Use Of Generalized Coordinates, Momenta and Force In Theoretical Mechanics

 When 3rd year physics students are first introduced to theoretical mechanics they are generally taken aback when confronted by the concepts of generalized coordinates, momenta and forces.  In this post I examine aspects of this and hope to show why such generalizations manifest practical use and power in solving problems in theoretical mechanics. 

We may write generalized coordinates as:

q ₁  , q ₂  ……..  q _n  

This would apply to some system composed of n particles with associated 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 over time, t with phase angle f):  

 x= r cos  (wt   + f)  and:  y = r sin (wt   +  f)

Example Problem: 

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

Solution:

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

And:

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 (p _q ,  p _f ) or linear ( p _x, p _y ,  p _z ):

 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 )

Given:  

¶ 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  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:

To derive the 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 q ‘² )

V =  mg r sin q    

Therefore:

L =   T  - V  =

½ m ( r’ ²    +   r ² q ‘² )  - 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 _r    =  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

dl / dq   =  -n

Suggested Problems:

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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