Lagrangian L = T − V

WHAT is the Lagrangian?

Note the minus sign. The total energy is $E = T + V$; the Lagrangian is $T - V$. They are not the same thing, and the difference is the whole point (see Mistakes).

WHY $T - V$? (Derivation from first principles)

We want to show that minimizing the action with $L = T-V$ reproduces Newton's law $F = ma$. If it does, the choice $T-V$ is justified, not arbitrary.

Step 1 — Define the action. $S = \int_{t_1}^{t_2} L(q,\dot q,t)\, dt$ Why this step? We need a single number measuring an entire path so we can ask "which path is best." Integrating a scalar over time gives one number per path.

Step 2 — Demand the action be stationary. Consider a true path $q(t)$ and a nearby varied path $q(t)+\delta q(t)$ with fixed endpoints ($\delta q(t_1)=\delta q(t_2)=0$). Stationarity means $\delta S = 0$ to first order. Why fixed endpoints? We compare paths between the same start and end events; the question is the route, not the destination.

Step 3 — Vary $S$. $\delta S = \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta\dot q\right)dt$ Why? First-order Taylor expansion of $L$ in its arguments.

Step 4 — Integrate the second term by parts. Using $\delta\dot q = \frac{d}{dt}\delta q$: $\int \frac{\partial L}{\partial \dot q}\frac{d}{dt}\delta q\,dt = \left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_1}^{t_2} - \int \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right)\delta q\,dt$ The boundary term vanishes because $\delta q=0$ at the ends. Why by parts? To pull every $\delta q$ outside its derivative so we can factor it.

Step 5 — Factor and apply the fundamental lemma. $\delta S = \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q}\right)\delta q\, dt = 0$ Since $\delta q$ is arbitrary, the bracket must vanish: $\boxed{\;\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0\;}$ This is the Euler–Lagrange equation.

Step 6 — Plug in $L=T-V$ for one particle in 1D. With $T=\tfrac12 m\dot x^2$ and $V=V(x)$: $\frac{\partial L}{\partial \dot x}=m\dot x,\qquad \frac{d}{dt}(m\dot x)=m\ddot x,\qquad \frac{\partial L}{\partial x}=-\frac{dV}{dx}=F$ So the Euler–Lagrange equation gives $m\ddot x - F = 0 \;\Longrightarrow\; F = m\ddot x.$ This recovers Newton exactly. That is why $L=T-V$: it is the unique combination whose stationary-action condition reproduces $ma = -\nabla V$. If we had chosen $T+V$, the velocity term and force term would carry the same sign and we'd get the wrong physics.

HOW to use it (recipe)

1. Choose generalized coordinates $q_i$ that exploit the system's symmetry/constraints.
2. Write $T$ and $V$ in those coordinates.
3. Form $L=T-V$.
4. For each $q_i$ apply the Euler–Lagrange equation.

Worked examples

Common mistakes (Steel-manned)

Active recall

Connections

• Euler–Lagrange Equation — the equation of motion derived above.
• Principle of Least Action — the variational principle $\delta S=0$ that justifies $L$.
• Generalized Coordinates and Constraints — why constraint forces vanish.
• Generalized Momentum and Conservation — $p_i=\partial L/\partial\dot q_i$; cyclic coordinates.
• Hamiltonian H = T + V — Legendre transform of $L$; energy reappears here.
• Newton's Second Law — the special case $L$ reduces to.

Concept Map

Hinglish (regional understanding)