# Boundary condition for a flow past a spherical obstacle

• happyparticle
happyparticle
Homework Statement
Consider the steady flow pattern produced when an impenetrable rigid spherical obstacle is placed in a uniformly flowing, incompressible, inviscid fluid.
Relevant Equations
##r(r,\theta, \phi) = V \hat{z} , r \to\infty##
##\vec{v} = - \nabla \Phi##
I'm trying to find how the author finds the boundary condition at ##r\to\infty## is ## \Phi(r,\theta, \phi) = - V r cos \theta##.

Using the spherical coordinates.

##- V \hat{z} = \nabla \Phi##

##- V ( cos \theta \hat{r} - sin \theta \hat{\theta}) = \frac{d \Phi}{dr}\hat{r} + 1/r \frac{d \Phi}{d \theta} \hat{\theta} + \frac{1}{r sin \theta} \frac{d \Phi}{ d \phi} \hat{\phi} ##

I'm not sure to understand why most of the terms vanishes.

happyparticle said:
I'm trying to find how the author finds the boundary condition at ##r\to\infty## is ## \Phi(r,\theta, \phi) = - V r cos \theta##.
What is the pattern of a uniformly flowing, incompressible, inviscid fluid before the obstacle is inserted into the flow? That's the same pattern the flow must approach at long distances (##r\rightarrow\infty##) after insertion of the object.

renormalize said:
What is the pattern of a uniformly flowing, incompressible, inviscid fluid before the obstacle is inserted into the flow? That's the same pattern the flow must approach at long distances (##r\rightarrow\infty##) after insertion of the object.
I understand that part. However is more the mathematical part that I don't really understand. Where ##\Phi(r,\theta, \phi) = - V r cos \theta## come from.

What are the velocity components far from the sphere?

happyparticle said:
I understand that part. However is more the mathematical part that I don't really understand. Where ##\Phi(r,\theta, \phi) = - V r cos \theta## come from.

The background velocity is $V\mathbf{e}_z$.

The calculation is $$\Phi = -\int \mathbf{v} \cdot d\mathbf{x}.$$ You can either do in cartesians, which is straightforward, or you can do in spherical polars; this however is much trickier, since the polar basis vectors are not constant, but vary with position, so that you cannot integrate component by component as you can with cartesians.

Last edited:
I think I'm even more confuse.

Using cartesians coordinates I have:

##-\vec{v} = \frac{d \Phi}{dx}\hat{x} + \frac{d \Phi}{dy}\hat{y} + \frac{d \Phi}{dz}\hat{z}##

##-V \hat{z} = \frac{d \Phi}{dx}\hat{x} + \frac{d \Phi}{dy}\hat{y} + \frac{d \Phi}{dz}\hat{z}##

It seems like you kept only the ##\hat{x}## component. Why?
Also, to have ##z = r cos \theta##, I must use polar coordinates.

What are the components of the far-field velocity in spherical coordinates?

Chestermiller said:
What are the components of the far-field velocity in spherical coordinates?
##V (cos \theta \hat{r} - sin \theta \hat{\theta})##

happyparticle said:
##V (cos \theta \hat{r} - sin \theta \hat{\theta})##
So, at large r, $$\frac{\partial\Phi}{\partial r}=V\cos{\theta}$$ and $$\frac{\partial \Phi}{\partial \theta}=-Vr\sin{\theta}$$

It seems like there is a issue with the sign. Is there an error on the page linked above?

happyparticle said:
It seems like there is a issue with the sign. Is there an error on the page linked above?
Oops. I assumed that v was equal to the gradient of phi rather than minus the gradient of phi. So just flip the signs in my previous post.

happyparticle
happyparticle said:
It seems like you kept only the ##\hat{x}## component. Why?

The spherical object has no preferred axes, but the background flow does have one: the direction of the flow. In spherical polar coodinates, that direction is usually aligned with the $\theta = 0$ ray, which is the positive $z$ axis.

We are only interested in the gradient of the potential, so we can take the constant of integration to be zero.

happyparticle
Thank you!

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