- #1

- 194

- 0

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

You should upgrade or use an alternative browser.

- Thread starter enricfemi
- Start date

- #1

- 194

- 0

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

Science Advisor

Homework Helper

Gold Member

- 3,656

- 203

In variation, delta x cannot be zero because you eventually divide by it.

- #3

- 194

- 0

could you concretely explain why delta q can be zero.

thank you very much!

thank you very much!

- #4

- 194

- 0

i think when x is changed ,q should be also change.

help ,help ,can anybody help me?

help ,help ,can anybody help me?

- #5

- 223

- 2

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.

Share:

- Replies
- 16

- Views
- 835

- Replies
- 5

- Views
- 558

- Replies
- 12

- Views
- 855

- Replies
- 1

- Views
- 628

- Replies
- 9

- Views
- 940

- Replies
- 6

- Views
- 921

- Replies
- 4

- Views
- 828

- Replies
- 2

- Views
- 631

- Replies
- 2

- Views
- 715

- Replies
- 22

- Views
- 1K