1. Nov 20, 2012

### vish22

Let's say we have a scalar function U in terms of r,theta and phi.
why cannot this be the gradient at any point P(r,theta,phi)-
partial of U wrt. r in the direction of r+partial of U wrt. theta in direction of (theta)+partial of U wrt. phi in the direction of (phi)?

2. Nov 20, 2012

### arildno

Do you see that in order for your terms to be of the same physical dimension, you need length factors in the denominators for the two angularly differentiated terms?

3. Nov 20, 2012

### vish22

I'm sorry but I'm not able to visualize what you are saying.
I thought that the components of the gradient are concerned with the rate of change of the potential in each direction?

Last edited: Nov 20, 2012
4. Nov 20, 2012

### arildno

Nope.
In general, you "divide" with the differential arc length associated with each variable, in the proper direction. "dx", "dy" are differential arc lengths, as is, for 2-D polar coordinates, "dr" and "rd(theta)"
Same for spherical coordinates in 3-D

5. Nov 20, 2012

### vish22

Ok,I think I got it.So its like the gradient components for the angular directions are the rates of changes of the potential U "in the direction of" theta with the arc length being rd(theta) and phi with the arc length bring rsin(theta)d(phi)-hence bringing in the denominator r and rsin(theta) respectively??

Last edited: Nov 20, 2012
6. Nov 20, 2012

### arildno

"If this is the case,can you please support your answer of the gradient components NOT being concerned with the rate of change in the basis vector's DIRECTION?"
It is.
You walk a little LENGTH along a curve in the variable's direction.

Thus, in 2-D polars, you go the little distance "dr" along the r-vector (i.e, some ray from the origin, on which the angular variable is constant), measuring thereby the rate of change of the function, by dividing its change by the length "dr".

Similarly, you walk a little rd(theta) length along a circle (where "r" is constant), in order to measure the rate of change when the angular value changes between the two positions of the circle.

7. Nov 20, 2012

### vish22

So the gradient in polar should ACTUALLY look like this?
∇U(r,α,β)=rδU/δr+αδU/(rδα)+βδU/(rsinαδβ)

But the custom is the take r and rsinα outside the brackets because they have constant magnitude in the differential lengths-rδα and rsinαδβ-in the direction of α and β respectively?

8. Nov 20, 2012

### arildno

Well, yes.
But REMEMBER.
Those factors are CONSTANTS with respect to the given variable, so they can perfectly well be drawn out from the brackets.

9. Nov 20, 2012

### vish22

Yeah I got it,thanks a lot really.That was big help for me.