Thursday, June 13, 2019

Applying Advanced Theoretical Mechanics To Circular And Planetary Motion (Part I)

It generally goes without saying that the most refined area of physics is theoretical mechanics. Its integration of advanced mathematics into practical problems has directly led to many other disciplines - from biology to economics - to try to copy it with mixed results.  In this post I want to look at how its concepts and approaches can be applied to familiar areas of astronomy I've covered in earlier, simpler post - mainly to do with circular motion, as well as planetary motion.

Consider a particle moving in a circle as shown below:
No photo description available.

 (Readers may want to examine this earlier post before going on:  )

We have the constraint on the motion: 

r  = (x2  +   y2)½

By looking at such constraints we can reduce the number of coordinates required to specify a mechanical system.

We may write generalized coordinates as:

q 1  , q 2  ……..  q n  

This would apply to some system composed of  n particles with position vectors:

r 1 ,   r 2  ……..  r n  

Such that we may write:     å n i   q i

In all such cases we require that the Jacobian determinant* J    0  else no legitimate set of generalized coordinates can be defined in a specific form. Consider the case of the above for x,y variables with stationary axis:  x =  r cos q,       y =  r sin q

And when the axis is moving:   x= r cos  (wt   + f)  and:  y = r sin (wt   +  f)

Show that (x, y) or (r, q) can be used as generalized coordinates  q 1 , q 2  ……..  q n   


Let   x  = x(q 1 , q 2 , t)

Then write:  x’  = dx/ dt =

 ( x/ q 1 q 1/ t  +( x/ q 2 q 2/t    +  …. x/ t

dx/ dt =

  ( x/ q 1q 1’   +  =  ( x/ q 2q 2’  +  ….…. x/ t


x’   = r’ cos q   -   r sin q  q’  

y’ =  r’  sin q  +   r’ cos q q’   

(Rem:  r’   =  dr/ dt ;   q’  =  dq/ dt )

So the coordinates work in either system, for circular motion

Generalized Momenta can be angular or linear:

 For generalized forces: consider a particle which has moved an incremental amount  d r   where:

d r    =   d x i   +   d  y j       +  d jk   

This is an actual displacement but so small that the forces don’t change, say for a vertical displacement. For N particles we have:

d W    =  å N i   (F i x d x +  F i y d y    + F i z d z )


 x/ q’  =  p q

this can be referred to generalized coordinates: q 1  , q 2  ……..  q n  

d x = ( x/ q 1) d q 1   +  ( x/ q 2) d q 2    +

( x/ q 3) d q 3   

Or in polar coordinates:

d x = cos q d r  - r sin d q

d y = sin q d r  - r cos d q

In general, we may write:  d W   =  Q k  d q k   

=   Q r   y r    +  Q q  d q    

Consider now the transformation of:

F   =  F  x  i +  F  y  j                                 To:

=   F r    +  F q  q

The generalized force is:

Q r       = -  V / r  =

 -  V / x   (-  x / r  )  -  V / y   (-  x / y  )

=   F x  ( x / r  )   +    F  y   ( y / r )

= F x   cos q   +  F  y   sin q   =   F r

We may now introduce the diagram below:

No photo description available.
To derive generalized angular force  Q q :

Q q   =   F x  ( x / q )   +    F  y   ( y / q )

=    F x   r sin q   +  F  y  r cos  q   =  r F q

Note from the diagram this is a component in the direction of   q ^  

It is of interest here to obtain the Lagrangian in polar coordinates.  We know:

x   = r cos q   and  y =  r  sin q 

So that:

x’   = r’ cos q   -   r sin q  q’  

y’ =  r’  sin q  +   r’ cos q q’  

The kinetic energy T is:

T  =  ½ m ( r’ 2    +   r 2 q2 )

V =  mg r sin q    


L =   T  - V  =

½ m ( r’ 2    +   r 2 q2 )  - mg r sin q 

It is also useful to consider the unit vectors associated with polar coordinates in central force problems: n pointing in direction of increasing r, and l, in the direction of increasing q .

The velocity components can then be written:

v   =  dr/ dt,       v q  =   r dq  /dt

Using the polar coordinate unit vectors (n, l) this can be rewritten as:

v  =   dr/ dt  n   +   r  dq  /dt l

For the change in the radial coordinate alone:

v  =   dr/ dt   = d(r n)/ dt = ( dr/ dt  n   +   r dn /dt)

Where the last derivative can be evaluated from:

dn /dt   =   dn / dq   (dq  /dt)

To evaluate  dn / dq     one makes use of the fact that by radian measure of angles:

 D n   =  D q    and in the limit as  D q   ®  0,  

D n ‖ / ‖D q ‖   =   1

So:   dn / dq   =  l

i.e.  As  D q   ®  0,  D n   becomes perpendicular to n and assumes the direction of  l .

In an analogous way we may obtain  the relation:

dl / dq   =  -n

Finally, we have:  dn /dt   =    dq  /dt l

And:  dl /dt   =   -  dq  /dt n


1.   Using  one or more of the preceding unit vector relations, obtain an expression for the acceleration in two dimensions of polar coordinates.

2.A particle  of mass m moves in a plane under the influence of a force F = - kr, directed toward the origin.  Sketch a polar coordinate system (r, q ) to describe the motion of the particle and thereby obtain the Lagrangian (L = T - V, i.e. difference in kinetic and potential energy).

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