# Thread: Derivative of a tensor

1. ## Derivative of a tensor

How do I go about evaluating:

$\nabla \cdot (T_{ij} \tau^i \hat x_j)$

if it helps I have $\tau = [y, -x, 0]'$ and T is symetric

I'll give it a go:

$\nabla \cdot (T_{ij} \tau^i \hat x_j)=\partial_j T_{ij} \tau ^i=\partial_j (yT_{xj} - xT_{yj}) = \partial_j \epsilon_{sk3}(T_{sj} r_k \hat x_3)$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k \hat x_3 +\frac{\partial r_k }{\partial x_j} T_{sj} \hat x_3 +\frac{\partial \hat x_3 }{\partial x_j} T_{sj} r_k\right]$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k \hat x_3 +\delta_{kj} T_{sj} \hat x_3 +\delta_{j3} T_{sj} r_k\right]$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k \hat x_3 + T_{sk} \hat x_3 +T_{s3} r_k\right]$ T is symetric so the middle term disapears

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k \hat x_3 + T_{s3} r_k\right]$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \hat z_i \epsilon_{isk} r_k\left[ \frac{\partial T_{sj}}{\partial x_j} \hat x_i + T_{si} \right]$ where z = [0 0 1]'

I can go no further and suspect I have made an error before this point.

I would really like to get an answer of the form:

$(\underline r \times \frac{\partial \underline S}{\partial t}) \cdot \hat z$ but that seems unlikely as I can see no way to conjure up the partial derivative w.r.t. t.

This problem is related to my earlier post:
Calculus of Variations - Noether's Theorem for fields

2. ## Re: Derivative of a tensor

Correcting some errors:
$\nabla \cdot (T_{ij} \tau^i \hat x_j)=\partial_j T_{ij} \tau ^i=\partial_j (yT_{xj} - xT_{yj}) = \partial_j \epsilon_{sk3}(T_{sj} r_k )$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k +\frac{\partial r_k }{\partial x_j} T_{sj} \right]$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k +\delta_{kj} T_{sj} \right]$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k + T_{sk} \right]$ T is symmetric so the second term disappears

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \epsilon_{sk3} \left[ \frac{\partial T_{sj}}{\partial x_j} r_k \right]$

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= \hat z_i \epsilon_{isk} r_k\left[ \frac{\partial T_{sj}}{\partial x_j} \right]$ where z = [0 0 1]'

$\nabla \cdot (T_{ij} \tau^i \hat x_j)= -(\underline r \times \nabla _j \cdot [T]) \right] \cdot \hat z$

I can go no further and suspect I have made an error before this point.

I would really like to get an answer of the form:

$(\underline r \times \frac{\partial \underline S}{\partial t}) \cdot \hat z$ but that seems unlikely as I can see no way to conjure up the partial derivative w.r.t. t.

Perhaps there is something special about the Maxwell Stress Tensor that allows me to do this?

This problem is related to my earlier post:
Calculus of Variations - Noether's Theorem for fields

3. ## Re: Derivative of a tensor

Can you be a little more specific about your T and S here so that we can help better? I think the negative sign in the end should not be there. And if S is the poynting vector and T is the stress tensor, then you have $f + \epsilon_o \mu_o \frac{\partial S}{\partial t} = \nabla.T$ where f is the force per unit volume for the charge distribution. If your f is 0, then you get exactly the form you are looking for. Let me know if this helped.