Recent content by tnho

To be a bit clearer "photon has zero rest mass, which is the mass we measured in its co-moving frame although it is not possible to do so. In general, photon do has mass which is due to its momentum, namely E=pc . With non-zero mass, there should have non-zero gravity created. Here i have a...

Yes. I think so. The sum of linear momentum of the particles is zero in the centre of mass frame in SR.

I bet it should be 0.5\rho g h^2 - \frac{k_B T}{A}\log(h). =P It is a very good example~thank you~

See if i have done anything wrong or not~ First do it in the frame where the target proton (marked 2 below) is stationary. P_1 = (E_1/c,p_1,0,0) \mbox{ ; } P_2 = (mc,0,0,0) By conservation of four-momentum, P_1 + P_2 = P_3 + P_4 +P_5 +P_6 (P_1 + P_2)^2 = (P_3 + P_4 +P_5 +P_6)^2 (P_1...

I think you may try to use four-momentum vector approach. This will never wrong~ and i think the answer is something like 7m_p c^2

It's a very useful trick~try to make good use of it~

\frac{1}{2}mv^2 = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}} then n=\frac{mvL}{\pi \hbar} put m=9\times 10^{-31}, v=3\times 10^{4}, L=48.5\times 10^{-9} \mbox{ and } \hbar = 1.055\times 10^{-34} then you can get n\approx 4 get it?

what is that "*" mean? complex conjugate (cc)?? or just multiplication? Use e^{x}=\sum_n x^n/n! and H|E\rangle = E|E\rangle then you can get that equation. I think that "*" is just a multiplication instead of a cc. :smile:

yes. I got 3.969 for using E_n=\frac{1}{2}mv^2. Besides, E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}} is a result from Schrodinger Equation which is non-relativistic. I am not sure if this form preserve in Dirac Equation or not.

the |phi> is the excited states in the new potential, right??

Homework Statement An alternating current electrical generator has a fixed internal impedance R_{g}+iX_{g} and is used to supply power to a passive load that has an impedance R_{g} +iX_{l}, where i = \sqrt{-1}, R_{g}\neq 0, and X_{g} \neq 0 . For maximum power transfer between the generator...

I have thought that before but not quite sure. It is true that oscillation angles do depend on energy. um..i don't think so as those experiments only measured the electron neutrinos but no other flavors.

You are right. Neutrinos DO change their flavors (or states). However, all experiments mentioned above were all observing the electron neutrinos with the same length traveled.

In the solar neutrino experiments, there were deficits in the observed solar neutrino flux when compared with the theoretical values. If i just focus on the Cl-37 and Ga-71 reactions to detect solar neutrinos, there is a question confuses me a lot. That is, from the result of experiment...

i think it would be much better if you can translate your .tex file into English...:cool: