# Determine Scalar Potential Function

1. Jan 4, 2010

### teeeeee

Hi,

I wonder if someone could help me.

I'm trying to find the potential function,$$\phi$$, of the field: v = y2z3i + 2xyz3j + 3xy2 z2k

So using v = $$\nabla\phi$$, I have found:

$$\frac{\partial \phi}{\partial x}$$ = y2z3x + F(y,z)

$$\frac{\partial \phi}{\partial y}$$ = y2z3x + G(x,z)

$$\frac{\partial \phi}{\partial z}$$ = y2z3x + H(x,y)

So my question is:

Does this mean that F,G and H are all zero... and therefore $$\phi$$ = y2z3x

or do I still need to include some constant, so that $$\phi$$ = y2z3x + C , for example.

Help will be much appreciated,
Thanks, teeeeee

2. Jan 4, 2010

### tiny-tim

Hi teeeeee!

(have a phi: φ and a curly d: ∂ and a del: ∇ )

If there are two solutions, φ and φ2, then ∇(φ - φ2) must be 0, so yes, φ - φ2 must be constant.

3. Jan 4, 2010

### teeeeee

I dont understand what you mean..

4. Jan 4, 2010

### teeeeee

So are F,G and H all zero? Or do I need to include a constant at the end? Or a function of all three variables at the end?

Thanks

5. Jan 4, 2010

### tiny-tim

If there are two different potentials, φ and φ2, such that ∇φ = v and ∇φ2 = v,

then ∇φ - ∇φ2= v - v = 0,

so ∇(φ - φ2) = 0,

so (φ - φ2) must be constant.

6. Jan 4, 2010

### teeeeee

How do I determine if there are 2 different potentials..?
All the information I have is in my first post.

7. Jan 4, 2010

### LCKurtz

As others have pointed out, there isn't any the potential because adding a constant gives another potential for the same field.

But I would suggest a better way to come to the solution:

Start with an unknown function $\phi(x,y,z)$ that is to be your potential function. Then you must have

$$\phi_x = y^2z^3$$

Integrating both sides of this with respect to x tells you

$$\phi(x,y,z) = xy^2z^3 + h(y,z)$$

since the "constant" of integration can have y and z in it. Next check the partial with respect to y:

$$\phi_y = 2xyz^3 + h_y(y,z) = 2xyz^3$$

This tells you that $h_y(y,z) = 0$ so h(y,z) = f(z), a pure function of z. So now we know:

$$\phi(x,y,z) = xy^2z^3 + f(z)$$

Now check the partial with respect to z:

$$\phi_z = 3xy^2z^2 + f'(z) = 3xy^2z^2$$

This says f'(z) = 0 so f(z) = C, a constant. This tells us

$$\phi(x,y,z) = xy^2z^3 + C$$

This takes the guess work out of it and also tells you any C works. The simplest answer is when C = 0.

8. Apr 29, 2010

### Urmi Roy

Hi everyone....I have been doing this kind of sums myself...but in college,we were taught a different method......they find out

$$\phi_x$$,$$\phi_y$$,$$\phi_z$$ individually,then integrate each..then we check for the common terms....when we collect all the common terms to get the final function.

This seems to be equivalent to the process described by LCKurtz,but if we apply a similar procedure (as taught in my college)to find out the harmonic conjugate of an analytic function,it doesn't work!!
(What is puzzling me is,inspite of using a similar procedure for a similar situation,we don't get required results.)

Say,for example we have u=1/2(log(x^2+y^2))...we have to find its harmonic conjugate(i.e v)
....now,we get ux,uy (the partial derivatives)...and we have vx=-uy ...(eqn1)and vy=ux....(eqn 2)....so if we integrate each of the equations,and then compare,we don't get the correct value of v!!
(The answer is v=tan inverse of (y/x), but if we integrate equation 1,we get tan inverse of (x/y)).

9. Apr 29, 2010

### LCKurtz

I'm sorry to hear that. That method, while it will work the easy ones, is not adequate for the more complex ones and can lead to misleading or wrong answers.

I think you get -arctan(x/y) don't you?

Hint: What happens if you differentiate arctan(x) + arctan(1/x)? What can you conclude?

10. Apr 30, 2010

### Urmi Roy

well,you get 0...which means the function arctan(x) + arctan(1/x) is constant (value is pi by 2).

But then in the sum,we get v=arctan(y/x) -arctan(x/y) ....which can't be simplified to give a proper answer.

11. Apr 30, 2010

### LCKurtz

It means that arctan(y/x) and -arctan(x/y) differ by a constant so they have the same partial derivatives. One answer is as good as the other.

12. Apr 30, 2010

### Urmi Roy

Oh... I get it! However,while working it out in the exam,I'd better be careful...perhaps if I follow the method that you originally mentioned,instead of integrating both equations,it might prevent the teacher getting confused,