# Why is that in the action principle, the Taylor's series is limited to the first order?

For the Hamilton's principle: $$\delta s =\int_{t_1}^{t_2}L(\mathbf {q+\delta q},\mathbf {\dot q+\delta \dot q},t) dt-\int_{t_1}^{t_2}L(\mathbf {q},\mathbf {\dot q},t) dt=0.\\$$

In the textbooks, using the Taylor's series:

$$L(\mathbf {q+\delta q},\mathbf {\dot q+\delta \dot q},t) =L(\mathbf {q},\mathbf {\dot q},t)+\frac{\partial L(\mathbf {q},\mathbf {\dot q},t)}{\partial \mathbf q} \bar \delta \mathbf q(t) +\frac{\partial L(\mathbf {q},\mathbf {\dot q},t)}{\partial \mathbf {\dot q}} \bar \delta \mathbf {\dot q(t)}+\frac{\partial}{\partial \mathbf q} (\frac{\partial L(\mathbf {q},\mathbf {\dot q},t)}{\partial \mathbf {\dot q}}\bar \delta \dot q)\bar \delta q + ...$$

But why is omitting the rest of the Taylor's series? I think it will propagate a lot of error in the derivation...

• The variations $\delta q$ must multiply a coefficient $\epsilon$ that you will eventually take the incremental limit on. Higher order derivatives will still present the multiplicative factor after the incremental ratio and will therefore vanish anyway. – gented Aug 28 '17 at 12:23

## 1 Answer

The Euler Lagrange equation is a differential equation resulting from the search for the extremum of a functional: this extremum is given by the first variation only.

This is similar to the condition for finding a point where a function $f$ is extremum: the condition $df/dx=0$ is on the first derivative only.

In both cases, one does not seek to approximate $L$ or $f$ by a series, but rather extract from the first term in the series either a differential equation to be satisfied by a function or an algebraic equation to be satisfied by a point.

Edit:

In addition, the order of the resulting Euler-Lagrange equation can be greater than 1 even if one considers only the first variation. For instance, in 1d, with a Lagrangian of the type $L=L(q,\dot{q},\ddot{q},t)$, the resulting EL equation is of second order: $$\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}+ \frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot{q}}=0.$$

Finally, the first variation (like the first derivative) only guarantees an extremum. To verify if this is a local max or min, one ought to use the second variation (like the second derivative) but it is often technically complicated and unnecessary: one can simply verify if any other construction gives a greater or smaller action or shorter or longer path.

• This is not the reason. The reason is that the variations $\delta q$ are multiplied by a coefficient that is set limit to zero afterwards. All terms in higher order derivatives still present the coefficient after the limit procedure and therefore vanish. – gented Aug 28 '17 at 12:21
• @GennaroTedesco What is multiplied by a coefficient, in the case shown above I didn't need to declarate a coefficient what is generally declated as $\epsilon$ – Joan J. Cáceres Aug 28 '17 at 12:33
• @GennaroTedesco This is equivalent, since the remaining coefficient is just proportional to the first variation (or the definition of the first derivative). – ZeroTheHero Aug 28 '17 at 12:34
• @JoanJ.Cáceres That is exactly the point. You want to define $\delta L$, which by definition is the limit of the incremental ratio $L(q+ \epsilon) - L(q)$ over $\epsilon$. Your notation is just shorthand notation for the whole thing. – gented Aug 28 '17 at 12:35