Severian

Correct, but if two space ships where moving towards each other at 2x10^8 m/s the speed of the second to someone in the first would not be 4x10^8 m/s.

Maxwell's equations are never true because they are classical. However, QM aside for a moment, Maxwell's equations are true in all inertial frames (in the classical sense) because they can be written in a covariant way ([math] \partial_{\mu} F^{\mu \nu}=0[/math]). This doesn't change by including gravity (although one should then add gravitational interactions too of course), so Maxwell's equations are true in all frames. There is an easy way to settle this. Imagine a frame S, where the the photon is travelling at speed c. Let us denot your photon rest frame as S'. Please give me a Lorentz transformation which transforms from S to S'. You will find that the time axis become parallel to the direction of the photon's flight and therefore S' is not a valid reference frame. Edit: One could I suppose argue that light is stationary at the event horizon of a black hole, but this is a limiting case, and not really applicable. Edit: I even found a paper on Maxwell's equations in non-inertial frames: http://ej.iop.org/links/q79/ju4Vl9q47pptrQXkOP059A/jav23i22p5169.pdf

http://www.amazon.co.uk/exec/obidos/ASIN/0201503972/qid%3D1109770995/026-2811211-6333217

Einstein didn't even believe in quantum mechanics, nevermind Quantum Field Theory! This is pure myth.

It's just wrong. That is not the definition of mass. You are trying to generalise the rest frame energy mass relation to something it isn't applicable for, because the photon has no rest frame. Let me put it this way, for a massive particle you are saying that E=mc^2, therefore why not regard this as the definition of mass and thus the mass of the photon is E/c^2. Right? But this is not the correct expression - you should really write [math]E^2=m^2c^4 +p^2c^2[/math]. Then we have [math]m=\sqrt{E^2/c^4 - p^2/c^2}[/math] Now, it is OK to use this as a definition of mass, but then you will find for the photon that E=pc and m=0. The photon has no mass.

I'm sorry, but this is just silly. You can't just make up a new definition of mass to allow you to say that photons have mass. Any 'definition' of mass which is not frame independent is completely useless and just confuses the issue. Although photons have no mass, they still have energy. As has already been pointed out numerous times [math]E^2=m^2 c^4 + p^2c^2.[/math] The 'm' in this equation should be (for the purposes of this thread at least) the definition of mass. For a photon m=0, so this becomes E=pc. Photons have energy corresponding to their momentum. Also note that I didn't say that photons have no 'rest' mass. Since photons have no rest frame, the concept of rest mass for a photon is not defined.

I am not sure why you make this assertion about conservation of energy. We already see violations of this in particle physics. It is just a consequence of the uncertainty principle: [math]\Delta E \Delta t > \frac{h}{2 \pi}[/math] So we can violate energy conservation as long as we do it quickly enough. We can 'borrow' an amount of energy [math]\Delta E[/math] as long as we give it back in time [math]\Delta t 2 \pi / h[/math]. We see this in particle physics all the time. For example, in e+e- collisions at say 50GeV, you will see the reaction [math]e^+e^- \to Z^*\to e^+e^-[/math] even although the Z boson has a mass of 91.2GeV. By rights we should not be able to make the Z with the energy available, but we can borrow energy for a short time to make up the difference. We say that the Z is 'virtual', and the * is added to distinguish this. In fact, the more energy we need to borrow (the more 'virtual' the Z is) the quicker we have to give it back, so the faster the decay back to electrons. To some extent, every particle we see is virtual, because a non-virtual ('on-shell') particle would never decay. Anyway, there is some speculation that the universe is just one big vacuum fluctuation. We have borrowed the energy for a short time, and will have to give it back soon or expect a visit from the bailiffs....

This comes about in quantum mechanics because of the uncertainty relation. That is, some properties of particles cannot be assigned simultaneously. The classic (no pun intended) example is position and momentum. If one measures the position of a particle, the act of measurement has fixed its position, and while position is defined, momentum is not defined. In the case of entanglement in the EPR 'paradox', which is what you are refering to, the two electrons are produced together in such a way that their spin is undefined. Since the spin of an electron (when defined) can be up or down only, it forms a nice binary switch for sending messages. However, while their individual spins are not defined, their sum is, so if electron A were spin up then electron B would be spin down (note this is not yet the case since neither has defined spin). It is only when spin of one of them is measured that a definite value of spin is taken, and since we know the relation between their spins, knowing one tells us the other. Although this is most definitely non-local (measuring the spin of the particle at one space-time point instantly 'changes' the value at another) it cannot be used to send signals faster than light. This is for two reasons: 1. we have no control over whether electron A chooses spin up or down - it is random; 2. we cannot measure the spin of B before the signal is sent, so cannot detect any 'change' in its behaviour. This is why the EPR 'paradox' is not a paradox at all. We are still constrained by having information flow as sublight speeds.

Well, it was 8 years ago! I actually personally know most of the names mentioned in this post. John's first ever paper was written with me in fact. Peskin is very impressive - I have a lot of respect for his physics. I have met him at a few conferences (and at CERN) and had dinner with him a few times, and he is a nice guy. Never met Susskind though.

Witten's latest work on twistors is really quite exciting I think. Some comments: This is just a wrong statement. They have not been used to calculate things that we couldn't calculate before (yet). They give much neater derivations though. This guy is just an idiot. First of all, twistors could potentially give us really amazing insights into the structure of the theory itself. Secondly, things like W+2 jets at the LHC really need to be calculated at NNLO if we was to do anything beyond simple discovery. Go ask Mangano - I know he will agree with me. (In fact, I am faily sure they all would, even Kunszt.) PS: The author of "The experimenter's wishlist" was an usher at my wedding....

I think I am going to have to retract my statement, since WMAP has only made a measurement of the mass density, not the total mass. In one sense it depends on how you define 'universe', but all matter is thought to have been created from energy at the Big bang. So unless the big bang was of infinite energy then there must be a finite amount of mass. I certainly can't prove this though....

Can anyone recommend a technical book on string theory? I mean, with lots of mathematical detail? I have a few (Wess & Bagger, Ramon etc) but have not been very impressed with them.

Since there is a finite mass in the universe, there are a finite number of planets, so the whole supposition is moot.

This is correct. As you say' date=' there are three unknowns, the accelaration (a), the radius ® and the angular velocity (omega, and related to that, v). Then there is one relation between them: [math'] a = \omega^2 r[/math] So if you want a for a specific omega, you must fix r, and if you want a for a specific r, you must fix omega. I was pointing out that the ring in Halo doesn't seem very realistic. So, for the fairground ride, lets say it was 9.81m across (for obvious reasons ), then for 1g [math]\omega=\sqrt{\frac{a}{r}}=\sqrt{\frac{9.81 ms^{-2}}{9.81m}}=1 s^{-1}[/math] In other words, you have to go round once a second, which corresponds to a speed of [math]v=\omega r = 9.81ms^{-1}[/math] I don't think that is unreasonable...

Well, the author claims that quarks have never been experimentally confirmed. They most certainly have.

But how can you distinguish them? You can't - every physical observable that both Bob and Bill can measure will give the same result (assuming they are in a box, or a lift or somesuch) so you cannot distinguish which one is in a gravitational field and which one is being 'slowed down'. Therefore there can be no difference in age. Of course, as I described it, it isn't a well defined experiment, because I should be comparing their ages in a particular frame - while that is OK at the end (where they are at rest relative to one another) it is not OK at the beginning (when they are not at rest relative to one another). Instead, we would have to have Bob on Earth and Bill somewhere far away (outside Earth's gravity well) but at rest in Bob's frame. Then Bill would accelerate away with 1g for a time T, and then decelerate at 1g (but still heading in the same direction) until he came to rest in Bob's frame again at time 2T. Then compare their ages in Bob's (and now Bill's) frame, and they wil be the same.

Think of it this way: you can't travel faster than the first graviton you send out (otherwise you would pass it) and since that is restricted to speeds < c then you can't travel faster than light.

Yes, this is a big problem with the big bang. Having said that, it would have been worse if the inhomogeneities were absent.