Severian

Yes' date=' that is right. I did mention this in my post too. In the case I derived, the medium was at rest with respect to the observer. However, as long as you quote the propagation speed with respect to the observer (rather then with respect to the medium), you can use one formula (the one I wrote down). The motion of the medium relative to the observer would of course change this speed. Agreed, but again if one quotes the velocity of propagation with respect to the observer, this should be the same formula as you would use for sub-light propagators. This is the same as the expression that I derived (the last equation of my post). The other equations can all be derived from this.

Science? You have GCSE's in 'Science'? What ever happened to Physics, Biology and Chemistry?

Yes it does. This is a weak decay. The neutron is made up of three quarks: one 'up' quark and two 'down' quarks. One of the down quarks decays to an up quark by emitting a 'W-boson' (which then decays to lepton and a neutrino). Since the proton is two up quarks and a down quarks, the neutron has decayed into a proton. The actual calculation is a bit beyond the scope of a web forum though....

Let me qualify that statment. The speed of light does not effect the qualitative description of what is happening' date=' although it does effect the quantitative description. Lets take a simple example: My friend and I are standing a distance [math']d[/math] apart and he throws a ball at me, with velocity [math]v[/math] at intervals of time [math]\Delta t[/math]. Asuming we can neglect air resitance, the ball takes time [math]d/v[/math] to reach me, so the first ball hits me at time [math]t_1=d/v[/math], and the second ball hits me at time [math]t_2=d/v + \Delta t[/math]. The frequency of balls is [math]\frac{1}{t_2-t_1} = \frac{1}{\Delta t}[/math] and is independent of [math]v[/math]. So far so good. Now, while still throwing the balls, my friend walks away from me with speed [math]u[/math] but still throws the ball at me with speed [math]v[/math] in my frame. (We are interested in the frequency in my frame, and since we will want to make the ball light later on, it makes sense to stay in my frame. Also, for sound, since the air is at rest, we should keep the same speed in the 'air's rest frame.) As before, the first ball (which he throws just as he starts to move backwards) hits me at time [math]t_1=d/v[/math]. By the time he throws the second ball, he is a distance [math]d+u \Delta t[/math] away, so the ball takes time [math]\frac{d+u \Delta t}{v}[/math] to cover the distance and hits me at time [math]t_2=\frac{d+u \Delta t}{v}+\Delta t[/math]. The frequency with which balls hit me is now [math]\frac{1}{t_2-t_1} = \frac{v}{u+v} \frac{1}{\Delta t}[/math]. So if he moves away from me, the frequency decreases. This is exactly the doppler effect for sound. notice that the expression is dependant on the speed of sound [math]v[/math]. My point was that this formula is also completely valid for the doppler effect with light. This is because all of the quantities are defined in my frame of reference and relativity only effects the comparison of quantities in different frames. To be fair though, we should point out (as I mentioned earlier) that if my friend is moving away at a relativistic speed, then [math]\Delta t[/math] is better to be defined in the 'ball thrower's" frame, since he is the one who decides when to throw. In this case, the time between his throws in my frame is time dilated to [math]\gamma \Delta t[/math], with [math]\gamma = \frac{1}{\sqrt{1-u^2/c^2}}[/math] so the frequency becomes: [math] \frac{v}{u+v} \frac{\sqrt{1-u^2/c^2}}{\Delta t}[/math]. This equation is now fully general and works for any situation. Further, if the 'ball' is made of light, ie. [math]v=c[/math] this becomes [math] \frac{c}{u+c} \frac{\sqrt{1-u^2/c^2}}{\Delta t} = \sqrt{\frac{c-u}{c+u}} \frac{1}{\Delta t}[/math] (I hope I haven't made any silly mistakes, since I did this on the fly without looking anything up....)

The ending of Time Enough for Love, when the hero has sex with his mother, was a bit much for me. Yuck!

What do you mean by "excitations are not always particles"? How are you defining a 'particle'? I would regard any localised field excitation as a particle. (Actually many people go further, and call non-local excitations particles too!) Can you give me an example of a localised excitation which is not a particle?

String theory is supposed (hoped) to provide a theory which merges quantum mechanics and gravity, so you don't need anything along side it! String Theory is a quantum theory, so it has to be consistant with the ideas of QM. The problem with String Theory is that it is extremely difficult to work with mathematically, so it is difficult to make testable predictions.

Very often people come to these fora with a belief that our current theories of physics, such as the Standard Model or relativity, are flawed and present some alternative of their own. On the whole, this is a fine attitude to take - we should always be skeptical, and it is good if people can think a little 'out of the box' and generate ideas which more standard thinkers may not have come up with. I have always thought that genius was not an ability to think 'better' than everyone else - it is an ability to think differently from everyone else. However, when coming up with a new theory it is important that it should be better than the old one. Therefore the first step of coming up with a new theory is a sufficient understanding of the old one. You have to make sure that your new theory does everything at least as well as the old theory, otherwise the old theory remains more attractive. This is very difficult mainly because our current theories are so spectacularly good in their predictions. Let me give an example: the magnetic moment of the electron. If we look at the energy (Hamiltonian) of an electron in an electromagnetic field, we find that there is a contribution from the interaction of the electron's angular momentum and the magnetic field. For an orbital angular momentum [math]L[/math], this is [math]\vec{\mu}_L \cdot \vec{B}[/math] with a magnetic moment [math]\vec{\mu}_L = - \frac{e \hbar}{2mc} \vec{L}[/math] (The charge of an electron is [math]-e[/math] and its mass is [math]m[/math].) However, if the particle has 'spin' (intrinsic angular momentum) [math]\vec{s}[/math], we also have a contribution to the magnetic moment of [math]\vec{\mu}_s = - g \frac{e \hbar}{2mc} \vec{s}[/math] [math]g[/math] is known as the gyromagnetic ratio, and its value depends on the theory. Since this can be measured in experiment very accurately, it is a good test of a theory to check if it predicts the correct gyromagnetic ratio. For example, simple QM (the Dirac equation in an em field) predicts a gyromagnetic ratio [math]g=2[/math]. Experiments shows that [math]g[/math] is very close to 2, so this is good news, but since experiment shows that it is not quite 2, the Dirac equation cannot be the whole answer. Quantum Field Theory, in the form of the Standard Model, predicts a deviation from 2. It is usual to write down the prediction for this deviation from 2 rather than the gyromagnetic ratio itself. For the SM this is: [math]\frac{g_{\rm th}-2}{2} = 1159652140(28) \times 10^{-12}[/math] The experimantal result is: [math]\frac{g_{\rm exp}-2}{2} = 1159652186.9(4.1) \times 10^{-12}[/math] (A note on errors: the numbers in brackets denote the error on the prediction/measurement at the same precision to which the value is specified. For example [math]1159652140(28)[/math] means [math]1159652140 \pm 28[/math] and [math]1159652186.9(4.1)[/math] means [math]1159652186.9 \pm 4.1[/math].) You can see that the theory predicts the correct experimental value to incredible precision (although the experimental error is still better than the theory one). If you want to persuade scientists that the Standard Model is wrong, then you have to explain why this is a coincidence or show that your new theory predicts [math]g-2[/math] to at least this accuracy.

That's not true. There is no a priori reason why maths should descibe the physical world. The fact that it does is very handy for us, otherwise we couldn't formulate our scientific laws. Indeed, the predictivity of the universe, ie. that it is governed by laws at all, is an assumption of science made at the outset. It is an assumption that has never been seen to fail but it is still an assumption (actually, I am not quite sure whether science would even spot a failure of predictivity - we would probably just put it down to having the theory a wee bit wrong). It could well be that there are some phenomena in the universe which cannot be descibed by mathematics and physical laws.

That is exactly what particles are. No-one is suggesting anything different for a graviton.

I think this is very interesting. This could be evidence for large scale structure in the universe, which could blow standard big bang models right out of the water. It is particularly dangerous for inflation because you would have expected these structures to be removed by inflation. It is most likely just showing a systematic error in the WMAP data though, since the structure seem to be correlated with the solar system's axis. I doubt that there really is large scale structure in the CMBR.

The standard idea is that it is expanding from the energy of the big bang itself. In inflation models the energy for a very rapid expansion is usually coming from a vacuum-expectation value of a scalar field.

Which bit? Pointing out that Metafizzics doesn't know any physics or pointing out the fallacy of his statements?

The only absurd thing around here is your belief that you know any physics. Since I can make measurements in different frames of reference and compare the results, relativistic invariance is hardly useless. I find it amusing that you support a Machian 'theory' which has the very property that you called 'absurd' in the first line of the very same post!

Yes, but the speed of light is not relevant here - it is the speed of the source with respect to the observer which is relevant.

You seem to have this ass-backwards. You are the one insisting that the medium is important - not me. Here is a nice explanation: http://archive.ncsa.uiuc.edu/Cyberia/Bima/doppler.html

The laws of physics are always independent of your reference frame, but what you measure may not be. If I measure your speed relative to me, I will get an answer dependent on my reference frame. If an ambulance is moving away from you, it has a lower frequency siren than if it was moving towards you - that is the whole point of the doppler effect.

That would depend on which direction they are moving and how fast. I think in most cases it is not a very big effect.

Write [math]f(x)=\sum_n a_n x^n[/math] [math]g(x)=\sum_m b_m x^m[/math] [math]h(y,z)=\sum_{k,l} c_{kl} y^k z^l[/math] Now stick your definitions for f(x) and g(x) into h(y,z) to give h(f(x),g(x)) and find the condition that it is zero for all x (ie. that all the coefficients vanish). This will give you the form of h(y,z) which you need.

You can't shoot them in the shops. Just give them a call or go online.

Try cutting off one of his fingers and see if it grows back.

But the important velocity difference is that between the source and the observer - not the signal propagation speed. As long as the observer isn't travelling at relativistic speeds relative to the source, it is exactly the same phenomena as it is for sound.