Recent content by theUndergrad

This is a straightforward application of the lorentz transformation for time. You know that time slows down due to relative motion and it takes some time for light to travel some distance, so for the special case that the relative velocity of Jupiter and Earth are 0, the amount of time should...

Really it's all about playing around with the unit circle (http://en.wikipedia.org/wiki/Unit_circle). The circle contains some simple angles (30,45,60, and 90 degrees, along with 90 degree shifts of all of these angles all of the way around the circle) represented in radians and the value of...

The phase difference is just the difference in the "angle" (ie the values inside the sinusoid). In this case the time is the same, and can be ignored. Also since you're only worried about the distance (not absolute position) you may as well set one of the particles at x1=0 and other at x2=200...

It looks like the best thing you can do is to assume that unpolarized => equally polarized in the x and y (or whatever coordinates you are using) direction. When you say light is unpolarized you are saying that it is randomly distributed. You can always break this down into polarization in x...

You're venturing into strange territory when you try to deal with things the size of atoms. Defining the "border" of a proton or neutron is tricky due to their representation as wave functions and the indeterminacy of their position. Also, virtual particles that mediate the forces are...

The eigenstates need to take the spin characteristic into account and will be of the form (harmonic oscillator)*(spin state). Even without the harmonic potential these particles wouldn't be truly "free" because they are constrained by the Pauli Exclusion Principle. Typically spin eigenvalues...

It's at that point the amplitude of the incoming waves continually sum to 0. However, just when two wave interact on a string (see this video ) the wave continue propagate through the interference and afterwards. That is why you can record the interference at a variety of distances (as long...

From there it's important to recognize that (1-a^2)=(1+a)(1-a) and then (1-2a-a^2)=(1-a)^2=(1-a)(1-a). Then it's just a matter of removing the common term. ------------------------------- theUndergrad http://www.theUndergraduateJournal.com

Since these spin-1/2 particles there are only two possible eigenstates for each of the particles: either S=(1,0)^{T} or (0,1)^{T}. Since these are non interacting particles, either one can be in either state, so you have 4 possible configurations for the system. Then it's just a matter of...

That Hamiltonian operator is looking at each interaction separately and making a statement that all of the interactions taken together describe the system completely. Omitting the inter-nueclon terms (which have little effect on the motion of the electrons), and assuming a stationary nucleus...

It's likely that the major reaction you were seeing was the combustion of hydrogen gas. Hydrogen combusts pretty violently, often explosively (which is why balloons are no longer made of hydrogen). While it is true that pure oxygen-hydrogen flames emit light in the ultraviolet, impurities from...

For this problem you need to work backwards. You have equations of kinematics, from which you can calculate the inital speed of the two interlocked cars (remember that for any force, including friction, F=m*a). After that you can used conservation of momentum (the fact that m1v1=m2v2) to...

I would go with trying to do a Fourier expansion of the position wave function with sine terms mostly because your box extends from 0 to a, (ie has odd symmetry). (Like you had mention doing before trying to transform the problem into momentum space.) Although changing to momentum space...

The real trick in physics and math is to do problems. Nearly everyone needs to do problems to learn the material. The number of problems you should do is really a personal thing that you have to figure out based on how easily you can do the problems you've done. If you book ranks the...

Your first method was on the right track, though it may be easier to see if you write u = f(x,y) and v=g(x,y), and then you follow through with differentiation getting du=(df/dx)dx+(df/dy)dy and dv=(dg/dx)dx+(dg/dy)dy, and then realizing that the derivatives of f and g are functions of x and y...