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Free Particle Action

by gulsen
Tags: action, free, particle
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gulsen
#1
Jun21-07, 12:31 PM
P: 218
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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smallphi
#2
Jun22-07, 12:40 AM
P: 443
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.
lightarrow
#3
Jun22-07, 10:50 AM
P: 1,521
Quote Quote by gulsen View Post
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.

hunt_mat
#4
Feb17-11, 05:06 PM
HW Helper
P: 1,583
Free Particle Action

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.
RedX
#5
Feb17-11, 10:18 PM
P: 969
Quote Quote by hunt_mat View Post
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).


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