# Curl and Force

#### Barley

I know that if the curl is zero then the Force is conservative, and that the curl is the cross product of the gradient and the Force function. What I'm having, I think, is a math problem;

Probably not able to get the partial derivatives of the function right...

Ok, first off, have a Force that is a function of position: F = ax, where a is a constant--I know the solution to the curl is a in the direction of z; not conservative. But, when I set up my cross product matrix I have for the partials in the z and z direction as zero, and the Fy and Fz too. Now this leaves me with only a and ax in the i column of my matrix-- that gives zero, so I'm doing something wrong.

Second I have a Force function that I know is conservative:

F = (yz^2)i + (xz^2 + 2)j + (2xyz -1)k

when I take the partials of the function I get,

Fx = z^2 + 2y
Fy = Z^2 + 2xy
Fz = 2yz + 2xz +2xy

Now, when I place these partials in the top row and place the Function components in the bottom row-- I do not get a zero cross product.

Please tell me what I'm doing wrong.

Thanks

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#### ehild

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Barley said:
I know that if the curl is zero then the Force is conservative, and that the curl is the cross product of the gradient and the Force function.

You know it wrong. The curl of a vextor F is the cross product of the "nabla vector" with F, that is

$$curl{\vec F}= \nabla \times \vec F =(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z})\vec i+ (\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial_x} )\vec j +(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial_y}) \vec k$$

ehild

#### Barley

I]The gradient is a vector operator denoted and sometimes also called Del or nabla. It is most often applied to a real function of three variables , and may be denoted ...

I got the above definition from mathworld. difference between the gradient and nabla vector aside I have the same result as echild for the curl.

What my problem is, I think is with the partials of the origional function. Notice if I put my result for the partials in to the curl eqn., I get 0 for the function F = ax, and non zero for the other function I mentioned in my post.

Any help is appreciated.

#### arildno

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First of all, your function F=ax is a scalar function; you cannot calculate the curl of it.
Secondly, using ehild's expression, you do get 0 curl.

#### Barley

F = ax is a scalar representing the force in on direction -- x only.

Such a force is not conservative and there's got to be a way to come up with the expression for a curl in the z direction. Now, this is probably no help but the gradient of a scalar is a vector. I know its simple but I can't get it.

I have a example of a particle traveling in a closed loop. When the particle governed by F = ax travels in the y direction only the work = 0 because the force is in the x direction, now when the particle moves in the x direction from the y axis to a poing xo, yo, the work on it is also zero because the work is perpencicular to the motion so work to move from (0,y) to (xo, yo) is zero. Last when the particle moves back to the origin from (xo, yo) there is an x and y component to the movement--taken as unit vecors both the x and y components are negative so work done is negative-- indicating a loss in KE. Having F = ax have a non zero curl represents that the force is not conservative. This meaning that all one directional motion is not conservative.

Did anyone see what I was doing wrong with my second function ?

thanks!

#### arildno

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Do you mean $$\vec{F}=ax\vec{i}$$ where $$\vec{i}$$ is the unit vector in the "x"-direction?
In that case, $$\vec{F}$$ is a conservative field, since it is the gradient of the scalar potential $$\phi=\frac{a}{2}x^{2}$$

#### Barley

I really appreciate your help but I'm certain that the result I'm looking for is that F = ax ( a fprce that is a function of position in this case) is not a conservative force. I know you can take the integral to get the potential, but only if as you said this Force is conservative.
I beginning to see how curl applies to fields and this is making things less clear to me.
I've sent a message to my professor and I'll post the reply. thanks again !

#### arildno

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You cannot take the curl of a scalar function!

#### Barley

Then its obvious I'm down the wrong path. I'll wait for the professor's response because its likely I've not got the statement of the problem correct.

I will post again when I have the problem and result correct.

#### arildno

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Barley said:
Then its obvious I'm down the wrong path. I'll wait for the professor's response because its likely I've not got the statement of the problem correct.

I will post again when I have the problem and result correct.

Indeed, I think this is the case with your first problem.

#### ehild

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Barley said:
F = ax is a scalar representing the force in on direction -- x only.
The force is always a vector quantity. Your F=ax might be

$$\vec F = ax \vec i$$.

Such a force is not conservative and there's got to be a way to come up with the expression for a curl in the z direction.
No. This is a conservative force.

Now, this is probably no help but the gradient of a scalar is a vector.
That is true, the gradient of a scalar is a vector, and the gradient of a function U=ax

$$grad (ax) = a \vec i$$

But the force is a vector, and your "ax" function might be potential instead of force.

I have a example of a particle traveling in a closed loop. When the particle governed by F = ax travels in the y direction only the work = 0 because the force is in the x direction, now when the particle moves in the x direction from the y axis to a poing xo, yo, the work on it is also zero because the work is perpencicular to the motion so work to move from (0,y) to (xo, yo) is zero. Last when the particle moves back to the origin from (xo, yo) there is an x and y component to the movement--taken as unit vecors both the x and y components are negative so work done is negative-- indicating a loss in KE. Having F = ax have a non zero curl represents that the force is not conservative. This meaning that all one directional motion is not conservative.
Wrong. The work is

$$W=\int_A^B\vec F\cdot \vec {dr}= \int_A^B(F_xdx+F_ydy)$$

when the motion takes place in the (x,y) plane.
As the force has got only x component the work is zero from (0,0) to (0,yo) but it is not zero from (0,yo) to (xo,yo) as the force and displacement are parallel.

$$W|_{(0,xo)}^{(xo,yo)}=\int_0^{xo}(axdx)=x_o^2/2$$

On the way back from (xo,yo) to (0,0)

$$W|_{(xo,yo)}^{(0,0)}= \int_{xo}^0 axdx =-x^2/2$$

The work along your closed path is zero. And the curl is also zero.

Did anyone see what I was doing wrong with my second function ?

Now, when I place these partials in the top row and place the Function components in the bottom row-- I do not get a zero cross product.
You should put the unit vectors i, j, k in the first row of the determinant, the components of the nabla vector (or gradient operator if you prefer that name) in the second row and the components of the force in the bottom row. You must get zero curl.

ehild

#### Barley

No, I'm having trouble with the second function too. I'm looking in a calc. book and trying to follow the text example.

Let me ask you this: If I have a function , F(x,y,z) = (x^3 y^2 z)i + (x^2 z)j +(x^2 y)k

Then the fx(partial with respect to x )is, fx = 3x^2 y^2 z +2xz + x^2xy ?
I have to take the partial of each component with respect to x, yes ?

If this were not true than the j term would have a fy = 0.

Thanks

#### arildno

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It seems you have very big troubles with understanding what vector notations mean, Barley!
Let's take the general vector function:
$$\vec{F}=F_{x}(x,y,z)\vec{i}+F_{y}(x,y,z)\vec{j}+F_{z}(x,y,z)\vec{k}$$
Now, $$F_{x},F_{y},F_{z}$$ are notations for the SCALAR vector components of $$\vec{F}$$; they do not signify partial derivatives of $$\vec{F}$$!!!
Those would be $$\frac{\partial\vec{F}}{\partial{x}},\frac{\partial\vec{F}}{\partial{y}},\frac{\partial\vec{F}}{\partial{z}}$$

Thus, for example, $$\frac{\partial{F}_{z}}{\partial{x}}$$ is the partial derivative of the scalar component function $$F_{z}$$ with respect to x.
Get it?

#### Barley

I was just going to post that I figured that out-- still your explaination has solidified it.

THANKS

#### arildno

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Go through ehild's setup, and verify for yourself that the curl of the given function in your second problem is, indeed, zero.

#### Barley

Yes I've verified that it's zero, and I must have something wrong with my problem statement. I know it's supposed to be the motion of a particle with only force only in the x direction. I think something like an object sliding across a table is not conservative. Energy is lost through friction. Sorry I can't explain it till I understand it.

How about this; a particle mass m has a speed v= a/x, where a is a constant and x represents displacement. The force of this function, from
F= ma, just take the integral of the velocity function and

F=malnx, thats mass x constant times the integral of the velocity function. Is this correct ?
Is this function conservative ?

Thanks

#### arildno

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Barley said:
Yes I've verified that it's zero, and I must have something wrong with my problem statement. I know it's supposed to be the motion of a particle with only force only in the x direction. I think something like an object sliding across a table is not conservative. Energy is lost through friction. Sorry I can't explain it till I understand it.
IF you model an object sliding along a table and take into account friction (and possibly, air resistance), then yes, energy would dissipate.

But why must you in every single problem take into account every one of the gazillion forces that act upon the object?
That is, to study idealized cases where you try to determine what would the motion be if we restrict our attention to just a small sub-set of those forces?

#### Barley

Error

I've got to do some work before I can post again. What I meant in the last post was F= -ma/x^2 and not the mass times the integral of the velocity function.

Thanks

#### Barley

Ahh F= (ax) jhat

F = (ax)jhat was the non consertive force function that was discussed in class. Starting to make sense now. Let's see what I can do before I get stuck again.

Kudos to arildno and eheld .

#### arildno

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There you have it!
First rule of maths:

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