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Generally, in path integral formalism a propagator between two states with definite position is computed, something like,$$K(x_1,t_1;x_0,t_0)=int_{x_0(t_0)}^{x_1(t_1)}mathcal{D}x(t)expleft(frac{i}{hbar}S[x(t)]right).$$However, if we want a propagator between two states, where the...
Asked on 06/10/2021
1 answerCan the fly-wheel lined with magnets in a heat-powered Stirling engine, be magnetically braked at such an angle in relation to the disk there by producing lateral motion in space?...
Asked on 06/10/2021 by Justintimeforfun
0 answerI was reading various reviews on neutrino physics but I couldn't understand the following to complete satisfaction. How is $theta_{12}$ identified with the Solar mixing angle and $Delta m^2_{21}$...
Asked on 06/10/2021
3 answerIn accelerated fluids, fluid in a container can orient itself in a direction due to acceleration. In that case, pressure at different heights (at the surface) is same (atm). Then...
Asked on 06/10/2021 by Shodai
3 answerI'm trying to do the following question from David Tong's problem sheets on string theory:A theory of a free scalar field has OPE $$partial X(z)partial X(w) = frac{alpha'}{2}frac{1}{(z-w)^2}+...$$ Consider...
Asked on 06/10/2021
1 answerIn the "Lecture notes on many-body theory" by Michele Fabrizio, it is stated: How we do show that the Fermi...
Asked on 06/10/2021
1 answerHow would one go about calculating forces that test objects feel using Feynman diagram methods? For example, say we have a massive object in GR so that the metric takes...
Asked on 06/10/2021
0 answerThe objective is to prove that the Lagrangian:$$L'=frac{2dot x+lambda x}{2Omega x}tan^{-1}(frac{2dot x+lambda x}{2Omega x})-frac{1}{2}ln(dot x^2+lambda dot{x } x + omega^2x^2), qquad Omega=sqrt{omega^2-lambda^2/4},$$is equivalent to the lagrangian of...
Asked on 06/10/2021
1 answerI've seen some claims that idempotency ($rho^2=rho$) is necessary and sufficient to guarantee the existence of some state $psi$ such that $rho=|psiranglelanglepsi|$, as well as claims on...
Asked on 06/10/2021
3 answerThere is the formula: $$ Deltatheta=-2frac{partial}{partial p_{theta}}left[int_{r_{min}}^{r_{max}}sqrt{2muleft(E-V_{eff}right)}drright] $$ That indicates the change in angle for the perihelion (in the case of the Earth and the Sun) in one radial...
Asked on 06/10/2021
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