# Why it doesn't matter to use velocity gradient or shear rate tensor to calculate wall shear stress?

The wall shear stress is defined based on Matyka et. al. eq. A.4:

$$\vec{\tau} = 2 \mu (\mathbf{S} \cdot \vec{n} - (\vec{n} \cdot \mathbf{S} \cdot \vec{n})\vec{n})$$

Where $$\mathbf{S} = \frac{1}{2} (\nabla \mathbf{u} + (\nabla \mathbf{u})^{T})$$ is shear rate tensor and $$\mu$$ is kinematic viscosity. There is one more popular definition of wall shear stress as:

$$\vec{\tau^{*}} = \mu \nabla \mathbf{u} \cdot \vec{n}$$

In the first look, these two definitions are really different, but I could simplify the Matyka's equation by remembering that wall shear stress is calculated at the wall and at the wall ($$\partial \Omega$$) we have no-slip boundary condition: $$\mathbf{u} = 0$$:

$$\vec{\tau} = \mu (\nabla \mathbf{u} \cdot \vec{n} + (\nabla \mathbf{u})^{T} \cdot \vec{n} - (\vec{n} \cdot (\nabla \mathbf{u} + (\nabla \mathbf{u})^{T}) \cdot \vec{n})\vec{n})$$

Now, we have:

$$(\nabla \mathbf{u})^{T} \cdot \vec{n} = \frac{\partial u_{j}}{\partial x_{i}} n_{j} \mathbf{e}_{i}$$

Here Einstein summation is used and $$\mathbf{e}_{i}$$ are unit vectors.

But:

$$\frac{\partial}{\partial x_{i}} (u_{j}n_{j}) = \frac{\partial u_{j}}{\partial x_{i}} n_{j} + \frac{\partial n_{j}}{\partial x_{i}} u_{j}$$

or:

$$(\nabla \mathbf{u})^{T} \cdot \vec{n} = \nabla (\mathbf{u} \cdot \vec{n}) - (\nabla \vec{n})^{T} \cdot \mathbf{u} = \nabla (\mathbf{u} \cdot \vec{n})$$

Because $$\mathbf{u} = 0$$ at the $$\partial \Omega$$ wall.

Also:

$$\vec{n} \cdot (\nabla \mathbf{u} + (\nabla \mathbf{u})^{T}) \cdot \vec{n} = \vec{n} \cdot \nabla \mathbf{u} \cdot \vec{n} + \vec{n} \cdot (\nabla \mathbf{u})^{T} \cdot \vec{n} = \vec{n} \cdot \nabla \mathbf{u} \cdot \vec{n} + \nabla (\mathbf{u} \cdot \vec{n}) \cdot \vec{n}$$

But:

$$\vec{n} \cdot \nabla \mathbf{u} \cdot \vec{n} = \frac{\partial u_{\alpha}}{\partial x_{\beta}} n_{\alpha} n_{\beta} = (\frac{\partial}{\partial x_{\beta}} (u_{\alpha} n_{\alpha}) - \frac{\partial n_{\alpha}}{\partial x_{\beta}} u_{\alpha})n_{\beta} = \vec{n} \cdot \nabla (\mathbf{u} \cdot \vec{n}) - (\vec{n} \cdot \nabla \vec{n}) \cdot \mathbf{u}$$

Again the last term is zero because of no-slip boundary condition so:

$$\vec{n} \cdot (\nabla \mathbf{u} + (\nabla \mathbf{u})^{T}) \cdot \vec{n} = 2 \nabla (\mathbf{u} \cdot \vec{n}) \cdot \vec{n}$$

So finally:

$$\vec{\tau} = \mu \nabla \mathbf{u} \cdot \vec{n} + \mu \nabla (\mathbf{u} \cdot \vec{n}) -2 \mu (\nabla (\mathbf{u} \cdot \vec{n})\cdot \vec{n})\vec{n}$$

So basically the difference of $$\vec{\tau}$$ and $$\vec{\tau^{*}}$$ is just:

$$\vec{\tau} - \vec{\tau^{*}} = \mu \nabla (\mathbf{u} \cdot \vec{n}) -2 \mu (\nabla (\mathbf{u} \cdot \vec{n})\cdot \vec{n})\vec{n}$$

I calculated the wall shear stress with both $$\vec{\tau}$$ and $$\vec{\tau^{*}}$$ and they are the same up to fourth digit, which means basically the term $$\mu \nabla (\mathbf{u} \cdot \vec{n}) -2 \mu (\nabla (\mathbf{u} \cdot \vec{n})\cdot \vec{n})\vec{n}$$ must be close to zero. Is there any reason that the gradient of normal component of velocity should be close to zero? I really appreciate some suggestions or insight here.

• Isn't it because u=0 at the wall? Thus the normal component is zero, and the gradient is also zero. – Xi Zou Feb 18 at 0:29
• $\mathbf{u} = 0$ is at the boundary but why $\nabla (\mathbf{u} \cdot \vec{n})$ must be equal to zero? I don’t see any apparent reason for that. – Alone Programmer Feb 18 at 1:35