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UPSC 2015 Maths Optional Paper 2 Q7c-ii — Step-by-Step Solution

10 marks · Section B

Lagrange's equations · Mechanics & Fluid Dynamics · asked 9× in 13 yrs · Read the full method →

Question

Find an equivalent Lagrangian that is not explicitly dependent on time.

Technique

Add a total time derivative $dF/dt$ to absorb the $e^{-\alpha t}$ terms. The clever $F=-\alpha b q^2 e^{-\alpha t}/2$ kills both offending terms simultaneously.

Solution

Starting point (from Q7(c)(i)):

Strategy. Lagrangians are defined modulo a total time derivative $\dfrac{dF}{dt}$ (where $F=F(q,t)$). And we can also do a point transformation $Q=g(q,t)$ to eliminate time-dependence.

Step 1 — Try point transformation $Q=qe^{-\alpha t/2}$? Or $Q=qe^{\beta t}$?

The time-dependence enters $L$ through factors $e^{-\alpha t}$. Try $Q=qe^{\alpha t/2}$ so $q=Qe^{-\alpha t/2}$.

Then $\dot q=\dot Q e^{-\alpha t/2}-\dfrac{\alpha}{2}Qe^{-\alpha t/2}=e^{-\alpha t/2}(\dot Q-\dfrac{\alpha}{2}Q)$.

Hmm — let me check the structure first.

The original $H$ has time-dependence only via $e^{-\alpha t}$. The factor $e^{-\alpha t}$ multiplies several terms.

Try $Q=qe^{-\alpha t/2}$, so $q=Qe^{\alpha t/2}$, $\dot q=\dot Q e^{\alpha t/2}+\dfrac{\alpha}{2}Qe^{\alpha t/2}=e^{\alpha t/2}(\dot Q+\dfrac{\alpha}{2}Q)$.

Then:

• $\dot q^2=e^{\alpha t}(\dot Q+\dfrac{\alpha}{2}Q)^2$.
• $q\dot q=Qe^{\alpha t/2}\cdot e^{\alpha t/2}(\dot Q+\dfrac{\alpha}{2}Q)=e^{\alpha t}Q(\dot Q+\dfrac{\alpha}{2}Q)$.
• $q^2=Q^2 e^{\alpha t}$.

Substitute into $L$:

• $\dfrac{\alpha\dot q^2}{2}=\dfrac{\alpha e^{\alpha t}}{2}(\dot Q+\dfrac{\alpha}{2}Q)^2$.
• $\alpha bq\dot q e^{-\alpha t}=\alpha b\cdot e^{\alpha t}Q(\dot Q+\dfrac{\alpha}{2}Q)\cdot e^{-\alpha t}=\alpha b Q(\dot Q+\dfrac{\alpha}{2}Q)$.
• $-\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}=-\dfrac{b\alpha^2 Q^2 e^{\alpha t}\cdot e^{-\alpha t}}{2}=-\dfrac{b\alpha^2 Q^2}{2}$.
• $-\dfrac{kq^2}{2}=-\dfrac{k Q^2 e^{\alpha t}}{2}$.

So

Still has $e^{\alpha t}$ factors. Doesn’t remove explicit time. Try the other direction:

Step 2 — Try $Q=qe^{\beta t}$ for general $\beta$

$q=Qe^{-\beta t}$, $\dot q=(\dot Q-\beta Q)e^{-\beta t}$.

$\dot q^2=(\dot Q-\beta Q)^2 e^{-2\beta t}$. $q\dot q=Q(\dot Q-\beta Q)e^{-2\beta t}$. $q^2=Q^2 e^{-2\beta t}$.

Substitute:

• $\dfrac{\alpha\dot q^2}{2}=\dfrac{\alpha}{2}(\dot Q-\beta Q)^2 e^{-2\beta t}$.
• $\alpha b q\dot q e^{-\alpha t}=\alpha b Q(\dot Q-\beta Q)e^{-(2\beta+\alpha)t}$.
• $-\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}=-\dfrac{b\alpha^2}{2}Q^2 e^{-(2\beta+\alpha)t}$.
• $-\dfrac{kq^2}{2}=-\dfrac{k}{2}Q^2 e^{-2\beta t}$.

For no explicit time-dependence, need the exponentials to cancel: requires $-2\beta=0$ and $-(2\beta+\alpha)=0$, which gives $\beta=0$ (trivial) or $\beta=-\alpha/2$ for the second, conflicting.

Alternatively, multiply $L$ by an overall function — but this changes the equations of motion unless it’s a constant.

Step 3 — Add a total time derivative

Lagrangians related by total $d/dt$ of a function give the same equations of motion. Try adding $\dfrac{dF}{dt}$ to $L$ for some $F(q,t)$ that absorbs the time dependence.

Try $F=-\dfrac{\alpha b q^2 e^{-\alpha t}}{2}$. Then

So $L+\dfrac{dF}{dt}$:

• $\dfrac{\alpha\dot q^2}{2}$ unchanged.
• $\alpha b q\dot q e^{-\alpha t}-\alpha b q\dot q e^{-\alpha t}=0$.
• $-\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}+\dfrac{\alpha^2 b q^2 e^{-\alpha t}}{2}=0$.
• $-kq^2/2$ unchanged.

So

No explicit time dependence!

Answer