Recent content by P3X-018

Then please enlighten us on how gravity interacts with massless particles. Are you sure light is assumed to be massless for that statement? "Bodies in free fall" already assumes mass.

You can't bend a trajectory without a force. The acceleration is just the result of the force between the bodies attracting each other. Gravity doesn't affect massless particles in the classical mechanics.

Both of these wavefunctions can't be normalized, since they still have the same functional form. I think that's what nrqed was referring to, so that they are equivalent, when you normalize them.

http://books.google.com/books?id=AEkfajgqldoC&printsec=frontcover&dq=Rodney+Loudon&hl=da#PPA54,M1" avoids this problem by taking the sum over frequencies first after squaring the amplitude, but he still mentions, after eq. (2.3.11), that the integral should be over the excitation amplitudes. How...

But it's supposed to be dimensionless like the probability, hence must have the same dimensions. EDIT: Ok to make it more clear, then (4.38) has the form c_b = \int_0^{\infty} f(\omega) \, d\omega while (4.39) has the form |c_b|^2 = \int_0^{\infty} |f(\omega)|^2 \, d\omega which...

Cyosis: Yes they are supposed to be dimensionless, but I just looked at the difference between the 2 expressions. Note that every term from (4.38) appears squared in (4.39) except for the d\omega term which has the same power in both, hence (4.39) is short by a factor of 1/sec.

At page 190 of Bransden & Joachain (see the page from http://books.google.com/books?id=i5IPWXDQlcIC&printsec=frontcover&dq=B.+H.+Bransden,+Charles+Jean+Joachain&hl=da#PPA190,M1"), there are 2 expressions for the transition probability, (4.38) and it's absolute value squared in (4.39). Is it...

I know that the coherence is one issue, but since the coherence length is so long due to the narrow linewidth of the laser, that the effects due to this can be ignored (coherence function being equated to 1), provided the optical path difference stays less than the coherence length. That's also...

By sending part of the initial beam through a resonator and filtering out a desired mode.

The beams come from the same source, and the optical path difference is much less than the coherence length of the laser. Though the 2 beams don't have exactly the same parameters (slightly different waist and waist location), and the two beams are different modes (one is TEM_00 the other...

If you have 2 laser beams, A and B, at right angle with each other, that you send towards a beamsplitter (BS), such that you get 2 output beams from the BS, one being the sum of the reflected part of A and transmitted part of B, and the other the transmitted part of A and reflected part of B...

I was also looking for a book on quantum optics. I have Loudon's text, but it doesn't seem to be going into big details with the topics covered (reasonable, given the size of it), though it does seem like an excellent starting point. So I was considering a book with a more advanced treatment...

I'm looking for a textbook on laser physics. I'm considering either Lasers by Siegman or same title by Eberly & Milonni. I want a textbook more directed to a physicist and not the more engineering style, which may lack rigour (from the physics point of view). I'm not very comfortable with books...

I kept looking at the wrong inner product to use CS on, that is that inner product of L_2, but I should look at the inner product of \ell_2 space. So by using that (replacing the minus sign with + in the afore mentioned equation for c_n(f)) \sum_{n\in\mathbb{Z}}|c_n(f)| \leq c_0(f) +...

[SOLVED] Fourier coefficients Homework Statement For f \in C^{2\pi}\cap C^1[-\pi,\pi] , I have to show that \sum_{n\in\mathbb{Z}}|c_n(f)| < \infty where c_n(f) is the Fourier coefficient of f; c_n(f) = (f, e_n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}\,dt f \in...