Derivatives of a radial function

  • Thread starter mmwave
  • Start date
211
2
I have a scalar function of position only, V(r) where r is the position vector (x, y,z). Since V depends only on position, is it true to say that

dV/dx = dV/dy = dV/dz ?
(these should be partial derivatives)

I am trying to show L = r x [nab] commutes with any radial function V(r) meaning that for any function f

r x [nab] ( V(r) * f) - V(r) * r x [nab] f = 0

most of the terms cancel out but I am left with

(y dV/dz - x dV/dz -zdV/dy + xdV/dy + z dVdx - y dV/dx) * f

or (y (dV/dz - dV/dx) + x( dV/dy - dVdz ) +z( dV/dx - dV/dy ) ) * f

If the partial derivatives cancel then I am done. If not I have no clue how to continue but I do know the final answer must be zero.
 
837
1
Originally posted by mmwave
I have a scalar function of position only, V(r) where r is the position vector (x, y,z). Since V depends only on position, is it true to say that

dV/dx = dV/dy = dV/dz ?
(these should be partial derivatives)
What, all the Cartesian components of the gradient of V should equal each other, for an arbitrary function V? If that's what you mean, then no.
 
837
1
By the way, L does not commute with V unless V is spherically symmetric, i.e. V ≡ V(r) where r is the distance from the origin (not the position vector).
 
211
2
Re: Re: derivatives of a radial function

Originally posted by Ambitwistor
By the way, L does not commute with V unless V is spherically symmetric, i.e. V ≡ V(r) where r is the distance from the origin (not the position vector).
Yes, thank you. That is correct and in fact I think the book means that V(r) is a central potential so the spherically symmetric is true in this case. Does that mean the answer is yes - the spherical symmetry means dV/dx = dV/dy = dV/dz ?
 
837
1
Re: Re: Re: derivatives of a radial function

Originally posted by mmwave
Yes, thank you. That is correct and in fact I think the book means that V(r) is a central potential so the spherically symmetric is true in this case. Does that mean the answer is yes - the spherical symmetry means dV/dx = dV/dy = dV/dz ?
No. (Consider a central potential at a point on the x axis. At that point, dV/dx will generally be nonzero in the x direction since it is the radial direction, but dV/dy = dV/dz = 0 since they are perpendicular to the radial direction.)

It's hard to incorporate the central constraint in Cartesian coordinates, which is why everyone does this calculation in spherical coordinates.
 
211
2
Thanks for keeping me from seizing the easy incorrect answer. When I took the physics out of the problem to simplify it I took out an essential element.

What I really had was
x dV/dy - y dV/dx and similar terms so you can have cancelation without having dV/dy = dVdx = 0. Saddly I didn't see the trick but using dV/dy = dV/dr * dr/dy and similar for x gets the terms to cancel. I hope I don't forget this trick.
 

Related Threads for: Derivatives of a radial function

Replies
4
Views
11K
  • Posted
Replies
2
Views
3K
Replies
2
Views
28K
Replies
1
Views
2K
Replies
3
Views
3K
Replies
5
Views
5K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top