Exercise 5.22 in the book Geometrical Methods of Mathematical Physics (by Bernard Schutz)

Question

I can't figure it out about the Exercise 5.22 in the book Geometrical Methods of Mathematical Physics (by Bernard Schutz).
Could any one give me a help? Thanks.

($\bar{V}$ means vector V. )

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Exercise 5.22 (on page 186)
The shear of a velocity field $\bar{V}$ is defined in Cartesian coordinates by the equation

$$ \sigma_{ij} = V_{i,j} + V_{j,i} - \frac{1}{3} \delta_{ij} \theta \qquad (5.83)$$

where $\theta$ is the expansion

$$ \theta = \nabla \cdot \bar{V} \qquad (5.84)$$

Show that in an arbitrary coordinate system

$$ \theta = \frac{1}{2} g^{ij} \mathcal{L}_{\bar{V}} g_{ij} \qquad (5.85) $$

$$ \sigma_{ij} = \mathcal{L}_{\bar{V}} g_{ij} - \frac{1}{3} \theta g_{ij} \qquad (5.86) $$

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I have tried the following steps:

$$ \sigma_{ij} = \mathcal{L}_{\bar{V}} g_{ij} - \frac{1}{3} \theta g_{ij} \qquad (5.86) \\ = V^k \frac{\partial}{\partial x^k}g_{ij} + g_{ik} \frac{\partial}{\partial x^j} V^k + g_{kj} \frac{\partial}{\partial x^i} V^k - \frac{1}{3} \nabla \cdot \bar{V} g_{ij} \qquad (by \ using \ a \ step \ of \ (3.37) \ on \ page \ 88)\\ = V^k \frac{\partial}{\partial x^k}g_{ij} + V_{i,j} + V_{j,i} - \frac{1}{3} \nabla \cdot \bar{V} g_{ij} \\ $$

But how does the last line equal the right side of (5.83)?

Answer 1

Unfortunately I don't have access to the book in question so I don't know your setting or context. But lets assume that the metric is symmetric, ie $g_{ij}=g_{ji}$ and $\det(g_{ij})\neq0$. Also, we need a connection which is assumed to be based on the metric, $\Gamma_{i\,\,j}^{\,\,h}=\overset{(g)}{\Gamma}{}_{i\,\,j}^{\,\,h}$. This construction gives us access to the musical isomorphism (or canonical isomorphism) between the tangent bundle and the cotangent bundle of a Riemannian manifold induced by its metric tensor, for example $$v_i=g_{ij}v^j\tag{1}$$

Also, the covariant derivative of any symmetric nonsingular type (0,2) tensor field vanishes identically whenever this covariant derivative is defined in terms of the connection whose coefficients are the Christoffel symbols defined with respect to the given tensor field which means

$$g_{ij|k}=0\tag{2}$$

Another useful property of Riemannian spaces (the special construction mentioned above) is

$$\frac{\partial g_{ij}}{\partial x^k}=\Gamma_{ijk}+\Gamma_{jik}\tag{3}$$

The Lie derivative of the metric is by definition

$$\mathcal{L}_v g_{ij}=\frac{\partial g_{ij}} {\partial x^k }v^k+g_{rj}\frac{\partial v^r} {\partial x^i }+g_{ir}\frac{\partial v^r} {\partial x^j }\tag{4}$$

Using equation (3) and the canonical isomorphism this can be expanded to

$$\mathcal{L}_v g_{ij}=(g_{rj}\Gamma_{i\,\,k}^{\,\,r}+g_{ri}\Gamma_{j\,\,k}^{\,\,r})v^k+g_{rj}\frac{\partial v^r} {\partial x^i }+g_{ri}\frac{\partial v^r} {\partial x^j }\tag{5}$$

Rearranging the terms we recognize the definition of the covariant derivative

$$\mathcal{L}_v g_{ij}=g_{rj}\left(\Gamma_{i\,\,k}^{\,\,r}v^k+\frac{\partial v^r} {\partial x^i }\right)+g_{ri}\left( \Gamma_{j\,\,k}^{\,\,r}v^k+\frac{\partial v^r} {\partial x^j }\right)=g_{rj}v^r_{|i}+g_{ri}v^r_{|j}\tag{6}$$

$$\mathcal{L}_v g_{ij}=v_{i|j}+v_{j|i}\tag{7}$$

The importance of (7) in practical calculations cannot be stressed enough.

In an arbitrary coordinate system (well, arbitrary within the limits of the Riemannian Geometry specified above) we generalize expressions from Cartesian coordinates by using the covariant derivative instead of the "usual" partial derivative and the metric $g_{ij}$ instead of $\delta_{ij}$. So for instance $\nabla \cdot v =v^k_{|k}=\theta$ and the generalized version of equation (5.83) is (using (7))

$$v_{i|j}+v_{j|i}-\frac{1}{3}g_{ij}\theta=\mathcal{L}_v g_{ij}-\frac{1}{3}\theta g_{ij}\tag{8}$$

Finally equation (5.85) is (again using (7))

$$\frac{1}{2}g^{ij}\mathcal{L}_vg_{ij}=\frac{1}{2}g^{ij}\left(v_{i|j}+v_{j|i}\right)=\frac12\left(v^j_{|j}+v^i_{|i}\right)=v^k_{|k}=\nabla \cdot v\tag{9}$$