Severian

It is both. The neutron is udd (on up-quark and two down-quarks). The proton is uud. The down-quark can decay into an up-quark by emitting a 'W-boson', and the W-boson can then decay to an electron and neutrino. So the down's decay looks like: [math]d \to W^{-} u \to e^- \bar \nu_e u[/math] The electron and neutrino escape but the up is sucked back in, making a uud state, a proton.

Lagrangian mechanics is fundamental to quantum field theory too. In fact, all of the modern high energy theories are described by Lagrangians (of more precisely, Lagrange densities). This is mainly because the Path Integral formalism of QFT weights each possible 'path' between initial and final conditions with the expenential of the action required for the path. The prefered path is then the one with the least action.

c=1

No, there are other scales in the universe set by the fundamental theories, so the scale is not arbitrary.

I'm afraid not. If the size of the universe was your only length scale and that length scale was infinite, then you would not be able to meaningfully measure distance at all. All measurements are in comparison to something else. In reality that 'something else' is set, in fundamental physics, by some other distance scale in the theory. For example, QCD has a fundamental energy scale of 1GeV (think of an energy as 1/distance), the energy where the force becomes strong enough to form condensates, binding quarks into hadrons (like the proton). It is interesting to note that it is the quantum corrections to the theory which introduce this energy scale - in the classical limit, QCD would have no associated distance scale.... Any theory which does not involve a distance scale, such as string theory, is termed 'conformal', but even then there is always some mechanism for generating a distance scale in some way (eg. in string theory, one has the string scale).

OK' date=' I'm going to be picky. Apologies in advance... Do you have any evidence for that statement? At all? The Planck length is the distance scale at which we [i']think[/i] gravity becomes strong, and will tear holes in space time. But this is conjecture over many orders of magnitude. We have never probed this small in experiment, and don't even know how to explain gravity quantum mechanically! The Plank energy scale is 1019GeV, while we have probed only up to 102GeV. I would be very surprised if our description of the three forces we do understand held up to an extrapolation of 100000000000000000 times in energy, nevermind gravity, the one we do not understand. I think it is time we stopped taking string theorists' comments as reality, without experimental data to back them up....

Classical physics is completely deterministic, so there are no effects of this nature. Only quantum mechanical effects have this, such as radioactive decay. This was Einstein's big objection to QM ("God does not play dice."). Interestingly, even using the radioactive decay to 'trigger' heads or tails according to some well specified rule (eg. if it decays with 3s then its heads, if not its tails) you would still be able to predict the probability of getting heads, so QM still has predictive power.

NASA are, largely speaking, a waste of money. They should take all the money off them and give it to particle physics instead! Seriously though, I don't think much of NASA's program: the only decent space projects are things like Planck and Cassini.

If we restrict ourselves to spin-up and spin-down' date=' then the electron is not always in a definite spin-state. It could be a superposition of a spin-up state and a spin-down state, and only when you measure the spin does it 'choose' to be spin-up or spin-down. Since the composition of 'spin-upness' or 'spin-downness' can change, it has the properties of a wave. All the time! A photon can turn into a 'virtual' electron-positron pair and then back into a photon. Swantson has already got this one. An electron travelling 'back in time' is just a positron travelling forward in time (this is a consequnce of the CPT theorem). by 'QFD' I suppose you mean quantum flavor dynamics, as in the weak interaction? The bosons are the particles which transfer the force from one particle to another, called the W and Z bosons. Think of this like two iceskaters throwing a ball between them. In throwing the ball they exchange momentum and push each other away; they create a force by exchanging an object. There is nothing very mysterious about these particles - the humble photon is a bosonic force carrier for electromagnetism, so they are just generalisations of the photon.

nm - got the wrong end of the stick there...

Sorry ed - I didn't realise your question was serious. Both. Is it not clear that they are the same?

Black body radiation? You mean the effect where futons get warmer when black people sit on them?

I think that is a nice point Ophiolite. Even if one could travel back in time to for example, see Christ on the cross, you would need to pick a spacial coordinate too. But where would Jersusalem in 33AD actually be? You would need to take into account the rotation of the Earth, the Earth going round the sun, the roatation of the galaxy and ultimately the Hubble flow of the universe! So even if you did somehow manage to invent a time machine, you would probably pop out in deep space....

'Derivative'. The word is 'derivative'. ....and you're welcome.

tut tut! You should do integrations yourself, not using maple.... [math]\int_0^{\infty} \frac{1}{\cosh^2 (\omega \theta)} d\theta = \frac{1}{\omega} \left\{ \tanh \omega \theta \right\}_0^{\infty} = \frac{1}{\omega}[/math] so [math]t=\frac{a^2}{h \omega}[/math] which , along with h=av gives the right answer.....

For the last bit, you should use [math]\frac{d \theta}{dt}=hu^2 = \frac{h}{a^2}\cosh^2 (\omega \theta)[/math]. The object hits O as [math]\theta \to \infty[/math] so we can use the above equation to write [math] t = \int_{0}^{\infty} \frac{a^2}{h \cosh^2(\omega \theta)} d\theta[/math]. Integrate this....

The Antiproton Source is made up of three parts. The first is the Target: Fermilab creates antiprotons by striking a nickel target with protons. Second is the Debuncher Ring: This triangular shaped ring captures the antiprotons coming off of the target. The third is the Accumulator: This is the storage ring for the antiprotons. http://www.fnal.gov

We don't have colliders which can probe string theory yet either.

Photons are not energy - they contain energy. In other words, their energy is a property of their state. In fact, every particle contains energy, whether a photon, an electron, quark, gluon etc. Gluons are the force carriers of the strong nuclear force, just as photons are the force carriers of electromagnetism.

OK, your general radial solution was: [math]u(\theta) = C e^{\omega \theta} +D e^{-\omega \theta} ~~~(1)[/math] Differentiating wrt [math]\theta[/math] gives: [math]u^{\prime}(\theta) = \omega ( C e^{\omega \theta} -D e^{-\omega \theta}) ~~~(2)[/math] Your boundary conditions are [math]u(0)=1/a[/math] and [math]u^{\prime}(0)=0[/math] since start a distance a out, with zero velocity parallel to OP. [math]u(0)=C+D=1/a[/math] [math]u^{\prime}(0)=C-D=0[/math] Thus C=D=1/2a. that was the mathsy was to do it, but you could think more physically. The general solution can be rewritten as: [math]u(\theta)=E \cosh (\omega \theta) + F \sinh (\omega \theta)[/math] where E and F are constants. Now since the force is a central force, pointing in towards the centre of the system, it doesn't care if you go round clockwise or anticlockwise. Whether you choose to define increasing theta in a clockwise or anticlockwise sense is entirely up to you, and therefore the answer can't depend on your choice. This symmetry is only broken by the choice of initial conditions (as to which way the object is moving). But this initial condition, having no radial part, cannot effect the radial equation (other than through the angular momentum). therefore you r answer must be symmetric in replacing theta with -theta. Since sinh is anti-symmetric about 0, its coefficient F=0. Thus E=1/a.

Well, you could impose your boundary condition on the velocity (or [math]\frac{du}{d \theta}[/math]) to give you the extra constraint, but it is fairly obvious that C=D from the symmetry of the problem. The boundary conditions don't specify whether your initial velocity is clockwise or anti-clockwise because the whole problem is necessarily symmetric under [math]\theta \to -\theta[/math].

The angular momentum (per unit mass) is h=va (since it is constant you just take this from initial conditions). The constraint [math]\lambda > v^2a^2[/math] is then just insiting that the object under the square root is positive. At time [math]t=0, \; \theta=0[/math] and [math]u=1/a[/math], so [math]1/a = C+D[/math]. the other 2 comes from the definition of cosh.

Lol! and I just pasted your equation! (that'll teach me to trust you!)

Write down a differential equation for r(t), using Newton's Second Law (F=ma). You can assume that motion is in a plane (since it is a central force) but should consider axial and radial motion (the equation for one of which is trivial but gives you an expression for angular momentum conservation). Convert (the non-trivial equation) to one concerning u(theta) (which makes it easier to slove) and solve it. If you feel ambitious, solve it for a central force of the form [math]\lambda r^{-3} +\kappa r^{-2}[/math]. Once you have the equation for u, you will see that this is the most general central force with an orbit equation which can be solved in the time frame of an exam (since the r-2 and r-3 are the only forces which will give a linear differential equation).

[math]0.510998918 \pm 0.000000044 {\rm ~MeV}/c^2[/math] (Yeah I know.... I'm a physics geek boy....)