# Excerpt from 3-D Schrodinger Eqaution...I'm stuck...

• Jul 26th 2008, 08:58 PM
Chris L T521
Excerpt from 3-D Schrodinger Eqaution...I'm stuck...
I'm solving the Three Dimensional Schrodinger Equation, and there is a little part that I'm stuck on...

I'm starting with this DE:

http://img.photobucket.com/albums/v4...e2143a3-12.jpg

and I have to show that if I change the independent variable from theta to z, using the substitution

http://img.photobucket.com/albums/v4...e2143a3-13.jpg

and the identity

http://img.photobucket.com/albums/v4...e2143a3-14.jpg

I should show that the DE has the form:

http://img.photobucket.com/albums/v4...e2143a3-15.jpg

My problem is going from:

http://img.photobucket.com/albums/v4...e2143a3-16.jpg

I'm probably overlooking something pretty minor, but I'd appreciate any help! (Sun)

--Chris
• Jul 26th 2008, 09:08 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
I'm solving the Three Dimensional Schrodinger Equation, and there is a little part that I'm stuck on...

I'm starting with this DE:

http://img.photobucket.com/albums/v4...e2143a3-12.jpg

and I have to show that if I change the independent variable from theta to z, using the substitution

http://img.photobucket.com/albums/v4...e2143a3-13.jpg

and the identity

http://img.photobucket.com/albums/v4...e2143a3-14.jpg

I should show that the DE has the form:

http://img.photobucket.com/albums/v4...e2143a3-15.jpg

My problem is going from:

http://img.photobucket.com/albums/v4...e2143a3-16.jpg

I'm probably overlooking something pretty minor, but I'd appreciate any help! (Sun)

--Chris

Which part are you having problems with? I got it, and we know that you are a thousand times better at DE's than me, so I must be overlooking something to. The 1-z^2 is obvious right? and the rest...so which part?
• Jul 26th 2008, 09:18 PM
Chris L T521
Quote:

Originally Posted by Mathstud28
Which part are you having problems with? I got it, and we know that you are a thousand times better at DE's than me, so I must be overlooking something to. The 1-z^2 is obvious right? and the rest...so which part?

Well...*ahem*...

http://img.photobucket.com/albums/v4...e2143a3-17.jpg

--Chris
• Jul 26th 2008, 10:39 PM
mr fantastic
Quote:

Originally Posted by Chris L T521

Instead of differentiating a second time wrt theta, differentiate wrt z and use the chain rule. Then you have d/dz [....] times dz/dtheta = d/dz [....] times (- sin theta).

And sin^2 theta = 1 - cos^2 theta = (1 - z^2) .....
• Jul 26th 2008, 10:57 PM
ThePerfectHacker
Quote:

Originally Posted by Chris L T521

No wonder you are confused. I am confused. Typical physics "math" garbage what does "change variable from cos@ to z" is even supposed to mean? It is stuff like this made me spend much time translating physics "math" into formal understanding. Anyway, I guess I know what you have to do - whether it actually makes any sense I live it to you to decide, because it makes no sense to me (eventhough I have experience in trying to understand what the problem is asking).

Open up the derivative,
(-1/sin@)*[cos@*(dT/d@)+sin@*(d^2T/d@^2)] ----- [1]
Now,
dT/d@ = (dT/dz)*(dz/d@)=(dT/dz)*(-sin @)
And,
d^2T/d@^2 = d/d@(dT/d@)=d/d@(- dT/dz*sin @)=
-d/d@(dT/dz)*sin@ - dT/dz*cos @= d^2T/dz^2 * sin @ - dT/dz*cos@
Go back to [1] and substitute (and divide out),
-cos@*dT/dz+d^2T/dz^2*sin@ - dT/dz*cos@
Use the fact that z=cos @ to get,
d^2T/dz^2 *sin@ - dT/dz*(1+z)
And that is supposed to give the thing you are trying to show.

I probably made a mistake somewhere but it is late and I cannot check it now. See if that mess above makes any sense?
• Jul 26th 2008, 11:13 PM
Chris L T521
Quote:

Originally Posted by mr fantastic
Instead of differentiating a second time wrt theta, differentiate wrt z and use the chain rule. Then you have d/dz [....] times dz/dtheta = d/dz [....] times (- sin theta).

And sin^2 theta = 1 - cos^2 theta = (1 - z^2) .....

Quote:

Originally Posted by ThePerfectHacker
No wonder you are confused. I am confused. Typical physics "math" garbage what does "change variable from cos@ to z" is even supposed to mean? It is stuff like this made me spend much time translating physics "math" into formal understanding. Anyway, I guess I know what you have to do - whether it actually makes any sense I live it to you to decide, because it makes no sense to me (eventhough I have experience in trying to understand what the problem is asking).

Open up the derivative,
(-1/sin@)*[cos@*(dT/d@)+sin@*(d^2T/d@^2)] ----- [1]
Now,
dT/d@ = (dT/dz)*(dz/d@)=(dT/dz)*(-sin @)
And,
d^2T/d@^2 = d/d@(dT/d@)=d/d@(- dT/dz*sin @)=
-d/d@(dT/dz)*sin@ - dT/dz*cos @= d^2T/dz^2 * sin @ - dT/dz*cos@
Go back to [1] and substitute (and divide out),
-cos@*dT/dz+d^2T/dz^2*sin@ - dT/dz*cos@
Use the fact that z=cos @ to get,
d^2T/dz^2 *sin@ - dT/dz*(1+z)
And that is supposed to give the thing you are trying to show.

I probably made a mistake somewhere but it is late and I cannot check it now. See if that mess above makes any sense?

It finally clicked. Thanks TPH and Mr. Fantastic! (Sun)

--Chris