A question on the action as a function of the co-ordinates in Landau 1

[dfj]

A question on "the action as a function of the co-ordinates" in Landau 1

Landau Mechanics, volume 1, third edition, page 138-139 (section 43)

I can understand equations (43.1) - (43.3), which talk about treating the ending point of the path as a variable.

Then, they continue to treat the time of the ending point as a variable. In deriving equation (43.4), they said that

From the definition of the action, its total time derivative along the path is
dS/dt=L.

Here t is the time of the ending point of the path, while the positions of the starting point and ending point are fixed, and the time of the starting point is also fixed.

However, I don't understand why they say dS/dt = L can be derived from the definition of the action.

A simple counter-example would be like this:

Suppose
$$L=\frac{m}{2}{\dot{x}}^2$$

Then, in treating the time of the ending point as a variable, we have

$$S(t) = \int_0^t Ldt'=\frac{m}{2t}{\Delta x}^2$$

Here $$\Delta x=x_2-x_1$$.

Apparently, dS/dt is not equal to L.

 

[qbert]

Remember - we're already assuming the particle obeys the equations of motion.
so the position at the final time also varies.

L = 1/2 m v^2

$$S = \int_{t_0}^t \frac{m}{2} \dot{x}^2 dt$$
the equation of motion is
$$\frac{d}{dt}\left( m \dot{x} \right) = 0$$
so that $\dot{x} = v$ is a constant.

that is S = (1/2 m v^2) (t - t₀)
and dS/dt = L as advertised.

-----------------------------------------
So what this is in general:
we vary the paths to pick the actual path - which we in principle could
solve. we plug this path (say x(t)) back into the action
$$S = \int_{t_0}^t L(x(t'), \dot{x}(t'), t') dt' = \int_{t_0}^t L(t') dt'$$
now since we "know" what x(t) is the integrand is only a function of t.

Then by the fundamental theorem of calc
$$\frac{dS}{dt} = L(t) = L(x(t), \dot{x}(t), t).$$

 

[dfj]

Thanks for your reply!

The thing is, on page 139 of this book, just below equation (43.3), the sentences are

The action may similarly be regarded as an explicit function of time, by considering paths starting at a given instant t1 and at a given point q1, and ending at a given point q2 at various times t2=t. ... ...

Apparently, what these words tells me is that the positions of both the starting and ending points are fixed, and the time of the starting point is also fixed, while the time of the ending point is kept as a variable.

So, in the simple example we used above, as the time of the ending point changes, the velocity v also changes (instead of being kept as a constant), in order that the starting point and ending point are not altered.

I hope this is not merely a English problem...

 

[qbert]

Let's see if i can make this clearer.

the derivative dS/dt = L is the full derivative of S as a function of t. And we
ALLOW the ending point to vary in time (such that the equation of motion are
satisfied). This is what I have above.

the point you are referring to is an alternate way to get to $\frac{\partial S}{\partial t}$.
Specifically, you must vary t but NOT allow the endpoint to change in time.
This calculation is not done in the book. What should you get if you hold the
end points fixed and vary the time?
$\frac{\partial S}{\partial t} = -H$
This is what you calculate in post 1.

 

[dfj]

Thanks! I think I understand it now:

The derivative dS/dt is defined along each real "world line". Each point of the world line has a unique t, and L is defined at each point. Understanding in this way, of course dS/dt=L. No calculation, even no thinking is needed, as far as the picture is there. $\partial S/\partial q$ also follows naturally from such a picture.

Maybe all the problems are caused by my misunderstanding of the word "given", and I kind of ignored the words "along the path" just above equation (43.4).