Why Must the Term dL/d(v^2) v.e Be Linear in v to Be a Total Time Derivative?

rc75
Messages
6
Reaction score
0
In section 4 of Landau and Lifgarbagez they derive the expression for the kinetic energy by expanding the Lagrangian around v+e. The resulting expression has a term which must be a total time derivative so that the equations of motion are unaffected. The text claims that the term dL/d(v^2) v.e must be linear in v to be a total time derivative, but I don't understand why this is.
 
Physics news on Phys.org
I just read that section.
I think it would have helped if they stated that the 2nd term is a total time derivative 'of a function of coordinates and time' ...
df(x,t) / dt = df/dx * dx/dt + df/dt (partial d's now)
Since f does not depend on the velocities, df/dx and df/dt don't, and the overall dependence of df/dt on v=dx/dt is linear.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top