# Homework Help: Complex Potential in Cartesians

1. May 31, 2009

### Firepanda

Relevant Equations

I know ∂ψ/∂y = ∂φ/∂x = u

and -∂ψ/∂x = ∂φ/∂y = v (Cauchy Riemann equations)

For the complex potential ω(z) = φ(x,y) + iψ(x,y)

u = ui + vj

and dω/dz = u - iv

Attempt 1

ω(z) = - ik log(z - z0)

Take z0 = 0

=> ω(z) = - ik log(x + iy)

=> ψ = -k log(x + iy)

=> ∂ψ/∂x = -k/(x + iy)

=> v = k/(x + iy)

and

∂ψ/∂y = -ik/(x + iy)

=> u = -ik/(x + iy)

so

u = -ik/(x + iy) i + k/(x + iy) j

Is this correct? I'm a little concerned as I have an i in my u component.

Perhaps I should have just done

Attempt 2

ω(z) = - ik log(z)

dω/dz = u - iv = - ik/z

where - ik/z = - ik/(x+iy) = -ik(x-iy)/(x2+y2)

= (yk - ikx)/(x2+y2)

=> u = yk/(x2+y2)

and v = - kx/(x2+y2)

=> u = yk/(x2+y2) i - kx/(x2+y2) j

Which is correct? :)

Should I have taken z0 = 0?

Last edited: May 31, 2009
2. May 31, 2009

### Firepanda

For above if Z0 = 0 isn't good enough I guess I can just sub in x = x - x0 and y = y - y0.

The rest of the question:

Here's my basic proof of the equilateral triangle:

With z = 1 etc. for the vertices.

But how to show the rotation?

Thanks.

3. Jun 1, 2009

### Hao

An error in Attempt one is in moving from
=> ω(z) = - ik log(x + iy)
to
=> ψ = -k log(x + iy)
The two potentials ψ and φ are the real/complex parts of ω.

As a result, ψ is not quite either one of them due to the log(x + iy).

Attempt 2 is correct.

Another way to work from Attempt one is to work out this problem in polar coordinates, and transform to cartesian coordinates using the standard transformations:
$$r^2 = x^2 + y^2$$
$$\theta = Tan^{-1}(x\y)$$

For the triangle question, I find it strange that the sources can rotate, but this question can still be solved if we assume that the vortices move with the velocity flow.

I believe that after working it out, the velocity flow at each vortex is in a tangential direction, and $$Period = 2 \pi / |v|$$.

To find the fourth vortex, we know that the flow at each vortex is tangential, so we introduce a fourth vortex that eliminates this tangential component.

4. Jun 1, 2009

### Firepanda

Ok thankyou.

So in my solution to the first part am I right in thinking I change all my x's and y's to x = x - x0 and y = y - y0 respectively? i.e for z = 1 the y0 = 0 and x0 = 1.

I've done all that, but finding the tangential velocities has me stumped.

5. Jun 1, 2009

### Hao

The thinking is correct - there is usually no other way to obtain the real and complex potentials if the real and complex parts are locked up in z.

To find velocities, we rely on superposition once more. Add the velocity flows for individual point vortexes to get the velocity flow for a more complex vortex distribution. By symmetry, once we find the velocity at one point vortex, we can be sure that it will be in the same relative direction and have the same speed for other point vortices.

Knowing that the velocity at a point vortex is tangential can also be drawn from a symmetry argument.

6. Jun 1, 2009

### Firepanda

Ok, I've probably misunderstood but I'll add the velocity flows for each point vortex

For z = 1

u = i yk/((x-1)2+y2) - j (x-1)k/((x-1)2+y2)

z = e2ipi/3

u = i (y-√3/2)k/((x+0.5)2+(y-√3/2)2) - j (x+0.5)k/((x+0.5)2+(y-√3/2)2)

z = z = e-2ipi/3

u = i (y+√3/2)k/((x+0.5)2+(y+√3/2)2) - j (x+0.5)k/((x+0.5)2+(y+√3/2)2)

So I add all the i components and add all the j components and hope it simplifies?

7. Jun 1, 2009

### Hao

Now that you have added them up, what is the velocity flow at the point z = 1?

8. Jun 1, 2009

### Firepanda

So now I just sub in all x = 1 and y = 0?

So

u = i [0 -(√3/2)k/3 + (√3/2)k/3] - j [0 + (1.5)k/3 + (1.5)k/3]

= i 0 - j k

Correct?

9. Jun 1, 2009

### Firepanda

The rest being

for z = e2ipi/3

u = i [(√3/2)k/3 + 0 + (√3)k/3] - j [(-1.5)k/3 + 0 + 0]

= i (√3/2)k + j k/2

for z = e-2ipi/3

u = i [(-√3/2)k/3 + (-√3)k/3 + 0] - j [(-1.5)k/3 + 0 + 0]

= i (-√3/2)k + j k/2

10. Jun 1, 2009

### Firepanda

So here are my tangential lines.

11. Jun 1, 2009

### Hao

You've gotten everything right - now, the vortices will always move in a tangential direction, which results in circular motion if we recalculate the velocity after every short time interval.

As a result, for a circle of radius 1, the time is 2*pi*1/speed