y', z'). The relation between the frames is

\mat{t'\cr x'\cr y'\cr z'\cr}= \mat{1 & 0 & 0 & 0\cr 0 & \cos\omega t & \sin\omega t & 0\cr 0 & -\sin\omega t & \cos\omega t & 0\cr 0 & 0 & 0 & 1\cr} \mat{t\cr x\cr y\cr z\cr} = \mat{t\cr x\cos\omega t+y\sin\omega t\cr -x\sin\omega t+y\cos\omega t\cr z\cr}

The inverse transformation can be calculated by simply inverting the matrix:

\mat{t\cr x\cr y\cr z\cr}= \mat{1 & 0 & 0 & 0\cr 0 & \cos\omega t' & -\sin\omega t' & 0\cr 0 & \sin\omega t' & \cos\omega t' & 0\cr 0 & 0 & 0 & 1\cr} \mat{t'\cr x'\cr y'\cr z'\cr}

so the transformation matrices are:

{\partial x'^\mu\over\partial x^\nu}= \mat{1 & 0 & 0 & 0\cr -x\omega\sin\omega t + y\omega\cos\omega t & \cos\omega t & \sin\omega t & 0\cr -x\omega\cos\omega t - y\omega\sin\omega t & -\sin\omega t & \cos\omega t & 0\cr 0 & 0 & 0 & 1\cr} ={\partial x'\over\partial x}

{\partial x^\nu\over\partial x'^\mu}= \mat{1 & 0 & 0 & 0\cr -x'\omega\sin\omega t' - y'\omega\cos\omega t' & \cos\omega t' & -\sin\omega t' & 0\cr x'\omega\cos\omega t' - y'\omega\sin\omega t' & \sin\omega t' & \cos\omega t' & 0\cr 0 & 0 & 0 & 1\cr} ={\partial x\over\partial x'}

The problem now is that Newtonian mechanics has a degenerated spacetime metrics (see later). Let’s pretend we have the following metrics in the x^\mu system:

g_{\mu\nu} = \mat{1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} =g

and

g'_{\alpha\beta} = {\partial x^\mu\over\partial x'^\alpha} {\partial x^\nu\over\partial x'^\beta} g_{\mu\nu} = \left({\partial x\over\partial x'}\right)^T g \left({\partial x\over\partial x'}\right) = \mat{1 + \omega^2 (x'^2+y'^2) & -\omega y' & \omega x' & 0\cr -\omega y' & 1 & 0 & 0\cr \omega x' & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} =g'

However, if we calculate with the correct special relativity metrics:

g_{\mu\nu} = \mat{-c^2 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} =g

and

g'_{\alpha\beta} = {\partial x^\mu\over\partial x'^\alpha} {\partial x^\nu\over\partial x'^\beta} g_{\mu\nu} = \left({\partial x\over\partial x'}\right)^T g \left({\partial x\over\partial x'}\right) = \mat{-c^2 + \omega^2 (x'^2+y'^2) & -\omega y' & \omega x' & 0\cr -\omega y' & 1 & 0 & 0\cr \omega x' & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} =g'

We get the same Christoffel symbols as with the \diag(1, 1, 1, 1) metrics, because only the derivatives of the metrics are important. Then the only nonzero Christoffel symbols are

\Gamma^1_{00}=-x'\omega^2

\Gamma^1_{02}=\Gamma^1_{20}=-\omega

\Gamma^2_{00}=-y'\omega^2

\Gamma^2_{01}=\Gamma^2_{10}=\omega

If we want to avoid dealing with metrics, it is possible to start with the Christoffel symbols in the x^\mu system:

\Gamma^\sigma_{\mu\nu}=0

and then transforming them to the x'^\mu system using the change of variable formula:

\Gamma'^\alpha{}_{\beta\gamma}= {\partial x^\mu\over\partial x'^\beta} {\partial x^\nu\over\partial x'^\gamma} \Gamma^\sigma{}_{\mu\nu} {\partial x'^\alpha\over\partial x^\sigma} + {\partial x'^\alpha\over\partial x^\sigma} {\partial^2 x^\sigma\over\partial x'^\beta\partial x'^\gamma} = {\partial x'^\alpha\over\partial x^\sigma} {\partial^2 x^\sigma\over\partial x'^\beta\partial x'^\gamma}

As an example, let’s calculate the coefficients above:

\Gamma'^2{}_{00}= {\partial x'^2\over\partial x^\sigma} {\partial^2 x^\sigma\over\partial x'^0\partial x'^0} = {\partial x'^2\over\partial x^\sigma} {\partial \over\partial x'^0} {\partial x^\sigma\over\partial x'^0} =

= \mat{-x\omega\cos\omega t - y\omega\sin\omega t & -\sin\omega t & \cos\omega t & 0\cr} {\partial \over\partial t'} \mat{1\cr -x'\omega\sin\omega t'-y'\omega\cos\omega t'\cr x'\omega\cos\omega t'-y'\omega\sin\omega t'\cr 0\cr} =

= \mat{-x\omega\cos\omega t - y\omega\sin\omega t & -\sin\omega t & \cos\omega t & 0\cr} \mat{0\cr -x'\omega^2\cos\omega t'+y'\omega^2\sin\omega t'\cr -x'\omega^2\sin\omega t'-y'\omega^2\cos\omega t'\cr 0\cr} =-y'\omega^2

\Gamma'^1{}_{00}=-x'\omega^2

\Gamma'^2{}_{01}=\Gamma'^2{}_{10}= {\partial x'^2\over\partial x^\sigma} {\partial^2 x^\sigma\over\partial x'^0\partial x'^1} = {\partial x'^2\over\partial x^\sigma} {\partial \over\partial x'^0} {\partial x^\sigma\over\partial x'^1} =

= \mat{-x\omega\cos\omega t - y\omega\sin\omega t & -\sin\omega t & \cos\omega t & 0\cr} {\partial \over\partial t'} \mat{0\cr \cos\omega t'\cr \sin\omega t'\cr 0\cr} =

= \mat{-x\omega\cos\omega t - y\omega\sin\omega t & -\sin\omega t & \cos\omega t & 0\cr} \mat{0\cr -\omega\sin\omega t'\cr \omega\cos\omega t'\cr 0\cr} =\omega

\Gamma'^1{}_{02}=\Gamma'^1{}_{20}=-\omega

So we got the same results.

Now let’s see what we have got. Later we’ll show, that the \Gamma^i_{00} coefficients are just \partial_i\phi in the Newtonian theory. E.g. in our case we have:

\Gamma'^1_{00} = -x'\omega^2 = \partial_x'\phi

\Gamma'^2_{00} = -y'\omega^2 = \partial_y'\phi

\Gamma'^3_{00} = 0 = \partial_z'\phi

from which:

\phi(t, x, y, z) = -\half(x'^2+y'^2)\omega^2 + C(t)

and the force acting on a test particle is then:

{\bf F} = -m\nabla\phi=m\,(x', y', 0)\,\omega^2=m{\bf r'}\omega^2

where we have defined {\bf r'} = (x', y', 0). This is just the centrifugal force. Also observe, that we could have read \phi directly from the metrics itself — just compare it to the Lorentzian metrics (with gravitation) in the next chapter.

The other two terms (\Gamma'^1_{02}, \Gamma'^2_{01} and the symmetric ones) don’t behave as a gravitational force, but rather only act when we are differentiating (e.g. only act on moving bodies). Below we show this is just the -2\boldsymbol\omega\times{\d{\bf r}\over\d t} term (responsible for the Coriolis acceleration).

Let’s write the full equations of geodesics:

{\d^2 x^0\over\d\lambda^2}=0

{\d^2 x^1\over\d\lambda^2}+\Gamma^1_{00}\left({\d x^0\over\d\lambda}\right)^2 +2\Gamma^1_{20}{\d x^2\over\d\lambda}{\d x^0\over\d\lambda}=0

{\d^2 x^2\over\d\lambda^2}+\Gamma^2_{00}\left({\d x^0\over\d\lambda}\right)^2 +2\Gamma^2_{10}{\d x^1\over\d\lambda}{\d x^0\over\d\lambda}=0

{\d^2 x^3\over\d\lambda^2}=0

This becomes:

{\d^2 x\over\d t^2}=x\omega^2 +2\omega{\d y\over\d t}

{\d^2 y\over\d t^2}=y\omega^2 -2\omega{\d x\over\d t}

{\d^2 z\over\d t^2}=0

we can define {\bf r} = (x, y, 0) and \boldsymbol\omega=(0,0,\omega). Then the above equations can be rewritten as:

{\d^2{\bf r}\over\d t^2}={\bf r}\omega^2-2\boldsymbol\omega\times{\d{\bf r}\over\d t}

So we get two fictituous forces, the centrifugal force and the Coriolis force.

Now imagine a static vector in the x^\mu system along the x axis, i.e.

V^\mu = \mat{1\cr 1\cr 0\cr 0\cr} = V

then

V'^\mu = {\partial x'^\mu\over\partial x^\alpha}V^\alpha= {\partial x'\over\partial x}V= \mat{1\cr -x\omega\sin\omega t + y\omega\cos\omega t + \cos\omega t \cr -x\omega\cos\omega t - y\omega\sin\omega t - \sin\omega t\cr 0\cr} = \mat{1\cr y'\omega + \cos\omega t' \cr -x'\omega - \sin\omega t'\cr 0\cr}=V'

In the last equality we transformed from x^\mu to x'^\mu using the relation between frames.

Differentiating any vector in the x^\mu coordinates is easy – it’s just a partial derivative (due to the Euclidean metrics). Let’s differentiate any vector in the x'^\mu coordinates with respect to time (since t=t', the time is the same in both coordinate systems):

\nabla_0V'^{\mu}=\partial_0V'^{\mu}+\Gamma^\mu_{0\alpha}V'^{\alpha}

\nabla_0 \mat{V'^0\cr V'^1 \cr V'^2\cr V'^3\cr} = \mat{\partial_0V'^0\cr \partial_0V'^1 +\Gamma^1_{00}V'^0+\Gamma^1_{02}V'^2\cr \partial_0V'^2 +\Gamma^2_{00}V'^0+\Gamma^2_{01}V'^1\cr \partial_0V'^3\cr} = \mat{\partial_0V'^0\cr \partial_0V'^1 -x'\omega^2V'^0-\omega V'^2\cr \partial_0V'^2 -y'\omega^2V'^0+\omega V'^1\cr \partial_0V'^3\cr}=

(3.40.2.3)= \partial_0 \mat{V'^0\cr V'^1 \cr V'^2\cr V'^3\cr} + \mat{ 0 & 0 & 0 & 0\cr -x'\omega^2 & 0 & -\omega & 0\cr -y'\omega^2 & \omega & 0 & 0\cr 0 & 0 & 0 & 0\cr } \mat{V'^0\cr V'^1 \cr V'^2\cr V'^3\cr}

For our particular (static) vector this yields:

\nabla_0 \mat{1\cr y'\omega + \cos\omega t' \cr -x'\omega - \sin\omega t'\cr 0\cr} = \mat{0\cr 0\cr 0\cr 0\cr}

as expected, because it was at rest in the x^\mu system. Let’s imagine a static vector in the x'^\mu system along the x' axis, i.e.

W'^\mu = \mat{1\cr 1\cr 0\cr 0\cr}

W^\mu = {\partial x^\mu\over\partial x'^\alpha}W'^\alpha= \mat{1\cr -x'\omega\sin\omega t'-y'\omega\cos\omega t'+\cos\omega t'\cr x'\omega\cos\omega t'-y'\omega\sin\omega t'+\sin\omega t'\cr 0\cr} = \mat{1\cr -y\omega+\cos\omega t\cr x\omega+\sin\omega t\cr 0\cr}

then

\nabla_0W'^\mu= \nabla_0 \mat{1\cr 1\cr 0\cr 0\cr} = \mat{0\cr -x'\omega^2\cr -y'\omega^2+\omega\cr 0\cr}

\nabla_0W^\mu= \partial_0 \mat{1\cr -y\omega+\cos\omega t\cr x\omega+\sin\omega t\cr 0\cr} = \mat{0\cr -\omega\sin\omega t\cr \omega\cos\omega t\cr 0\cr} = \mat{ 0 & 0 & 0 & 0\cr 0 & 0 & -\omega & 0\cr 0 & \omega & 0 & 0\cr 0 & 0 & 0 & 0\cr } \mat{0\cr \cos\omega t\cr \sin\omega t\cr 0\cr} = \boldsymbol\omega\times{\bf W}

Similarly

\nabla_0\nabla_0W'^\mu= = \mat{0\cr -y'\omega^3-\omega^2\cr -x'\omega^3\cr 0\cr}

\nabla_0\nabla_0W^\mu= = \mat{0\cr -\omega^2\cos\omega t\cr -\omega^2\sin\omega t\cr 0\cr}

How can one prove the relation:

(3.40.2.4){\d{\bf A}\over\d t} = \boldsymbol\omega \times {\bf A} + {\d'{\bf A}\over\d t}

that is used for example to derive the Coriolis acceleration etc.? We need to write it components to understand what it really means:

\nabla_0 \mat{A'^0\cr A'^1\cr A'^2\cr A'^3\cr} = \mat{ 0 & 0 & 0 & 0\cr 0 & 0 & -\omega & 0\cr 0 & \omega & 0 & 0\cr 0 & 0 & 0 & 0\cr } \mat{A'^0\cr A'^1\cr A'^2\cr A'^3\cr} + \partial_0 \mat{A'^0\cr A'^1\cr A'^2\cr A'^3\cr}

Comparing to the covariant derivative above, it’s clear that they are equal (provided that x'=0 and y'=0, i.e. we are at the center of rotation).

Let’s show the derivation by Goldstein. The change in a time dt of a general vector {\bf G} as seen by an observer in the body system of axes will differ from the corresponding change as seen by an observer in the space system:

(d{\bf G})_{\rm space} = (d{\bf G})_{\rm body}+(d{\bf G})_{\rm rot}

Now consider a vector fixed in the rigid body. Then (d{\bf G})_{\rm body}=0 and

(d{\bf G})_{\rm rot} = (d{\bf G})_{\rm space} = d\boldsymbol\Omega \times {\bf G}

For an arbitrary vector, the change relative to the space axes is the sum of the two effects:

(d{\bf G})_{\rm space} = (d{\bf G})_{\rm body}+d\boldsymbol\Omega \times {\bf G}

A more rigorous derivation of the last equation follows from:

G_i = a_{ji}G'_j

dG_i = a_{ji}dG'_j + da_{ji}G'_j

Let’s make the space and body instantaneously coincident at time t, then a_{ji} = \delta_{ji} and da_{ji}=-\epsilon_{ijk}d\Omega_k=\epsilon_{ikj}d\Omega_k, so we get the same equation as earlier:

dG_i = dG'_i + \epsilon_{ikj}d\Omega_kG'_j

Anyhow, introducing \boldsymbol\omega by:

\boldsymbol\omega = {d\boldsymbol\Omega\over dt}

we get

\left({d{\bf G}\over dt}\right)_{\rm space} = \left({d{\bf G}\over dt}\right)_{\rm body} + \boldsymbol\omega \times {\bf G}

Linear Elasticity Equations in Cylindrical Coordinates

Authors: Pavel Solin & Lenka Dubcova

In this paper we derive the weak formulation of linear elasticity equations suitable for the finite element discretization of axisymmetric 3D problems.

Original equations in Cartesian coordinates

Let’s start with some notations: By \bfu = (u_1, u_2, u_3)^T we denote the displacement vector in 3D Cartesian coordinates, and by \epsilon the tensor of small deformations,

\epsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right),\ \ \ \ 1 \le i,j \le 3.

The stress tensor \sigma has the form

(3.40.2.5)\sigma_{ij} = \lambda \delta_{ij}\mbox{div}\bfu + 2\mu \epsilon_{ij},\ \ \ \ 1 \le i,j \le 3,

where

\mbox{div}\bfu = \sum_{k=1}^3 \frac{\partial u_k}{\partial x_k} = \sum_{k=1}^3 \epsilon_{kk} = \mbox{Tr}(\epsilon).

The symbols \lambda and \mu are the Lam'e constants and \delta_{ij} is the Kronecker symbol (\delta_{ij} = 1 if i = j and \delta_{ij} = 0 otherwise). The equilibrium equations have the form

(3.40.2.6)\sum_{j=1}^3 \frac{\partial \sigma_{ij}}{\partial x_j} + f_i = 0,\ \ \ \ 1 \le i \le 3,

where (f_1, f_2, f_3)^T is the vector of internal forces (such as gravity).

The boundary conditions for linear elasticity are given by

\begin{eqnarray*} u_i &=& \hat{u}_i \quad \mbox{on} \ \Gamma_1\\ \sum_{j=1}^3 \sigma_{ij} n_j &=& g_i \quad \mbox{on} \ \Gamma_2, \end{eqnarray*}

where g_i are surface forces.

Weak formulation

Multiplying by test functions and integrating over the domain \Omega we obtain

(3.40.2.7)- \int_{\Omega}\sum_{j=1}^3 \frac{\partial \sigma_{ij}}{\partial x_j}v_i = \int_{\Omega}f_i\ v_i,\ \ \ \ 1 \le i \le 3.

Using Green’s theorem and the boundary conditions

\int_{\Omega}\sum_{j=1}^3 \sigma_{ij} \frac{\partial v_i}{\partial x_j} - \int_{\partial \Omega} \sum_{j=1}^3 \sigma_{ij} n_j v_i = \int_{\Omega}f_i\ v_i,\ \ \ \ 1 \le i \le 3.

Thus

(3.40.2.8)\int_{\Omega}\sum_{j=1}^3 \sigma_{ij} \frac{\partial v_i}{\partial x_j} - \int_{\Gamma_2} g_i v_i = \int_{\Omega}f_i\ v_i,\ \ \ \ 1 \le i \le 3.

Let us write the equations (3.40.2.8) in detail using relation (3.40.2.5)

(3.40.2.9)\begin{eqnarray*} \int_{\Omega} \left[\lambda \mbox{div}u + 2 \mu \frac{\partial u_1}{\partial x_1}\right] \frac{\partial v_1}{\partial x_1} + \mu \left(\frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1}\right)\frac{\partial v_1}{\partial x_2} + \mu \left(\frac{\partial u_1}{\partial x_3} + \frac{\partial u_3}{\partial x_1}\right)\frac{\partial v_1}{\partial x_3} - \int_{\Gamma_2} g_1 v_1 &=& \int_{\Omega}f_1\ v_1,\nonumber\\ \int_{\Omega} \mu \left(\frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1}\right)\frac{\partial v_2}{\partial x_1} + \left[\lambda \mbox{div}u + 2 \mu \frac{\partial u_2}{\partial x_2}\right] \frac{\partial v_2}{\partial x_2} + \mu \left(\frac{\partial u_2}{\partial x_3} + \frac{\partial u_3}{\partial x_2}\right)\frac{\partial v_2}{\partial x_3} - \int_{\Gamma_2} g_2 v_2 &=& \int_{\Omega}f_2\ v_2,\\ \int_{\Omega} \mu \left(\frac{\partial u_1}{\partial x_3} + \frac{\partial u_3}{\partial x_1}\right)\frac{\partial v_3}{\partial x_1} + \mu \left(\frac{\partial u_2}{\partial x_3} + \frac{\partial u_3}{\partial x_2}\right)\frac{\partial v_3}{\partial x_2} + \left[\lambda \mbox{div}u + 2 \mu \frac{\partial u_3}{\partial x_3}\right] \frac{\partial v_3}{\partial x_3} - \int_{\Gamma_2} g_3 v_3 &=& \int_{\Omega}f_3\ v_3.\nonumber \end{eqnarray*}

Elementary transformation relations

First let us show how the partial derivatives of a scalar function g are transformed from Cartesian coordinates x_1, x_2, x_3 to cylindrical coordinates r, \phi, z. Note that

x_1(r, \phi) = r \cos \phi, \ \ \ x_2(r, \phi) = r \sin \phi, \ \ \ x_3(z) = z.

Since

g(x_1, x_2, x_3) = g(x_1(r, \phi), x_2(r, \phi), x_3(z)),

it is

\begin{eqnarray*} \frac{\partial g}{\partial r} &=& \frac{\partial g}{\partial x_1}\cos\phi + \frac{\partial g}{\partial x_2}\sin\phi,\nonumber \\ \frac{\partial g}{\partial \phi} &=& \frac{\partial g}{\partial x_1}(-r\sin\phi) + \frac{\partial g}{\partial x_2}r\cos\phi,\nonumber\\ \frac{\partial g}{\partial z} &=&\frac{\partial g}{\partial x_3}.\nonumber \end{eqnarray*}

From here we obtain

(3.40.2.10)\begin{eqnarray*} \frac{\partial g}{\partial x_1} &=& \frac{\partial g}{\partial r}\cos\phi - \frac{1}{r}\frac{\partial g}{\partial \phi}\sin\phi,\nonumber \\ \frac{\partial g}{\partial x_2} &=& \frac{\partial g}{\partial r}\sin\phi + \frac{1}{r}\frac{\partial g}{\partial \phi}\cos\phi, \\ \frac{\partial g}{\partial x_3} &=& \frac{\partial g}{\partial z}. \nonumber \end{eqnarray*}

The relations between displacement components in Cartesian and cylindrical coordinates are

(3.40.2.11)\begin{eqnarray*} u_1 &=& u_r \cos \phi, \nonumber \\ u_2 &=& u_r \sin \phi, \nonumber \\ u_3 &=& u_z. \nonumber \end{eqnarray*}

The same relations hold for surface forces g_i and volume forces f_i.

Applying (3.40.2.10) to u_1, we obtain

\begin{eqnarray*} \frac{\partial u_1}{\partial x_1} &=& \frac{\partial u_1}{\partial r}\cos\phi - \frac{1}{r}\frac{\partial u_1}{\partial \phi}\sin\phi,\nonumber \\ \frac{\partial u_1}{\partial x_2} &=& \frac{\partial u_1}{\partial r}\sin\phi + \frac{1}{r}\frac{\partial u_1}{\partial \phi}\cos\phi, \nonumber \\ \frac{\partial u_1}{\partial x_3} &=& \frac{\partial u_1}{\partial z}. \nonumber \end{eqnarray*}

Using (3.40.2.11) and the fact that u_r does not depend on \phi, this yields

\begin{eqnarray*} \frac{\partial u_1}{\partial x_1} &=& \frac{\partial u_r}{\partial r}\cos^2\phi + \frac{1}{r} u_r\sin^2\phi,\nonumber \\ \frac{\partial u_1}{\partial x_2} &=& \frac{\partial u_r}{\partial r}\cos\phi\sin\phi - \frac{1}{r}u_r \cos\phi\sin\phi, \nonumber \\ \frac{\partial u_1}{\partial x_3} &=& \frac{\partial u_r}{\partial z}\cos\phi. \nonumber \end{eqnarray*}

Analogously, for u_2 we calculate

\begin{eqnarray*} \frac{\partial u_2}{\partial x_1} &=& \frac{\partial u_r}{\partial r}\cos\phi\sin\phi - \frac{1}{r}u_r \cos\phi\sin\phi, \nonumber \\ \frac{\partial u_2}{\partial x_2} &=& \frac{\partial u_r}{\partial r}\sin^2\phi + \frac{1}{r} u_r\cos^2\phi,\nonumber \\ \frac{\partial u_2}{\partial x_3} &=& \frac{\partial u_r}{\partial z}\sin\phi. \nonumber \end{eqnarray*}

For u_3, using that it does not depend on \phi, we have

\begin{eqnarray*} \frac{\partial u_3}{\partial x_1} &=& \frac{\partial u_z}{\partial r}\cos\phi, \nonumber \\ \frac{\partial u_3}{\partial x_2} &=& \frac{\partial u_z}{\partial r}\sin\phi,\nonumber \\ \frac{\partial u_3}{\partial x_3} &=& \frac{\partial u_z}{\partial z}. \nonumber \end{eqnarray*}

For further reference, transform also \mbox{div}u into cylindrical coordinates

\begin{eqnarray*} \mbox{div}u &=& \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3} = \\ &=& \frac{\partial u_r}{\partial r}\cos^2\phi + \frac{1}{r} u_r\sin^2\phi + \frac{\partial u_r}{\partial r}\sin^2\phi + \frac{1}{r} u_r\cos^2\phi + \frac{\partial u_z}{\partial z} = \\ &=& \frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z} \end{eqnarray*}

Axisymmetric formulation

Assuming that the domain \Omega is axisymmetric, we can begin to transform the integrals in (3.40.2.9) to cylindrical coordinates. Recall that the Jacobian of the transformation is J(r, \phi, z) = r. The first equation in (3.40.2.9) has the form:

\begin{eqnarray*} &&\int_{\Omega} r \left[\lambda (\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z}) + 2 \mu (\frac{\partial u_r}{\partial r}\cos^2\phi + \frac{1}{r} u_r\sin^2\phi)\right] (\frac{\partial v_r}{\partial r}\cos^2\phi + \frac{1}{r} v_r\sin^2\phi) + \\ &&r 2 \mu \left(\frac{\partial u_r}{\partial r}\cos\phi\sin\phi - \frac{1}{r}u_r \cos\phi\sin\phi\right)\left(\frac{\partial v_r}{\partial r}\cos\phi\sin\phi - \frac{1}{r}v_r \cos\phi\sin\phi\right) + \\ &&r \mu \left(\frac{\partial u_r}{\partial z}\cos\phi + \frac{\partial u_z}{\partial r}\cos\phi\right)\frac{\partial v_r}{\partial z}\cos\phi - \int_{\Gamma_2} r g_r v_r {\cos}^2 \phi = \int_{\Omega} r f_r\ v_r \cos^2 \phi, \end{eqnarray*}

The second equation in (3.40.2.9) has the form:

\begin{eqnarray*} &&\int_{\Omega} r 2\mu \left(\frac{\partial u_r}{\partial r}\cos\phi\sin\phi - \frac{1}{r}u_r \cos\phi\sin\phi\right)\left(\frac{\partial v_r}{\partial r}\cos\phi\sin\phi - \frac{1}{r}v_r \cos\phi\sin\phi\right) +\\ && r \left[\lambda (\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z}) + 2 \mu (\frac{\partial u_r}{\partial r}\sin^2\phi + \frac{1}{r} u_r\cos^2\phi)\right] (\frac{\partial v_r}{\partial r}\sin^2\phi + \frac{1}{r} v_r\cos^2\phi) + \\ && r \mu \left(\frac{\partial u_r}{\partial z}\sin\phi + \frac{\partial u_z}{\partial r}\sin\phi\right)(\frac{\partial v_r}{\partial z}\sin\phi) - \int_{\Gamma_2}r g_r v_r \sin^2 \phi = \int_{\Omega}r f_r\ v_r \sin^2 \phi,\\ \end{eqnarray*}

Adding these two equations together we get

\begin{eqnarray*} &&\int_{\Omega} r \lambda (\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z}) (\frac{\partial v_r}{\partial r} + \frac{1}{r} v_r) + \\ &&\int_{\Omega} r \mu \left[ 2 \left(\frac{\partial u_r}{\partial r}\frac{\partial v_r}{\partial r}\cos^4\phi + \frac{1}{r} u_r \frac{\partial v_r}{\partial r}\sin^2 \phi \cos^2 \phi + \frac{1}{r}\frac{\partial u_r}{\partial r} v_r\sin^2 \phi \cos^2 \phi + \frac{1}{r^2} u_r v_r\sin^4\phi\right) +\right.\\ &&\qquad\ \left.2 \left(\frac{\partial u_r}{\partial r}\frac{\partial v_r}{\partial r}\sin^4\phi + \frac{1}{r} u_r \frac{\partial v_r}{\partial r}\sin^2 \phi \cos^2 \phi + \frac{1}{r}\frac{\partial u_r}{\partial r} v_r\sin^2 \phi \cos^2 \phi + \frac{1}{r^2} u_r v_r\cos^4\phi\right) + \right. \\ && \left. 4 \left( \frac{\partial u_r}{\partial r}\frac{\partial v_r}{\partial r}\cos^2\phi\sin^2\phi - \frac{1}{r}u_r\frac{\partial v_r}{\partial r} \cos^2\phi\sin^2\phi - \frac{1}{r} \frac{\partial u_r}{\partial r} v_r \cos^2\phi\sin^2\phi + \frac{1}{r^2}u_r v_r \cos^2\phi\sin^2\phi\right)\right. + \\ && \left. \left(\frac{\partial u_r}{\partial z}\frac{\partial v_r}{\partial z} + \frac{\partial u_z}{\partial r}\frac{\partial v_r}{\partial z}\right)\right] - \int_{\Gamma_2} g_r v_r r= \int_{\Omega}f_r\ v_r r \end{eqnarray*}

This can be simplified to

\int_{\Omega} r \lambda (\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z}) (\frac{\partial v_r}{\partial r} + \frac{1}{r} v_r) + \int_{\Omega} r \mu \left[ 2 \left(\frac{\partial u_r}{\partial r}\frac{\partial v_r}{\partial r} + \frac{1}{r^2} u_r v_r\right) + \left(\frac{\partial u_r}{\partial z}\frac{\partial v_r}{\partial z} + \frac{\partial u_z}{\partial r}\frac{\partial v_r}{\partial z}\right)\right]

- \int_{\Gamma_2} g_r v_r r = \int_{\Omega}f_r\ v_r r

Finally, the third equation in (3.40.2.9) has the form

\begin{eqnarray*} && \int_{\Omega} r \mu \left(\frac{\partial u_r}{\partial z}\cos\phi + \frac{\partial u_z}{\partial r}\cos\phi\right)\frac{\partial v_z}{\partial r}\cos\phi + r \mu \left(\frac{\partial u_r}{\partial z}\sin\phi + \frac{\partial u_z}{\partial r}\sin\phi\right)\frac{\partial v_z}{\partial r}\sin\phi + \\ && r \left[\lambda (\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z} ) + 2 \mu \frac{\partial u_z}{\partial z}\right] \frac{\partial v_z}{\partial z} - \int_{\Gamma_2} g_z v_z r = \int_{\Omega}f_z\ v_z r. \end{eqnarray*}

This gives us

\begin{eqnarray*} && \int_{\Omega} r \mu \left(\frac{\partial u_r}{\partial z}\frac{\partial v_z}{\partial r} + \frac{\partial u_z}{\partial r}\frac{\partial v_z}{\partial r} + 2 \frac{\partial u_z}{\partial z} \frac{\partial v_z}{\partial z}\right) + r \lambda \left(\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z} \right)\frac{\partial v_z}{\partial z} - \int_{\Gamma_2} g_z v_z r = \int_{\Omega}f_z\ v_z r. \end{eqnarray*}

Since the integrands do not depend on \phi, we can simplify this to integral over \Omega_0, where \Omega_0 is the intersection of the domain \Omega with the x^+_1x_3 half-plane. Dividing both equations by 2\pi we get

\int_{\Omega_0} r \lambda (\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z}) (\frac{\partial v_r}{\partial r} + \frac{1}{r} v_r) + \int_{\Omega_0} r \mu \left[ 2 \left(\frac{\partial u_r}{\partial r}\frac{\partial v_r}{\partial r} + \frac{1}{r^2} u_r v_r\right) + \left(\frac{\partial u_r}{\partial z}\frac{\partial v_r}{\partial z} + \frac{\partial u_z}{\partial r}\frac{\partial v_r}{\partial z}\right)\right]

- \int_{\Gamma_2} g_r v_r r = \int_{\Omega_0}f_r\ v_r r

\int_{\Omega_0} r \mu \left(\frac{\partial u_r}{\partial z}\frac{\partial v_z}{\partial r} + \frac{\partial u_z}{\partial r}\frac{\partial v_z}{\partial r} + 2 \frac{\partial u_z}{\partial z} \frac{\partial v_z}{\partial z}\right) + r \lambda \left(\frac{\partial u_r}{\partial r} + \frac{1}{r} u_r + \frac{\partial u_z}{\partial z} \right)\frac{\partial v_z}{\partial z} - \int_{\Gamma_2} g_z v_z r = \int_{\Omega_0}f_z\ v_z r.

Coordinate Independent Way

Let’s write the elasticity equations in the cartesian coordinates again:

\sigma_{ij} = \lambda \delta_{ij}\partial_k u^k + \mu (\partial_j u_i + \partial_i u_j) \partial_j \sigma^{ij} + f^i = 0

Those only work in the cartesian coordinates, so we first write them in a coordinate independent way:

\sigma^{ij} = \lambda g^{ij}\nabla_k u^k + \mu (\nabla^j u^i + \nabla^i u^j) \nabla_j \sigma^{ij} + f^i = 0

so:

\nabla_j \left( \lambda g^{ij}\nabla_k u^k + \mu (\nabla^j u^i + \nabla^i u^j) \right) + f^i = 0

The weak formulation is then (do not sum over i):

-\int \nabla_j \left( \lambda g^{ij}\nabla_k u^k + \mu (\nabla^j u^i + \nabla^i u^j) \right) v^i \sqrt{|g|}\d^3x = \int f^i v^i\sqrt{|g|}\d^3x

We apply the integration by parts:

\int \left( \lambda g^{ij}\nabla_k u^k + \mu (\nabla^j u^i + \nabla^i u^j) \right) \nabla_j v^i \sqrt{|g|}\d^3x = \int f^i v^i\sqrt{|g|}\d^3x

This is the weak formulation valid in any coordinates. Using the cylindrical coordinates (see above) we get:

{\bf x} = (\rho, \phi, z) \d^3 x = \d\rho\, \d \phi\, \d z g^{ij} = \mat{1 & 0 & 0\cr 0 & 1\over\rho^2 & 0\cr 0 & 0 & 1\cr} \sqrt{|g|} = \sqrt{|\det g_{ij}|} = \rho \nabla_k u^k = {1\over\sqrt{|g|}}\partial_k (\sqrt{|g|} u^k) = {1\over\rho}\partial_k (\rho u^k) = = {1\over\rho} u^\rho + \partial_\rho u^\rho + \partial_\phi u^\phi + \partial_z u^z (\nabla^j u^z + \nabla^z u^j) \nabla_j v^z = (g^{jk}\nabla_k u^z + g^{zk}\nabla_k u^j) \nabla_j v^z = (\partial_\rho u^z + \partial_z u^\rho) \partial_\rho v^z + (\partial_z u^z + \partial_z u^z) \partial_z v^z = = (\partial_\rho u^z + \partial_z u^\rho) \partial_\rho v^z + 2\partial_z u^z \partial_z v^z g^{\rho j}\nabla_j v^\rho = g^{\rho\rho}\nabla_\rho v^\rho = \partial_\rho v^\rho + \Gamma^\rho_{k\rho} v^k = \partial_\rho v^\rho + {1\over\rho} v^\phi g^{\phi j}\nabla_j v^\phi = g^{\phi\phi}\nabla_\phi v^\phi = {1\over\rho^2}(\partial_\phi v^\phi + \Gamma^\phi_{k\phi} v^k )= {1\over\rho^2}(\partial_\phi v^\phi + {1\over\rho} v^\rho )= g^{zj}\nabla_j v^z = g^{zz}\nabla_z v^z = \partial_z v^z + \Gamma^z_{kz} v^k = \partial_z v^z \int \left( \lambda g^{ij}\left( {1\over\rho} u^\rho + \partial_\rho u^\rho + \partial_\phi u^\phi + \partial_z u^z \right) + \mu (\nabla^j u^i + \nabla^i u^j) \right) \nabla_j v^i \rho \,\d\rho\, \d \phi\, \d z = \int f^i v^i \rho \,\d\rho\, \d \phi\, \d z

for i=1, 2, 3 we get:

\int \lambda \left( {1\over\rho} u^\rho + \partial_\rho u^\rho + \partial_\phi u^\phi + \partial_z u^z \right)\left(\partial_\rho v^\rho + {1\over\rho} v^\phi\right)\rho + \mu \left(2\partial_\rho u^\rho \partial_\rho v^\rho + (\partial_z u^\rho +\partial_\rho u^z)\partial_z v^\rho\right) \rho \,\d\rho\, \d \phi\, \d z = \int f^\rho v^\rho \rho \,\d\rho\, \d \phi\, \d z \int \lambda \left( {1\over\rho} u^\rho + \partial_\rho u^\rho + \partial_\phi u^\phi + \partial_z u^z \right){1\over\rho^2}\left(\partial_\phi v^\phi + {1\over\rho} v^\rho \right)\rho + \mu \left(2\partial_\rho u^\rho \partial_\rho v^\rho + (\partial_z u^\rho +\partial_\rho u^z)\partial_z v^\rho\right) \rho \,\d\rho\, \d \phi\, \d z = \int f^\phi v^\phi \rho \,\d\rho\, \d \phi\, \d z \int \lambda \left( {1\over\rho} u^\rho + \partial_\rho u^\rho + \partial_\phi u^\phi + \partial_z u^z \right)\partial_z v^z\rho + \mu \left((\partial_\rho u^z + \partial_z u^\rho) \partial_\rho v^z + 2\partial_z u^z \partial_z v^z\right) \rho \,\d\rho\, \d \phi\, \d z = \int f^z v^z \rho \,\d\rho\, \d \phi\, \d z