Derivatives of a radial function

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)

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.

Ambitwistor

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.

Ambitwistor

By the way, L does not commute with V unless V is spherically symmetric, i.e. V &equiv; V(r) where r is the distance from the origin (not the position vector).

mmwave

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 &equiv; 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 ?

Ambitwistor

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.

mmwave

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.

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