Physics+ Work Lemmas: UNIZOR.COM - Classic Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Field Work Lemmas

In the previous lecture we have introduced the concepts of a field and field intensity force that is equal to a gradient of the field potential.
Also, we have proven that, dealing with such force, the work of this field intensity force along any trajectory of an object moving in the field depends only on the field potential at the beginning and at the end of a trajectory and is independent of a path between these two points.

There is a converse theorem that states that, if the work performed by some force on an object depends on the object's position in the beginning and at the end of its movement and does not depend on a trajectory between these points, then this force can be represented as a gradient of some scalar function, the field potential.

This lecture presents certain auxiliary theorems (lemmas) that will help to prove the above mentioned theorem in the next lecture.


Lemma A

This lemma, in short, is about comparing the work performed by a force, when an object moves along the same trajectory in two opposite directions.

More rigorously, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.
The coordinate components of the vector of force F at point (x,y,z) are
||F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)||.

An object is moving along certain trajectory from point A to point B within this area, while the force F acts on it and performs certain work.
The coordinate components of the position vector r are
||x,y,z||.
The coordinate components of the infinitesimal increment, the differential, of the position vector dr are
||dx,dy,dz||.

Compare the work done by the force F along the object's trajectory from point A to point B with the work done when an object moves from point B to point A along the same trajectory in the opposite direction.

Solution A

Recall the definition of work performed by force F on an object moving along a trajectory described by position vector r from starting point A to finishing point B:
W_[AB] = ∫_[AB]dW = ∫_[AB]F·dr
The above integral is an infinite sum of infinitesimal work increments dW(x,y,z) performed by force F(x,y,z) on an infinitesimal interval of a trajectory r(x,y,z) from point (x,y,z) to point (x+dx,y+dy,z+dz), where F·dr is a scalar product of two vectors, so
dW(x,y,z) = F(x,y,z)·dr(x,y,z) =
= F_x·dx+F_y·dy+F_z·dz

Assume, at some moment of time our object, moving from A to B, is at position (x,y,z).
During an infinitesimal increment of time it's new position will be (x+dx,y+dy,z+dz) and the force will do an infinitesimal amount of work
dW = F_x·dx+F_y·dy+F_z·dz.

Now assume that the object moves in the opposite direction from B to A along the same trajectory.
Being at the same position (x,y,z) at some moment in time, the infinitesimal increments dx, dy and dz will have signs opposite to those when an object moved from A to B, while the force vector will be the same.

Therefore, the differential of work dW will also be of an opposite sign, and subsequent integration will result in total work done by the force on trajectory from B to A to have the same magnitude but opposite sign comparing with object movement from A to B.

Answer A
W_[AB]=∫_[AB]F·dr =
= −∫_[BA]F·dr = −W_[BA]


Lemma B

As in Lemma A, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.

An object is moving along certain trajectory within this area and the force F(x,y,z) acts on it, performing some work along its trajectory.
The object's position at time t is r(t)=||x(t),y(t),z(t)||.

It's given that the work of this force on an object moving along any trajectory between any pair of points A and B depends only on a choice of these two endpoints and does not depend on a choice of trajectory between them.

Prove that the work this force performs on an object moving along a closed trajectory, when points A and B coincide, equals to zero.

Proof B

In our case of a closed trajectory the endpoints A and B coincide. So, we will deal only with point A.

Choose any closed trajectory, its starting and ending point A and point M on it that does not coincide with point A.
Now we have two different paths from A to M, let's call them path #1 and path #2.

If an object moves along a closed trajectory, it moves from point A to point M along path #1 and then moves from point M to point A along path #2.

According to the condition of the problem, a work W_1[AM] performed by a force acting on our object along a path #1 from A to M should be equal to a work W_2[AM] performed by a force acting on our object along a path #2 from A to M:
W_1[AM] = W_2[AM]

As has been proven in the Lemma 1, during the second part of the trajectory, when object moves from M to A along path #2, the work of a force is of the same magnitude as if an object moved from A to M along the same path #2 but with an opposite sign:
W_2[MA] = −W_2[AM]

Therefore, the total work along path #1 from A to M followed by moving from M to A along path #2 equals to
W_1[AM]2[MA] = W_1[AM]+W_2[MA] =
= W_1[AM] − W_2[AM] =
= W_1[AM] − W_1[AM] = 0


Lemma C

This lemma is a converse to Lemma B.
As in Lemma A, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.

An object is moving along certain trajectory within this area, and the force F(x,y,z) acts on it, performing some work.
The object's position at time t is r(t)=||x(t),y(t),z(t)||.

It's given that the work of this force on an object moving along any closed trajectory that starts and ends at the same point equals to zero.

Prove that the work this force performs on an object moving from any fixed point A to any fixed point B does not depend on trajectory between these points.

Proof C

Choose any two paths from point A to point B - path #1 and path #2.
We will use the symbols introduced in Lemma 2.
As stated in the condition of this lemma,
W_1[AB]2[BA] =
= W_1[AB] + W_2[BA] = 0
At the same time
W_2[BA] = −W_2[AB]
Therefore,
0 = W_1[AB] + W_2[BA] =
= W_1[AB] − W_2[AB]
from which follows that
W_1[AB] = W_2[AB]

As we see, a statement that the work performed by a force on an object moving along any closed trajectory equals to zero is equivalent to a statement that the work performed by a force on an object moving from any point A to any point B does not depend on a trajectory an object moves between these points.


Lemma D

As in Lemma A, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.
It's given that the work of this force on an object moving along any trajectory between any pair of points depends only on a choice of these two endpoints and does not depend on a choice of trajectory between them.

Assume, there are three points in the area where the force is acting, A, B and C.
Consider amounts of work the force performs on an object during its movements between these points
W_[AB], W_[AC], W_[BC].

Prove that
W_[BC] = W_[AC] − W_[AB].

Proof D

From Lemma B above follows that the amount of work performed by a force on an object moving along a closed trajectory from A to B to C and back to A equals to zero
W_[AB] + W_[BC] + W_[CA] = 0

From Lemma A above follows that, if object moves in the opposite direction along the same trajectory, the amount of work performed by a force is the same in magnitude but opposite in sign to the amount of work performed on a directly moving object
W_[AC] = −W_[CA]

Therefore,
to B to C and back to A equals to zero
W_[AB] + W_[BC] − W_[AC] = 0
from which follows that
W_[BC] = W_[AC] − W_[AB].

Coincidentally, this equality reminds the rule about a difference between two vectors
BC = AC − AB