Path Integral Derivation Question

In summary, the conversation is about the derivation of a path integral formulation for finding the length of a curve on a given interval. The process involves breaking the curve into small portions and summing the distances between them. The problem arises when trying to evaluate the sum as the interval tends to zero. The formula for arclength by "infinitesimal reasoning" is used, and it is argued that it approximates the integral of the square root of 1 + the derivative of the function squared. However, there is concern that this may not always work, as shown in the example of y = x. Ultimately, there is uncertainty about how to continue with the derivation.
  • #1
gordonj005
56
0
Hello, this will be my first post on the physics forum, so i wanted to make it decent :P

I've been trying recently to derive for myself a path integral formulation (not quantum mechanical or anything feynman like but for finding the length of a curve on a given interval). Heres my attempt at deriving it for a simple curve, but its a very time consuming tedius process to actually calculate. I know real path integrals involve Reinman sums, but after trying to learn them on my own, I decided to give it a shot on my own, here goes:

f(x) = x^2, from x1 = 3 to x2 = 4 using a small interval dx = 0.01

distance between two points on a function is defined as:
d = √(x2 - x1)^2 + (f(x2) - f(x1))^2

so breaking the curve down into infentesibly small portions, (or in this case our small interval dx), and adding the distances together, we should get a reasonably good approximation for the distance on the curve... right?

so in this case, between x1 = 3 and x2 = 4, we have several peices of the curve, ex, the distance between 3 - 3.01, 3.01 - 3.02, 3.02 - 3.03...3.99 - 4.0. I've worked it out so that the number of these peices is defined by (x2 - x1)*(1/dx). so in this case we have:

1*100 , so 100 individual peices we have to sum up. now this is where I've run into some problems. I've defined the sum as:

Ʃ(i = 1 to 100) = √((3 + i*dx) - (3 + (i-1)*dx))^2 + (f(3 + i*dx) - f(3 + (i-1)*dx))^2

which gives a reasonable approximation. now my problem is, i want to evaluate this sum as dx tends to zero, so we would have an exact answer. so we could write this as:

lim(dx -> 0) Ʃ(i = 1 to ∞) = √((3 + i*dx) - (3 + (i-1)*dx))^2 + (f(3 + i*dx) - f(3 + (i-1)*dx))^2

but i am stuck here, i feel like i should take an integral, but i don't know how and that probably isn't correct (or even doable). which leads me to my question, what now?

thank you very much

EDIT: using 0.01 i got ~6.94, and continuing to decrease it it seems to converge at ~7.074
 

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  • #2
I don't know much about path integrals, but for deriving the formula for arclength by "infinitesimal reasoning" the procedure in calculus is:

[tex] \sum_n \sqrt{ dx^2 + (f(x+(n+1) dx) - f(x+ n dx))^2} [/tex]

which is how you began. Then divide inside the radical by [itex] dx^2 [/itex] and multiply on the outside by [itex] dx [/itex]

[tex] = \sum_n \sqrt{ 1 + (\frac {f(x+(n+1)*dx) - f(x+ n dx)}{dx})^2} dx [/tex]

Argue that this approximates

[tex] \sum_n \sqrt{ 1 + f'(x+n dx) } dx [/tex]

Argue that this is a Riemann sum to approximate the integral

[tex] \int \sqrt{1 + f'(x)} dx [/tex]
 
  • #3
Stephen Tashi said:
Argue that this approximates

1) [tex] \sum_n \sqrt{ 1 + f'(x+n dx) } dx [/tex]

Argue that this is a Riemann sum to approximate the integral

2) [tex] \int \sqrt{1 + f'(x)} dx [/tex]

to get expression 1) you're noticing that [itex]\frac{f(x + (n+1)dx) - f(x + ndx)}{dx}[/itex] is just [itex]f'(x + ndx)[/itex], i feel like that makes sense but I'm not sure why? both these steps make sense to me but I can't seem to reason them out. At any rate I'm going to reveiw integrals maybe that will help clear this up. Thanks again for your time
 
  • #4
[tex]dl=\sqrt{dx^2+dy^2}=dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}[/tex]
[tex]l=\int dl=\int \sqrt{dx^2+dy^2}=\int dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}[/tex]
 
  • #5
lurflurf said:
[tex]dl=\sqrt{dx^2+dy^2}=dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}[/tex]
[tex]l=\int dl=\int \sqrt{dx^2+dy^2}=\int dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}[/tex]

AH! Perfect, that makes perfect sense. Thank you very much lurflurf.
 
  • #6
After thinking about it, there's a problem.

take :
[tex]y = x[/tex]

this means:
[tex]\frac{dy}{dx} = 1 [/tex]

which means you'll be taking the integral of zero, so you will have no path lenth. also, when taking the definite integral between two points, you can run into imaginary path lengths, there's something wrong
 
  • #7
gordonj005 said:
After thinking about it, there's a problem.

take :
[tex]y = x[/tex]

this means:
[tex]\frac{dy}{dx} = 1 [/tex]

which means you'll be taking the integral of zero, so you will have no path lenth. also, when taking the definite integral between two points, you can run into imaginary path lengths, there's something wrong

I'm not sure what you were just looking at that got you confused, but the formula lurflurf posted has 1 + (y')^2 in the square root, not 1 - (y')^2. There's no problem.
 
  • #8
Of course, nevermind. Thanks again everyone.
 
  • #9
No, that is a "+" inside the square root not a "-".
 

1. What is a path integral derivation?

A path integral derivation is a mathematical technique used to solve problems involving quantum mechanics and statistical mechanics. It involves breaking down a complex system into smaller, simpler parts and integrating over all possible paths that the system could take. This allows for a more complete and accurate understanding of the system's behavior.

2. How is a path integral derivation different from other mathematical techniques?

Unlike other mathematical techniques, a path integral derivation takes into account all possible paths that a system could take, rather than just the most likely or most probable path. This allows for a more comprehensive and accurate analysis of the system.

3. What are some applications of path integral derivation?

Path integral derivation has various applications in physics, such as in quantum field theory, quantum electrodynamics, and statistical mechanics. It can also be used in other fields, including economics and finance, to analyze complex systems with many interacting variables.

4. What are the limitations of path integral derivation?

One limitation of path integral derivation is that it can be mathematically complex and difficult to perform for systems with many variables. It also requires knowledge of the underlying physical laws and principles governing the system, which may not always be known. Additionally, the accuracy of the results depends on the chosen path integral approximation.

5. How is the path integral derivation used in real-world experiments?

In real-world experiments, the path integral derivation is often used to calculate the probability of a system's outcome and to predict the behavior of the system under different conditions. It can also be used to analyze data and make predictions about future events. However, due to its complexity, it is not always feasible to use in every experiment and is often used in conjunction with other mathematical techniques.

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