A question about Analytical Mechanics

  • #1
194
0
i am studing Analytical Mechanics in these days.it is really amazing. but a question pazzles me .it seems:

to generalized coordinates q,δq is arbitrary,it can equal 0;while variation in the x-coordinate, δx is not.

i just cann't understand
 
  • #2
In variation, delta x cannot be zero because you eventually divide by it.
 
  • #3
could you concretely explain why delta q can be zero.

thank you very much!
 
  • #4
i think when x is changed ,q should be also change.

help ,help ,can anybody help me?
 
  • #5
It depends on what your generalized coordinates are.

The derivation with variations like [tex]\delta q_i [/tex] depends on the variation possibly being anything at all. Lagrange's equations arise from the fact that you have a product of two functions in an integral that equals zero. Since the variation in the generalized coordinates could, in general, be anything, the other function has to be uniformly zero. Weinstock's book in Dover edition covers this point quite nicely.
 

Suggested for: A question about Analytical Mechanics

Replies
16
Views
835
Replies
12
Views
855
Replies
1
Views
628
Replies
9
Views
940
Replies
6
Views
921
Replies
4
Views
828
Replies
2
Views
715
Replies
22
Views
1K
Back
Top