## Monday, September 19, 2022

### More Field Problems: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

More Field Problems

Problem 1

Prove that a curl of two-dimensional conservative vector field equals to zero.

Solution

(a) From the definition of the conservative field (the integral of the vector field along any path depends only on the end points) follows that an integral around any closed path equals to zero.
(b) A curl is defined as a limit of the ratio of an integral around a closed path (which is zero for conservative vector field, as stated above) to the area enclosed by this path. The result of this division is zero.

Problem 2

Prove that a two-dimensional conservative field
F(x,y) = {P(x,y), Q(x,y)} =
= P(x,y)·i + Q(x,y)·j

is a gradient of some two-dimensional scalar field f(x,y) called its potential.
In other word, prove that there exists function f(x,y) such that
P(x,y) = f(x,y)/x and
Q(x,y) = f(x,y)/y.

Solution

Before offering a solution, let's try to use some knowledge about conservative fields that we have learned before.

Recall a concept of potential energy of a gravitation or an electrostatic field.
By definition, it's the work performed by or against a field by moving a unit of mass or a unit of electrical charge from infinitely remote location to a specific location of interest in space, thus making this potential energy a function of a point in space where gravitational or electrostatic forces act.
Validity of a concept of potential energy is based on its independence on the path taken from infinity to a point of interest. Does not it remind the conservative field? It sure does.

That means, a potential energy is a scalar field, a scalar function of a specific location in space. It does not depend on trajectory of movement.

The work participating in the definition of a potential energy is, roughly speaking, a product of a force by distance. More precisely, it's an integral of this product along a trajectory we move a unit of mass or a unit of electric charge, because the force is, generally speaking, variable in magnitude and direction. The force itself constitutes a vector field in space, and the integral of a product of this force by distance along a trajectory (that is, a work performed by or against this vector field by moving a unit of mass or a unit of electric charge) corresponds to a definition of a circulation of this vector field of force along some trajectory of movement.

As an example, let's find the work of moving a unit of mass in the gravitational field. For simplicity, we'll analyze this in one-dimensional case. Let's choose a point {x,0,0} as a point of interest and calculate the work to move a unit of mass along the X-axis from infinity to this point. Let the source of gravity be a mass M at the origin of coordinates.

According to the Newton's Law of Gravity, the force of gravity for a unit mass at distance s from the origin of coordinates is
F(s) = G·M/s²
Therefore, the potential energy at point {x,0,0} is
[∞,x]F(s)·ds = [∞,x]G·M/s²·ds
The indefinite integral of 1/s² is −1/s.
Therefore, our potential energy at point {x,0,0} is
U(x) = −G·M/s|x = −G·M/x

The potential energy U(x) is a scalar field.
What's interesting is that
dU(x)/dx = G·M/x² = F(x).
So, a scalar field of potential energy for a gravitational field plays the role of potential, gradient of which is the vector field of gravitational force.

Our guess then is, that in case of a given conservative vector field we can find its potential, gradient of which is our vector field, and this potential is just amount of work performed by or against the vector field of force by moving along some (actually, any) trajectory from some fixed location to a point of interest.
We can choose any trajectory because our vector field is conservative.

Consider a two-dimensional case and point of interest {x,y}.
The given conservative vector field is defined at all points as
F(x,y) = {P(x,y); Q(x,y)}

Let's calculate a work needed to move an object in this field from origin of coordinate to point {x,y}.
Since the field is conservative, we choose an easy path:
A{0,0} → B{x,0} → C{x,y}
and calculate the work separately on an interval AB, then on BC and then add them together.

WAB = [0,x]P(u,0)du
WBC = [0,y]Q(x,v)dv
Total amount work is W(x,y)=WAB+WBC
Can this scalar field W(x,y) be the needed potential, gradient of which is our vector field F(x,y)?

Let's check by calculating the gradient
If we are correct in the assumption that work by or against the conservative vector field along some trajectory to a point of interest gives a potential of this field, the gradient of work W(x,y) (that is, partial derivatives by each argument) should deliver the vectors of the field F(x,y).
In other words, we should check that
W(x,y)/x = P(x,y) and
W(x,y)/y = Q(x,y)

Here are the calculations.
W(x,y)/x = WAB(x,y)/x +
+
WBC(x,y)/x =
= (
/x)
[0,x]P(u,0)du +
+ (
/x)
[0,y]Q(x,v)dv

The first expression is a derivative by an upper limit of integration, which reverses the integration and results in
(/x)[0,x]P(u,0)du = P(x,0)

The second expression allows to bring differentiation under the integral, since differentiation and integration are for different arguments
(/x)[0,y]Q(x,v)dv =
=
[0,y][(Q(x,v)/x]dv

From the definition of the conservative field (the integral of the vector field along any path depends only on the end points) follows that an integral around any closed path equals to zero.
A curl is defined as a limit of the ratio of an integral around a closed path (which is zero for conservative vector field, as stated above) to the area enclosed by this path. The result of this division is zero.
The curl in two dimensional case was calculated as
curl F(x,y) =
=
Q(x,y)/x − P(x,y)/y

Since curl is zero for a conservative field,
Q(x,y)/x = P(x,y)/y
and the expression for the second integral above can be changed to
(/x)[0,y]Q(x,v)dv =
=
[0,y][(Q(x,v)/x]dv =
=
[0,y][(P(x,v)/v]dv
Integral by v and differentiation by v reverse each other and the result is the original function
[0,y][(P(x,v)/v]dv =
= P(x,v)|y0 = P(x,y) − P(x,0)

Adding this to the first of two integrals above, that we found to be equal to P(x,0), the result will be P(x,y), which corresponds to our expectations.

Absolutely analogous calculations for W(x,y)/x will yield Q(x,y) as a result.
These results prove that
and that proves the existence and construction of the scalar field W(x,y), gradient of which is our conservative vector field F(x,y).

Problem 3

Prove that a curl of a gradient of a three-dimensional scalar field is zero.

Solution

The solution in three-dimensional space is based on the formula for a curl and an expression of a gradient of a scalar field.
Let vector field F(x,y,z) be a gradient of a scalar field f(x,y,z):
Fx(x,y,z) = f(x,y,z)/x
Fy(x,y,z) = f(x,y,z)/y
Fz(x,y,z) = f(x,y,z)/z

A curl of F(x,y,z) is a vector with components
curl F(x,y,z) = ∇⨯F(x,y,z) =
=
{Fz(x,y,z)/y−Fy(x,y,z)/z;
Fx(x,y,z)/z−Fz(x,y,z)/x;
Fy(x,y,z)/x−Fx(x,y,z)/y
}

Substituting Fx, Fy, Fz with their expressions in terms of partial derivatives of f(x,y,z), we obtain
Fz(x,y,z)/y−Fy(x,y,z)/z =
=
2f(x,y,z)/zy −
2f(x,y,z)/yz = 0

because a mixed partial second derivative by two different arguments does not depend on the order of differentiation (assuming the participating function is twice continuously differentiable)

Analogously,
Fx(x,y,z)/z−Fz(x,y,z)/x =
=
2f(x,y,z)/xz −
2f(x,y,z)/zx = 0

Finally,
Fy(x,y,z)/x−Fx(x,y,z)/y =
=
2f(x,y,z)/yx −
2f(x,y,z)/xy = 0

Since a conservative vector field can be represented as a gradient of some scalar field (its potential), we have proven that
curl grad f(...) = 0 or
∇⨯(∇f(...)) = 0
for any scalar vector field f(...).