Recent content by grindfreak

Thanks for the very intelligent post AnTIFreeze3! I'm definitely still considering grad school for physics (I have to go back to school for something regardless), but I still have time to weigh my options before applying anywhere (hence the reason for this post). I just really wanted to get...

jbrussell93 - Well, if you follow the academic route, then yes you would probably be able to do what you were talking about. Plus a lot of the geologists/geophysicists/petrophysicists collaborate with oil companies on certain projects as well. But no, I'm really not interested in doing it...

jbrussell93 - It really depends on how high you go I suppose. If you're going to be going to grad school for petrophysics or geophysics you stand a chance of going straight into a desk job. However you'll probably be doing what I'm doing right now, since it's pretty much the standard...

Well, now is most definitely the time to get into the oil business if you're thinking about it. Probably the worst aspect of my job is the hours. I work, on average, 12 hour shifts from 6 in the morning and 6 in the evening. There is no time off in general, unless you ask for it. I'm also...

As in occupation? Right now I work in the oilfield doing field geological work(and in the near future possibly some more advanced things like mass spectrometry) for oil companies. Not really related to anything I've mentioned above I know, but I guess I should get more specific about my...

As far as math is concerned I started out in calc III (I passed the AP test for both calc I and II) and took differential equations, linear algebra and modern algebra as well. As far as statistical coursework is concerned, I don't have any formal coursework per se (not the most promising answer...

So right now I'm working a job that I dislike. Of course it's only temporary, the goal being to allow my wife to finish school and save up enough money in order to allow me to enter graduate school with a minimum amount of debt. Up until this point I was bound and determined to get a Ph.D in...

In what world does Δp ->∞ mean that v->c? Δp = √(<p2> - <p>2) where <p> = ∫ψ*pψdp for a wave function in the p basis as seen in any basic quantum physics text. This is the standard deviation, it has nothing to do with the momentum or the velocity approaching a certain value.

Although I am not currently skilled in QFT, I do know that one can have infinite uncertainty of a variable (meaning we are approaching infinite precision for the other). In my mind this does not mean that the particle is moving with a relativistic speed, it just means that the uncertainty is of...

Yes I noticed this, most texts use: c_{j+1}=\frac{2(j+l+1-n}{(j+1)(j+2(l+1))}c_{j}\ I believe.

Hmmm, I'm not sure about moving backwards since, like you've stated, every bj is zero due to the fact that every numerator of the recursion is zero. The one exception being bj=l where both the denominator and the numerator are zero. Since this is the case, we end up with an indeterminate form...

Well you've already found three solutions from the second equation(x = 0, x=-y and x=y), what happens when you plug these into the first equation and solve?

Here's how it should integrate. Since the initial values for both velocity and displacement are 0, the right integral looks like: \int_{0}^{s}ads^{'}=k\int_{0}^{s}(s^{'})^{n}ds^{'}=\frac{ks^{n+1}}{n+1} The right integral is: \int_{0}^{v}v^{'}dv' = \frac{1}{2}v^{2} Now don't be confused...

Anytime man, I appreciate the feedback.

I've been using Fedora 16 for a bit and have really enjoyed it. I'd recommend trying anyway.