# Consequences of tensor form due to various symmetries

I'm reading a paper on 2D hydrodynamics, specifically on the drag of a rod in a 2D fluid. It is a low-Reynolds number regime, therefore linear hydrodynamics and the velocity of the rod can be expressed via:

$$v_i^{rod} = \mu_{ij}F_j$$

then it says: By in-plane rotational symmetry combined with the $\hat{n} → -\hat{n}$ symmetry of the rod, the mobility tensor must take the form:

$$\mu_{ij} =\mu_\parallel\hat{n}_i\hat{n}_j + \mu_\perp(\delta_{ij}-\hat{n_i}\hat{n_j})$$

Here, $\mu_\parallel$ and $\mu_\perp$ are the mobilities for motion parallel and perpendicular to the rod's long axis, respectively.

I'm having a hard time connecting the dots with this logic, can anyone show me how this follows?

• So far, it makes sense that the perpendicular mobility takes a jth component of the force and causes an ith component of the velocity of the rod, for i not equal to j, however there is still a parallel mobility term for this case, why is that? Or am I wrong? Commented Apr 4, 2014 at 2:15
• That's wrong. I am writing my answer now. I was eating before. Commented Apr 4, 2014 at 2:16

We are asking how the rod moves through the viscous medium if we apply a force to it. Since the Reynolds number is so low, inertial forces must be small, and the externally applied force must be balanced by a viscous force.

Also since we are in the low Reynold's number limit, the viscous force is linear in the velocity of the object. Thus there must be a linear relationship between the force being applied to the object and the resulting velocity of the object.

Now we will determine some properties of the linear relationship using symmetries of the rod. Let's work in coordinates where the direction along the long axis of the rod is $\hat{n}$. Let's call the other direction $\hat{y}$. Now let's find what the coordinates of $\mu$ might be in the $n,y$ coordinate system.

First consider pulling the rod so the force is directed along the long axis of the rod (in the $\hat{n}$ direction). What will the resulting velocity be? By symmetry, we know that the velocity cannot move in the $\hat{y}$ direction. Thus we conclude the final velocity is in the $\hat{n}$ direction. Then the final velocity in this case is given by $\mu_\parallel \hat{n}F$.

Now consider pulling the rod perpendicular to the long axis. Similar as before symmetry prevents the rod's velocity to have a component along its length. Thus the velocity must be purely in the $\hat{y}$ direction. Let's say the velocity is $\mu_\perp \hat{y}F$.

Now if we have components of the force in both directions, we know by linearity that the resulting velocity must be $$\vec{v} = \mu_\parallel \hat{n}F_n+\mu_\perp \hat{y}F_y =(\mu_\parallel \hat{n}\otimes \hat{n} +\mu_\perp \hat{y} \otimes \hat{y}) \vec{F}.$$ This tells us that $\mu$ is the expression in paretheses: $$\mu = \mu_\parallel \hat{n}\otimes \hat{n} +\mu_\perp \hat{y} \otimes \hat{y}.$$ It is a diagonal tensor in this coordinate system.

Now that I have found $\mu$ I should show that it is the same as the expression you give in the question. We agree on the first term of $\mu$, but the second term is different. It seems I have to show that $1-\hat{n}\otimes \hat{n} = \hat{y} \otimes \hat{y}$, but this is true because, since $\hat{n}$ and $\hat{y}$ form an orthonormal basis, then by the resolution of identity formula we have $1=\hat{n}\otimes \hat{n}+\hat{y} \otimes \hat{y}$, so our answers do agree.