# Tag Info

0

"does the shift directly depend on being above or below resonance?" Phase of reflected light will depend on which side of resonance laser is(check out cavity phase response). "I would have expected you to mix the output signal with the initial laser signal, such that the shift in phase could be determined and hence accounted for." You could in theory do ...

0

Q^2 is the Mandelstam variable t, i.e. the four momentum transfer squared where the s channel is the x axis in the feynman diagram. where p1 p2 are incoming. In the same link it is seen that at the relativistic limit : The dot product for t (Q^2) is p transverse to the incoming beam direction of p1 , p2 are incoming. etc. Is this Q the same if ...

0

If you have a multimeter you can do it easily. Think of the capacitor as three resistors in series: the first lead, the dielectric, and the other lead. Resistors in series add resistance to combine, so we will add those three: $$R_{eq} = 2R_{leads} + R_{dielectric}$$ We can assume the leads to have a resistance of almost 0 since they conducting metals. So ...

3

The very flexibility that you mention in your post is a bit of a problem in an experimental context. In order to understand the signal that a Cf-Be calibration source would generate in your detector and tease useful information out of it, you're going to have to model all three channels that generate neutrons and the gammas that escape the source. This means ...

3

As you suggested in your comment, the $\mu^-$ and $\mu^+$ that stop in matter do not have the same lifetimes. The $\mu^+$ come to rest between the atoms of your stopper (eg: scintillator?) and decay into $\nu_{\mu}e^+\nu_e$ with the standard 2.2 usec lifetime. However, the $\mu^-$ get captured into Bohr orbits about the stopper nuclei. The $\mu^-$ then ...

1

I know that you are explicitly asked about not equipment related answers. But when I learned something from experimental physics then that you should always consider equipment flaws. I could imagine a scenario where the events on which you trigger to start/stop the clock have different rise times depending on where they take place in the scintillator, ...

3

One can calculate a lot more in quantum field theory if one goes beyond asymptotic computations into thermal field theory. I recommend that you look at the book ''Nonequilibrium Quantum Field Theory'' by Calzetta and Hu.

2

The small number of "conceptually independent types of processes and calculations" is exactly a symptom of the theory's being fundamental! Even in classical physics, all calculations could have been mathematically reduced to the calculation of the final state that evolves from an initial state (or a state that is stationary etc.). In quantum mechanics, this ...

Top 50 recent answers are included