Question

Peculiar choice of representation of action.

Answer 1

So just to get the stuff out of the way for those who don't know,

\[S = \int_{t_a}^{t_b} L(\dot x, x, t) dt\]
S is the action, and the Euler-Lagrange equation is:
\[\frac{\partial L}{\partial x} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot x} \right) =0\]

So now given \(L = \frac{m}{2} \dot x^2\), I am expected to show that:

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

However I don't understand why this particular form is useful or important. The way I solve it, x is a function of t, so I end up with S as solely a function of t in the end. I can algebraically turn one into the other, but I feel like I've missed the point somehow.

Answer 2

My way of solving it is this:

I solve the Euler-Lagrange equation since I have L to get:
\[x(t) = \frac{C}{m}t + K\]

so \[\int \frac{m}{2} (\frac{C}{m})^2 dt = \frac{C^2}{2m}(t_b-t_a)\]
Similarly, just plugging in my x(t) into what they want, gives the same thing.

Anywho, any kind of reasoning for why they want that special form would be nice.

Answer 3

My best guess is perhaps there's some cute way of solving this in general:

\[\int_a^b (f'(x))^2 dx\]

Answer 4

for fun, gets us nowhere I think,

\[\int (y')^2 dx = yy' -\int yy''dx\]

Answer 5

i was wondering why the form they want pops out more naturally (least i think it does) if we switch \(L(t, x, \dot x) \to \bar L (x, t, t') \)

\(S = \dfrac{m}{2} ~ \int\limits_{x1}^{x2} \dfrac{1}{t'^2} ~ t' ~ dx = \dfrac{m}{2} ~ \int\limits_{x1}^{x2} \dfrac{1}{t'} ~ dx\)

so applying the full E-L to this \(\dfrac{d}{dx } \left( -\dfrac{1}{t'^2} \right) = 0 \implies t' = const = alpha \) or \(t_2 - t_1 = \alpha (x_2 - x_1) \)

so the action is \(S = \dfrac{m}{2} ~ \int\limits_{x1}^{x2} \dfrac{1}{\alpha} ~ dx = \dfrac{m}{2} ~ \dfrac{(x_2 - x_1)}{\alpha} = \dfrac{m}{2} ~ \dfrac{(x_2 - x_1)}{\frac{(t_2 - t_1)}{(x_2 - x_1)}} = \dots\)


clutching at straws - the solution posted is totally by the book, and it's all swings and roundabouts - but, i think we are now guaranteed invariance in x rather than t. so wondering if this is to do with invariance itself?

is that really telling us about conservation of linear momentum?

dunno.

@Michele_Laino

Answer 6

Fascinating I didn't think of that substitution flipping the dependence like that, cool!
\[L(t,x(t),\tfrac{dx(t)}{dt}) = \bar L(x, t(x), \tfrac{dt(x)}{dx})\]
Specifically I hadn't really considered this being useful in that way before, I had more only used this when thinking of taking derivatives of matrices of several functions to get inverses, but I guess these are like 1x1 matrices multiplied to make a 1x1 identity matrix xD
\[\frac{dx(t)}{dt} *\frac{dt(x)}{dx} = 1\]

Thanks, your comment about invariance is good too, it seems to fit better with the book I'm reading since I'm summing over paths which I hadn't really considered need to be invariant, while the time doesn't necessarily I think.