# Shifting polar functions vertically

## Main Question or Discussion Point

Hi PF!

I have a function that looks like this $$f(r,\theta) = \sinh (\omega \log (r))\cos(\omega(\theta - \beta))$$

You'll notice $f$ is harmonic and satisfies the BC's $f_\theta(\theta = \pm \beta) = 0$. Essentially $f$ has no flux into the wall defined at $\theta = \pm \beta$. So we can imagine the domain is a sort of groove with it's vertex centered at the origin. However, this groove actually needs to be centered at a different vertical location, $(h,0)$.

I calculate the new function to be $$f(r,\theta) = \sinh \left(\frac{\omega}{2} \log (r^2-2hr\cos\theta + h^2)\right)\cos\left[\omega\left(\arctan \left(\frac{r\sin\theta}{r \cos \theta-h}\right)- \beta\right)\right]$$
but it's so messy. Any clean-up ideas or different approaches?

Last edited:

haruspex
Homework Helper
Gold Member
I don't get the bit about fθ(.,-β) being zero. Have I misunderstood the notation?

I don't get the bit about fθ(.,-β) being zero. Have I misunderstood the notation?
Sorry, what I mean is $$\left.\frac{\partial f}{\partial \theta} \right|_{\theta = \pm \beta} = 0$$
The vertex has angle $2\beta$.

haruspex
Homework Helper
Gold Member
Sorry, what I mean is $$\left.\frac{\partial f}{\partial \theta} \right|_{\theta = \pm \beta} = 0$$
The vertex has angle $2\beta$.
Yes, that's what I thought you meant, but you seem to be saying this follows from
$f(r,\theta) = \sinh (\omega \log (r))\cos(\omega(\theta - \beta))$
I get $\left.\frac{\partial f}{\partial \theta} \right|_{\theta = -\beta} = -\omega\sinh (\omega \log (r))\sin(\omega( - 2\beta))$

Yes, that's what I thought you meant, but you seem to be saying this follows from

I get $\left.\frac{\partial f}{\partial \theta} \right|_{\theta = -\beta} = -\omega\sinh (\omega \log (r))\sin(\omega( - 2\beta))$
Oh shoot, sorry, I forgot to say $\omega = k \pi / \beta : k\in \mathbb N$.

FactChecker
Gold Member
Is there a reason that you can't just introduce new variables for the shift and leave the original equations essentially as-is? I think it would be much more clear. If I were to document or program that problem, I think that is how I would do it.

Is there a reason that you can't just introduce new variables for the shift and leave the original equations essentially as-is? I think it would be much more clear. If I were to document or program that problem, I think that is how I would do it.
Can you elaborate?

See, this is one component to a very large problem I'm working on. The issue is, I am in two different coordinate systems that do not have the same center. Obviously in order to have them both work I need to choose one of the two coordinate systems. I know this one is by far easier (namely, I need to shift the vertex to $(h,0)$).

I also think that when I shift the coordinate system, the BC's are no longer valid at $\theta = \beta$. Instead, the boundaries are now valid along a line not passing through the origin. Specifically, the upper boundary line is parameterized as $$r = h \csc(\beta - \theta) \sin\beta$$
What do you think? Am I making an error by thinking I can simply shift $f$ vertically, evaluate $f_\theta$ along the shifted line, and still get a good answer? I tried doing that and I'm not getting 0.

Last edited:
FactChecker
My suggestion is just a cosmetic difference which I think would make it easier to understand what is going on. I would define the shifted coordinate $(r',\theta') = (\sqrt {r^2-2hr\cos \theta + h^2}, \frac {r \sin \theta}{r \cos \theta - h})$ and then use $f(r', \theta')$.