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

1. ## 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:

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

and the identity

I should show that the DE has the form:

My problem is going from:

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

--Chris

2. 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:

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

and the identity

I should show that the DE has the form:

My problem is going from:

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

--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?

3. 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*...

--Chris

4. Originally Posted by Chris L T521
Well...*ahem*...

--Chris
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) .....

5. Originally Posted by Chris L T521
My problem is going from:

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?

6. 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) .....
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!

--Chris