Severian

The state of the art collider being built at CERN is the Large Hadron Collider (LHC) which will collide protons together at an energy of about 14TeV. It should turn on (if there is no delay) in 2007. However, since prtons are not funtamental particles, it is really the collisions of gluons inside the proton which are important, and these collisions are at about 1TeV. A 10PeV collider would need radically different technology. First of all, it would need to be straight since bending a particle round in a circle is an acceleration, realeasing breaking-radiation, which reduces the enrgy of the particle and places a cap on how energetic you can make it. But if it is straight, you can't pass it round and round to gather energy - you only have one shot through the accelerator to achive the 10PeV. So we would have to think of some radically new way of accelerating the particles since we couldn't just 'scale-up' the ones we have.

Quantum Mechanics is not studied much anymore, basically because it is incomplete. Quantum Field Theory (where the fields themseleves are also quantised) on the other hand is very widely studied.

since you can reformulate all of physics in a way which doesn't use complex numbers at all (although it doesn't look nice) I don't see how there can be anything 'non-physical' about them being used in physics. Just think of every 'number' as being a 2d vector in some vector-space. The first entry is the 'real' part while the second is the 'imaginary' part, and the magnitude of the vector is the modulus. This should tell you that the imaginary entry is just as physical as the real one.

To actually observe gravity waves is extremely difficult, since gravity is so weak. You have to wait for a massive gravitational event (like a supernova) and try and detect the tiny gravitational influence of this on something on Earth. Generally speaking Gravitational Wave Detectors have very long arms with large masses on either end - a gravitational wave will move the masses differently (since it is a 'wave' and will have different magnitudes in different places) which, if you are careful enough, you can detect. There should be two arms (at least) at right angles to each other to cover both directions on Earth. There are pretty much two schools of thought of how to build these things. The US machines are the biggest (of course) but tend not to be very sophisticated, whereas the European machines are smaller but more high tech (in the way which they remove noise). However, there is now a new initiative to apply European technology (form GEO 600) to an American machine (LIGO), so hopefully we might see something in the next decade or so.... PS: Gravitons are only theoretically postulated - they have never been observed, and are not likely to be in my lifetime.

Since we don't have a working theory of quantum gravity we can't say for sure. If we were to formulate a quantum gravity it is most likely that the strength of the gravitational interaction would change with energy. Extrapolating to high energies, the Planck scale is the energy to which is is thought that gravity will become strong. The Planck length is the length associated with this energy. People say that one cannot have a length smaller than the Planck length because in order to probe such a length you need so much energy that gravity becomes strong enough to 'foam' space-time - i.e it breaks it up into lot os little black holes. But this is all speculation. There is no reason to believe that the gravity coupling strength behaves in this way - there may be some other physical effect which prevents gravity from becoming this strong. Since we have no scientific experiments or observations to probe the physics of this high scale it is rather unscientific to speculate on the physics there.

http://map.gsfc.nasa.gov/m_uni/uni_101bbtest3.html

Incidentaly, the CMBR is not vacuum energy - it is just the photons which were expelled from the perfect blackbody at the surface of last scattering when the universe became opaque. They have been travelling for 14 billion years or so. (If you look at the CMBR's spectrum you will see that it is a perfect blackbody spectrum.) Vacuum energy occurs when a particle's ground state is not at zero energy, so for example the Higgs boson contributes vacuum energy. Also the cosmological constant would provide similar effects to vacuum energy (although technically it is not vacuum energy). Vacuum energy is needed for inflation (which may be why you are getting confused).

Well, it fails on a theoretical level. By the time we get down to a size where quantum effects are noticable, gravity is so weak that we can't do any experiments on it (yet). But the classical viewpoint of GR is incompatible with the rest of physics - for example, in classical GR one can know the position and momentum of a particle at the same time, which is inconsistant with QM.

Actually we know that GR is wrong. That is, it does not correctly describe gravity at very small distances since it is a classical theory, not a quantum one. However, it is still a very good description of gravity at macroscopic distances so we should not throw it away. We also know that our current theories of QG are wrong since they are either incomplete or have bad properties (like infinities cropping up all over the place). So we don't really understand gravity very well yet. However, it is pretty certain that the correct theory of gravity will contain a graviton which mediates the force (so your case B). But this final theory has to look like GR at large distance scales: so if we screw up our eyes it should look like case C. The fact that it can look like both at once is a consequnce of particle-wave duality.

I think you may be getting confused between the distinction of GR and Quantum gravity. In GR, there is no graviton since it is a clasical field theory (just as in classical electromagnetism there is no photon - only light waves). One has to quantize the theory to get particles and quantum gravity then has a graviton (and lots of problems!). For classical GR, choice C would be correct (if you remove the word graviton) so that the Earth has a constant gravitational field around it. This is analagous to the electric field round a point charge in classical electromagnetism. In QG choice B is correct if you modify it to be any body with energy (not just mass). In fact this analagous to Quantum Electrodynamics where photons are exchanged between charged bodies. In the covariant formulation of QED you can't even define which way the photon travels, ie. did it leave charge A and travel to charge B or the other way round (since photons are their own anti-particle).

Or you could use time for the 4th dimension. So a unit hypercube would be a unit cube which existed only for one unit of time. It would not exist, then suddenly flick into existence, last a unit of time and disappear again. (I will leave the Minkowski metric to one side...)

Since it is supposed to be the UK lottery, they might have at least gone to the effort to write the prize money in pounds rather than dollars.

I thought the world ended in 2000? No? I want my money back....

Ironically gravity is the weakest force that we know of. The other three forces, electromagnetism, the strong nuclear force and the weak nuclear force, are all much much stronger than gravity. However, the strength of these forces changes with energy/distance and it is postulates (but unproven) that gravity will become strong on very short distance scales or at very high energy.

Yes. Since the graviton has energy, it exerts gravity itself. This is not the same as quantum electrodynamics, where the photon is neutral and doesn't feel the electromagnetic force. (But the gluon of QCD is colored, and feels the strong force.)

Why do you say this? A simple formation of an electron positron pair is the creation of matter (together with antimatter). On the discussion at hand, I find myself agreeing with Martin (on all points).

I don't see how GR can be regarded as harder than QM since we really don't understand QM at all (what makes the wavefunction collapse? how do we define an operator as separate from the system?). Even if you are willing to sweep a lot under the carpet (in the name of testability) then GR is just a classical field theory while QM has a lot of new ideas like anti-commutation of operators. And GR is certainly a lot less complicated than Quantum Field Theory. In fact it is turning GR into a QFT which has caused so many problems.

Well, the 'i' arises because the wave is decribed by an expenential of a complex number. So a field may look like [math]\psi(x,t) = \psi_0 exp(i(kx-\omega t))[/math] for example. Remember that [math]e^{ix} = \cos x + i \sin x[/math] so this form is really just saying that the field is 'waving'. Now the momentum of this wave is [math]\hbar k[/math] and we can extract the k from the field only by differentiating with respect to x: [math] \frac{\partial}{\partial x} \psi (x,t) = \psi_0 exp(i(kx-\omega t)) ik = \psi (x,t) ik[/math] So to get the momentum [math]\hbar k[/math] the operation we must perform on the field is the differentiation and then multiply by [math]\hbar[/math] and finally multiply by -i to get rid of the extra i. Therefore the momentum operator (which extracts the mometum from the field) is [math]\hat p \equiv -i \hbar \frac{\partial}{\partial x}[/math] and [math]\hat p \psi(x,t) = \hbar k \psi(x,t)[/math] as desired.

The Schrodinger Equation is just conservation of energy in a quantum mechanical form. Classical (non-relativistic) conservation of energy is: Kinetic energy + potential energy = total energy Kinetic energy = 1/2 m v2 = [math]\frac{p^2}{2m}[/math] where p=mv is the momentum. Potential energy depends on the form of the potential V(x) So classically [math]\frac{p^2}{2m}+ V(x) = E[/math] In qunatum mechanics, the momentum of a field [math]\psi(x)[/math] is given by [math]-i \hbar \frac{\partial}{\partial x} \psi(x)[/math] or in other words the momentum operator is [math]\hat p = -i \hbar \frac{\partial}{\partial x}[/math]. Also the energy operator is [math]\hat E = i \hbar \frac{\partial}{\partial t}[/math]. So the Schrodinger equation is just [math](\frac{\hat p^2}{2m}+V(x))\psi(x) = \hat E \psi(x)[/math] or [math]-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+V(x)\psi(x) = i \hbar \frac{\partial}{\partial t} \psi(x)[/math].

We don't. That is why we call it the Copenhagen interpretation. But it really isn't a scientific question - it is a question of philosophy, since it can't be tested. The evolution of the wavefunction in QM can be tested of course. It is just the mechanism of collapse which cannot.

The Big Freeze is more commonly known as 'The Heat Death of The Universe'.

None and never.

Sorry to break it to you, but apparently Episode III does have a battle with Wookies in the trees of Kashyyyk....