problems related to gradient of scalar field

[PG1995]

Hi

Could you please help with these two problems, Problem #1 and Problem #2? Thank you.

Regards
PG

 

[steveB]

For Q1

I would read the word "then" as "then let", which would mean the same as "suppose" in this context.

Yes you do make sense.

Note that this proof is very essential to understanding potential theory in electromagnetism. We know that voltage potential is related to electric field through the gradient operator as follows.

[latex] \vec E = -\nabla V[/latex]

Hence, electric field direction is always perpendicular to surfaces of constant potential.

 

[steveB]

For Q2

The direction AP is just the direction from the point A to the point P, and this is the directio of the vector R formes by subtractiing the position vector to the point A from the position vector to the point P.


One way to interpret is to compare to the one dimensional case. Here, the gradient reduces to d/dx, and the vector R is (x-a). Hence, it is clear that the derivative of x-a is simply 1 at all points in the space. In 2 or 3 dimensions, this still works, only the gradient is a vector with more than one component. Still, the change in the position with respect to the position is always 1, even in this vector form. This works relative to any point in space because derivatives of constants (e.g. a, b and c) is zero.

 

[MrAl]

Hi,

I agree that the wording in these papers we are seeing here lately is anything but choice. I'd like to see where this stuff came from.

 

[PG1995]

Thank you.

Re: Post #3

One way to interpret is to compare to the one dimensional case. Here, the gradient reduces to d/dx, and the vector R is (x-a). Hence, it is clear that the derivative of x-a is simply 1 at all points in the space. In 2 or 3 dimensions, this still works, only the gradient is a vector with more than one component. Still, the change in the position with respect to the position is always 1, even in this vector form. This works relative to any point in space because derivatives of constants (e.g. a, b and c) is zero.

Here is a follow-on question. My original question from the post #1 was this.

@MrAl: Here is the book I'm using.

Regards
PG

 

[steveB]

For your follow on question, I think it is fair to say that A should be fixed. In fact, the text states that explicitly in the first sentence.

 

[PG1995]

For your follow on question, I think it is fair to say that A should be fixed. In fact, the text states that explicitly in the first sentence.

Yes, the text is saying that if you have one point fixed and the other non-fixed then the magnitude of the grad R is always going to be unity. So, I didn't miss it. But, here too, the grad R having magnitude of unity seems somewhat counter intuitive. Just think that the point P has been moved little upward while the A being fixed. According to the text the magnitude of the grad R is still going to be unity as was before (when the P has not been moved upward)! So, where am I having it wrong? Kindly guide me. Thank you.

Regards
PG

 

[steveB]

So, where am I having it wrong? Kindly guide me. Thank you.

My guess is that you just aren't visualizing the correct picture. Perhaps you are thinking of the distance change as a percentage of the distance you are already away by. If you are 10 m away and move 1 m, then the percentage is bigger than if you are 100 m away and move 1 m. However, in both cases you have moved a distance of 1 m. So, the gradient of the separation-distance (remember distance is a scalar and not a vector) is a vector, and the magnitude of that vector is independent of the separation distance, since gradient deals with actual changes and not relative changes. Gradient points in the direction of greatest change, which is the direction away along the line connecting the points. We evaluate the magnitude of the gradient in that direction by traversing an infinitesimal distance in that direction and looking at the change in the function (in this case, the function is the scalar distance R), for that change in the coordinates. A unit magnitude displacement in the coordinate vector along that direction results in a distance change equal to the hypothetical displacement we made to evaluate the gradient. Can you see the picture now?

 

[PG1995]

Once again, my apologies for asking you this again but seriously I don't it.

Let's give it another try. This is about Problem #2 from the post #1. Please have a look here. Thanks a lot for your patience and help.

Regards
PG

 

[steveB]

So in this description you say that gradient is the directional derivative which is not true. The directional derivative equals the gradient dotted with a unit vector in the direction you wish to evaluate the rate of change.

Perhaps this is the cause of the confusion?

 

[PG1995]

So in this description you say that gradient is the directional derivative which is not true.

Yes, it was incorrect. The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is that rate of increase. Thanks.

Regards
PG