Recent content by Mr Boom

and the answer is... So I found the solution described as f(r) -> f(x): exp(-r2/2σ2) -> (2π)1/2σ*exp(-x2/2σ2) The added constant does not change the the value of sigma so, which doesn't make sense to me. When viewing the line itegral, shouldn't the distribution appear more narrow?

Hello. I'm wondering if anyone has a table of transforms showing the result of an Abel transform on a Gaussian distribution. I have been unable to find the solution to this. Many thanks for any help. I'm reconstructing an an image from a picture that fits a Gaussian very well, hence I'm hoping...

Hello. I'm reviewing background information on Raman scattering and I've noticed that the polarizability scales with v+1 for Stokes transitions and v for anti-Stokes. Why is this and what assumptions are used in the derivation? An explanation or pointing me to a good reference would be...

OK, that makes sense. I'd like to try to do this problem on my own. Any recommendations on how to start this program? I've plotted the static field lines and I need to superimpose the field due to the resistor?

I was thinking of spheres, yes, but I was trying to just use a 2D example. This was my question. I'm wondering how the field lines look visually outside the resistor by allowing current to pass. I realize there will also be a magnetic component. Will the lines straighten or become more arched?

Let's say I have a circle filled with positive charges. Some distance away I have an identical circle filled with negative charges. Since the distribution is uniform in the circles, I can just use the center of the circles as points and calculate the field lines between the two. So far so good...

Energy is put into some kind of gain medium by an external source (electrical discharge, pump laser, etc.). The energy being put into that gain medium drives atoms to a higher electrical state. This energy is released in the form of photons, much like a light bulb. The difference here is that a...

The units of S can be photons, counts, or electrons depending if you put those conversions in. This is a two photon cross section so the tabulated values I see are cm4*s. Not sure where the seconds come from unless it is supposed to be multiplied by a power rather than energy.

Yes, the fraction has no units. It's just a large number. Thanks. Any idea about the x-section units?

Hello. I was hoping someone could help me understand some units. In several papers, I have seen the two photon LIF signal written as: S = n*(Ω/4π)*T*G*σ*(E/hv)^2*η*a where the units are n (density of measured atom): #/cm3 Ω/4π (solid angle fraction): unitless T (optical transmission...

OK, that makes sense. I think my confusion was that I was comparing the nitrogen and oxygen term symbols with known transition lines and for oxygen (triplet P) there are three close P line corresponding to different J values. These three lines don't represent the triplet of the ground state but...

So the term symbol for the nitrogen atom is 4S3/2 This means L=0, S=3/2, hence J=3/2. My question is that if S=3/2 than this is a quartet, meaning there are four levels, yet I only ever see one state listed on energy diagrams? What gives? Obviously, I'm still trying to learn term symbols Are...

I believe that you are correct. In this case, the word "heat" does not indicate the motion of molecules because the bulb itself is a vacuum. I really doubt that the small amount of matter in that bulb heats up and transfers energy to the glass which then gets conducted out. The efficiency just...

I am reading up on Paschen's Law and came across some papers measuring breakdown voltage between two identical spherical electrodes. I've read in a couple places that the electric field is uniform when the gap distance is no larger than the radii. Why is that? It seem to me that the field...

You might want to check the supply voltage to the laser to make sure it isn't changing, hence changing your wavelength and power a bit. I'm also not sure how the polarization ratio in these lasers changes as they warm up. Are you monitoring it after you pass it through your optics?