Meir Achuz

This force is never exponential in z. It can be calculated explicitly for any distance. It is always expandable in a power series in z. I just gave the leading terms. It is not a question of logic, but of explicit calculaton.

At large z, the forced falls off like 1/r^4 (not r^3). It is the force between two magnetic dipoles

The formula I gave is for two touching magnets. It will be approximately true as long as the distance z between the magnet faces satifies z<< r, where r is a radius of the face. As the magents get further apart, the force starts to fall off like 1/z, then 1/z^2, and eventually (when z>>r and z>>L, the length of the magnet) like 1/z^3.

If the system is rigidly attached, nothing will move because there is lno esternal force. The B field will just try to bend the wire. A bending stress will be set up in the wire.

"stable" and "unstable" sometimes have to be quallified. All elements above Pb are unstable, sometimes with very long lifetimes. Often, when talking about these heavier elements, they are not called "unstable" if they have long enough lifetime. ("Enough" depends on the circumstances.) For instance, U238, with a lifetime over 10^9 years is often called stable, and only rarely called unstable. Some transuranium elements have long lifetimes. Plutonium 244 has a lifetime of about 10^8 years, which is too stable for safety.

The force will be in dynes.

The force needed to push the two magnets together is easy. F=2 pi M^2 A, where M is the strength of each magnet in gauss, and A is the cross-sectional area of the tube. Calculating how the force decreases as the magnets move apart is a bit messy, until they get so far apart that you can make some approximations. To calculate the force when they are still close together requires using a Legendre polynomial expansion of the field from the end of one magnet.

E=-Z^2*m*alpha^2/(2*n^2), where Z=6 for carbon, alpha=1/137 m=511 keV n=1,2,3. This gives E=-490/n^2 eV for Carbon.

Temperature is related to the random motion of the particles of a substance, not the coherent motion of the entire object.

For bound states, the classical region is when the level of excitation is so large that a change in N of one unit seems continuous. Just where this occurs depends on what accuracy you want. For instance, for hydrogen E is proportional to me^4/n^2. When n gets large, this looks classical, since changes in n look continuous.

"Is my maths correct? Let's say each bar magnet has 1 Tesla. Then 10 bar magnets put together would have 10x0.1 = 1 Tesla Is it right? Can I get 1 Tesla?" I guess you meant each magnet has 0.1 T. But, no. Placing magnets end to end do not add up the field. 200 magnets, each of 0.1T, placed end to end would give 0.1 T.

Newton's gravitational law and Coulomb's law have the same form ~1/r^2. Therefore Gauss's for the E field translates directly into a Gauss's law for the gravitational force field. Poor Newton predated Gauss by over a hundred years, so he had to work harder without G's law.

I should have mentioned that the magnetic moment of a solenoid is much easier: \mu=NIA/c, where A is the cross-sectional area of the space inside the coils. If you don't like Gaussian units let 1/c-->mu0/4pi.

The magnetic field on the z axis of a solenoid (in Gaussian units) is: B=[2\pi(N/L)I/c][sin A_2-sin A_1], where A_i is the angle between a line perpendicular to the axis of the solenoid at one end and a point on the z axis. If you write this in terms of z and expand, you can use Legendre polyomials to find the field off the axis and the dipole moment of the solenoid.

The PM can be modeled as a solenoid with a current carrying wire wrapped around it. In fact, this is just what happens to the molecular currents that cause the magnetization. They cancel in the interior of a bar magnet, but circle the outer surface.

QM has a large number of tractable problems that have definite, testable experimental and practical consequences. GR has very few of any of the above. Even some seemingly simple phenomena are still intractable in GR. The GR equations are deceptively simple. Their solution in most cases of interest have still not been worked out.

You have to calculate resonant frequencies of the particular molecules. Absorption is high near a resonance. I guess this is done by molecular chemists, which I am not.

Swansont: In any given theory (even classical mechanics) you have to determine a complete set of variables. Then "not random" (easier to define than random) means that knowing the value of each of these variables at one time, you could tell the value at any future time (assuming causality). For CM, those variables are the momentum and position of a particle (doing one particle for now). If you chose momentum and temperature, it would look like a random theory. Just look at the 5 day weather forecast. That does not mean CM is random, but just that you picked the wrong variables. In QM, momentum and energy are the wrong variables. If you pick the wave function at one time for all space, you can uniquely predict the wave function at a future time. That is what I call "not random".

QM is not random. If the wave function is given at one time, it is uniquely determined at a later time. Position and momentum are not suitable variables for getting a definite future prediction, but that is their fault, not QM's.

The standard convention is to use Latin letters for sum from 1 to 3 (just space), and Greek indices for the sum over 0,1,2,3, where x^0=ct.

When N in a bound state formula becomes large enough, the QM result can agree with the CM result. The Moon is in a QM state, but N is so large, we can't notice this discreteness. The use of a QM formalism called "coherent states" facilitate the classical look of high N quantum states. So QM does apply in the Macro world, but is usually unrecognizable. Superconductivity is an example of a macro QM state.

Your difficulty is in significant figures. Since the input numbers are given to 3 sf, your answer is only accurate to 3 sf. All the numbers after the decimal are meaningless. Your answer is 535 photons.

That answer is typical of Mathematica. It "solves" your relatively simple integral in terms of a complicated function of a more complicated (elliptic) integral. Your integral can be readily evaluated by expanding the sqrt in a binomial series. The non-integral power gives no difficulty in summing a binomial series.

Dave, I know Latex but don't know how to get it to typeset on this forum. I tried [latex]\phi[/latex], which worked on another site, but not here. Please advise. Thanks.

[math] \mu_0[/math] and [math]\epsilon_0 [/math] have no physical meaning. They have nothing to do with any property of free space or anything else. They are parameters introduced in SI units due to the fact that SI units are internally incompatible. [math] \mu_0/4\pi [/math] is exactly [math] 10^{-7} [/math] in SI units. It is not a number to be measured and cannot fluctuate. It is as significant as the number [math] 10^3 [/math] for the number of meters in a kilometer. The numerical value of [math] \epsilon_0 [/math] is determined by the equation you wrote. The number c first appeared in physics as the ratio [math]q'/q[/math] between the natural unit for electric charge in magnetic interactions (as [math] {\bf F=q'v\times B} [/math]) and in electric interactions (as [math] F=q^2/r^2[/math]). This ratio was first measured in 1856 as [math] 3.1\times 10^{10}[/math] cm/sec. It was later shown (in 1864) by Maxwell to also be the speed at which light traveled. Einstein then showed in 1905 that space and time were just two different directions in a symmetric "space-time", much like distance and altitude on a topographic map. The constant c is now understood to be just a conversion constant between cm and seconds, much like the number 5,280 is used in the US. c just defines the meter in terms of the second. The meter is now defined as the distance light travels in 299,792,458 seconds. c is no longer a number to be measured in any experiment. So the three constants in that equation are all just conversion constants between different units. They have no physical meaning, and cannot be used for FTL (whatever that means).