Tom Mattson

The apple exists in every frame. Seriously, I have no idea of where you're coming up with this. Why don't you try what I already suggested: Explicitly show your reasoning, starting from the Lorentz transformation. No, just one. I don't know why you would think that SR suggests otherwise.

OK' date=' now for the next obvious question: On what grounds? Yes. And I've already explained why your analogy with the steel tape fails.

When you factor the numerator of that function and cancel the common factor, you aren't "simplifying" the function. You are obtaining a new function that agrees with the original function at all but a single point. z=1 is not in the domain of the original function, and so f(1) does not equal 2.

You think that it has been misread by the army of highly trained physicists' date=' postdocs, and grad students who are all salivating to become famous by being the one to devise the repeatable experiment that proves the theory wrong? But as I am sure you know, a tape made from a different metal will expand a different distance under the same temperature gradient. What I am telling you is that all clocks will show the same time lag, regardless of the mechanism by which they work. And the lag is precisely what is predicted by relativity.

No' date=' that's not it. Time dilation in relativity never results in anything moving [i']backwards[/i] in time, which is what you were suggesting. "Time frame" is not my term, so I'm not going to answer for it. I prefer "inertial frame". And it is perfectly sensible to suggest that an object can be in an infinite number of inertial frames, because there are an infinite number of states of motion that other observers may be in, relative to that object.

What's the difference? There are no indicators of time other than clocks. And if all clocks--regardless of the mechanism by which they work--all show the exact same lag in elapsed time when subjected to "twin-paradox" type experiments, then in what sense can it be said that time dilation has not occured? Or should we suppose that the clocks are all conspiring to play a trick on us?

Yes. Now the question is, "How did you come up with it?"

This debate has come up before. I don't think that the nonexistence of space is a consequence of relativity, although it sure is consistent with it. Why do people keep saying this? Oh yeah, it's because they have no clue as to what relativity actually says.

Oops, never mind. The solutions to the above equation will indeed make E=0, but E is not the energy of the particle anymore. E*=E+V is the energy, and that is not zero.

That's not right. The norm of a particle's 4-momentum is the mass of the particle. [math] p^{\mu}=(E,\mathbf{p}) [/math] [math] p_{\mu}=(E,-\mathbf{p}) [/math] Taking their inner product yields: [math] p^{\mu}p_{\mu}=E^2-p^2=m^2 [/math] where [math]p[/math] is the norm of the 3-momentum.

It is justification for statements that begin with, "SR predicts that..." I can't even count the number of arguments against relativity I've seen by people who think they've found a paradox in the theory, when in truth the person simply does not know what the theory says.

Damn it everyone, the man's name is Hawking!!!

Just a few other small comments. To really get to "what relativity is" you should go back to its postulates. Special Relativity is founded on the two beliefs that: 1. The laws of physics are the same for any two inertial observers. 2. The speed of light is independent of the motion of its source. What you state above is a derived consequence of this foundation.

If you want to learn relativity then the place to begin is the following link: http://www.physics.nyu.edu/hogg/sr/sr.pdf The first 4 chapters assume knowledge of basic physics and precalculus mathematics. Chapter 5 assumes that you know how to do simple matrix operations, and Chapter 6 uses a minimal amount basic calculus.

Your reasoning does indeed go south for this reason. As we learn from quantum mechanics, [math]p= h / \lambda[/math], regardless of whether a particle has mass or not. It's funny you bring this up because just minutes ago I was reading over a beautiful presentation of just this very thing in a free e-book: http://www.geocities.com/zcphysicsms/chap3.htm

Just let the particle be sujected to any potential, and do a transformation on it like I described in Post #5. The two potentials (V and V*) give rise to precisely the same dynamics. Yes, an atom can be said to have zero total energy. As I keep saying, you can add any constant you like to the potential function and it does not make one lick of difference to the physics of the atom. All you have to do is define that constant to be such that the total energy is zero. Yes.

Huh? I thought that the ZPE was the minimum energy a particle can have. The spectrum for the SHO is [math]E_n=(n+1/2)\hbar \omega[/math]. The ZPE corresponds to n=0, which is a finite energy.

Yes. As I said you can always transform the potential to "move the zero" to any point. What makes you think that a particle with zero energy would be totally stationary? Take a particle of mass m. Let's its KE have the value mc2, and its PE have the value 2mc2. What's its total energy? Zero. Is it stationary? No. Could you tell me what physics courses you've taken? It might help.

See post #5. You can always transform a potential to make the total energy of a particle zero at any arbitrary point, by adding an appropriate constant. Remember it's not the potentials that are physically meaningful, it's the forces. And F=-∇V. So if I replace V(x) in post #5 with V*(x), I find that the exact same force is derivable from the new potential, because the gradient of a constant is zero. All this leads up to the idea of a gauge transformation. Have you heard of those?

The Zero Point Energy is just the minimum energy that a bound quantum mechanical particle can have. But that ZPE can be set to any real-numbered value, including zero.

Not one bit of this makes any sense.

While it's true that a massive particle is guaranteed to have an energy associated with its mass (E=mc2), it is also true that one is free to adjust a potential energy function by any number one wishes. That is because it is only potential energy gradients that are physically meaningful. So for instance say a particle of mass m is subject to a force that can be derived from a potential V(x), such that V(x0=0. And say further that the particle is at rest at x=x0. So it's total energy is it's rest mass energy, right? Not necessarily. You can just as easily use the potential V*(x)=V(x)-mc2. After all, it has the same gradient (and therefore gives rise to the same force) as the original V(x). So now, using V*(x), we say that the particle has "zero energy". That's why merely stating the energy of a system without also specifying a datum is physically ill-defined.

The requirement of discrete energy levels alone is not sufficient to guarantee that the atom won't collapse. The specific form of the Coulomb potential must be just so in order for that to be ensured. If the Coulomb potential were, say, an inverse square potential then it would you would get discrete energy levels, but the eigenvalues of the Hamiltonian would not be bounded from below, and you'd still get atomic collapse. But since the Coulomb potential varies as 1/r, there is a finite lower bound on the spectrum of energy eigenvalues. Of course you're free to put the zero of energy anywhere you like, but convention you are choosing here is not common. It's far more common (in nonrelativistic QM) to associate zero energy with a free electron (that is, an electron that is infinitely far from the nucleus). Under the normal convention electron energy is taken to go to neg. infinity as the electron collapses to the nucleus. But as I said the atomic Hamiltonian is bounded from below, so this does not happen. No, it is not.