Is a symmetric Lagrangian leads to a symmetric Stress-Energy Momentum?

centry57
Messages
8
Reaction score
0
Is a symmetric Lagrangian leads to a symmetric Stress-Energy Momentum ?
 
Physics news on Phys.org
A "symmetric energy momentum tensor" obeys T_{\mu\nu}=T_{\nu\mu}.

A Lagrangian is a scalar, with no indices.

So what does one have to do with another?
 
I was adoubt if a symmetric stress-energy tensor 's lagrangian is symmetry .

Since {\cal L}= - \frac{1}{16\pi}F^{\mu\nu} F_{\mu\nu} is symmetry on \mu &\nu,the corresponding Stress-Energy Tensor \Theta^{\mu}\,_{\nu} = - \frac{1}{4 \pi} F^{\mu \alpha} \partial_{\nu}A_{\alpha} + \frac{1}{16\pi} \delta^{\mu}_{\nu} F^{\alpha\beta}F_{\alpha\beta} is also symmetry.

Is this the special one ?
 
Last edited:
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