Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: Landau & Lifschitz, Mechanics, Principle of Least Action
Replies: 10   Last Post: Jan 19, 2014 8:24 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Hetware

Posts: 148
Registered: 4/13/13
Re: Landau & Lifschitz, Mechanics, Principle of Least Action
Posted: Jan 18, 2014 7:30 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 1/17/2014 8:11 PM, Hetware wrote:
> On 1/17/2014 5:52 PM, Hetware wrote:
>> On 1/17/2014 2:28 AM, William Elliot wrote:
>>> On Wed, 15 Jan 2014, Hetware wrote:
>>>

>>>> On page 3 of Landau & Lifschitz's Mechanics it is stated that
>>>>
>>>> delta q-dot = d(delta q)/dt.

>>>
>>>> This fact is not demonstrated, it is asserted. q-dot is a vector
>>>> /tangent to/
>>>> the particle trajectory. delta-q is a displacement /of/ that
>>>> trajectory. I
>>>> see no a priori reason to believe the two are interchangeable.

>>>
>>> pho = position vector = q
>>> tau = tangent vector = q-dot
>>> tau = dpho/dt
>>>
>>> delta tau(t) = tau(t + delta t) - tau(t)
>>> = dpho(t + delta t)/dt - dpho(t)/dt
>>> = d(pho(t + delta t) - pho(t))/dt
>>> = d(delta pho(t))/dt
>>>

>>
>> deltaq (same as delta q) is a variation of the trajectory, not an
>> infinitesimal displacement along the trajectory. The variation of the
>> Lagrangian is written as
>>
>> deltaL = L(q(t) + deltaq(t), q'(t) + deltaq'(t), t)-L(q(t), q'(t), t)
>>
>> where q-dot has been re-witten as q'(t), etc. Note specifically that t
>> is NOT varied. That is to say, q(t), deltaq(t), q'(t) and deltaq'(t)
>> are all evaluated at t.
>>
>> deltaq'(t) = d(deltaq(t))/dt = lim[(deltaq(t+h) - deltaq(t))/h, h->0].
>>
>> deltaq(t) is an arbitrary smooth vector field parametrized which is
>> added to q(t).

>
> Sorry for changing notation again. It's hard to communicate in ASCII.
>
> It seems the piece that I'm missing is the definition of delta(q'[t]).
> L&L only explicitly define delta_q[t]. So taking the limit as
> Delta_t->0 of the following gives dq/dt + d/dt(delta_q)
>
> (q[t+Delta_t] + delta_q[t+Delta_t] - q[t] - delta_q[t])/Delta_t
> = (q[t+Delta_t]- q[t])/Delta_t
> + (delta_q[t+Delta_t] - delta_q[t])/Delta_t
>
> So d/dt(delta_q) appears to be implicitly defined as delta(dq/dt).
>
> It appears to be implied by the first expression on page 2. This is the
> 2nd edition. Mine is the the 3rd, but the content under discussion is
> the same.
>
> https://ia601205.us.archive.org/11/items/Mechanics_541/LandauLifshitz-Mechanics.pdf
>


Sorry again. I should have said page 3. I saw section 2 and got confused.




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.