Generalized Coordinates and Invariant Form for

Equations of Motion
Enrique Segura
September 27, 2019

Goldstein 1.15 | Symon 9.24. Show by direct calculation that for some L(q 1 , . . . , q f , q˙1 , . . . , q˙f ; t)
which satisfies Lagrange’s equation:

d ∂Lq ∂Lq
− k =0
dt ∂ q˙k ∂q

If we introduce new coordinates (s1 , . . . , sf ), where:

q k = fk (s1 , . . . , sf ; t), k = 1, . . . , f

then,
d ∂Ls ∂Ls
− j =0
dt ∂ s˙j ∂s
is obtained with:

L(q 1 , . . . , q f , q˙1 , . . . , q˙f ; t) = L(s1 , . . . , sf , s˙1 , . . . , s˙f ; t)

by substitution for fk (s1 , . . . , sf ; t) for q k where j = 1, . . . , f .

Solution. Let’s take the simplest case we can to be able to build this result; namely only use s, q
to build this system:
s = s(q, t), q = q(s, t) (1)
Our goal is show that in order for the L(s, ṡ, t) = L(q, q̇, t) then we must convert:

d ∂Lq ∂Lq d ∂Ls ∂Ls

− ←→ − (2)
dt ∂ q̇ ∂q dt ∂ ṡ ∂s
So let’s start with inspecting our Lq :

Lq = L(q, q̇, t) = L(q(s, t), q̇(s, t), t) (3)

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According to the Euler-Lagrangian equation, this equation:

d ∂Lq ∂Lq
− (4)
dt ∂ q̇ ∂q

will change due to q = q(s, t).

By chain rule, we can find by inspection:

∂Lq ∂Lq ∂s
= (5)
∂q ∂s ∂q
∂Lq ∂Lq ∂ ṡ
= (6)
∂ q̇ ∂ ṡ ∂ q̇
From this inspection, we have to show:

∂ ṡ ∂s
= (7)
∂ q̇ ∂q
For that, let’s start with s and calculate its full derivative:
∂s ∂s
ds = dq + dt (8)
∂q ∂t
Dividing by dt yields
ds ∂s dq ∂s
= + (9)
dt ∂q dt ∂t
which can expressed as
∂x ∂x
ṡ = q̇ + (10)
∂q ∂t
By inspecting we can yield the following observation:

ṡ = ṡ(q̇, t) (11)

Having this form, we can then find the following:

∂ ṡ ∂x ∂ ∂x
= q̇ + (12)
∂ q̇ ∂q ∂ q̇ ∂t
Since the second partial is a function of t, this term vanishes leading to:

∂ ṡ ∂s
= (13)
∂ q̇ ∂q

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 Having arrived at this result, we go back and work with the lagrangian of interest:
d ∂Lq ∂Lq
− =0 (14)
dt ∂ q̇ ∂q
d ∂Lq ∂ ṡ ∂Lq ∂s
( )− =0 (15)
dt ∂ ṡ ∂ q̇ ∂s ∂q
∂Lq ∂ ṡ ∂Lq ∂s
( )= dt (16)
∂ ṡ ∂ q̇ ∂s ∂q
And by the substitution:
∂Lq ∂ ṡ ∂Lq ∂ ṡ
( )= dt (17)
∂ ṡ ∂ q̇ ∂s ∂ q̇

Just for reference:

Lq = L(q, q̇, t) = L(q(s, t), q̇(s, t), t)

Ls = L(s, ṡ, t) = L(s(q, t), ṡ(q, t), t)

Using our substitution we find:

∂Lq ∂Lq
= dt (18)
∂ ṡ ∂s
which brings:
d ∂Lq ∂Lq
( )− =0 (19)
dt ∂ ṡ ∂s
This means that Lq is a function s, which means Lq = L(s, ṡ, t), which means Lq is actually
Ls . Thus,
d ∂Ls ∂Ls
( )− =0 (20)
dt ∂ ṡ ∂s
Thus, we find that for the simplest case:
L(q, q̇, t) = L(s, ṡ, t) (21)
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Solution. Now having shown this result for this case, we expand it for two coordinates systems
represented by:
q k = (q 1 , . . . , q f ; q˙1 , . . . , q̇ f ; t) (22)
sl = (s1 , . . . , sf ; s˙1 , . . . , ṡf ; t) (23)
(24)
If there is f k that maps q k ←→ sl such that:
q k = q k (s1 , . . . , sf ; t) (25)
sl = sl (q 1 , . . . , q f ; t) (26)

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 Then by finding:
∂ q˙k ∂q k
= (27)
∂ ṡl ∂sl
this can be use to yield:

d ∂Lq ∂Lq
( )− k =0 (28)
dt ∂ q˙k ∂q
d ∂Lq ∂ q˙k ∂Lq ∂q k
( )− k l =0 (29)
dt ∂ q˙k ∂ ṡl ∂q ∂s
d ∂Ls ∂Ls
( )− =0 (30)
dt ∂ ṡ l ∂sl
And thus showing that indeed:

L(q 1 , . . . , q f , q˙1 , . . . , q˙f ; t) = L(s1 , . . . , sf , s˙1 , . . . , s˙f ; t) (31)

Or, as Goldstein puts it : “...the form of Lagrange’s equations is invariant under a point
transformation.”
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