Severian

That is not a prediction of String theory - that is a prediction of a certain boundary condition which is sometimes used in some string theories. The boundary condition doesn't need to have anything to do with string theory though, and in fact if this is true then we would expect to see signs of it at the LHC, since the Plank mass would only be a few TeV.

"Stone walls do not a prism make, nor iron bars a diffraction grating." "Quasars are far out!" "Absolute zero is cool." Resistance begins at ohm.

If God has made the world a perfect mechanism, He has at least conceded so much to our imperfect intellects that in order to predict little parts of it, we need not solve innumberable differential equations, but can use dice with fair success. -- Max Born The electron is not as simple as it looks. -- (William) Lawrence Bragg, British Physicist(1890-1971) "The second law of thermodynamics holds, I think, the supreme position among the laws of nature. If someone points out to you that your pet theory of the Universe is in disagreement with Maxwell's equations - then so much the worse for Maxwell's equations. If it is found to be contradicted by observation - well, those experimentalists do bungle things up sometimes. but if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing to do but to collapse in deepest humiliation." -- Arthur S. Eddington (British Astrophysicist, 1882-1944) in The nature of the Physical World (1928) If it's all right with Dirac, it's all right with me. -- Enrico Fermi, on being told that there was direct experimental evidence that helium-3 nuclei obey Fermi-Dirac statistics. Physics is to Math what Sex is to Masturbation --Richard Feynman Physics is not a religion. If it were, we'd have a much easier time raising money. - Leon Lederman

Maybe you are the same person....

If they were as poor as they sounded, they would have qualified for a government grant which would have paid their living expenses while they studied. (And they would not have been charged any fees.)

You have got to be kidding me! University is waaaay cheaper in the UK than in the US. And I think someone has been having navynuke on, unless 'a while back' means just after WWII...

What do you mean by 'at once'? Simultaneity depends on the reference frame. Two events which are simultaneous in a particular reference frame may not be in a different reference frame. So photons (or any other particle) cannot be "in two places at once". (I suspect what you mean is that if you were in the photon's rest frame, all other time intervals between external events would appear infinitely time dilated. But since the photon has no rest frame, the point is moot....)

"foot pounds"! *shudder* I heard someone claim recently that the English army at Agincourt would have defeated a similar sized Napoleonic army. His argument was that the lowbongs had better range and accuracy than muskets. The only reason that muskets were used was because it was much easier to train troops how to use them, but the English army at Agincourt had a huge number of soldiers who had been trained to fire a bow from childhood. I don't know how true this is...

Oh... I hadn't noticed the post was so old, and had just been resurrected. Sorry.

There are plenty of measured quantities that need precision better than the number of digits on your calculator.... ... this one springs to mind.

Whether or not a quantum number is additive or multiplicitive is completely semantic. If I have multiplicative quantum numbers of two particles A and B being [math]q_A[/math] and [math]q_B[/math], then the quantum number of the combined state is [math]q_Aq_B[/math] if the quantum number is conserved. But I could have defined [math]Q_i=\log(q_i)[/math] as my quantum number. Then the combined object has quantum number [math]\log(q_Aq_B)=\log q_A+\log q_B = Q_A+Q_B[/math]. In other words, [math]Q_i[/math] are additive. The only important thing is whether or not the quantum number is conserved.

They annihilate because the resulting system (usually a photon) has a smaller rest mass. They are very happy to annihilate to photons, which has a zero mass, but in principle they will annihilate to any available lighter state which has the right quantum numbers. Since particles and antiparticles have opposite quantum numbers, they have to annihilate into something which is neutral and colorless, just like the photon (but in principle a very heavy particle-antiparticle pair could annihilate into something else, like a Z boson).

Ah - I see what you mean. Yes, you are right - dark matter can't be charged. This is actually a constraint on SUSY models because quite often the lightest SUSY particle of a model is a stau, the SUSY partner of a tau, which is charged.

It is easier to do this using work and energy. Say the distance dropped is [math]h[/math], and the downwards force is [math]F=mg[/math] then the work done by the force is [math]W=Fh=mgh[/math] This is the energy which is converted from potential energy to kinetic energy, so the change in kinetic energy is [math]\Delta K = \frac{1}{2}mv^2 = mgh[/math] So after dropping a height [math]h[/math] the downward speed is [math]v=\sqrt{2gh}[/math] You can then use [math]v=gt[/math] to find [math]t=v/g=\sqrt{\frac{2h}{g}}[/math]

I am a theoretical particle physicist.

They have different spin and are much heavier, so this wouldn't happen. In fact, the interactions look rather different when you include the kinematics - what Atheist meant (I think) was that they couple to the same particles as their SM partners, with related strengths. So for example, they have the same electromagnetic interactions (so I don't understand your last comment). Also, only the lightest SUSY particle is stable. This is probably a neutralino, which is a mixture of the SUSY partners of the Higgs boson and the photon. We don't see universes built out of Higgs bosons and photons do we...?

What was wrong with typing [math]R_{\alpha \beta}-\frac{1}{2}g_{\alpha \beta}R+\lambda g_{\alpha \beta} = \frac{8 \pi \kappa}{c^2} T_{\alpha \beta}[/math]?

I am fairly sure this would be strange quarks, since there was a worry that they would destroy the universe or something.

OK. A photon (the wiggly line on the left) splits into an electron positron pair (the solide circle - remember time is left to right). The positron interacts with a magnetic field (the photons coming down - there needs to be two to conserve momentum) and finally the virtual electron-positron pair recombine back into a photon.

My GPS in my car can tell if you turn around and I am pretty sure that it is not just sensing how many turns you make, so I think it has a compass. It must have some whay of knowing which way I am pointing because it still knows which way it is pointing if I take a ferry. Also, it cannot be using the GPS satellite to tell it (by having multiple sensors) because GPS is only accurate to 10m or so (and my car isn't that long!).

Here are some graphs. First, here is an example of a virtual electron interacting in a magnetic field: and here is your second process, which is basically compton scattering: (time is left to right)

Both processes are allowed. Unless I am mistunderstanding you, the second process is a special case of the first. And the second case is one part of Compton scattering, which is a very well studied process. (I could draw you the Feynman diagrams if I were at work, but I am home right now. Maybe later...) Maybe this might help: http://www.wolframscience.com/nksonline/page-1060a-text?firstview=1

I think this discussion is rather semantic at present, since no-one can think of how a 'complex observable' would be manifest. When you make a measurement you are given a real number of a series of real numbers. Even if somehow you managed to measure a 'complex observable' the apparatus is not going to suddenly spit out a great big 'i' indicating that it is complex - you will either get one real number (the imaginary part of the complex number) or you will get 2 (the real and imaginary parts) (more likely you would get a magnitude and a phase). Actually we already have this in QM - we can measure the size of a cross-section and the phase between different processes. But while this may look like we are measuring a complex number, we are not - we are separately measuring the magnitude and phase, which are both real numbers. Furthermore, in QFT (or QM) one does not make the statement that all measurable quantities must be represented by Hermitian operators. One instead says that observables that we know how to measure such as momentum, energy, production rates, must be real and therefore represented by Hermitian operators. Since we can measure these things. they are undisputably real, so this must be true. I challenge you to come up with a 'complex observable' and I will show you that it is not a complex observable, but simply two real observables.

The problem is not creating matter (which is very easy). The problem is creating matter without creating anti-matter.