Lagrangian L = T − V: Question bank — Physics deep dive

Before we start, a reminder of the symbols so nothing is used unearned:

$T$ = kinetic energy (the "moving" energy, $\frac{1}{2} m v^{2}$ for a point mass).
$V$ = potential energy (the "stored" energy, e.g. $m g y$ or $\frac{1}{2} k x^{2}$ ).
$L = T - V$ = the Lagrangian, one scalar number at each instant.
$q$ = a generalized coordinate: any independent number that pins down the configuration (a position $x$ , or just an angle $θ$ ).
$\overset{q}{˙}$ = its time-rate-of-change (generalized velocity), read "q-dot".
$\overset{q}{¨}$ = the second time-derivative, i.e. acceleration: $\overset{q}{¨} = \frac{d ^{2} q}{d t ^{2}}$ (two dots = differentiate with respect to time twice). So $\ddot{θ}$ is angular acceleration, $\overset{x}{¨}$ is ordinary acceleration.
$δ$ = the variation symbol: $δ q$ means a tiny "wiggle" you add to a path (a small trial change), and $δ S$ is the resulting first-order change in the action. "Stationary" means $δ S = 0$ : no first-order change under any small wiggle.
$S = \int_{t_{1}}^{t_{2}} L d t$ = the action, one number for a whole path.

Four pictures anchor the whole page — glance at them before you start so the words below have something to point at.

Picture 1 — the action is an area. $L$ is just a number that changes as the system moves; plot $L$ against time and the area under that curve is the action $S$ . Two different paths give two different curves, hence two different shaded areas.

Picture 2 — varying the path. The real path (solid) is surrounded by wiggled "trial" paths (dashed) that share the same two endpoints. Nature picks the one whose action doesn't change under tiny wiggles, i.e. $δ S = 0$ .

Picture 3 — stationary is not the same as minimum. Slide along the family of paths and watch $S$ : at the true path the slope is flat, but flat can be a valley bottom (minimum) or a saddle. "Least action" is a nickname, not a law.

Picture 4 — the pendulum coordinate. One angle $θ$ from the vertical fixes the bob completely, even though it lives in a 2D plane. The rod is the constraint that erased a degree of freedom.

True or false — justify

L = T + V

False.

L = T - V

; the plus-sign object is the total energy

E = T + V

, which plays a different role. Only the minus sign makes the equations of motion come out with the correct force sign.

The action

S

is a number attached to a single instant.

False. Look at Picture 1:

S = \int L d t

is the area under the whole $L$ -vs- $t$ curve between two fixed events, so

S

is one number per candidate path, not per instant.

Nature always minimizes the action.

False in general. Picture 3 shows the point: nature makes the action stationary (

δ S = 0

), which is a flat slope — a saddle or maximum has a flat slope too, so "least action" is a historical name, not always literally a minimum.

The Euler–Lagrange equation must be solved once for the whole system at once.

False. It holds independently for each generalized coordinate

q_{i}

: one equation per degree of freedom, obtained by varying that coordinate alone while holding the others fixed.

Inside

L (q, \overset{q}{˙}, t)

, the variables

q

and

\overset{q}{˙}

are treated as independent when taking partial derivatives.

True. When forming

\partial L / \partial q

and

\partial L / \partial \overset{q}{˙}

you freeze the other; only after building

\frac{d}{d t} \frac{\partial L}{\partial q ˙}

does the chain rule reconnect them.

The Lagrangian method still needs you to draw tension and normal forces.

False. Look at Picture 4: because

θ

runs along the circle the bob is allowed to move on, the rod's tension (which points along the rod, perpendicular to the motion) does no work and never enters

L

For any smooth function

f (q, t)

, the Lagrangians

L

and

L^{'} = L + \frac{df ( q , t )}{d t}

give the same equations of motion.

True. Here

f (q, t)

is any differentiable function of the coordinate and time (a free "gauge" choice, not a physical energy). Its integral over the path is

\int \frac{df}{d t} d t = f (q (t_{2}), t_{2}) - f (q (t_{1}), t_{1})

, which is fixed by the endpoints, so it adds a constant to

S

and cannot change where

δ S = 0

T

and

V

are both zero the Euler–Lagrange equation is meaningless.

False but degenerate.

L = 0

gives

0 = 0

trivially; there is simply no dynamics — a free, forceless, potential-free "system" is consistent, just empty.

The Lagrangian is always

T - V

even for a charged particle in a magnetic field.

Roughly false as stated. Velocity-dependent (magnetic) forces need a generalized potential

U (q, \overset{q}{˙})

, and

L = T - U

; the plain "

T

minus a position-only

V

" form is the special case of conservative forces.

A Lagrangian whose kinetic part is

\frac{1}{2} m \overset{q}{˙}^{2}

can always be solved for

\overset{q}{¨}

False in general. If the coefficient of

\overset{q}{˙}^{2}

(the "mass matrix") degenerates to zero the Lagrangian is singular:

\partial^{2} L / \partial \overset{q}{˙}^{2} = 0

and you cannot invert to get

\overset{q}{¨}

. Such singular Lagrangians (common in field theory / with constraints) need the special machinery of constrained dynamics.

Spot the error

"Since

E = T + V

and

L = T - V

, then

L = E - 2 V

, so knowing

L

and

V

gives energy for free."

The algebra is fine but the identity hides a condition:

E

is conserved only when

\partial L / \partial t = 0

(no explicit time in

L

). Concretely, the conserved quantity is

H = \overset{q}{˙} \frac{\partial L}{\partial q ˙} - L

, and

\frac{d H}{d t} = - \frac{\partial L}{\partial t}

; if

\partial L / \partial t \neq = 0

then

H

(hence

E

) drifts and the "free energy" claim is empty.

"I formed

L

, then set

\frac{\partial L}{\partial q} = 0

to find the motion."

Error: that is not the equation of motion. You must use the full

\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = 0

; dropping the first term throws away the entire dynamics (it would only give equilibrium points).

"For the pendulum I put the string tension into

V

so gravity plus tension are both accounted for."

Error: tension is an ideal constraint force that does no work; it must not enter

V

. Only genuine potentials (gravity here) belong in

V

. Choosing

θ

already removes the tension.

"When varying the action, the boundary term

[\partial L / \partial \overset{q}{˙} δ q]

(left over after integrating by parts) vanishes because

\partial L / \partial \overset{q}{˙} = 0

Error: the momentum is not zero. The boundary term dies because

δ q = 0

at both endpoints (fixed-endpoint condition, exactly the shared endpoints in Picture 2), not because the momentum factor vanishes.

"Because

\overset{q}{˙}

is the derivative of

q

, when I differentiate

L

with respect to

q

I must also account for how

\overset{q}{˙}

changes."

Error: during the partial derivatives

q

and

\overset{q}{˙}

are independent slots. Coupling them prematurely produces spurious extra terms; the linkage is applied later via

\frac{d}{d t}

"I used

L = T + V

and got

m \overset{x}{¨} + F = 0

; that's basically Newton, just rearranged."

Error: it is the wrong sign. Newton gives

m \overset{x}{¨} = F

, i.e.

m \overset{x}{¨} - F = 0

. The

T + V

version predicts anti-physics (force pushing the wrong way), which is exactly why

T - V

is mandatory.

"There's friction on my block, so I'll just add a friction term to

V

and use

L = T - V

Error: friction is dissipative and cannot come from any potential

V (q)

(it depends on velocity direction and drains energy). It enters instead as a generalized force

Q_{nc}

(the "nc" = non-conservative: any force, like friction or drag, not derivable from a potential) on the right-hand side:

\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = Q_{nc}

. Plain

L = T - V

only covers conservative and velocity-dependent-potential forces.

"I'll model a rolling-without-slipping constraint by just writing it into

V

Error: rolling-without-slipping is often non-holonomic — a relation between velocities that cannot be integrated into a coordinate relation, so it cannot be absorbed into coordinates or

V

. It must be imposed with Lagrange multipliers, unlike the holonomic pendulum rod which can be baked into

θ

Why questions

Why do we demand fixed endpoints when varying the path?

Because (Picture 2) we compare different routes between the same start and end events; letting endpoints move would ask a different question and would leave a nonzero boundary term that spoils the derivation.

Why integrate

L

over time to form the action instead of just examining

L

at one moment?

One instant cannot decide a whole path. Integrating produces a single scalar (the shaded area of Picture 1) per candidate path, which is exactly what you need to compare paths and pick the stationary one.

Why does the minus sign in

T - V

matter physically?

Because

\frac{\partial L}{\partial q} = - \frac{d V}{d q} = F

; the minus turns "stored energy" into the correct force direction, so the Euler–Lagrange result matches

F = m \overset{q}{¨}

. Flip it and the force sign flips.

Why can a single angle

θ

fully describe a pendulum that lives in 2D space?

The rigid rod is a constraint that removes one degree of freedom; the bob is stuck on a circle (Picture 4), so one coordinate (the angle) suffices — this is the payoff of generalized coordinates.

Why does a coordinate that does not appear in

L

(only its

\overset{q}{˙}

appears) signal a conservation law?

\frac{\partial L}{\partial q} = 0

then

\frac{d}{d t} \frac{\partial L}{\partial q ˙} = 0

, so the generalized momentum

p = \frac{\partial L}{\partial q ˙}

is constant — see Generalized Momentum and Conservation.

Why does an explicit time dependence in

L

break energy conservation?

Think of it as symmetry: if

L

looks identical at every instant (no explicit

t

), sliding the whole experiment forward in time changes nothing, and that time-translation symmetry is energy conservation. But if someone is actively cranking a parameter (say

L

depends on

t

through a moving support), the "same experiment tomorrow" is a different experiment — the symmetry is broken. Quantitatively

\frac{d H}{d t} = - \frac{\partial L}{\partial t}

, so the energy leaks at exactly the rate the clock is written into

L

Why does the Lagrangian method reduce to Newton's Second Law for a single particle in 1D?

Plugging

L = \frac{1}{2} m \overset{x}{˙}^{2} - V (x)

into Euler–Lagrange yields

m \overset{x}{¨} = - \frac{d V}{d x} = F

exactly; Newton is the special case, and matching it is how

L = T - V

was justified.

Why is the Hamiltonian built from the Lagrangian rather than replacing it?

H

is the Legendre transform of

L

, swapping

\overset{q}{˙}

for momentum

p

; it re-expresses the same physics with energy as the star quantity, but the underlying content still comes from

L

Edge cases

What happens to the Euler–Lagrange equation for a free particle (

V = 0

\frac{\partial L}{\partial q} = 0

, so

\frac{d}{d t} (m \overset{q}{˙}) = 0

: the momentum is conserved and the particle moves in a straight line at constant speed — the stationary path is literally the straight one.

What is the motion at the pendulum's stable equilibrium

θ = 0

\ddot{θ} = - \frac{g}{ℓ} sin θ

(recall

\ddot{θ} = d^{2} θ / d t^{2}

) gives

\ddot{θ} = 0

θ = 0

, so it sits still if placed there; nearby it oscillates since the restoring term

sin θ

pushes back toward zero.

What about the pendulum's inverted position

θ = π

sin π = 0

\ddot{θ} = 0

: it is also an equilibrium, but unstable — any tiny displacement grows because the restoring sign flips, so a real bob won't stay there.

Does the method work when

L

has explicit time dependence, e.g. a driven oscillator?

Yes, Euler–Lagrange still holds; but then energy is generally not conserved because

\frac{d H}{d t} = - \partial L / \partial t \neq = 0

— the time-translation symmetry that would guarantee conservation is broken.

For small angles, does the pendulum equation collapse to something simpler?

Yes:

sin θ \approx θ

gives

\ddot{θ} \approx - \frac{g}{ℓ} θ

, simple harmonic motion with

ω = g / ℓ

— a limiting case recovering the SHM form.

What if two candidate paths give the same action value?

The action being stationary is about vanishing first-order change

δ S = 0

, not uniqueness of value; degenerate actions can occur (e.g. symmetric paths) but each true path still satisfies Euler–Lagrange.

What if the kinetic term is degenerate (a singular Lagrangian)?

Then

\overset{q}{¨}

cannot be solved for from Euler–Lagrange, and hidden constraints appear instead of ordinary equations of motion — the standard

L = T - V

recipe silently assumes a non-degenerate (invertible) kinetic term.

Connections

Euler–Lagrange Equation — the equation every "spot the error" item leans on.
Principle of Least Action — clarifies "stationary, not minimum."
Generalized Coordinates and Constraints — why tension never enters $V$ ; holonomic vs non-holonomic.
Generalized Momentum and Conservation — the cyclic-coordinate items.
Hamiltonian H = T + V — the energy counterpart; $H = \overset{q}{˙} \partial L / \partial \overset{q}{˙} - L$ .
Newton's Second Law — the 1D limiting case.