Verify Integration and differentiation of a vector.

[yungman]

I want to verify simple integration and differentiation of a vector and verify that the direction of the derivative and integral of a vector is not the same direction of the original vector. Let:

\vec A = \hat x A_x + \hat y A_y + \hat z A_z

1) Differentiation:

\frac {d \vec A}{dx} = \hat x \frac {d A_x}{dx} \;+\; \hat y \frac {d A_y}{dx} \;+\; \hat z \frac {d A_z}{dx}


2) Integration:

\int\int\int \vec A \;dxdydz \;\;= \;\; \hat x \int\int\int A_x \;dxdydz \;\;+\;\; \hat y \int\int\int A_y \;dxdydz \;\;+\;\; \hat z \int\int\int A_z \;dxdydz

eg. If

\vec A = \hat x x \;+\; \hat y y \;+\; \hat z z \;\Rightarrow\; \int\int\int \vec A \;dxdydz \;\;=\;\; \hat x (\frac 1 2 x^2yz +C_1)\;\;+\;\; \hat y (\frac 1 2 xy^2z +C_2)\;\;+\;\; \hat z (\frac 1 2 xyz^2 +C_3)

For simplification, I did not perform a true volume integral that have limits on x, y and z.

Therefore the direction of the derivative and integral of a vector is not the same direction as the original vector.

 

[tiny-tim]

hi yungman!

\frac {d \vec A}{dx} = \hat x \frac {d A_x}{dx} \;+\; \hat y \frac {d A_y}{dy} \;+\; \hat z \frac {d A_z}{dz}

no, \frac {d \vec A}{dx} = \hat x \frac {d A_x}{dx} \;+\; \hat y \frac {d A_y}{dx} \;+\; \hat z \frac {d A_z}{dx}

but anyway i don't understand why you expect a triple integral to be the inverse of a single derivative

 

[yungman]

hi yungman!


no, \frac {d \vec A}{dx} = \hat x \frac {d A_x}{dx} \;+\; \hat y \frac {d A_y}{dx} \;+\; \hat z \frac {d A_z}{dx}

but anyway i don't understand why you expect a triple integral to be the inverse of a single derivative

Hi Tiny-Tim, thanks, I was just typing too fast. I corrected that already.

I am not expecting triple integral to be the inverse of a single derivative, I just want to verify the direction of the derivative and integral of a vector is not the same as the direction of the original vector A.

 

[Redbelly98]

For the triple integral, in general, the vector can have different directions at different points within the integration volume. So it doesn't make sense to talk about the direction of the vector. The integral will give a weighted average of the directions the vector has within the integration volume.

As for the derivative (as defined in Post #1), consider A = (x, x, 0). What is the direction of A? What is the direction of the derivative?

 

[yungman]

For the triple integral, in general, the vector can have different directions at different points within the integration volume. So it doesn't make sense to talk about the direction of the vector. The integral will give a weighted average of the directions the vector has within the integration volume.

As for the derivative (as defined in Post #1), consider A = (x, x, 0). What is the direction of A? What is the direction of the derivative?

Thanks for the reply, I am referring to the vector A has different direction than the volume ingtegral of vector A at ANY point.

The reason I ask was mainly because of the discussion of finding vector magnetic potential for a distribution of current. I want to understand why vector A is not necessary the same direction as vector J in this equation:

1) \vec A \;=\; \frac {\mu_0}{4\pi}\int_{v'} \frac {\vec J}{|\vec r - \vec r_0|} dv'




2)If A = (x, x, 0).

\frac{\partial \vec A}{\partial x} = (1,0,0)

So the direction of A is not necessary same as the direction of it's derivative.

 

[Redbelly98]

Thanks for the reply, I am referring to the vector A has different direction than the volume ingtegral of vector A at ANY point.

The reason I ask was mainly because of the discussion of finding vector magnetic potential for a distribution of current. I want to understand why vector A is not necessary the same direction as vector J in this equation:

1) \vec A \;=\; \frac {\mu_0}{4\pi}\int_{v'} \frac {\vec J}{|\vec r - \vec r_0|} dv'

Hmmm, my suspicion is that if J is continuous, then some J within the integration volume would have the same direction as A. But I could be wrong. (EDIT: I believe I am wrong.)

If J need not be continuous, then I can come up with a fairly simple counterexample:

You're obviously talking about the magnetic vector potential. Consider a square loop of current in the xy plane, oriented so that J always points in the (±1,0,0) or (0,±1,0) direction:

[PLAIN]http://www.lei.ucl.ac.be/~matagne/ELECMAGN/SEM05/S05F64.GIF[/INDENT][/INDENT]

However, in the vicinity of the corners of the square, A will point at roughly 45 degrees to the J's, or the (±1,±1,0) directions.

2)If A = (x, x, 0).

\frac{\partial \vec A}{\partial x} = (1,0,0)

So the direction of A is not necessary same as the direction of it's derivative.

Correct.​

 

[yungman]

Hmmm, my suspicion is that if J is continuous, then some J within the integration volume would have the same direction as A. But I could be wrong. (EDIT: I believe I am wrong.)

If J need not be continuous, then I can come up with a fairly simple counterexample:

You're obviously talking about the magnetic vector potential. Consider a square loop of current in the xy plane, oriented so that J always points in the (±1,0,0) or (0,±1,0) direction:

[PLAIN]http://www.lei.ucl.ac.be/~matagne/ELECMAGN/SEM05/S05F64.GIF[/INDENT][/INDENT]

However, in the vicinity of the corners of the square, A will point at roughly 45 degrees to the J's, or the (±1,±1,0) directions.


Correct.​

Thanks for your time.

From what we have so far, I think the integral of a vector is not "usually" in the same direction of the original vector. When you talk about some special cases, yes, they cound be in the same direction, but the main point is that they are "usually" not in the same direction and one cannot count on any directional relation.

All these are just to proof that there is no easy way to find the direction of the vector magnetic potential just by the given B. One can only derive A from B only in some cases that have particular symetry where you can apply something like the Gauss law or Amperes' law like int E and B resp.​