Landau's derivation of the Langangian expression

1. Jul 4, 2010

lewis198

I'm going over the Landau's Mechanics, and can't get over two hurdles.

The first is the following(I'm not good with latex here, so please bear with me):

Landau asserts that the action S' between t1 and t2 of a Lagrangian L such that L is a function of (q+dq) and (q'+dq') minus the action S between t1 and t2 of a Lagrangian L such that L is a function of q and q', the gives a difference, that, expanded in powers of dq and dq' equals zero. Also, dq(t1)=dq(t2)=0.

OK, for this to be the case, S' must equal S' plus some residue R. R must then be formed of integrals of terms that are multiples of dq and dq' so that when integrated (definite) give 0.

Landau therefore expands S'-S 'in powers of' dq ad dq'. Here's my problem: How can you expand L in powers of dq and dq'? Talyor expansion allows for expansion in terms of q and q', and if I substitute q for q+dq I do in fact get the terms in dq and dq', but I also have annoying terms like 2q*dq and the like- I can see that q and q^2 terms will vanish due to S expanded.

So the question still stands: How can you expand L in powers of dq and dq'? is the above way satisfactory?

2. Jul 4, 2010

zzzoak

Consider q and q' as two independent variables.

Let think that you have function
f(x,y) and use Talyor expansion for this function with two independent variables x and y.

Last edited: Jul 4, 2010
3. Jul 5, 2010

lewis198

I understand that, but here we have the variables 'x'=(q+dq) and 'y'=(q'+dq') and we want an expansion for dq and dq'.

4. Jul 5, 2010

zzzoak

f(x+dx,y+dy)-f(x,y)=(df/dx)dx+(df/dy)dy
where dx=dq and dy=dq'.

5. Jul 5, 2010

Dickfore

One way to derive the Euler-Lagrange equations is the following. Assume the extremal curve is $\bar{q}(t)$. We define a one-parameter family of trial functions:

$$q(t; \alpha) = \bar{q}(t) + \alpha \, \eta(t)$$

where $\eta(t)$ are arbitrary smooth functions which satisfy the boundary conditions:

$$\eta(t_{1}) = \eta(t_{2}) = 0$$

so that the trial functions automatically satistfy the boundary conditions:

$$q(t_{1}; \alpha) = \bar{q}(t_{1}) = q_{1}$$

$$q(t_{2}; \alpha) = \bar{q}(t_{2}) = q_{2}$$

It is obvious from our construction that among all the trial functions, the extermum is achieved for $\alpha = 0$, because $q(t; 0) = \bar{q}(t)$ then. But, if we insert the trial function in the expression for the action:

$$F(\alpha) = S[q(t; \alpha)] = \int_{t_{1}}^{t_{2}}{L(t, \bar{q} + \alpha \, \eta, \dot{\bar{q}} + \alpha \, \dot{\bar{q}} + \eta \, \dot{\eta}) dt}$$

it becomes an explicit function of the parameter $\alpha$. Using the necessary condition of extermum, we must have:

$$F'(0) = 0$$

If we try to calculate the derivative of $F(\alpha)$ with respect to $\alpha$, we will need to use the rules of finding parametric derivatives:

$$F'(0) = \int_{t_{1}}^{t_{2}}{\left\{ \frac{\partial L}{\partial q}(t, \bar{q}, \dot{\bar{q}}) \, \eta(t) + \frac{\partial L}{\partial \dot{q}}(t, \bar{q}, \dot{\bar{q}}) \, \dot{\eta}(t) \right\} \, dt}$$

If we try to express the $\eta(t)$ in terms of the trial functions and the extremal curve, we get:

$$\eta(t) = q(t; \alpha = 1) - \bar{q}(t) \equiv \delta q(t)$$

Notice that the variation of a function of time is again a function of time. We can differentiate it:

$$\frac{d}{d t} \delta q(t) = \dot{\eta}(t) = \dot{q}(t; \alpha = 1) - \dot{\bar{q}}(t) = \delta \frac{dq(t)}{d t}$$

This is the famous rule of exchanging variation and differentiation with respect to the argument of the function. If we also define:

$$F'(0) \approx S[q(t; \alpha = 1)] - S[\bar{q}(t)] \equiv \delta S[\bar{q}(t)]$$

Then, the above derivative may be rewritten as:

$$\delta S[q(t)] = \int_{t_{0}}^{t_{1}}{\left[\frac{\partial L}{\partial q}(t, q, \dot{q}) \, \delta q(t) + \frac{\partial L}{\partial \dot{q}}(t, q, \dot{q}) \, \delta \dot{q}(t)\right] \, dt}$$