Johnny5

Yes. But you could be more clear about what you mean. I don't believe that the speed of electric field propagation is a frame independent quantity. But if you don't understand me that's ok. But I did understand the analogy just the same. An electric field accelerates a particle, and thats as if someone blows on the particle to accelerate it, but the 'wind' isn't air, the 'wind' is the electric field. It's a good analogy. I'm just not sure what an electric field is though.

That's ok with me.

Yes, I realized that as soon as I went away. But I am right about the angle. That has to be measured by some experiment.

Isn't that a matter for experiment to decide? I gotta go grab something to eat, I'll be back later. Thanks for everything Martin.

I've read enough to understand this much. You have water in the reactor. So speeds are being defined in the rest frame of the reactor. So in this frame, we have photons moving through the water. They will have the following speed in the reactor frame: c/n Where c=299792458 m/s, and n = 1.33 (the index of refraction of water) But apparently, there are electrically charged particles moving through the water as well. And these charged particles have a speed through the water which exeeds the speed of photons through the water. Which means that if the charged particles are accelerated to c/n in the reactor frame, and then surpass c/n in the reactor frame, there is a mach cone formed, similiar to when a jet fighter reaches the speed of sound in air. So thats what you meant when you said v> c/n. V is the speed of the electrons through the water c/n is the speed of some photons through the water. Alright I understand the diagram in the wikipedia article somewhat. The red line represents the velocity vector of some electron which is moving through the water, which is at rest in the reactor frame. Now, as this beam electron moves in this frame, there is an associated magnetic field with it. I guess that B field influences the atomic electrons in the atoms of the water. And apparently if the speed of the beam electron exceeds c/n, then there is some kind of shock wave thing happening, Mach cone thing, which leads to all that blue cerenkov radiation which a human eye can see. The blue lines in the diagram represent the blue photons. I followed the geometry, and I see that [math] cos \phi = \frac{1}{n \beta} [/math] Where b is v/c. I assume that beta is the Cerenkov angle. So we can write: [math] cos \phi = \frac{1}{n \beta} = \frac{c}{nv} [/math] c/n = speed of light through dielectric medium. So we can write the formula as follows: [math] cos \phi = \frac{1}{n \beta} = \frac{v_l}{v} [/math] Where Vl represents the speed of light in a medium, in this case water. [math] cos \frac{\pi}{6} = \frac{v_l}{v} = \frac{.75 c}{v} [/math] [math] cos \frac{\pi}{6} = \frac{\sqrt {3}}{2} = .866025... [/math] As can be checked with a handheld calculator. Hence: [math] .866 = \frac{.75 c}{v} [/math] It therefore follows that: [math] v = \frac{.75 c}{.866} [/math] And c = 299792458 meters per second' date=' so [math'] v = \frac{.75 (299792458)}{.866} = 259627884 [/math] It was supposed to come out faster than c. The only thing I can figure is that phi = 60, not 30. I guess you meant q = 30, where: [math] 180 = 90 + \phi + \theta [/math] If I use 60, instead of 30, i get: [math] cos 60 = \frac{1}{2} = \frac{.75 c}{v} [/math]

You have to go slow with me. Sorry. I am doing my best to comprehend a universe you know. The angle the light goes off at? I can tell the speed of the particles' date=' from the angle the light goes off at? Ok it has a name Cerenkov angle. Let me google it and see what happens. I am reading this now. I am especially intrigued by this part here: I need that part explained to me.

The wind is blowing at c. Jacques, what do you mean?

What is the wind inside a particle accelerator though? I've been to a 900 MeV one, you have big evacuated tubes, there's no air inside.

I think, though I'm sure someone will correct me if I'm wrong, that recently there has been some divergence from the mass changes predicted by the theory of relativity, in particle accelerator experiments.

Long ago, I was talking with Tom Mattson about something, and he asked me what I meant, and I never did finish explaining to him what I meant. Part of what I said was, that I operationally defined the binary relation 'before,' on the set of moments in time. Since I didn't finish then, I will now. An operational definition is a non-verbal experiment which somehow assigns meaning to some term. The term here is 'before' A little thought, and you will see that if you are an adherent to "absolute time" and "absolute simultaneity," that before is a binary relation on the set of moments in time. At any rate... Suppose you go to the moon, and drop two objects of different mass, say two spheres of equal radii, from the same height above the surface of the moon, simultaneously in the rest frame of the moon. You will discover that they hit the ground simultaneously. Thus, you can use that experiment to operationally define the term 'simultaneous' if you wish. It is your sensory perception which lets you know that you released them simultaneously. You can use an experiment to operationally define 'before' instead. Suppose you release one object in the rest frame of the moon. It will fall to the surface of the moon, and strike the ground. The moment in time you released the object is before the moment in time at which the object struck the ground. Thus, an experiment serves to operationally define the term before. You can then link that operational definition to a clock in the rest frame of the moon. In other words, a clock maps numbers onto moments in time in a frame. But logically, the term 'before' is undefined. So you have this kind of thing going on: Undefined term: before Now you can construct a logical definition of simultaneity: Definition: For any moment in time X, and any moment in time Y: X simultaneous to Y if and only if (not(X before Y) and not(Y before X)) PS: Galileo pioneered free-fall experimentation, his experiments led him to definitions of speed, and acceleration. They can be found here: Dialogue Concerning Two New Sciences by Galileo In texts and figures on the third day pp: 153-160 Galileo defines uniform motion. In subsequent pages Galileo goes on to discuss accelerated motion.

Good thank you.

To any object in the universe, there is a center of mass of that object. You can attach a frame to that object, so that the center of inertia of that object is always located at the origin of the frame. This is what I mean by "rest frame" of something. So for example, the rest frame of the moon I was talking about, is a three dimensional coordinate system, whose origin is always the center of inertia of the moon. wherever the moon goes, so does the coordinate system. Likewise for the rest frame of the rocket. Somewhere inside the rocket, is the center of mass of the rocket. Let that point be the origin of a three dimensional rectangular coordinate system. Wherever the rocket goes, the coordinate system goes, the coordinate system is the rest frame of the rocket, in other words, the rocket is at rest in the coordinate system, even if the rocket is accelerating in someone else's frame. PS: You didnt bold anything, and I don't understand what you mean by "constantly changing rest frames as it accelerates."

Exactly. But in the rest frame of the ship, the moon is accelerating. And In the rest frame of the moon, the ship is accelerating. The rest frame of the moon was stipulated to be an inertial reference frame. So since, in the rocket frame, the moon is accelerating without an external applied force (no action reaction pair) it necessarily follows that the rest frame of the rocket isn't an inertial reference frame. But still, the center of inertia of the moon is accelerating in the rest frame of the ship. Yes. Just switch back and forth between the frames, and you know what I'm taking about.

Can you be more specific. I am thinking of a missile moving through the air. I realize that the nosecone of the missile pushes the air away from where it was, as the missile moves, and then after the missile goes a ways futher, the air which it displaced "falls back" to where it was before the missile passed through. Yes I get this very much, it's intuitive. So right when the missile reaches Mach 1, there is a Mach cone. Do you know anything about that YT2095?

What is v?

Ok' date=' I've read number four and I don't get it. You say "faster than light" but "not faster than c" I just don't understand. I am reading the article on Cerenkov radiation as well. For whatever it is worth, I am trying to get it. [b']Do you mean that the speed of the beta particles in the rest frame of the reactor are travelling faster than the speed of light in the water?[/b]

Let me ask you something then Bob. Suppose you are in a rocket, passing by a moon, which is at rest in deep space. And your ship's engines are on. You look out your spaceship's window. Is the moon accelerating?

I was hoping for that script L, that's on the site you gave the link to. I want to be consistent in symbolism. [math] \mathcal{F} [/math] [math] \mathcal{L} [/math] Ok so, here then is the definition of the Gamma function: [math] \Gamma (n+1) \equiv \int_{x=0}^{x= \infty} x^n e^{-x} dx [/math] And here is the definition of the Laplace transform of f(t): [math] \mathcal{L} [ f(t) ] \equiv \int_{t=0}^{t= \infty} f(t) e^{-St} dt[/math] Let x become St in the definition of the Gamma function. Hence [math] \Gamma (n+1) = \int_{St=0}^{St= \infty} (St)^n e^{-St} d(St) [/math] Let S be finite, and nonzero, therefore we can divide both sides of the upper and lower limits of integration by S to obtain: [math] \Gamma (n+1) = \int_{t=0}^{t= \infty} (St)^n e^{-St} d(St) [/math] Let S be constant in time, or abstractly let d(St) = S dt, hence: [math] \Gamma (n+1) = \int_{t=0}^{t= \infty} (St)^n e^{-St} Sdt [/math] Therefore: [math] \Gamma (n+1) = \int_{t=0}^{t= \infty} S^{n+1} t^n e^{-St} dt [/math] Since S is constant in time, and n is constant in time, we can pull Sn+1 outside of the integral to obtain: [math] \Gamma (n+1) = S^{n+1} \int_{t=0}^{t= \infty} t^n e^{-St} dt [/math] Since S is nonzero, we can divide both sides of the equation above by Sn+1 to obtain: [math] \frac{\Gamma (n+1)}{S^{n+1} } = \int_{t=0}^{t= \infty} t^n e^{-St} dt [/math] We can now recognize the RHS as the Laplace transform of tn, therefore: [math] \frac{\Gamma (n+1)}{S^{n+1} } = \mathcal{L} [ t^n ] [/math] This is exactly the relation given by Dave, only with n instead of p.

Alright, I am starting to run through this myself. We begin with Fourier's heat equation which is this: [math] \alpha^2 \nabla^2 U = \frac{\partial U}{\partial t} [/math] In the equation above, U denotes temperature. So in general, the temperature of the parallelepiped can change in time, the body could heat up, or cool down. But here we are going to discuss stationary states, meaning that the temperature cannot change in time, which means this: [math] \frac{\partial U}{\partial t} = 0 [/math] Which means that the value of U only depends upon the spatial variables, x,y,z. That is: U(x,y,z) You then give the boundary conditions of the parallelepiped, as well as a neat little picture. In your opening post you say this: