Solving Free Particle Action with Feynman & Gibbs

In summary, the conversation discusses the solution to Exercise 1-1 in Feynman & Gibbs, which asks for the action of a free particle. The participants discuss different approaches, including using the Euler-Lagrange equations and substituting a known solution for the path. The final equation for the action is derived and its derivatives are shown to equal momentum and negative energy.
  • #1
gulsen
217
0
I've recently started Feynman & Gibbs. I was sure exercises will be fun, but i can't enjoy myself when i fail solving the first one! Exercise 1-1 says: show that free particle action is

[tex]\frac{m}{2} \frac{x_b^2 - x_a^2}{t_b-t_a}[/tex]

I tried finding anti-derivative of [itex]\dot x^2[/itex], ended up with [itex]x\dot x - \int x d(\dot x)[/itex] via integration by parts. Couldn't do much about the integral. Of course, I know the solution, and can show it using Euler-Lagrange equations, [itex]x=vt[/itex] where [itex]v[/itex] is a constant (taking [itex]x_0=0[/itex]). I can "solve" the question by substituting [itex]x[/itex] in the action integral

[tex]S = \int_{t_a}^{t_b} \frac{m}{2}v^2 dt[/tex]

but i suppose this counts as cheating --kinda solving backwards.

Any ideas on how this kind of stuff can be solved?
 
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  • #2
It's not cheating. The action have different values on different paths. The principle of least action claims that the actual path will extremize the action.

They want you to find the path that extremizes the action by using the usual Euler-Lagrange equations. You know the answer should be motion of constant velocity, x = xa + vt, where v = const = (xb-xa)/(tb-ta). Then you plug that in and evaluate the action on that path.
 
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  • #3
gulsen said:
I've recently started Feynman & Gibbs. I was sure exercises will be fun, but i can't enjoy myself when i fail solving the first one! Exercise 1-1 says: show that free particle action is

[tex]\frac{m}{2} \frac{x_b^2 - x_a^2}{t_b-t_a}[/tex]
Shouldn't it be:

[tex]\frac{m}{2} \frac{(x_b - x_a)^2}{t_b-t_a}[/tex]
...
Any ideas on how this kind of stuff can be solved?
The exercise asks you to find the action S. How S is defined? It's defined as

[tex]S = \int_{t_a}^{t_b} \L dt[/tex]

where L = T - V; T = kinetic energy and V = potential energy;

In this case, V = 0 and T = [tex] \frac{m}{2}v^2 [/tex] and to write kinetic energy you surely don't need Eulero-Lagrange equations, I hope, kinetic energy is defined in that way!

So:

[tex]S = \int_{t_a}^{t_b} \frac{m}{2}v^2 dt[/tex]

Where is the problem? Always start from definitions.
 
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  • #4
This looks wrong but the equation of motion here is [tex]\ddot{x}=0[/tex], so:

[tex]
S=\frac{m}{2}\int_{t_{a}}^{t_{b}}\dot{x}^{2}dt=\left[ x\dot{x}\right]_{t_{a}}^{t_{b}}-\int_{t_{a}}^{t_{b}}x\ddot{x}dt
[/tex]

Now in second integral is zero by the equation of motion. Not too sure where to go from here.
 
  • #5
hunt_mat said:
This looks wrong but the equation of motion here is [tex]\ddot{x}=0[/tex], so:

[tex]
S=\frac{m}{2}\int_{t_{a}}^{t_{b}}\dot{x}^{2}dt=\left[ x\dot{x}\right]_{t_{a}}^{t_{b}}-\int_{t_{a}}^{t_{b}}x\ddot{x}dt
[/tex]

Now in second integral is zero by the equation of motion. Not too sure where to go from here.

Remember that [tex]\dot{x} [/tex] is the same at the initial time and final time, since [tex]\ddot{x}=0 [/tex].

So you get [tex]\left[ x\dot{x}\right]_{t_{a}}^{t_{b}}=\dot{x}(x_b-x_a)=\frac{(x_b-x_a)^2}{t_b-t_a} [/tex]

So you can check that the derivative of the action:

[tex]S=\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a} [/tex]

with respect to x_b equals momentum, and with respect to time equals negative energy (which in this case is just kinetic).
 

1. How does Feynman & Gibbs solve free particle action?

Feynman & Gibbs use the path integral approach to solve free particle action, which involves summing over all possible paths a particle can take from one point to another. This is done by assigning a probability amplitude to each path and then summing them together.

2. What is the significance of solving free particle action?

Solving free particle action allows us to understand the behavior and dynamics of particles in a given system. This has applications in various fields such as quantum mechanics, statistical mechanics, and condensed matter physics.

3. How does the Feynman & Gibbs approach differ from other methods?

The Feynman & Gibbs approach is based on the path integral formulation, which is a more intuitive and visual way of understanding particle behavior compared to other methods such as the Schrödinger equation. It also allows for the inclusion of quantum effects and interactions with other particles.

4. What are the limitations of solving free particle action with Feynman & Gibbs?

The path integral approach can become computationally challenging when dealing with complex systems or interactions between multiple particles. It also requires a good understanding of mathematical concepts such as integrals and complex numbers.

5. How is the Feynman & Gibbs approach used in practical applications?

The Feynman & Gibbs approach is used in various practical applications, such as calculating the probability of a particle's path in a given system, predicting particle behavior in quantum systems, and simulating complex physical processes in fields such as quantum computing and particle physics.

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