Math 4 Teens - Path Integral Problems

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

Path Integral Problems

Problem 1

How to determine whether a sufficiently differentiable two-dimensional vector-function is conservative?

Solution

Let F(x,y)={F_x(x,y);F_y(x,y)} be some two-dimensional vector-function with differentiable components.
Assume, it is conservative.
Then there exist scalar function f(x,y) such that
F_x(x,y) = ∂f(x,y)/∂x
F_y(x,y) = ∂f(x,y)/∂y

Since all functions above are sufficiently differentiable, we can partially differentiate the first equation by y and the second - by x.
The results are
∂F_x(x,y)/∂y = ∂[f(x,y)/∂x]/∂y =
= ∂²f(x,y)/∂y∂x
∂F_y(x,y)/∂x = ∂[f(x,y)/∂y]/∂x =
= ∂²f(x,y)/∂x∂y

But for sufficiently smooth functions of multiple arguments the order of partial differentiation by different arguments is irrelevant.
∂²f(x,y)/∂y∂x = ∂²f(x,y)/∂x∂y
Therefore,
∂F_x(x,y)/∂y = ∂F_y(x,y)/∂x

The above equality is a necessary condition for representation of the components of our vector-function F(x,y) as partial derivatives of some scalar function f(x,y).
If we check this equality, and it's not held, our vector-function is not conservative.

Notice, we have only determined that the above equality of mixed partial derivatives is a necessary condition.
It can be proven that, if components F_x(x,y) and F_y(x,y) have continuous first partial derivatives throughout a simply connected domain (no holes, no separate parts), then this equality is also sufficient and guarantees the conservativeness.


Problem 2

Consider a two-dimensional vector-function
F(x,y)={F_x(x,y);F_y(x,y)} = {x·y,x+y}

Check whether this vector-function is conservative, and, if it is not, demonstrate this by calculating the path integral of F from point (0,0) to point (1,2) along two different paths and showing the results differ.

Solution

Step 1 - Test for conservativeness

Checking the necessary condition for a vector-function to be conservative, as determined in Problem 1.
∂F_x(x,y)/∂y = ∂(x·y)/∂y = x
∂F_y(x,y)/∂x = ∂(x+y)/∂x = 1

Since these partial derivatives are different, the necessary condition for conservativeness fails.
No scalar function f(x,y) exists, whose partial derivatives ∂f(x,y)∂x and ∂f(x,y)∂y correspondingly equal to components of vector-function F(x,y) - F_x(x,y)=x·y and F_y(x,y)=x+y.
So F(x,y) is not conservative.

Step 2 - Path 1: (0,0)→(1,2) straight

Parameterize this path as
x(t)=t, y(t)=2t,   t∈[0,1]
so dx=dt, dy=2dt.

The path integral is
∫_path1F·dr = ∫_path1(F_x·dx + F_y·dy) =
= ∫_path1(x·y·dx + (x+y)·dy) =
= ∫_[0,1][t·(2t)·dt+(t+2t)·2dt] =
= ∫_[0,1][2t²+6t]·dt =
= (2t³/3 + 3t²)|¹₀ = 11/3

Step 3 — Path 2: (0,0)→(1,0)→(1,2)

This path has two segments.

Segment A:
from (0,0) to (1,0), along y=0.
Parameterize x(t)=t, y(t)=0, t∈[0,1]
so dx=dt, dy=0.
∫_path2AF·dr = ∫_path2A(F_x·dx + F_y·dy) =
= ∫_path2A(x·y·dx + (x+y)·dy) =
= ∫_[0,1][t·0·dt+(t+0)·0] = 0

Segment B:
from (1,0) to (1,2), along x=1.
Parameterize x(t)=1, y(t)=2t, t∈[0,1]
so dx=0, dy=2dt.
∫_path2BF·dr = ∫_path2B(F_x·dx + F_y·dy) =
= ∫_path2B(x·y·dx + (x+y)·dy) =
= ∫_[0,1][1·2t·0+(1+2t)·2dt] =
= ∫_[0,1](1+2t)·2dt =
= (2t + 2t²)|¹₀ = 4

Total for Path 2 is Path2A+Path2B=0+4=4, which is different from Path 1 that was 11/3.

Step 4 — Conclusion

The straight Path 1 gives the value of the path integral for our vector-function equal to 11/3, while two-segmented Path 2 gives 4, for the same start and end points. Since the value of the path integral depends on the path taken and not only on the endpoints, F is confirmed to be non-conservative, which is consistent with the failed partial-derivative test in Step 1.


Problem 3

Consider a three-dimensional Cartesian system of coordinates with the Sun, as a point-mass M, at its origin.
Let the location of a point-mass m (a planet) be
r={x,y,z}
The vector F representing the force on a planet caused by the Sun's gravitation, which, according to the Newtons Law of Gravitation, equals in magnitude to
|F(r)| = G·M·m/|r|²
and it's directed from a planet towards the Sun (that is, opposite to r).

Check that this force is conservative with a scalar function, whose partial derivatives equal to components of force F={F_x,F_y,F_z}, being proportional to 1/|r|.

Solution

Our first task is to express the gravitational force in terms of Cartesian coordinates.
The magnitude of F is
|F(r)| = G·M·m/|r|² =
= G·M·m/(x²+y²+z²)

The direction of the gravitational force is opposite to vector r={x,y,z}.
So, vector F can be represented as
F(r) = |F(r)|·ρ
Where ρ is a unit vector directed opposite to position vector r.

The unit vector that has the direction opposite to r={x,y,z} is
ρ={−x,−y,−z}/|r| and
|r| = √(x²+y²+z²)

That makes the components of force F={F_x,F_y,F_z} to be
F_x= −G·M·m·x/(x²+y²+z²)^3/2
F_y= −G·M·m·y/(x²+y²+z²)^3/2
F_z= −G·M·m·z/(x²+y²+z²)^3/2

Consider now a scalar function
f(r) = f(x,y,z) = 1/|r| = 1/√(x²+y²+z²). Its partial derivatives are
∂f(x,y,z)∂x = (−½)·(x²+y²+z²)^−3/2·2x =
= −x/(x²+y²+z²)^3/2
∂f(x,y,z)∂y = (−½)·(x²+y²+z²)^−3/2·2y =
= −y/(x²+y²+z²)^3/2
∂f(x,y,z)∂z = (−½)·(x²+y²+z²)^−3/2·2z =
= −z/(x²+y²+z²)^3/2

As we see, all we need to equalize partial derivatives of f(x,y,z) with components of force F is to multiply f(x,y,z) by constant G·M·m.

That proves that gravitational force F is conservative.

Remark on Potential Energy

The scalar function we used above, multiplied by the constant G·M·m, is
φ(r) = G·M·m/|r|, with
F = {∂φ/∂x,∂φ/∂y,∂φ/∂z}.

In physics, the potential energy U(r) of an object at position r relative to a conservative force F(r) is defined as external work needed to bring this object from infinity to its current position.
The gravitation is an attractive force, so it helps to bring an object from infinity to r.
Hence, the external work is negative.
To match this convention, we set
U(r) = −φ(r) = −G·M·m/|r|.

This is the familiar gravitational potential energy: negative (since gravity is attractive), approaching zero as |r|→+∞, and growing more negative as the planet approaches the Sun.