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Offline DoctorBeaver

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« Reply #25 on: 30/07/2007 20:52:34 »
Heh.  I assumed since the question started off "I'm trying to solve the Schrodinger equation in 3-D for the Hydrogen atom..." that he wanted a technical answer.  ;)

I wasn't complaining. It was meant to convey the fact that you totally lost me. Your answer seems very comprehensive and I wish I could understand it.
 

Offline Mr Andrew

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« Reply #26 on: 01/08/2007 02:34:14 »
Thanks for the help, and yes, I did want a technical answer.  I'm right beside the Doctor doing mental push-ups but I will work it out eventually.  I'm probably in way over my head with this but it's summer and I've nothing better to do.

I understand where you got the scaling factors from but how do they fit into the equation for the gradient?  It looks like you have the derivative with respect to r added to itself there.  Is it because the gradient only act on positions?
 

Offline JP

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« Reply #27 on: 01/08/2007 15:19:26 »
The gradient is supposed to "live" in position space, since it tells you in what spatial direction the function is increasing fastest. 

I'm not sure I follow your question about adding the derivative with respect to r to itself, but I probably should define the notation I used.  Bold text was indicating unit direction vectors, and what the gradient will do is tell you how much a function changes with respect to one of its arguments, and then you have to multiply that by a unit vector so that it tells you in what direction it's changing.  Hence you get a term
r∂/∂r
which means "find out how much the function is changing as we change r (i.e. take the derivative), and then make this value point in the direction it's changing (i.e. multiply it by the r unit vector).  You don't need to scale it since it's already measuring how the function changes in space, since r is a spatial coordinate.

If you have a function in spherical coordinates, you'll want to do the same things in the θ and φ directions.  The problem is that these aren't measuring spatial coordinates.  You start off using the same form for your derivative terms θ∂/∂θ, where the boldface indicates a unit vector pointing either latitudinally or longitudinally from r.  However, you couldn't simply add these terms to r∂/∂r since they're measuring angles, not spatial coordinates [you can check units here: the unit vectors are unitless, ∂/∂r is 1/length, and the ∂/∂θ and ∂/∂φ are 1/angle (which is unitless)].  To get these angular derivatives to measure how a function is changing in spatial position, you need to throw those scaling factors in.

Here's a couple of links that explain things as well (they show where the scaling factors come from, though they're pretty math-intense):
http://mathworld.wolfram.com/Gradient.html
http://mathworld.wolfram.com/Laplacian.html
http://mathworld.wolfram.com/CurvilinearCoordinates.html
 

Offline Mr Andrew

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« Reply #28 on: 01/08/2007 20:43:45 »
Yes, thank you.  I am well aquainted with wolfram already and the explainations on their site are helpful, sometimes, but rarely do they include the origin of things such as these scaling factors.  They explain what they are but not where they come from.

As for my question about adding the ∂/∂r to itself three times.  It is in reference to the fact that (1/r) ∂/∂θ ≡ ∂/∂r ≡ (1/r sinθ) ∂/∂φ.  Ah, hold on... I feel an epiphany coming!  Grad[v] = vd/dr = rd/dr + θd/dr + φd/dr and the only way to make any sense of the last two terms is to make the substitutions dr = rdθ and dr = r sinθ dφ so that the unit vectors and independent variables match!  Does that make sense or am I losing my mind?
 

Offline JP

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« Reply #29 on: 01/08/2007 22:02:22 »
I've dug up another way of explaining this that should be clearer.  I think you're almost there, but to be honest I'm getting confused on notation at this point.  However, what I'll give you now should be a pretty easy-to-follow explanation.  First, let s measure a spatial position.  (r is now reserved for a unit vector pointing radially, not a spatial position).

If you want to know how s varies when you change the spherical variables a bit, you get:

ds=rdr+θ r dθ+φ r sinθ dφ

Now take a function of the spatial coordinate: f(s)
df=dr(∂f/∂r)+dθ(∂f/∂θ)+dφ(∂f/∂φ)
is a measure of how much f changes as the spherical coordinates change.  This has to be equivalent to how much f changes as you move a little bit in space.  The dot product ds•Grad[f] (the directional derivative) measures how f changes in space as you move by a small amount ds, so:

df=ds•Grad[f].

You already know how df and ds relate to the spherical coordinates, so you can solve for how Grad[f] must relate to them.
Match up terms in dr and likewise for dθ and dφ (you can do this since you could set dr, dθ and dφ to zero independently of each other):

dr (∂f/∂r) = dr r•Grad[f]

so the r component of Grad[f] should just be (∂f/∂r).

Matching up the other two parts:

dθ (∂f/∂θ) = r dθ θ•Grad[f]

so the θ component of Grad[f] must equal (1/r)(∂f/∂θ),

dφ (∂f/∂φ) = r sinθ dφ φ•Grad[f]  →  1/(r sinθ) (∂f/∂φ) =  φ•Grad[f]

 

Offline Soul Surfer

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Quantum Physics
« Reply #30 on: 01/08/2007 23:44:03 »
I agree that changing the co-ordinates can be very hard work because it was one of the tings we did during my physics degree course many years ago.  There are lots of steps and the equations got so big that each step tended to cover a whole page of a notebook.  confocal elliptical co-ordinates were the worst!
 

Offline ramphysix

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« Reply #31 on: 11/08/2007 12:02:48 »
you can find lots of good sources about quantum on the following link..
newbielink:http://www.whatusearch.net/Science/Physics/Quantum_Mechanics/ [nonactive]
 

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« Reply #31 on: 11/08/2007 12:02:48 »

 

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