Fluid dynamics - Proving the Lorentz reciprocal theorem using Cartesian tensor methods

Question

How can Cartesian tensor methods be used to prove this identity? I tried starting with the continuity equation but can't seem to get anywhere. Thank you in advance for any help.

Answer 1

For Stokes flow the velocity $\mathbf{u}$ and stress tensor $\mathbf{T}$ satisfy the continuity equation $\nabla \cdot \mathbf{u} = 0$ and momentum equation $\nabla \cdot \mathbf{T} = \mathbf{0}.$ We assume here that the body force is conservative (eg, gravity) and has been absorbed into the isotropic part of the stress tensor.

In terms of Cartesian components, we have

$$\tag{1}\partial_i u_i = 0,$$

and

$$\tag{2} \partial_j T_{ij}=0,$$

where Cartesian components are introduced with the Einstein convention (summation over repeated indexes).

The stress tensor and velocity are related by the constitutive equation

$$\tag{3} T_{ij} = -p \,\delta{ij} + 2 \mu D_{ij},$$

where $p$ is the pressure, $\mu$ is the viscosity, $\delta_{ij} = 1 \, (i = j); \, = 0 \, (i \neq j)$ is the Kronecker delta, and $D_{ij}$ are components of the rate-of strain tensor given by

$$\tag{4} D_{ij} = \frac{1}{2} (\partial_j u_i + \partial_i u_j).$$

Now suppose we have a second velocity $\mathbf{u}'$ and stress tensor $\mathbf{T}'$ satisfying

$$\tag{1'} \partial_i u_i' = 0,$$ $$ \tag{2'} \partial_j T_{ij}' =0,$$ $$\tag{3'} T_{ij}' = -p' \,\delta{ij} + 2 \mu' D_{ij}',$$ $$\tag{4'} D_{ij}' = \frac{1}{2} (\partial_j u_i' + \partial_i u_j').$$

We assume here that the two flow fields are functions defined on the same domain $V \subset \mathbb{R}^3$ with boundary $S$.

Multiplying $T_{ij}$ by $D_{ij}'$ we obtain

$$T_{ij}D_{ij}' = -p\delta_{ij} D_{ij}' + 2 \mu D_{ij} D_{ij}' = 2 \mu D_{ij} D_{ij}' ,$$

since $\delta_{ij}D_{ij}' = D_{ii}' = \partial_i u_i' = 0$, using the continuity equation $(1')$.

Similarly, interchanging unprimed and primed variables, we have

$$T_{ij}'D_{ij} = 2 \mu' D_{ij}' D_{ij}. $$

Thus,

$$\tag{5}\mu' \, T_{ij} \, D_{ij}' = \mu \, T_{ij}' \, D_{ij}.$$

Since the stress tensor is symmetric, we have $T_{ij} = T_{ji}$, $T_{ij}' = T_{ji}'$ and

$$\tag{6}T_{ij} \, D_{ij}' = \frac{1}{2} T_{ij} \, \partial_j u_i' + \frac{1}{2} T_{ij} \, \partial _i u_j' = T_{ij} \, \partial_j u_i' = \partial_j \, (T_{ij} \, u_i') - u_i' \partial_j \, T_{ij}, \\ T_{ij} \, 'D_{ij} = \frac{1}{2} T_{ij}' \, \partial_j u_i + \frac{1}{2} T_{ij}' \, \partial _i u_j = T_{ij}' \, \partial_j \, u_i = \partial_j \, (T_{ij}' \, u_i) - u_i \, \partial_j T_{ij}'.$$

From equations $(2)$ and $(2')$ we have $\partial_j T_{ij} = \partial_j \, T_{ij}'= 0$ and $(6)$ reduces to

$$\tag{7}T_{ij}\, D_{ij}' = \partial_j \, (T_{ij} \, u_i'), \\ T_{ij}' \, D_{ij} = \partial_j \, (T_{ij}' \, u_i) $$

Substituting the results from $(7)$ into $(5)$ we get

$$ \mu'\partial_j \,(T_{ij} \, u_i') = \mu\partial_j \, (T_{ij}' \,u_i),$$

and in general tensor form

$$\mu' \nabla \cdot (\mathbf{T} \cdot \mathbf{u}') = \mu \nabla \cdot (\mathbf{T}' \cdot \mathbf{u}).$$

Integrating over $V$ we have

$$\mu' \int_V \nabla \cdot (\mathbf{T} \cdot \mathbf{u}') \, dV = \mu \int_V \nabla \cdot (\mathbf{T}' \cdot \mathbf{u}) \, dV$$

Applying the divergence theorem it follows that

$$\mu' \int_S \mathbf{n} \cdot (\mathbf{T} \cdot \mathbf{u}') \, dS = \mu \int_S \mathbf{n} \cdot (\mathbf{T}' \cdot \mathbf{u}) \, dS.$$

Your result is the special case where $\mu = \mu'.$