elfmotat

It seems like you're generalizing "amateurs" to be essentially people who watched a couple string theory documentaries and decided they would come up with their own theory of everything. I would consider an amateur to be someone who's built up enough of a background in physics that producing new research isn't wholly unreasonable. I plan on (hopefully in the not-to-distant future) being a member of the latter group. I may be approaching this from an atypical vantage point, however; I know of only one other autodidact who frequents forums like these and who has a knowledge-base similar to mine. People generally seek out others who have similar interests and autodidacts do not have the luxury of being constantly surrounded by them (unlike academics), so one would think that science-based forums would attract many such people like myself. I suppose my assumption is either wrong for one reason or another, or autodidacts who have done a fair amount of self-study are rare.

Energy is measured in units of Joules. We also know that 1 Joule = 1 kilogram*meter2 / second2. Mass is measured in units of kilograms. "c" is a velocity, which is measured in units of meter/second. c2 is therefore measured in units of meters2 / second2. If we multiply mass by c2, we get units of (kilograms)*(meters2 / second2) = Joules, so we know that the equation is dimensionally correct because mc2 has units of energy.

What's your real goal here? Are you trying to make theoretical physics your full-time job? Or do you just want to have the ability to read modern papers and perhaps make some contribution to the field with no monetary compensation? Or do you just want to learn most of modern physics for knowledge's sake? If you're looking to make physics your day job, you will almost certainly need a PhD. Getting your PhD will likely consume the majority of a decade, will drain your bank account, and it will not be a cake-walk. After you get your PhD your next step will most likely be getting a Post-Doc, which pay horribly and apparently aren't very fun. You'll be lucky to be eventually hired by a University, at which point you'll have to lecture undergrad students - an activity that can also consume a lot of your time. If you don't plan on being paid, it's not really too difficult to learn physics on your own. For example I taught myself basic General Relativity with Schutz's textbook when I was eighteen, and I have never taken a formal class in the subject (though I may have a slight advantage because I'm an engineering student). Nonetheless, I currently consider myself relatively (pun semi-intended) well-versed in GR. I also taught myself classical field theory, non-relativistic Quantum Mechanics, and I'm working on Quantum Field Theory at the moment. From my first intro physics class in High School (where I learned only algebra-based basics like Newton's 2nd Law, conservation of energy, Newtonian gravitation, etc.) until now, it has only been ~three years. So you certainly can learn what you need in a relatively short period of time if you're (1) genuinely curious enough to keep learning (this is probably the most important), (2) have the time to spare, and (3) are willing to put in the effort. Since you left school early, I assume you're probably not so versed in math. I would go through the algebra, trigonometry, and calculus videos on this site first: https://www.khanacademy.org/ . Make sure you understand all of the concepts (do practice problem - passive learning isn't enough) before you move on. Once you've learned some basic calculus, you're ready to start learning some introductory physics. Watch Khan's videos, and then find a (calculus-based) textbook on classical mechanics.

I simply don't have the patience to continue catering to your nonsense, so this will likely be my last response (unless for some reason I can't help myself). You "answered" the exercise by essentially rewording the question. Obviously the events can't disagree between the two frames - your job was to figure out why they don't disagree. You were unable to do so because you don't really understand SR. The plate certainly is tilted in the meter stick's rest frame. That's just good old SR - it falls right out of the Lorentz transformation. The acceleration your rocketship needs to produce in order to hover at a constant r-value asymptotically approaches infinity as you approach the event horizon, as I explained. I can't really make out the logic of your "rebuttal." An object free-falling into a BH and an object accelerating away from a BH are not equivalent no matter how much you bang your fists and exclaim that they are. Length contraction and time dilation are phenomena which are derived from a special case of GR. They don't really generalize to curved space. Claiming GR is incorrect because LC and TD aren't well-defined is like claiming arithmetic is incorrect because division by zero isn't defined. Whether or not objects cross the event horizon in finite time is coordinate-independent. Schwarzschild coordinates do not cover the entire manifold because they are singular at the event horizon. That means you can't use Schwarzschild coordinates to describe objects crossing the horizon. Coordinate singularities are not physically meaningful, they just restrict the domain of a particular coordinate system. This will be my fourth or so time repeating myself. Your statements about singularities are nonsensical. We don't know how singularities behave by definition. You changed your argument about geodesic motion from "geodesic worldlines and curved worldlines are equivalent" to "free-falling objects experience tidal forces, Nana-Nana Boo-Boo!" Regardless of whether or not tidal acceleration constitutes proper acceleration, that was pretty sneaky of you. You for some reason think that coordinate acceleration and proper acceleration are equivalent, which is quite clearly absurd. Acceleration isn't relative the same way as velocity because we can measure it on an accelerometer.

Do you understand what the word "defined" means?

Thanks for the input guys. I asked this question on another site as well, and I was referred here: http://en.m.wikipedia.org/wiki/Hyperfunction . So it seems I wasn't too far off.

It doesn't have infinite mass. Some black holes have masses only a few times that of our Sun, though we have detected others that can be up to billions of times the mass of the Sun. Once a spherical object of mass M is crushed down to a ball with radius [math]r_c=\frac{9GM}{4c^2}[/math], it has no choice but to collapse into a black hole because resisting collapse would require infinite internal pressure. Once collapse is finished, the remaining black hole will have an event horizon at radius [math]r_s=\frac{2GM}{c^2}[/math]. Once anything falls inside the event horizon, it can never escape (because to do so would require infinite energy). After it falls in, it continues to fall toward the singularity. What happens at the singularity is unknown with current physics. To answer the question in the title, the acceleration required to "hover" outside of a black hole (i.e. the acceleration that your rocket-ship needs to produce) is given by: [math]a=\frac{GM}{r^2 \sqrt{1-r_s/r}}[/math] As you can see from the equation and from a plot of the equation below, as you approach the event horizon the acceleration required to hover outside of the BH approaches infinity.

The EM field couples to a rank-1 tensor (the four-current vector) and its field equations are linear, so your theory looks like it is able to reproduce classical electromagnetism. Unfortunately, however, your theory cannot possibly reproduce General Relativity. It is tempting to assume (in analogy with EM) that the field equations for gravity are linear and that the gravitational field couples to a rank-1 tensor with components (mass density, x-momentum density, y-momentum density, z-momentum density). It turns out that this assumption is incorrect. The field equations for gravity are nonlinear and the gravitational field actually couples to a rank-2 tensor (the stress-energy tensor). I think the way to correct this in the context of your theory would be to let "informations" be emitted by both mass-density and pressure. Then you have to find some way of letting the informations of gravity self-interact.

Your analysis is correct. Electrons are only "allowed" to occupy a discrete (i.e. not continuous) number of energy levels, E1, E2, E3, etc. They aren't allowed to have energies in between any of those values. This is definitely very different from the way classical objects behave, like your bowling ball example.

Does the Dirac Delta function have a residue? I feel like it should, considering the close parallels between delta function integral identities and theorems in complex analysis, namely the Cauchy integral formula and residue theory. I'm unsure exactly how they tie together (if they do).

So I was looking through Wald when I noticed his definition of the stress-energy for an arbitrary matter field: [math]T_{ab}=-\frac{\alpha_M}{8\pi} \frac{1}{ \sqrt{-g}} \frac{\delta S_M}{\delta g^{ab}}[/math] where [math]S_M[/math] is the action for the particular type of matter field being considered, and [math]\alpha_M[/math] is some constant that determines the form of the Lagrangian for the coupled Einstein-matter field equations: [math]\mathcal{L}=R\sqrt{-g}+\alpha_M \mathcal{L}_M[/math] For example, for a Klein-Gordon field we take [math]\alpha_{KG}=16\pi [/math], and for an EM field we take [math]\alpha_{EM}=4[/math]. Now, my question is whether or not there is some prescription for finding the value of [math]\alpha_M[/math]. How do we know that the constant takes on those particular values? How could I go about finding [math]\alpha_{M}[/math] for an arbitrary [math]\mathcal{L}_M[/math]?

It's not really wrong. My point was to demonstrate CPT symmetry, although I agree I didn't really convey it unambiguously. The scenario I was imagining in my head was a charge in a static external electric field. I used Coulomb's Law in my post because I thought it would be easier to imagine like/opposite charges repelling/attracting than it would be to imagine a charge moving in one direction or the other in an electric field, which is slightly more abstract. The unintentional result was that the whole thing became convoluted.

Yes, you're right. I'm mixing ideas now - I was thinking of the fact that positrons can be considered holes in a negative energy Dirac sea.

Edit: Nevermind.

A negative energy particle with negative proper time is equivalent to a positive energy particle with positive proper time.

Only one charge's sign is reversed. It's really only an analogy to help demonstrate the idea, so don't take the Coulomb's Law thing too seriously. What's actually going on has to do with the time reversal operator in QFT.

The idea that antimatter is just matter with negative proper time isn't really too strange if you think about it. To simply demonstrate the idea, consider Coulomb's law between two like charges of equal magnitude: [math]m\frac{dv}{dt}=kqq/r^2[/math] If we "reverse time" where [math]t\rightarrow -t[/math] (you can think of it as playing a movie backwards), then we get: [math]-m\frac{dv}{dt}=kqq/r^2[/math] But this is exactly the same law that governs the force between two opposite charges of equal magnitude: [math]-m\frac{dv}{dt}=kqq/r^2 ~\Leftrightarrow ~m\frac{dv}{dt}=kq(-q)/r^2[/math]. With the "movie analogy" this is obvious: if we record what happens when we place two like charges near each other and then run the movie backwards, it looks exactly like what we would find if we place two opposite charges near each other (well, over small distances at least).

Ah, I see what you were saying there. I was referring to the fact that you literally stick the operator between the wave function and its conjugate to calculate the corresponding observable: [math]\langle x \rangle = \int_{-\infty}^{\infty}\Psi^*(\hat{X})\Psi dx=\int_{-\infty}^{\infty}x | \Psi |^2 dx[/math] [math]\langle p \rangle = \int_{-\infty}^{\infty}\Psi^*(\hat{P})\Psi dx=-i \hbar \int_{-\infty}^{\infty}\Psi^*\frac{\partial \Psi}{\partial x} dx[/math] [math]\langle E \rangle = \int_{-\infty}^{\infty}\Psi^*(\hat{H})\Psi dx=i \hbar \int_{-\infty}^{\infty}\Psi^*\frac{\partial \Psi}{\partial t} dx[/math] I don't know what you're trying to say here. If you can find a solution in polar coordinates then you can automatically find a solution in Cartesian coordinates. This is a mathematical fact - there's no arguing it. One coordinate system may be more useful than another when solving the Schrodinger equation, but that's physically meaningless. Nature doesn't care how you decide to label points. The OP wanted a basic explanation of QM, so I gave him one. What is "it?" What are you talking about? I explained it the only way I understand it: mathematically. If it were possible to paint an intuitive picture of what's going on in the OP's head, I might be inclined to try that approach. But as things stand there is no intuitive explanation of QM, nor is there even a consensus on what its standard interpretation should be. As far as anyone's concerned, the math is the theory. The OP said he had a strong knowledge of physics in other topics, so I figured my post shouldn't be too far of a stretch for him. You started rattling off random stuff about electron orbitals and how Cartesian coordinates aren't good. It looks like word salad to me.

I have literally no idea what you're talking about. That looks like nonsensical word-salad to me. First of all, if you can find a solution in polar coordinates then you automatically have a solution in Cartesian coordinates since they only differ by a coordinate transformation. I'm also not sure what this has to do with my post, or how it's relevant to the conversation in any way.

I'm a bit tired so I'll try to do this with minimal effort (which is probably bad). Essentially the fundamental equation of quantum mechanics is the Schrodinger equation: [math]i\hbar \frac{\partial \Psi }{\partial t}=-\frac{\hbar^2}{2m}\nabla^2 \Psi +V\Psi [/math] This equation gives you a relationship between the potential energy function in space, and the "wave function" for the particle in question. If you specify a potential V and solve the Schrodinger equation for [math]\Psi[/math], you essentially obtain the probability distribution for that particle. If we restrict our scenario to one dimension, then the probability that the particle can be found between the points x=a and x=b is given by: [math]P(a\leq x\leq b)=\int_a^b|\Psi|^2dx[/math] where [math]|\Psi|^2=\Psi^*\Psi [/math] and [math]\Psi^*[/math] is the complex conjugate of [math]\Psi[/math]. Using the wave function we can also determine the expectation values for various observables by "sandwiching" their corresponding operator between the wave function. For example, the expectation value of position (where we "expect" it to be - an average of its position) is given by: [math]\langle x \rangle = \int_{-\infty}^\infty \Psi^*x\Psi dx [/math] The momentum operator is determined by the fact that it should be the generator of spacial displacements (in accordance with Noether's Theorem): [math]e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|x \rangle=|x+a\rangle[/math] [math]\Rightarrow ~~\mathbf{P}=-i\hbar \nabla[/math] So the expectation value of momentum is given by: [math]\langle p \rangle =\int_{-\infty}^\infty \Psi^*(-i\hbar \frac{\partial}{\partial x})\Psi dx[/math] We can build up most other observables from the position and momentum observables. I have no idea whether or not this was helpful, but oh well.

If you take a longer trip, the time dilation accumulates. For example say you take a 1 year long trip in your spaceship at 99% the speed of light then ~7 years will pass on Earth. A 2 year long trip will for you is about a 14 year long period for Earth, etc. I suppose it really depends on what you mean by "teleport." We can't use our theories to describe scenarios which violate those very theories. The closest thing to "teleporting" that is allowed by GR (allowed so far as we can tell thus far) I suppose would be a wormhole. If you move one mouth (call it end #1) of a wormhole at high velocity and then bring it back, the end 1 will be time-dilated with respect to end 2. This can result in some wacky phenomena. For example if you entered at end 1, then you would exit at end 2 before you even entered at end 1. This is all assuming that stable wormholes are possible in the first place. If I had to guess I would say that they aren't, though this is by no means proven and certainly isn't an area of expertise for me. My perspective is a practical one: time travel to the past opens up a slew of causality paradoxes which need to be resolved (Grandfather Paradox, etc.). Well in this case you're mixing gravitational time dilation (time "slows" when you're near massive bodies) with kinetic time dilation (time "slows" as your velocity increases). In general, if you want to calculate the time that someone on Earth experiences relative to an orbiting observer, you use: [math]T=T_\oplus k\sqrt{1-\frac{r_s}{r}-\frac{v^2}{c^2}}[/math] where [math]T[/math] is the time experienced by the satellite, [math]T_\oplus[/math] is the time experienced on the surface of the Earth, [math]r_s=2GM_\oplus/c^2\approx 0.008869~ meters[/math] is the Schwarzschild radius of Earth, [math]r[/math] is the radial coordinate (which is approximately equal to the distance from the center of the Earth to the satellite), [math]v[/math] is the tangential velocity of the satellite, and [math]k=1/\sqrt{1-r_s/r_\oplus}\approx 1.000000000695[/math] is a constant. If you want to incorporate radial velocity then things get a lot more complicated.

. In that case I think Khan Academy (Cap'n's link) should be perfect.

Why shouldn't it be able to? There's no fundamental reason to think that physics should be local. Say a particle decays into two electrons traveling in opposite directions. We have it setup so that, in one frame, one electron is detected before the other. Say we find that this electron is spin-up, and then we go and see what the other detector had to say about the state of the other electron. We (always) find that the second electron was spin-down. But relativity tells us that an instantaneous "interaction" (I use that word loosely) will result in the order of events being reversed in some frames. We can find a frame in which the spin-down electron is detected first. Since the events aren't causally related, there is no contradiction and relativity is not violated. QM is a mathematical theory, and as such is necessarily "logical." It may be unfamiliar or unintuitive, but it's still logical.

They were - about a century ago. Figure what out? What about entanglement do you think we don't understand?