Recent content by Psi^2

Because it is approximation. In realty, there is no such a thing as a free particle.

I agree.

You simply solve Schrödinger equation for free particle, and look at the result. No boundary conditions, no quantization ;)

Energy doesn’t have to be quantized. For example, energy of the free particle (in quantum mechanics) is continuous.

Huh, this can be explained in couple ways. I don't know did you do derivation of Kepler laws - when you do that, you can see that objects have elliptic (or almost circular) orbits because they had a initial velocity, and therefore angular momentum and in the combination with gravitational law...

You are right. Earth is not sphere, but we do that kind of approximation so we can easily do calculation. When we consider Earth as a sphere, we can show that we can replace that problem with a point mass problem. If you know how to integrate, you can investigate what difference in force would...

P.S. I have some technical issues here.. you can copy these formula code into this application, and you will see it: http://www.codecogs.com/latex/eqneditor.php

Hello. You don't need to know what kind of process is between A and B on picture 2. You simply calculate work by integrating, or simpler, by calculating surface under the AB line. It is easy to calculate, because you have one right angled triangle and one rectangle. (i will do approximation...

No, it would be F=mg in the direction of falling, and some air resistance force in the opposite direction (if you taking air friction into account). Whatever you use, you have to have the same results. No. When it reaches terminal velocity, it have constant speed due the equality of F=mg...

My first step would be: for first reservoir (sphere) use Maxwell – Boltzmann distribution for velocity in spherical system, and then you say that only these particles who have certain condition (angle theta, angle phi) will go into a rectangular container. I don’t know, it could help.

You probably think on two nablas - differetiates with respect to the coordinates of particle 1 and 2. This is because you have to masses at a fixed separation R.

I don't have an answer for your question, but I’m curious (because I don’t know anything about black holes): what kind of potential are you using in Schrödinger equation for system of two micro black holes?

Yes, you need, because Boyle's and Charle's Law are derived form ideal gas formula (pV=nRT=NkT) where temperatures are given in Kelvins. If you put Celsius you will have different dimensions.

Second picture is correct. So, for the pressure in point 1 you will take h1, and for the pressure in pont 2 you will take (h1+h2). O course, pressure in the point 2 will be greater than pressure in the point 1, and you will have resultant force upward. Now you have to calculate weight and you...

Read it, it’s in Griffiths book. You cannot normalize wave function of a free particle unless you do transform to integral over the continuous variable, which is k.