# 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:

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