Severian

Good question. Historically this was to avoid solutions which had negative energy. If you can derive the relativistic form (as I suggested) you will see what I mean....

If you have a positron and electron annihilate to form a photon in the collider, the photon will usually decay pretty quickly. This is because the photon (usually) borrows energy using Heisenberg's uncertainty principle and must give it back rather quickly. However, what it decays into is not fixed. As long as it has enough energy to create the new particles (and usually in a collider it has lots of energy), it can decay into any charged particle-antiparticle pair. For example, it can decay into a quark-antiquark pair. However, the collider doesn't just collide one elctron with one positron. It collides lots of them together (in bunches). Often none of them will hit (there is a lot of space in between ) but sometimes two will collide. No - it is left pretty much to chance. There is no way to direct what will come out, so you can in principle create 'exotic' matter if you have enough energy. However, you may be able to influence which particles are produced a little by using polarized beams or by tuning the energy. For example, the LEP collider ran for a long time with the centre-of-mass energy tuned to the mass of the Z-particle - this guaranteed lots of Z's to study. At the LHC, one of the aims will be to produce supersymmetric particles, the lightest of which is a good dark matter candidate. Anti-matter will have the opposite spin of its particle partner, but they can both be left or right handed, so this isn't a definition or distinction of antimatter. I can produce a left-handed particle with a right-handed antiparticle, but I can also produce a right-handed particle with a left-handed antiparticle. You may be confused by the weak interaction, which only couples to left-handed objects.

I especially like this bit: Do you think they really did this study, or are they just making it up? Can the general public really be this stupid?

No!No! Dreadful book! Please don't buy it!

Hmm... it was working a minute ago. Edit: Weird - I just edited it but didn't change anything and now it is working! Edit 2: or not... seems to be a site problem rather than me.

A plane wave of momentum [math]\vec{p}[/math] and energy [math]E[/math]is [math]\psi = N e^{-i(Et-\vec{p}\cdot\vec{x})}[/math]. You can extract the momentum and energy with operators [math]\hat \vec{p}[/math] and [math]\hat E[/math] such that [math]\hat \vec{p}\psi = \vec{p} \psi[/math] and [math]\hat E \psi = E \psi[/math], if you make the choice of operators: [math]\hat \vec{p} \equiv - i \hbar \vec{\nabla}[/math] and [math]\hat E \equiv i \hbar \frac{\partial}{\partial t}[/math]. Now, we know from classical mechanics that the total energy is the sum of the kinetic and potential energies. Also, the kinetic energy (in a non-relativistic system) is [math]E_K = \frac{1}{2} m \vec{v}^2 = \frac{1}{2m} \vec{p}^2[/math]. In other words, the total energy is: [math]E = \frac{1}{2m} \vec{p}^2 + V[/math]. Simply replacing the energy and momentum by the right operators for a plane wave gives the Schrodinger Equation: [math]i \frac{\partial \psi}{\partial t} = \left( - \frac{\hbar^2}{2m} \vec{\nabla}^2 + V \right) \psi [/math]. Using the same principals, can you derive the relativistic version (the Klien-Gordon Equation)?

This is also not true. Whether of not the universe is infinite is an unscientific question which science can never answer (it could conceivably prove it was finite (if it were) but not the other way round). Since we can only ever see to our horizon distance, we can never know what is beyond our horizon therefore, from a scientific point of view, what lies beyond the horizon is an ill-defined question. Of course, scientists use a model where the universe is infinite because it is easier to formulate the mathematics, and there is no consequence to observable quantities, but they do not (should not!) use this model to make statements about physics beyond our horizon. On a similar note, people keep making statements about the universe starting with a sigularity. Can you support this claim with evidence? I am sure you cannot. In most inflationary models, one cannot recover any information from before about the last 60 e-folds, so what happened before this time is unknowable in pinciple. What happened before becomes (almost) as unscientific a question as what lies beyond our horizon.

The problem with this thread is that no-one has bothere to define what the Big Bang actually is. There are rather a lot of different definitions, and if you want to ask if the Big Bang happened then you must define what you mean. For example, is the 'Big Bang' the point where space and time were created? If so, then we can never answer the question. If time was created at the instant, then we cannot probe earlier times because there were no earlier times, but we cannot definitively say there are no earlier times because there may simply have been a machanism for removing all the information from the system. If this is your definition, then your question is unsceintific and we need to move this thread to Philosophy and Religion. Alternatively, one can define the Big Bang as the moment reheating starts after cold inflation. Then the 'Big Bang' is not the point of creation of space-time, and we can definitely ask if it happened or not. Since the jury is still out on inflation (never mind anything else) we cannot definitively answer this yet. Then there are Ekpyrotic models, where the big bang is the collision of two 'branes' (manifolds moving in a higher dimensinal space). I personally don't have much time for this model, and this definition is certainly not confirmed. Irrespective of the definition though, I think the answer to the question is: We don't know.

In out current theory, the quarks are point like objects, so their mass is just set by a parameter in the theory and is not dynamically generated. However, if we ever find out that the quarks are not fundamental, then their constituents will have motion which contributes to the mass. So you are correct in your 'prediction'. Most particle physicists think there will be something more fundamantal (strings?) so this will probably one day be proven correct. Interestingly enough, this is also the way that a lot of physicists believe things will turn out. An object having an 'inherent mass' is not very attractive from a theoretical perspective because it is a parameter of the theory which you cannot predict. So probably masses 'come form someting else'. Also interesting is the fact that String Theory is 'conformal'. This is a technical term which basically means that it has no inherent mass scale. So string theory would fit your bill - a theory where mass, as you put it, is all motion.

174GeV or so. Above that the electroweak symmetry is unbroken and the natural states are the pure SU(2) and U(1) (hypercharge) gauge bosons. The photon is a mixture of SU(2) and U(1) fields.

This is correct, but it isn't the analogous picture for the bgi bang. People tend to thing of the big bang as a big explosion, because of the name, but an explosion in real life is localized to a small area, and then spreads out. This isn't how the big bang (is thought to have) started - the big bang's explosion is not localised to one place - everywhere explodes at once! It has to be this way, because space-time was supposed to be created at the big bang. If it wasn't created everywhere at once, there would be no space for the light from a localized explosion to move into anyway.

No - it is not new. There have been lots of attempts to do this, but none have shown any real pattern (although a few, such as Regge trajectories, have been helpful).

More importantly I think, if there were no cosmological flow (which is what I think you mean by static) then any redshifts you observe would be local fluctuations. The object would be as likely to be moving towards you as away, so you would see as many blue-shifts as red-shifts. The fact that everything is redshifted is evidence of an expanding universe. Everything is moving away from us.

Yes, it is a direct product. Basically all particles must be representations of these symmetry groups, but the symmetry groups are essentially unconnected. For example, you can perform a SU(3) transformation on the system and get the same physics back, without needing to do anything with the SU(2) and U(1). Technically, saying it is SU(3)xSU(2)xU(1) is true only above energies of about 174GeV. Below this scale the Higgs mechanism breaks the electroweak part of the symmetry down to U(1), ie. SU(2)xU(1) -> U(1), where the last U(1) which is a result of the breaking is electromagnetism, and is different from the U(1) in 'SU(2)xU(1)'.

Let me explain this yet again. It depends on your definition of mass. The correct definition of mass is the 'm' in E2= m2 c4+p2c2. This mass does not change with velocity and is the same in all inertial frames. It does not approach infinity as we go towards light speed. The bad definition of mass, is [math]|\vec{p}|/|\vec{v}|[/math] (the magnitude of momentum divided by the magnitudes of velocity). This is sometimes used as the definition of mass out of a desire to preserve the Newtonian equation p=mv. If one uses this definition then mass is velocity dependent and will appraoch infinity as we appraoch the speed of light. However, this is a bad definition because it is frame dependent - the mass of the particle depends on who is looking at it! Since [math]p=m_{\rm bad}v=\frac{1}{\sqrt{1-v^2/c^2}}m_{\rm good} v[/math] you can see that it is really the momentum which is becoming infinite as we approach the speed of light. Therefore you need to put an infinite amount of energy and momentum into the particle to make it go at the speed of light, and this is why we can never accelerate any massive object to the speed of light.

It will exceed the speed of light. This is a known result. More commonly this is described as a spotlight being swept over low clouds - the beam will 'move' over the clouds faster than the speed of light when the nagle approaches zero. However, there is no information flow so this does not violate Special Relativity.

Yes, a PhD is all about research. They do supervised research so that they can learn how to do unsupervised research. I would never trust a grad student's calculation without checking it though (no offense to grad students). Usually if there is a mistake in a calculation it is really obvious - you find an answer with the wrong properties. But even if a mistake goes unnoticed, it doesn't really matter. A false calculation will never agree with the data (unless the mistake is very small) so you will find a discrepancy with data, and the first thing you do (or someone else does!) is check the calculation.

Heh - no. A PhD is research done by the superviser under particularly arduous circumstances.

Time dilation is directly seen by the decay of particles in the upper atmosphere. Muons are produced by cosmic rays hitting the upper atmosphere, but since muons decay very very fast, you might expect them to decay before reaching the Earth. However, they do reach the Earth: they are travelling so fast that they suffer time dilation, making them live long enough to reach us before decaying. Without SR, they would decay long before they reach us.

Well, simple Feynman diagrams calculations are done by our final year undergraduates, so it doesn't require that much training to get started. To work in the field and publish papers you really need to do a PhD. To get a permanent faculty job you need to have done a few 2-year postdocs. I think perseverance if more important that hyper-intelligence. No physicist understands everything, so we are all still learning.

OK - I see what you problem is now. The difficulty is that the doppler shift looks like it has different causes in different frames. You are correct that in your new frame (moving towards the source) the light will still be moving at a speed c towards you. So it looks like nothing has changed: previously you had the light moving towards you at relative speed c, and now you still do! However, now you are moving fast all the distance scales will suffer a length contraction, so the distance to the next wavepeak becomes smaller and you will reach it more quickly. Hence the frequency goes up and the light is blue-shifted as before.

You will see it right away. The important thing in this case how frequently you pass through the wave peaks. Since you are travelling towards them you will hit them more frequently.

To really understand the Standard Model, you really need to understand Quantum Field Theory. The best book on QFT that I have seen is called "An Indtroduction to Quantum Field Theory" by Peskin and Schroeder. It starts off quite advanced though. What level are you at? Would you understand this: http://www.phys.ualberta.ca/~gingrich/phys512/latex2html/node1.html or is that too advanced?