Math4Teens - Path Integral Properties

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

Path Integral Properties

The properties of path integrals below follow from the definition of path integral in terms of regular integral using the parameterization of the path
γ = γ(s) = {x(s),y(s),z(s)} for s∈[0,S]
and corresponding properties of regular integrals.

Path Integrals Properties
of Scalar Functions

1. ∫_γ [F(x,y,z)+G(x,y,z)] dγ =
= ∫_γ F(x,y,z) dγ + ∫_γ G(x,y,z) dγ

With parameterization of path γ functions F and G become functions of one argument s
F(x(s),y(s),z(s)) = f(s)
G(x(s),y(s),z(s)) = g(s)
Hence,
∫_γ[F(x,y,z)+G(x,y,z)] dγ =
= ∫_[0,S][f(s)+g(s)] ds
and we can refer to the additive property of the regular integrals.


2. ∫_γ c·F(x,y,z) dγ = c·∫_γ F(x,y,z) dγ
for the same reasons as above.


3. Going along a path in an opposite direction changes the sign of an integral.
Indeed, assume that path γ is from A to B and is parameterized by s∈[0,S], which implies
γ(0)=A, γ(S)=B
Then
∫_AB F(x,y,z) dγ = ∫_[0,S]F(x(s),y(s),z(s))ds
∫_BA F(x,y,z) dγ = ∫_[S,0]F(x(s),y(s),z(s))ds
and, by properties of regular definite integrals, the latter is opposite in sign to the former.

4. Assume, path γ from A to B is divided by point M into two parts:
α from A to M and
β from M to B.
So, we can say γ=α+β.
Then we can state the following additive property of path integrals:
∫_γ F(x,y,z) dγ =
= ∫_α F(x,y,z) dα + ∫_β F(x,y,z) dβ
Here is the proof of this statement.
Consider the parameterization of path γ={x(s),y(s),z(s)}, where s∈[0,S].
Using this, point A corresponds to s=0, point B corresponds to s=S.
Assume, point M corresponds to s=s_M.
This allows to parameterize both parts of an original path:
α={x(s),y(s),z(s)}, where s∈[0,s_M]
β={x(s),y(s),z(s)}, where s∈[s_M,S]
Then, by definition of a path integral,
∫_γ F(x,y,z) dγ=∫_[0,S]F(x(s),y(s),z(s))ds
∫_α F(x,y,z) dα=∫_[0,sM]F(x(s),y(s),z(s))ds
∫_β F(x,y,z) dβ=∫_[sM,S]F(x(s),y(s),z(s))ds
From these definitions and the corresponding additive property of regular integrals follow the additive property of path integrals.


Path Integrals Properties
of Vector Functions

5. In physics we sometimes deal with gravitational, electrostatic and some other types of forces that have an important property called conservative.
Conservative force F={F_x,F_y,F_z}, as a vector, can be represented as partial derivatives of some scalar function f(x,y,z) by corresponding coordinates, which means:
F_x(x,y,z) = ∂f(x,y,z)/∂x
F_y(x,y,z) = ∂f(x,y,z)/∂y
F_z(x,y,z) = ∂f(x,y,z)/∂z
In physics concepts of potential and potential energy are based on such function.
Let's prove that path integral of a vector-function that represents a conservative force depends only on the endpoints of a path and does not depend on path itself between these endpoints.
In other words, no matter which path we choose between two fixed endpoints, path integral along it will be the same.
Given two endpoints in space A and B and a parametrically defined path AB that depends on parameter s∈[0,S]
r = r(s) = {x(s),y(s),z(s)}
with A={x(0),y(0),z(0)}
and B={x(S),y(S),z(S)}.
Our task is to integrate a vector-function that represents a conservative force F={F_x,F_y,F_z} along path AB.
This integral equals to
∫_AB[F(r)·dr] =
under integral we have a scalar (dot) product of two vectors
F={F_x,F_y,F_z} and dr={dx,dy,dz}
= ∫_[0,S][F_x(x(s),y(s),z(s))·dx(s) +
+ F_y(x(s),y(s),z(s))·dy(s) +
+ F_z(x(s),y(s),z(s))·dz(s)]
Assuming our vector-function represents a conservative force, that is
F_x = ∂f/∂x
F_y = ∂f/∂y
F_z = ∂f/∂z
we can rewrite this path integral as
∫_AB[F(r)·dr] =
= ∫_[0,S]{[∂f(x(s),y(s),z(s))/∂x]·dx(s) +
+ [∂f(x(s),y(s),z(s))/∂y]·dy(s) +
+ [∂f(x(s),y(s),z(s))/∂z·dz(s)]}
Notice that the expression in {...} is a full differential of function f(x,y,z).
Therefore,
∫_AB[F(r)·dr] = ∫_[0,S]df(x(s),y(s),z(s))
According to the rules of integration, this equals to
f(x(S),y(S),z(S))−f(x(0),y(0),z(0))
As we see, the path integral of a conservative vector-function depends only on the values of this scalar function f(x) at endpoints of a path.


6. If a vector-function F represents a conservative force, then the path integral along any closed loop path is zero.
Assume, path γ is a closed loop and is parameterized by s∈[0,S], which implied that A=s(0)=s(S)=B.
According to property #5 above, the values of a path integral along this path is
f(x(S),y(S),z(S))−f(x(0),y(0),z(0))
But points A=s(0)={x(0),y(0),z(0)} and B=s(S)={x(S),y(S),z(S)} coincide, which makes the expression above to be zero.
Alternative proof might be as follows. Let's choose any two points on a closed loop of this path, M and N.
Consider two different ways of moving from M to N along two different routes within path γ that we conditionally call "left" and "right".
Since path integral from M to N for conservative vector-function F is independent of path chosen, but only on the endpoints,
∫_[MleftN]F(γ)dγ = ∫_[MrightN]F(γ)dγ
Full loop along path γ can be represented as a sum of MleftN+NrightM.
But changing the direction of integration from MrightN to NrightM changes the sign of a path integral. ∫_[NrightM]F(γ)dγ = −∫_[MrightN]F(γ)dγ
Full path integral is equal to
∫_[MleftN]F(γ)dγ + ∫_[NrightM]F(γ)dγ =
= ∫_[MleftN]F(γ)dγ −∫_[MrightN]F(γ)dγ
As we see, we are subtracting a value from an equal value.
Therefore, integration along a full loop MleftNrightM results in zero.