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What is the physical interpretation of $$ \int_{t_1}^{t_2} (T -V) dt $$ where, $T$ is Kinetic Energy and $V$ is potential energy.
How does it give trajectory?

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Possible duplicate: (closed) and – Qmechanic Aug 26 '12 at 6:50
up vote 2 down vote accepted

The quantity $$ S= \int_{t_1}^{t_2} (T -V) dt $$ is known as the classical action. There exists a physical law (called the "principle of least action") which says that the true path an object takes is that which minimizes $S$.

Check that it's true. I'll throw a ball straight up. When the ball leaves my hand its kinetic energy $T$ is high, and since nature prefers to minimize the integral $S$, the potential energy of the ball $V$ rises quickly to minimize the integrand $T-V$. The principle of least action, then, explains why balls go up when you throw them.

So why don't baseballs keep going into the stratosphere to make $T-V$ as small as possible? They would need a lot of kinetic energy to do that! So much that it would outweigh the additional negative contribution from $-V$. It turns out that the true path is somewhere in-between rising high and going fast, which is what we observe. (Balls slow down as they go up.)

Beyond this qualitative argument one may use Variational Calculus to derive Newton's laws from the principle of least action.

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Plus, the term T-V, is just the classical Lagrangian, and if you know that, using just what Alec S said, you can obtain the equations of motion for the given system :) – dingo_d Aug 26 '12 at 6:45

The only real physical interpretation of this quantity is in quantum mechanics. This is the phase of a contribution from a path that goes from $t_1$ to $t_2$ along the path $x(t)$. The two terms then are relatively clear, when you exponentiate this to make a phase and make time a lattice:

$$ e^{i \int_{t_1}^{t_2} (T - V)} = \prod_{t_1<t<t_2} e^{i {m(x(t+\epsilon)-x(t))^2\over 2\epsilon}}e^{-i \epsilon V(x(t))} $$

Where the product is over all t's between $t_1$ and $t_2$ in $\epsilon$ size steps.

The first term gives you the phase for free particle propagation from $x(t)$ to $x(t+\epsilon)$. The second term gives an extra phase rotation for the potential energy at the position $x(t)$. The two phases add up, and you add the phases over all the paths to get the total quantum propagation.

The classical path is then the place where the phase is stationary, so that the paths tend to add together with the same phase, rather than cancel due to interference. This is the place where a first-order variation in the path makes no change in the action.

To find this, you can shift $x(t)$ to $x(t)+\delta x(t)$, and find the leading variation

$$ e^S \int_t ( m \dot{x} {d\over dt}(\delta x) - V'(x) \delta x) dt $$

Integrating the first term by parts, you find

$$ m\ddot{x} = - V'(x) $$

or that the particle obeys Newton's law. The same derivation works in the path integral formalism to demonstrate Ehrenfest's theorem, and the Heisenberg equation of motion. This is because the path integral is invariant under shifts of the integration variable $x(t)$ by a constant amount $\delta x(t)$, even if that constant is different from time to time.

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