Jaezyl Comon Elumba

Classical Mechanics

September 20, 2020

1 Lagrange Equation
From the D’ Alembert’s Principle,
X ~(a)
= (Fi − P~i )· δ~ri = 0
i

This equation expresses the equilibrium of both the static and dynamic case.

Now, the position of the ith particle expressed in terms of the generalized
coordinates is

r~i = r~i (q1 , q2 , ..., qn ; t)

Thus, the velocity vi can be expressed in terms of q˙k by the formula

d~
ri X ∂~
ri ∂~
ri
~v = = q˙k +
dt ∂qk ∂t
We may define the virtual displacement δ~
ri as
X ∂~
ri
δ~
ri = δqi
j
∂t

Note that δt is not involve here, because in virtual displacement we held the
time as fixed.

1
 In terms of the generalized coordinates, the virtual work of the Fi becomes
X X ∂~
ri
F~i · δ~
ri = F~i · δqj
i i,j
∂qj

We can express the jth components generalized force Qi as

X ∂~
ri
Qj = F~i ·
i,j
∂qj

We turn now to the momentum componet which can be written as

X X
p~˙i · δ~
ri = mi r~¨i · δ~
ri
i
X ∂~
ri
= mi r~¨i
i,j
∂qj

Evaluating the expression

d ˙ ∂~ ri ∂~
ri d ∂~ri
i i·
(m~r ) = mi r~¨i · + mi r~˙i · ( )
dt ∂qj ∂qj dt ∂qj

What we need is
∂~
ri
mi r~¨i
∂qj
Expanding the momentum component in terms of i for a given j, we have
X ∂~
r i
X d ∂~
r i d ∂~
ri

mi r~¨i · = (mi r~˙i · ) − mi r~˙i · ( )
i
∂q j i
dt ∂q i dt ∂qi

If we look at the last term

d ∂~ri ∂ d~ri ∂ v~i
( )= =
dt ∂qj ∂qj dt ∂qj

Note that
X ∂~
ri ∂~
ri
v~i = q˙k +
k
∂qk qt

2
Substituting the value of v~i we now have
( )
d ∂~ ri ∂ X ∂~ ri ∂~ri
( )= · q˙k +
dt ∂qj ∂qj k
∂q k ∂t
X ∂ 2 r~i ∂ 2 r~i
= q˙k +
k
∂qj ∂qk ∂t

If we go back to the main problem

X ∂~
ri
X d ∂ v
~ i ∂ v
~i

mi r~¨i · = [mi v~i · ] − mi v~i ·
i
∂qj i
dt ∂ q˙j ∂qj

If you recall, there is an expression

∂ 1 ∂vi
( mi vi 2 ) = mi vi ·
∂ q˙j 2 ∂ q˙j
∂ v~i
= mi v~i ·
∂ q˙j
We also have  
∂ 1 2 ∂ v~i
mi vi = mi v~i ·
∂qi 2 ∂qi
Gather the last 2 equations we have
     
X
¨ ∂~
ri X d ∂ ∂
mi r~i · = Ti − Ti
i
∂q j i
dt ∂ q˙i ∂q i
PP P
Since there is a in the momentum, we still have a j written as
! (     )
X X ∂~ri
X X d ∂~r i d ∂~
r i
mi r~¨i · δqj = mi r~˙i · − mi r~˙i · δqj
j i
∂q j j i
dt ∂q j dt ∂q j

We can rewrite this further as

( )
X X d  ∂ v~i

∂ v~i
= mi v~i · − mi v~i · δqj
j i
dt ∂ q˙j ∂q j

Taking out the j factor, we have

( )
X d ∂ X 1 
∂ X1
= mi vi 2 − mi vi 2 δqj
j
dt ∂ q˙j i
2 ∂q j i
2

3
Rewriting the momentum term,
X ∂~
ri X  d ∂T ∂T

¨
mi r~i · δqj = − δqj
i,j
∂q j j
dt ∂ q˙j ∂q j

Putting everything together in D’ Alembert’s Principle, we have

X
F~i − p~˙i · δ~


ri = 0
i
!
X ∂~
ri ∂~
ri
F~i · − mi r~¨i · δqj = 0
i,j
∂qj ∂qj
We define the first term for the generalized force, we have now
( " #)
X d ∂T ∂T
Qj − − δqj = 0
j
dt ∂ q˙j ∂q j

If you have holonomic constraints, there are actually k constraints.

Suppose you have k constraints from 3N coordinates, we now have 3N −k

independent j - equations.
What we get now is  
d ∂T ∂T
− − Qj = 0
dt ∂ q˙j ∂qj
This equation is the general result.
However, the special case is when the forces F are derivable from a scalar
potential V
F~i = −∇
~ iV
Rewrite the Qj
X ∂~ri
Qj = F~i ·
i
∂qj

~ i V · ∂~
ri
X
=− ∇
i
∂qj

If we take the dot product of the last term along the direction of i, the result
would be
∂V
=−
∂qj

4
Everything can be written now as
 
d ∂T ∂T ∂V
− + =0
dt ∂ q˙j ∂qj ∂qj

Remember that V does not depend on q˙j

  
d ∂
T −V − ∂(T − V ) over∂qj = 0
dt ∂ q˙j

Define L = T − V , we get the Lagrange Equation

 
d ∂L ∂L
− =0
dt ∂ q˙j ∂qj