Elementary derivation of the perturbation equations of celestial
003 Derivation of Lagrange equations from D'Alembertcomplex.gmu.edu/ · Derivation of Lagrange...
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PHYS 705: Classical Mechanics Derivation of Lagrange Equations from D’Alembert’s Principle 1
Transcript of 003 Derivation of Lagrange equations from D'Alembertcomplex.gmu.edu/ · Derivation of Lagrange...
PHYS 705: Classical MechanicsDerivation of Lagrange Equations from D’Alembert’s Principle
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We can write down, when we include dynamics,
D’Alembert’s Principle
Following a similar argument for the virtual displacement to be consistent
with constraints, i.e, (no virtual work for fi)
( ) 0ai i i
i F p r
This is the D’Alembert’s Principle.
0i ii
f r
Again, since the coordinates (and the virtual variations) are not necessary
independent. This does not implies, . ( ) 0ai i F p
We now need to look into changing variables to a set of independent
generalized coordinates so that we can write and set
the independent coefficients in the sum to zero.
? 0jjj
q ? 0
j
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Derivation of Lagrange Equations
( ) 0ai i i
i F p rBreak into two pieces:
(Index convention: i goes over # particles and j over generalized coords)
1. ( ) (1)ai i
iF r
Assume that we have a set of n=3N-K independent generalized coordinates
and the coordinate transformation,
1 2, , , ,i i nq q q tr r
From chain rule, we have
ii j
j j
rr (note: since it is a virtual disp)0i tt
r
jq
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Derivation of Lagrange Equations
( ) ( ) ( )a a ai ii i i j i j
i i j ij j j
q qq q
r rF r F F
This links the variations in ri to qj, substituting it into expression (1), we have,
(Note: Qj needs not have the dimensions of force but must have
dimensions of work.)
Definingas the “generalized forces”
( )a ij i
i j
rF
We can then write,
( ) (1')ai i j j
i j
Q q F r
j jQ q
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Derivation of Lagrange Equations
(2)i ii
p r
Now, we look at the second piece involving :
2.
ip
(don’t forget the “-” sign in the original Eq)
(2 )
i i ii
ii i j
i j j
ii i j
i j j
m
m qq
m q aq
r r
rr
rr
(mass is assumed to be constant)
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Derivation of Lagrange Equations
Let, go backward a bit. Consider the following time derivative:
i i ii i i i i i
j j j
d dm m mdt q dt q q
r r rr r r
Rearranging, the last term (from the previous page) can be written as,
(2 )i i
ji i i i i i
j
i
j
ddt q
d m m bq dt q
m
rr v v rr where ii
ddt
rv
Now, consider the blue and red terms in detail,
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Derivation of Lagrange Equations
Since we have , applying chain rule, we have
i
jqr
i i ii k
k k
d qdt q t
r r rv
blue term:
1 2, , , ,i i nq q q tr r
Taking the partial of above expression with respect to , we havejq
j j
i i
rv
red term:i
j
ddt q
r(switching derivative order)
Is it ok? Check …
(note: ri does not depend on )jq
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i i
j j
dq dt q
r v
explicit check
i i ik
kj k j j
d qdt q q q t q
r r r
Check! The two terms are the same.
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1 2, , , ,i i nq q q tr r
i i ik
kj j k
i ik
k j k j
i ik
k k j j
d qq dt q q t
qq q q t
qq q t q
r r r
r r
r r
LHS:
RHS:
Derivation of Lagrange Equations
Putting these two terms back into Eq. (2b):
With this, we finally have the following for expression (2):
iii i i i i i
j
i
j j
dm m mq dt q q
rr v v vv
ii i i i j
i i j j
m qq
rp r r
(reminder: i sums over # particles and j sums over generalized coords)
ii i i i
i
ji
ii
j j
dm m dmq d dt qt q
rr v rr v
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(2 )i ii i i j
i ji
j j
cd m mdt q
v vv v
Derivation of Lagrange Equations
We are almost there but not quite done yet. Consider taking the qj derivative of
the Kinetic Energy,
Similarly, we can do the same manipulations on T wrt to ,
21 12 2
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i i i i ii ij j
i ii i i
i j j
ii i
i j
m mq q
mq q
mq
v v v
v vv v
vv
j
Tq
jq
j
Tq
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ii i i i
i ij j
m mq q
vv v
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Derivation of Lagrange Equations
Substituting these two expressions into Eq. (2c), we have:
Finally, reconstructing the two terms in the D’Alembert’s Principle, we have:
i ii i i i i i j
i j i i
jj
j j
j j
T Tq
d m m q
q
dt q q
d qdt
v vp r v v
( ) 0ai i i
i
F p r
0j jj j j
d T TQ qdt q q
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Derivation of Euler-Lagrange Equations
Now, since all the are assumed to be independent variations, the
individual bracketed terms in the sum must vanish independently,
There are 3N-K of these differential equations for 3N-K qj and the solution of these
equations gives the equations of motion in terms of the generalized coords
without the need to explicitly knowing the constraint forces.
(3)jj j
d T T Qdt q q
jq
Also, note the advantage of this equation as a set of scalar equations (with T)
instead of the original 2nd law which is a vector equation in terms of forces.
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ˆ ˆ ˆ ˆˆ ˆi i i
i i i i j
i i i
i i j i j i j
jj
U x y zx y z q
x y zU U Ux q y q z q
UQq
i j k i j k
( )a i ij i i
i ij j
Q Uq q
r rF
E-L Equation for Conservative ForcesCase 1:
( )1 2, , , ,a
i i NU t F r r r (note: U not depend on velocities)
derivable from a scalar potential( )aiF
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E-L Equation for Conservative Forces
Notice that since U does not depends on the generalized velocity , we
are free to subtract U from T in the first term,
Putting this expression into the RHS of Eq. (3), we have,
0
jj j j
j j
d T T UQdt q q q
T Ud Tdt q q
jq
0j j
T T Udq
Udt q
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E-L Equation for Conservative ForcesWe now define the Lagrangian function L = T – U and the desired
Euler-Lagrange’s Equation is:
0j j
d L Ldt q q
will be a different Lagrangian giving the same EOM.
' , , , , dGL q q t L q q tdt
Note: there is no unique choice of L which gives a particular set of
equations of motion. If G(q, t) is a differentiable function of the
generalized coordinate, then
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E-L Equation for Velocity Dependent PotentialsCase 2: U is velocity-dependent, i.e., ( , , )j jU q q t
jj j
U d UQq dt q
In this case, we redefine the generalized force as,
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Now, substitute this into , we then have,jj j
d T T Qdt q q
jQ
j j j j
d T T U d Udt q q q dt q
E-L Equation for Velocity Dependent Potentials
Combing terms using L = T – U, we again have the same Lagrange’s Equation,
0j j
d L Ldt q q
This is the case that applies to EM forces on moving charges q with velocity v,
U q A v where is the scalar potential
and A is the vector potential And, 212
L mv q A v
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0j j
T U T Uddt q q
E-L Equation for General ForcesCase 3 (General): Applied forces CANNOT be derived from a potential
One can still write down the Lagrange’s Equation in general as,
Here,
- L contains the potential from conservative forces as before and
- Qj represents the forces not arising from a conservative potential
jj j
d L L Qdt q q
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E-L Equation for Dissipative Forces
, ,f x x y y z zk v k v k v F
Example (dissipative friction):
For this case, one can define the Rayleigh’s dissipation function:
Then, the friction force for the ith particle can be written as,
2 2 212 i i i i i ix x y y z z
ik v k v k v
, ,, ,i i i
f i ix y zv v v
vF
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E-L Equation for Dissipative Forces
Plugging this into the component of the generalized force for the
force of friction, we can get,
To see this, plug in our earlier relation : , we have
,i
j f ii j j
Qq q
rF
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i i
j jq q
r v
,i
j f ii j
vF
,i
ii j jq q
vv
E-L Equation for Dissipative Forces
jj j j
d L L Qdt q q q
Then, the Lagrange’ Equation for the case with dissipation becomes,
- Both scalar function L and must be specified to get EOM.
- L will contains the potential derivable from all conservative forces as
previously.
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Simple Applications of the Lagrangian FormulationA particle moving under an applied force F in Cartesian Coordinates :
In 3D, r = (x, y, z) and there will be three diff eqs for the EOMs.
Then, the x-equation is given by,
xd L L Fdt x x
The Lagrangian is given by, 2 2 212
L T m x y z
This gives,
0 xd mx Fdt
similarly for y & z
m r F
ix i x
i
Q Fx
rF(Note: )
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Simple Applications of the Lagrangian FormulationLet redo the calculation in Cylindrical Coords with the same applied force F:
The coordinates are: r = (r, , z)
From before, we have 2 2 212
T m x y z
Expressing T in Cylindrical Coords:
Tranformation: , , , ,x y z r z
x
y
z
rcos sin cossin cos sin
x r x r ry r y r rz z z z
2
2 22 2 2
2 2 22 2
cos
si
2 sin cos
2 cos sin
si
)
n
s n2
co
(mT rr
rr
r
r
r
zr
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Simple Applications of the Lagrangian FormulationCombining and canceling like-color terms, we have
It is constructive to consider the following alternative way to get to this expression,
2 2 22( ) (*)2
rT rm z
Let try to express the speed in T in cylindrical coordinates,
x
y
z
rˆzz
ˆrr
ˆ ˆr z r r zStart with the position vector r,
Taking the time derivative,
ˆ ˆ ˆd dr r zdt dt
r rv r z
Note: the directional vectors change in time as the particle moves
ˆ r
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Simple Applications of the Lagrangian FormulationTo examine on how these directional vectors changes, consider the following
infinitesimal change,
ˆ ˆˆˆ
ˆˆ ˆ ˆ
dd d dt
dd ddt
r θr θ
θθ r r
Back to the v vector,
Notice that,
ˆ ˆˆ ˆˆ ˆd dr r z r r zdt dt
r rv r z θ r z
x
y ( ')tr
( )tr
't t dt dr̂
ˆ 'rˆdr
θ̂
ˆ 'θ
ˆdθ
So, 22 2 2v r r z and 2 2 2 2 2( )2 2m mT v r r z
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Simple Applications of the Lagrangian FormulationNow, let calculate the generalized force in cylindrical coordinates,
r:
Since, , we haverQr
rF ˆ ˆr z r r z ˆ
r
r r
So, ˆr rQ F F r
:Q
rF
ˆ ˆˆr r r
r θr θ (recall previous page)
So, ˆQ r rF F θ (this looks like torque)
z:ˆz zQ F
z
rF F z
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Simple Applications of the Lagrangian FormulationThe EOM is then given by:
2r r
d T T F mr m r Fdt r r
:
Recall,
(this is )
z :
r :
2 2 2 2( )2mT r r z
2d T T drF mr rFdt dt
dL Ndt
2 2mr m rr rF 2m r r F
z z zd T T dF mz F mz Fdt z z dt
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Simple Applications of the Lagrangian FormulationPutting the components of F together,
2
ˆˆ ˆˆˆ ˆ2 (*)
r zF F F
m r r m r r mz
F r θ z
F r θ z
Is that the same that we have gotten previously in Cartesian Coords?mF r
Check:
ˆ ˆ ˆ=d d r r zdt dt
vr θ r z
ˆˆ ˆ ˆ ˆ
ˆ ˆˆ ˆ ˆ ˆ
d dr r r r zdt dt
d dr r r r zdt dt
rr θ θ r z
θ rθ θ r z
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ˆ ˆ ˆr r z v r θ r z
Simple Applications of the Lagrangian FormulationUsing the directional vectors relations that we had earlier,
2 ˆˆ ˆ2m r r r r z m F r θ z r
ˆ ˆ ˆˆ ˆ ˆr r r r z θ θ θrr zr
ˆ ˆ ˆˆ,d ddt dt
θ rr θ
2 ˆˆ ˆ2r r r r z r r θ z
So the EOM in (*) on the previous page is indeed , i.e.,mF r
Collecting all terms in the same direction,
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