The Lagrangian is Not Unique

Lagrangians are not in a one-to-one relationship with physical systems—many Lagrangians can be used to describe the same physical system. In this section we will demonstrate this by show-ing that the addition to the Lagrangian of a “total time deriva-tive” of a function of the coordinates and time does not change the paths of stationary action or the equations of motion deduced from the action principle.

Total time derivatives

Let’s first explain what we mean by a “total time derivative.” Let F be a function of time and coordinates. Then the time derivative of F along a pathq is

D(F ◦Γ[q]) = (DF ◦Γ[q])DΓ[q]. (1.108)

Because F only depends on time and coordinates:

DF◦Γ[q] = [∂₀F◦Γ[q], ∂₁F ◦Γ[q]]. (1.109) So we only need the first two components of DΓ[q],

(DΓ[q])(t) =¡

1, Dq(t), D²q(t), . . .¢

, (1.110)

to form the product

D(F ◦Γ[q]) =∂₀F◦Γ[q] + (∂₁F ◦Γ[q])Dq

= (∂₀F + (∂₁F) ˙Q)◦Γ[q], (1.111) where ˙Q = I₂ is a selector function:⁷² c = ˙Q(a, b, c), so Dq = Q˙ ◦Γ[q]. The function

D_tF =∂₀F + (∂₁F) ˙Q (1.112)

is called the total time derivative of F; it is a function of three arguments: the time, the generalized coordinates, and the gener-alized velocities.

In general, thetotal time derivative of a local-tuple functionF is that functionD_tF that when composed with a local-tuple path is the time derivative of the composition of the function F with the same local-tuple path:

D_tF◦Γ[q] =D(F◦Γ[q]). (1.113)

The total time derivativeDtF is explicitly given by D_tF(t, q, v, a, . . .) =∂₀F(t, q, v, a, . . .)

+∂₁F(t, q, v, a, . . .)v

+∂₂F(t, q, v, a, . . .)a+· · ·, (1.114)

72Components of a tuple structure, such as the value of Γ[q](t) can be selected with selector functions: I_igets the element with indexifrom the tuple.

where we take as many terms as needed to exhaust the arguments of F.

Exercise 1.25: Properties of D_t

The total time derivativeD_tF is not the derivative of the function F. Nevertheless, the total time derivative shares many properties with the derivative. Demonstrate thatD_thas the following properties for local-tuple functions F and G, number c, and a function H with domain containing the range ofG.

a. D_t(F+G) =D_tF+D_tG b. D_t(cF) =cD_tF

c. D_t(F G) =F D_tG+ (D_tF)G d. D_t(H◦G) = (DH◦G)D_tG.

Adding total time derivatives to Lagrangians

Consider two LagrangiansL andL⁰ that differ by the addition of a total time derivative of a function F that depends only on the time and the coordinates

L⁰ =L+DtF. (1.115)

The corresponding action integral is S⁰[q](t₁, t₂) =

Z _t2

t₁ L⁰◦Γ[q]

= Z _t2

t1

(L+DtF)◦Γ[q]

= Z t₂

t₁ L◦Γ[q] + Z t₂

t₁ D(F◦Γ[q])

=S[q](t₁, t₂) + (F ◦Γ[q])|^t_t²₁. (1.116) The variational principle states that the action integral along a realizable trajectory is stationary with respect to variations of the trajectory that leave the configuration at the endpoints fixed. The action integralsS[q](t₁, t₂) andS⁰[q](t₁, t₂) differ by a term

(F ◦Γ[q])|^t_t²₁ =F(t₂, q(t₂))−F(t₁, q(t₁)) (1.117) that depends only on the coordinates and time at the endpoints and these are not allowed to vary. Thus, ifS[q](t₁, t₂) is stationary

for a path, then S⁰[q](t₁, t₂) will also be stationary. So either Lagrangian can be used to distinguish the realizable paths.

The addition of a total time derivative to a Lagrangian does not affect whether the action is critical for a given path. So if we have two Lagrangians that differ by a total time derivative the corresponding Lagrange equations are equivalent in that the same paths satisfy each. Moreover, the additional terms introduced into the action by the total time derivative only appear in the endpoint condition and thus do not affect the Lagrange equations derived from the variation of the action, so the Lagrange equations are the same. So the Lagrange equations are not changed by the addition of a total time derivative to a Lagrangian.

Exercise 1.26: Lagrange equations for total time derivatives LetF(t, q) be a function of tandqonly, with total time derivative

D_tF=∂₀F+∂_qFQ.˙ (1.118)

Show explicitly that the Lagrange equations forD_tFare identically zero, and thus that the addition of D_tF to a Lagrangian does not affect the Lagrange equations.

The driven pendulum provides a nice illustration of adding total time derivatives to Lagrangians. The equation of motion for the driven pendulum (see section 1.6.2),

ml²D²θ(t) +ml(g+D²y_s(t)) sinθ(t) = 0, (1.119) has an interesting and suggestive interpretation: it is the same as the equation of motion of an undriven pendulum, except that the acceleration of gravity g is augmented by the acceleration of the pivot D²ys. This intuitive interpretation was not apparent in the Lagrangian derived as the difference of the kinetic and potential energies in section 1.6.2. However, we can write an alternate La-grangian that has the same equations of motion that is as easy to interpret as the equation of motion:

L⁰(t, θ,θ) =˙ ¹₂ml²θ˙²+ml(g+D²ys(t)) cosθ. (1.120) With this Lagrangian it is apparent that the effect of the acceler-ating pivot is to modify the acceleration of gravity. Note, however, that it is not the difference of the kinetic and potential energies.

Let’s compare the two Lagrangians for the driven pendulum. The

difference ∆L=L−L⁰ is

∆L(t, θ,θ) =˙ ¹₂m(Dy_s(t))²+mlDy_s(t) ˙θsinθ

−gmy_s(t)−mlD²y_s(t) cosθ. (1.121) The two terms in ∆Lthat do not depend on either θor ˙θ do not affect the equations of motion. The remaining two terms are the total time derivative of the function F(t, θ) = −mlDy_s(t) cosθ, which does not depend on ˙θ. The addition of such terms to a Lagrangian does not affect the equations of motion.

Identification of total time derivatives

If the local-tuple functionG, with arguments (t, q, v), is the total time derivative of a function F, with arguments (t, q), then G must have certain properties.

From equation (1.112), we see that G must be linear in the generalized velocities

G(t, q, v) =G₁(t, q, v) v+G₂(t, q, v) (1.122) where neither G₁ nor G₂ depend on the generalized velocities:

∂₂G₁=∂₂G₂ = 0.

IfGis the total time derivative ofF then G₁ =∂₁F and G₂ =

∂₀F, so

∂₀G₁=∂₀∂₁F

∂₁G₂=∂₁∂₀F. (1.123)

The partial derivative with respect to the time argument does not have structure, so ∂₀∂₁F = ∂₀∂₁F. So if G is the total time derivative of F then

∂₀G₁=∂₁G₀. (1.124)

Furthermore,G₁ =∂₁F, so

∂₁G₁=∂₁∂₁F. (1.125)

If there is more than one degree of freedom these partials are actually structures of partial derivatives with respect to each co-ordinate. The partial derivatives with respect to two different coordinates must be the same independent of the order of the differentiation. So ∂₁G₁ must be symmetric.

Note that we have not shown that these conditions are sufficient for determining that a function is a total time derivative, only that they are necessary.

Exercise 1.27: Identifying total time derivatives

For each of the following functions, either show that it is not a total time derivative or produce a function from which it can be derived.

a. G(t, x, v_x) =mv_x b. G(t, x, v_x) =mv_xcost c. G(t, x, v_x) =v_xcost−xsint d. G(t, x, v_x) =v_xcost+xsint

e. G(t;x, y;v_x, v_y) = 2(xv_x+yv_y) cost−(x²+y²) sint

f. G(t;x, y;v_x, v_y) = 2(xv_x+yv_y) cost−(x²+y²) sint+y³v_x+xv_y