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Topic: Landau & Lifschitz, Mechanics, position dependence of kinetic energy,
T?

Replies: 3   Last Post: Jan 20, 2014 10:15 PM

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 Hetware Posts: 148 Registered: 4/13/13
Re: Landau & Lifschitz, Mechanics, position dependence of kinetic
energy, T?

Posted: Jan 20, 2014 11:05 AM

On 1/20/2014 3:11 AM, William Elliot wrote:
> On Sun, 19 Jan 2014, Hetware wrote:
>>
>> The Lagrangian is given as [Einstein summation convention assumed]
>>
>> L = 1/2 a[q]_ik q'_i q'_k - U[q] (5.5)

>
> L = sum_jk 1/2 * a_jk(q) q'_j q'_k - U(q)

'Einstein summation convention assume'

http://mathworld.wolfram.com/EinsteinSummation.html
" The convention that repeated indices are implicitly summed over. This
can greatly simplify and shorten equations involving tensors. For
example, using Einstein summation,
a_ia_i=sum_(i)a_ia_i

and
a_(ik)a_(ij)=sum_(i)a_(ik)a_(ij).

The convention was introduced by Einstein (1916, sec. 5), who later
jested to a friend, "I have made a great discovery in mathematics; I
have suppressed the summation sign every time that the summation must be
made over an index which occurs twice..." (Kollros 1956; Pais 1982, p.
216). "

I try to reserve it's application to the contraction of contragredient
entities, but I'm being a bit lax here. Also, I am using [] to enclose
the arguments of functions. It's a convention I am trying to adapt in
order to disambiguate f(x+y) meaning 'f*x+f*y' from f(x+y) meaning 'f is
a function taking x+y as arguments'. I am not the only person who has
I'm going to be in trouble.

>> "where the a_ik are functions of the coordinates only. The kinetic energy in
>> generalized coordinates is still a quadratic function of the velocities, but
>> it may depend on the coordinates also".
>>
>> It's not clear what this really means. Every point q of the generalized

>
> q is the position vector; q'_j are the velocity vectors in the j-th
> 3direction.
>
> L(q) = (1/2)sum_jk a_jk(q) dq/dx_j dq/dx_k - U(q)
>
> Which for rectangular coordinate three space,
> a_jk is Kroniker's delta_jk and L = mv^2 / 2.
> since v = sqr((v_x)^2 + (v_y)^2 + (v_z)^2),
> assuming potiential energy U(q) = 0.

That part I got.

I understand that in order to calculate T in terms of generalized
coordinates, I need both q and q'.

My apprehension is regarding the notion that T=T[x'] is /not/ a function
of the Cartesian 3N-component position vector x, but T=[q,q'] /is/ a
function of /both/ the generalized /position/ and generalized /velocity/.

When I asked John Archibald Wheeler what a tensor is he replied "A
tensor is a geometric object independent of coordinates."

Since a vector is a tensor, it would seem to follow that the Cartesian
representation x should be synonymous with q.

How, then, can it be the case that T=T[q,q']=T[x']. It seems to imply
that q and q' can be expressed in terms of x' alone.

>> coordinates corresponds to a point X = {x_i,y_i,z_i} in Cartesian coordinates.
>> That means to me that U[x]=U[q[x]]. That is to say U of a given state is
>> invariant under a change of coordinates. Since the Lagrangian is also an
>> invariant, it seems T must be an invariant. IOW, I expect
>>
>> 1/2 a[q]_ik q'_i q'_k = 1/2 m_a(x_a^2 + y_a^2 + z_a^2).
>>
>> Clearly the a[q]_ik are dependent on the generalized coordinates, but is the
>> /magnitude/ of the kinetic energy coordinate-dependent?
>>

Date Subject Author
1/19/14 Hetware
1/20/14 William Elliot
1/20/14 Hetware
1/20/14 William Elliot