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Gradient is only the vector pointing at the steepest slope. Or it can be said as the change in the function due to a small change in x, in the direction of x + the change in the function due to a small change in y, in the direction of y, + the change in the function due to a small change in z, in the direction of z.

In the spherical co-ordinate system, what's wrong with saying that the gradient is the change in the function due to a small change in r, in the direction of r + the change in the function due to a small change in theta, in the direction of theta + the change in the function due to a small change in phi, in the direction of phi.

Wouldn't that point in the direction of steepest ascent as well?

(P.S- I know what the actual formula is, and I also understand how it's derived but I just don't understand why it's done that way!)

underdog
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  • The key issue is: how do you measure which direction has the "steepest" slope? You choose the direction in which the function value decreases the most when you move a small fixed distance in that direction. Now, how much distance do you move if you change $r$ by a small amount, if you change $\theta$ by a small amount, or if you change $\phi$ by a small amount? Is it the same for all three? –  Dec 03 '19 at 16:06
  • Taking the derivative of a function with respect to $r$ gives you units of the function divided by distance, which derivatives with respect to angles give you just the units of the angles. If you then converted back to the Cartesian coordinate system you would get a clash of units because they would not have all the same units. So if we defined spherical gradient in your way, it wouldn't agree with the Cartesian way. The result is the gradient starts at a formal coordinate system independent definition. – Kraigolas Dec 03 '19 at 16:14
  • You claim in your question to know this definition, because you say you know how to derive the formula. But the gradient wasn't defined without a coordinate system and from there it just so happens to give a nice result in Cartesian, but the Cartesian definition doesn't define gradient – Kraigolas Dec 03 '19 at 16:15
  • See if this question (about polar coordinates in the plane) helps: https://math.stackexchange.com/questions/1867392/intuitive-way-to-understand-polar-coordinate-gradient – Hans Lundmark Dec 03 '19 at 16:24
  • @HansLundmark That was a very good explanation in 2 dimensions. But if I consider the same example in 3D. A slight change in θ does result in a change of rdθ given Φ is constant. But for a slight change is Φ the change should also be rdΦ given θ is constant. Which is not what we see in the real formula(The term rsinθ) appears. – underdog Dec 03 '19 at 17:17
  • No, if you change $\phi$ by $d\phi$ (keeping $r$ and $\theta$ constant), the point moves along a circle of radius $r \sin \theta$. (The radius is the line segment straight out from the $z$ axis to the point, not the line segment from the origin to the point.) So the point does move a distance $r \sin \theta , d\phi$. – Hans Lundmark Dec 03 '19 at 17:29
  • An extreme case is if the point lies on the $z$ axis ($\theta=0$ or $\theta=\pi$). Then the point doesn't move at all if $\phi$ changes, and this agrees with the fact that $r \sin\theta=0$ in this case. – Hans Lundmark Dec 03 '19 at 17:31
  • @HasnLundmark I just realized we meant opposite things by theta and phi. Now it's clear. Thanks a lot mate. – underdog Dec 03 '19 at 17:51
  • Yes, there are two opposite conventions, unfortunately... But anyway, you're welcome! I'll vote to close this question as a duplicate then, since you seem to have got the answer you wanted. – Hans Lundmark Dec 03 '19 at 18:49
  • Possible duplicate of Intuitive way to understand Polar Coordinate Gradient – Hans Lundmark Dec 03 '19 at 18:49

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