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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 adapted this convention. Now, when I start talking about commutators, 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 jth > 3direction. > > 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 3Ncomponent 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 coordinatedependent? >>



