Principle of least action with transformed time variable

Question

I have the following action $$S=\int L(t,a,\dot{a},\ddot{a}) \ dt,$$ $$L=3\ a^2\ \ddot{a}+3\ a\ (\dot{a})^2+3\ a.$$ I wish to find the function $a(t)$ which minimizes the action.

I use the Euler-Lagrange equation: \begin{eqnarray} \frac{\delta S}{\delta a}&=&\frac{\partial L}{\partial a}-\frac{d}{dt}\frac{\partial L}{\partial \dot{a}} \\ &=& 6\ a\ \ddot{a} + 3\ (\dot{a})^2 + 3 - \frac{d}{dt}(6\ a\ \dot{a}) \\ &=& 6\ a\ \ddot{a} + 3\ (\dot{a})^2 + 3 - (6\ (\dot{a})^2+6\ a\ \ddot{a}) \\ &=& -3\ (\dot{a})^2 + 3 \\ &=& 0 \end{eqnarray} Therefore $$a(t)=t.$$

If I change the time variable to $\eta$ where $d\eta=dt/a$ then the above solution is given by $$\int d\eta = \int \frac{dt}{t}$$ $$\eta = \log t$$ $$e^\eta=t$$ $$a(\eta)=e^\eta.$$

I should be able to derive the solution in terms of $\eta$ directly by varying the action expressed in terms of $\eta$. In order to derive such an expression I need to be able to convert derivatives of $a$ in terms of time $t$, $\dot{a}$, into derivatives of $a$ in terms of time $\eta$, $a'$. $$\dot{a}=\frac{da}{dt}=\frac{d\eta}{dt}\frac{da}{d\eta}=\frac{da/d\eta}{a}=\frac{a'}{a},$$ $$\ddot{a}=\frac{d\dot{a}}{dt}=\frac{d}{dt}(a'a^{-1})=\frac{1}{a}\frac{d}{d\eta}(a' a^{-1})=\frac{1}{a}(a''a^{-1}-a'a^{-2}a')=\frac{a''}{a^2}-\frac{(a')^2}{a^3}.$$

Therefore the Lagrangian $L$ in the new time variable $\eta$ is given by \begin{eqnarray} L &=& 3\ a^2\ \ddot{a}+3\ a\ (\dot{a})^2+3\ a \\ &=& 3\ a^2 \ \Big(\frac{a''}{a^2} - \frac{(a')^2}{a^3}\Big)+3\ a\ \Big(\frac{a'}{a}\Big)^2+3\ a \\ &=& 3\ a'' + 3\ a. \end{eqnarray} The action $S$ is then given by \begin{eqnarray} S &=& \int L(t,a,\dot{a},\ddot{a}) \ dt \\ &=& \int L(\eta,a,a'')\ a\ d\eta \\ &=& \int (3\ a \ a'' + 3\ a^2)\ d\eta \end{eqnarray} I now use the Principle of least action with the new time variable $\eta$ \begin{eqnarray} \frac{\delta S}{\delta a}&=&\frac{\partial L}{\partial a}-\frac{d}{d\eta}\frac{\partial L}{\partial a'} \\ &=& 3\ a'' + 6\ a \\ &=& 0 \end{eqnarray} Thus I get the equation $$a''=-2\ a$$ I was expecting the equation $$a''=a$$ which has the solution $$a(\eta)=e^\eta.$$

What's gone wrong?

Answer 1

When you have a double time-derivative as in $\ddot a$, the correct Euler-Lagrange equation is $$ \frac{\partial L}{\partial a}-\frac{d}{dt}\frac{\partial L}{\partial \dot a}+\frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot a}=0\, . $$ The generalization to higher order derivatives is called (in some texts) the Euler operator (although this is not unique).

A very good reference for this is Peter Olver's Applications of Lie Groups to Differential Equations (Graduate Texts in Mathematics) by Springer.