*Notes to a video lecture on UNIZOR.COM*

__Potential Theorem__

As you recall, we have defined a

**field**as an area (a subset) of points in our three-dimensional space with a vector of force

*called*

**F**(x,y,z)**field intensity**defined at each point of this area and a real scalar function

*U(x,y,z)*called

**potential**defined at all these points, when the following equation between a force and a potential at each point

*P(x,y,z)*is held:

*= −∇*

**F**(x,y,z)*U(x,y,z)*

where symbol ∇ signifies gradient of a function

*U(x,y,z)*- a vector of function's partial derivatives by each coordinate

∇

*U(x,y,z) =*

= ||∂U/∂x, ∂U/∂y, ∂U/∂z||

= ||∂U/∂x, ∂U/∂y, ∂U/∂z||

We have also proven that the work performed by such a field intensity function on an object moving along some trajectory depends only on the endpoints of the object's movement and is independent of a path chosen between these two endpoints.

Thus, independence of work of the trajectory is a necessary condition of the existence of the field

**potential**function

*U(x,y,z)*, whose gradient with a minus sign equals to the field intensity force

*.*

**F**(x,y,z)In this lecture we will prove that the condition of work being independent of a trajectory between the endpoints of an object's movement is also a sufficient condition for the field force to be a gradient of some scalar function - the field

**potential**.

*Theorem*

A vector of force

*is defined at each point*

**F**(x,y,z)*P(x,y,z)*of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.

An object, acted upon by this force, moves within this area.

It's given that the work performed by this force on an object moving along some trajectory between any two points depends only on positions of these endpoints and is independent of the trajectory between them.

Prove that there exists a scalar function of position

*U(x,y,z)*called

**potential**such that the vector of force equals to a negative gradient of this potential, that is

*= −∇*

**F**(x,y,z)*U(x,y,z) =*

= −||∂U/∂x, ∂U/∂y, ∂U/∂z||

= −||∂U/∂x, ∂U/∂y, ∂U/∂z||

Since vector

*at point*

**F**(x,y,z)*P(x,y,z)*can be expressed in coordinate form as

*||F*

_{x}(x,y,z),F_{y}(x,y,z),F_{z}(x,y,z)||the above statement can be formulated in coordinate form as

*F*

_{x}(x,y,z) = −∂U(x,y,z)/∂x*F*

_{y}(x,y,z) = −∂U(x,y,z)/∂y*F*

_{z}(x,y,z) = −∂U(x,y,z)/∂z*Proof*

As a proof, we will explicitly define a potential function and prove that it satisfies the required equalities.

Choose arbitrarily some fixed point

*A(x*in the area where our force is defined, that we will use as the beginning of some trajectory.

_{0},y_{0},z_{0})Choose any other point

*B(x,y,z)*there, where we will explicitly define the scalar function

*U(x,y,z)*(the

**potential**) that satisfies the conditions of the problem.

Let's chose any particular path from point

*A(x*to point

_{0},y_{0},z_{0})*B(x,y,z)*and define the function

*U(x,y,z)*(the field potential) for point

*B(x,y,z)*as the work of force

*performed during an object's movement along a chosen path from*

**F***A(x*to

_{0},y_{0},z_{0})*B(x,y,z)*with a negative sign.

This definition of a potential is quite legitimate since the work performed by force

*does not depend on a path chosen, but depends only on position of points*

**F***A*and

*B*.

So, by definition, for any point

*B(x,y,z)*where the field is defined

*U(B) = U(x,y,z) = −W*

_{[AB]}

Let's chose a point

*C(x+dx,y+dy,z+dz)*on infinitesimal distance from point

*B(x,y,z)*.

Let

*d*be a vector of displacement from

**r***B*to

*C*:

*d*

**r**= ||dx,dy,dz||According to the definition of work, when the field intensity force

*acts on an object that moves from point*

**F***B(x,y,z)*to an infinitesimally close to it point

*C(x+dx,y+dy,z+dz)*, the infinitesimal amount of work performed by the force is equal to

*dW*

_{[BC]}=

+ F

**F**(x,y,z)**·**d**r =**

=F=

_{x}(x,y,z)·dx + F_{y}(x,y,z)·dy ++ F

_{z}(x,y,z)·dzAt the same time, using the

*Lemma D*of the previous lecture

*Work Lemmas*, this same amount of work equals to

*dW*

_{[BC]}=

*W*

_{[AC]}−

*W*

_{[AB]}

where fixed point

*A(x*was arbitrarily chosen above.

_{0},y_{0},z_{0})Expressions

*W*

_{[AB]}and

*W*

_{[AC]}represent the amount of work the field force performs, if an object moves from point

*A*to

*B*and from points

*A*to

*C*correspondingly. These same amounts were used to define a potential at these point

*W*

_{[AB]}=

*−U(x,y,z)*

*W*

_{[AC]}=

*−U(x+dx,y+dy,z+dz)*.

Therefore,

*dW*

_{[BC]}=

*−U(x+dx,y+dy,z+dz)*+

+

*U(x,y,z) = −dU(x,y,z)*

where

*dU(x,y,z)*is a full differential (infinitesimal increment) of function

*U(x,y,z)*on an interval from

*B(x,y,z)*to

*C(x+dx,y+dy,z+dz)*.

As known from Calculus, the full differential of a function can be expressed in terms of partial derivatives and differentials of arguments (see

*Partial Derivatives - Basic Properties*lecture of the

*Calculus*chapter in the course

*Math 4 Teens*on

*UNIZOR.COM*).

Therefore,

*dW*

_{[BC]}=

*−dU(x,y,z) =*

= −(∂U(x,y,z)/∂x)·dx −

− (∂U(x,y,z)/∂y)·dy −

− (∂U(x,y,z)/∂z)·dz

= −(∂U(x,y,z)/∂x)·dx −

− (∂U(x,y,z)/∂y)·dy −

− (∂U(x,y,z)/∂z)·dz

Comparing this expression of

*dW*

_{[BC]}with the one in terms of the field force components above, we come to an equation

*F*

+ F

= −(∂U(x,y,z)/∂x)·dx −

− (∂U(x,y,z)/∂y)·dy −

− (∂U(x,y,z)/∂z)·dz

_{x}(x,y,z)·dx + F_{y}(x,y,z)·dy ++ F

_{z}(x,y,z)·dz == −(∂U(x,y,z)/∂x)·dx −

− (∂U(x,y,z)/∂y)·dy −

− (∂U(x,y,z)/∂z)·dz

While

*dx*,

*dy*and

*dz*are infinitesimal increments of position along some infinitesimal displacement for each coordinate, the direction of this displacement can be chosen freely.

Choosing infinitesimal

*dx*and

*dy=0, dz=0*leads to an equality

*F*

_{x}(x,y,z) = −∂U(x,y,z)/∂xSimilarly, leaving only

*dy*or

*dz*as infinitesimal increments and setting displacement along other coordinates to

*0*, we obtain the equalities

*F*

_{y}(x,y,z) = −∂U(x,y,z)/∂y*F*

_{z}(x,y,z) = −∂U(x,y,z)/∂zEnd of Proof

The question now arises, is the field potential

*U(x,y,z)*uniquely defined by the field force intensity?

The answer is NO, since we have chosen point

*A(x*as, basically, any fixed reference point where the movement of an object begins.

_{0},y_{0},z_{0})We can choose any other point

*A'(x*as the beginning, and the work

_{1},y_{1},z_{1})*W(x,y,z)*will be different. More precisely, it will differ by the amount of work the force performs moving an object from

*A*to

*A'*.

This means that our field potential is not uniquely defined by the field force, only its partial derivatives are, since they must correspond to force components. This is similar to the fact that, given a derivative of a function, the function is defined as an integral from a derivative plus some freely chosen constant.

Traditionally, for gravitational or electrostatic fields, as the staring point, physicists choose a point infinitely far from the source of the field force. There the force is equal to zero. With this convention the field potential

*U(x,y,z)*is fully defined and equals to the negative work performed by the force

*to move an object from an infinitely far point to any point*

**F**(x,y,z)*P(x,y,z)*.

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