Lagrangian Mechanics

[Will]

Hello,
I am reading "Quantum Mechanics and Path Integrals" and the first few chapters are a 'recap' (at least it would be for the age of people it's aimed at) of classical mechanics. There is a section on the Lagrangian and principle of least action etc and the derivation of Lagrange's equation is fine and I understand all that (at least I think I do). Then there is a set of problems. The first if "For a free particle L=(m/2)(x(dot))^2 show that the classical action is (m/2)((x(subscript b) - x(subscript a))^2/(t (subscript b) - t(subscript a))) . So, take the partial derivative of L with respect to x(dot) and get m(x(dot)), fine. Then with respect to just x, I went about this by saying that (for d read partial differential) (dL/dx)=(dl/dx(dot)) x (dx(dot)/dx) so then m(x(dot)) x d(x(dot))/dx = dL/dx so m(x(dot)) x x(dot dot) = dl/dx (not sure about this part, I did d(x(dot))/dx by another method and got x(dot dot)/x(dot), unsure which (if either) is correct)) then sub this into Lagrange's equation: (d is no longer partial) d/dt(mx(dot)) - mx(dot)x(dot dot) = 0 then I'm not entirely sure where to go from here, or even if the above is relevant and correct.

Thanks,
Will

EDIT: this particular problem is now solved! See post 5

 

[Krikkit]

If I am reading this right(which was not easy in this script), your second partial should just = 0.

 

[Will]

You mean dL/(dx(dot)) ? I thought that at first but surely because x(dot) is a function of x that isn't the case?
I'll see if I can type this up in Word's equation editor tomorrow and paste that into here.

 

[Ares]

Hi Will,

If I remember this correctly, you treat the coordinates and their derivatives as separate entities in the Euler-Lagrange equation. So (*=dot):

dL/dx* = mx*
dL/dx = 0
mx** = 0

Then:
S = INTEGRAL (L dt) ta < t < tb

 

[Will]

Thanks for the help guys! I think I got it:

x*=xdot
ta = initial time (t subscript a, only partially translated from MS word to OF)
xa = initial position
tb = final time
xb = final position

L=(m/2)x*

∂L/∂x* = mx*
∂L/∂x = 0

(d/dt)(mx*) = 0 = mx** => x** =0 => x(t) = A+Bt

xa = A+Bta
xb = A+Btb
A=xa - Bta
(xb – xa)/(tb – ta) = B

S=∫ L dt (t between ta and tb)
=∫ (m/2)(B^2) dt
=(m/2)(B^2) (∫dt) = (m/2)(B^2) (tb – ta)
=(m/2)((xb – xa)/(tb – ta))^2(tb – ta)
=(m/2)((xb – xa)^2/(tb – ta))
QED

That's just the first problem in the book though, so I will probably need help with others too.....