elfmotat

Pedophilia already is legal. Acting on it isn't.

But those aren't equal. They're opposite in sign! Unless you've defined [math]\Phi_{alt}=-\Phi [/math]. Regardless, it seems pointless. You've written down a bunch of symbols but haven't bothered to define any of them. What are [math]\eta[/math], [math]\Gamma[/math], and [math]T_{\mu \nu}^{alt}[/math]? Also you've written down the EFE's wrong. They should be: [math]R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R+\Lambda g_{\mu \nu}=G_{\mu \nu}+\Lambda g_{\mu \nu}=\kappa T_{\mu \nu}[/math] Also, from what I can see, you're defining that equation to be the field equations. What is there to derive? Are you trying to quantize it?

Nobody "experiences" time dilation. Observer's clocks run at different rates, and everyone agrees how much time is ticked off on each observer's clock.

It's not an illusion. (The word "illusion" in this context is rather ill-defined anyway.) Time dilation is a phenomenon with measurable consequences.

The whole "conversion" thing is rather contrived. All E=mc2 says is that mass is a form of energy. The equation falls directly out of Special Relativity, so if you really want to understand where it comes from you're probably going to have to read an SR textbook. You could use the following line of reasoning to get an intuition for it: Suppose mass is a form of energy. If this is true, then there should be some universal constant relating mass to mass-energy. By dimensional analysis, this constant should have units of [length]2[time]-2. Since [math]c[/math] is a universal constant with dimensions of [length][time]-1, it seems natural that this constant should be proportional to [math]c^2[/math] This doesn't tell you exactly what the constant should be, i.e. it could be [math]\frac{1}{2} c^2[/math], or [math]2 c^2[/math], etc. The correct constant comes from SR.

Interesting question. I don't think I've ever seen anything like this worked out before. I suppose there would be two ways to approach it: either by considering how voltage and current are effected, or considering how resistivity and volume are affected. The effect on resistivity seems like a complicated problem. Of course, resistance is related to resistivity by area and length, both of which could be affected by a Lorentz boost. For the other approach, you could consider the EMF induced by a changing magnetic field. A boost will change the value of the magnetic field, and thus the EMF. The current will also change with a boost. I might work out a specific scenario later.

[math]\delta S=0[/math] doesn't guarantee that the action is minimal, or even maximal, just that it's stationary. Take, for example, the curve y=x3 : the derivative at x=0 is zero, but it's neither a minimum nor a maximum (it's a saddle point). Intuitively we know that geodesics (usually, if not always) correspond to minimal path-length because it's not possible to draw a path with maximum length - you can always draw a crazier, more squiggly path that's longer. I can't, off the top of my head, think of any situations where the length of a geodesic would be a saddle point. The fact that geodesics correspond to curves of stationary length comes from the fact that varying the Lagrangian [math]L=\frac{ds}{d\lambda}[/math] yields the geodesic equation. You can find this worked out here: http://www.mth.uct.ac.za/omei/gr/chap6/node6.html . **Note that in the link I provided, they use the Lagrangian [math]L=\frac{d\tau}{d\lambda}[/math] instead of [math]L=\frac{ds}{d\lambda}[/math]. This corresponds, (usually if not always) to curves of maximum proper time, because [math]ds^2=-d\tau^2[/math]. This should make some intuitive sense because, for example in the Twin Paradox, the twin that goes away and comes back has always aged less than the twin who was sitting around in his inertial reference frame. This has to do with the orientation of the curve. For timelike curves in GR, there is really only one possible orientation because you can't travel backwards in time. For your sphere example, which is a spacelike geodesic, there are two possible orientations. One of them corresponds to the shortest path, and the other doesn't. The geodesic equation doesn't know which two points you're trying to find shortest distance between, so both orientations are possible. I.e. if you're trying to find the shortest path connecting A and B, you use orientation #1. If you then want to find the shortest path between B and A, you need to use orientation #2.

Marijuana.

Great book. It's very entertaining.

It seems to be a popular trend on science boards like this, where people naively decide one day that they're going to come up with "the next big idea," or "the theory of everything," etc., without having even a basic knowledge of the topics they want to talk about. The result is an oversupply of nonsense threads like this.

Why do you keep calling him Dr. Strauss?

But that's simply not true. An electron, obeying classical physics and orbiting around a nucleus, will give off electromagnetic radiation by the Larmor formula. Since it is constantly radiating away energy, it will eventually spiral into the nucleus. Thus, stable classical atoms cannot exist. Classical theory doesn't explain a host of other phenomenon, such as the photoelectric effect, Compton scattering, etc. (there's too many to mention). Your mind is stuck back in the 19th century. Haven't you heard of Quantum Electrodynamics? You're about half a century behind on the times. But it's not. That's a fact, not an opinion.

Read the whole thread pwagen posted, because you're essentially asking the same question that he was. Understanding what "expanding space" means requires knowledge of General Relativity, or at least some experience in working with a metric. The best way to demonstrate why galaxies are accelerating away from us without using any GR is with the Newtonian approximation with nonzero Cosmological Constant: [math]F=m \left (- \frac{GM}{r^2}+\frac{c^2 \Lambda r}{3} \right )[/math] The second term acts as a repulsive force so that when you get far enough away from a gravitating mass, gravity actually becomes repulsive. Up close, gravity is still attractive because the first term dominates the second. In Newtonian gravity the space is static and unchanging, so you don't need to worry about expansion. Keep in mind that this is just an approximation to GR.

I haven't looked into it at all (other than having having watched your popular youtube video), so these questions might have a simple answers. At first glance, the "zero-energy universe" idea seems rather ill-defined in GR. How exactly are you defining "total energy of the universe?" Under some simplifying conditions I suppose the Komar, ADM, and Bondi masses could be considered "total energy." But those are all undefined for cosmological spacetimes like the FLRW spacetime. More fundamentally, the FLRW spacetime and similar spacetimes do not possess a timelike Killing vector, so global energy conservation is not possible. The energy density of the gravitational field isn't defined at all in GR, so it would seem to be problematic to take its contribution into account as well. Looking forward to a response if my question is picked.

You're mixing the times and distances of two equally valid frames. It seems like you think 7 ly is the "real" distance, but both distances are equally valid in their respective frames.

No, he only traveled 1 ly in a little more than a year. When he stops he's in a different inertial reference frame. You're mixing the times and distances of two different frames.

I have no idea what that's supposed to mean. In the spaceship frame the distance is 1 ly, and you measure time at its regular rate. In the Earth frame the distance is 7 ly and they notice that your clock is running slow.

Since we know that the equation of motion for a test particle is the geodesic equation, the action should yield the geodesic equation. We also know geodesics correspond to paths of shortest length, as can easily be demonstrated by measuring distances on, for example, the surface of a sphere. The path along a great-circle of the sphere is the shortest path connecting two points. Therefore when you minimize the action, it should correspond to the path of shortest length. This implies that the action should be: [math]S=\int ds=\int \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}} d\lambda[/math] So the Lagrangian is just: [math]L=\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}[/math]. Similarly, you can also show that the Lagrangian [math]L=g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}[/math] also yields the geodesic equation when plugged into the Euler-Lagrange equations. This is the Lagrangian that your author used.

There are two important questions that need to be answered before we can give sensible recommendations: How much physics do you already know? How much math do you already know?

Semi-related: in the STM extension of Kaluza-Klein theory, all matter is a consequence of the geometry of extra dimensions.

I'm talking about the proper acceleration (i.e. what you would measure with an accelerometer) an observer experiences if he maintains a constant [math]r[/math] value: [math]a=\frac{GM}{r^2 \sqrt{1-2GM/c^2r}}[/math] On Earth we maintain a constant [math]r[/math] value because the ground holds us in place. If you plug in the mass and radius of Earth, you get ~9.8 m/s2. Notice that when [math]r\gg 2GM/c^2[/math] the square root term is approximately one, so it reduces to the familiar Newtonian value: [math]a=\frac{GM}{r^2 \sqrt{1-2GM/c^2r}}\approx \frac{GM}{r^2}[/math] The square-root term only has a significant effect when you're near the Schwarzschild radius, and it causes proper acceleration to blow up to infinity at the event horizon. You only experience that acceleration if you "hold station." It should intuitively make sense, because once you cross the horizon the radial coordinate becomes timelike and you must always fall towards the singularity. I.e. no amount of acceleration will keep you from falling in. If you free-fall into the BH then you experience nothing strange as you cross the event horizon.

Any mathematical model is purely abstract. Even in our simplest physical model, Newtonian mechanics, forces are represented by vectors which live in an abstract vector space. Objecting to a theory because it's abstract is rather absurd. As you said, our theories have great predictive power. That's the measure of a good theory, not whether or not it conforms to your naive intuitions. That's a rather strange viewpoint. Our brains evolved to intuitively understand a slow, macroscopic world. That is where all of our intuition comes from. Why would you expect fast-moving things, very small things, very large things, etc., to conform to the way our brains evolved? When you learn about physics after ~1900, you have to train yourself to develop an intuition for the math because it's generally not possible to "picture" what's going on. To quote Feynman: "...and then there's the kind of thing which you don't understand. Meaning 'I don't believe it, it's crazy, it's the kind of thing I won't accept.' I hope you'll come along with me and you'll have to accept it because it's the way nature works. If you want to know the way nature works, we looked at it, carefully, that's the way it works. You don't like it? Go somewhere else! To another universe! Where the rules are simpler, philosophically more pleasing, more psychologically easy. I can't help it! OK! If I'm going to tell you honestly what the world looks like to human beings who have struggled as hard as they can to understand it, I can only tell you what it looks like. And I cannot make it any simpler, I'm not going to do this, I'm not going to simplify it, and I'm not going to fake it. I'm not going to tell you it's something like a ball bearing inside a spring, it isn't. So I'm going to tell you what it really is like, and if you don't like it, that's too bad."

No it's not. In relativity there's no such thing as a gravitational force. Special Relativity does not cover gravitation, and applying Newtonian formulas will only lead to confusion. In General Relativity, gravity is a consequence of geometry. If by "inertial mass" we're talking about the coefficient relating three-force to three-acceleration, then that's not true. By this definition "inertial mass" is actually a matrix. If an object with rest mass m is moving along the x-direction, then: [math]f^i=M^i_{~j} a^j[/math] where: [math]M^i_{~j}=\begin{bmatrix} \gamma^3 m & 0 & 0\\ 0 & \gamma m & 0\\ 0 & 0 & \gamma m \end{bmatrix}[/math]

No, what he missed was that F=mg is not a valid equation in relativity, as has been pointed out multiple times not. The strength of a string has nothing to do with its relativistic mass, similar to how in Newtonian physics the strength of a string has nothing to do with its kinetic energy.

Right. If you were to set up a ring of photon detectors of radius [math]R[/math] around a (cylindrical) mirror, give the ring some angular velocity [math]\omega[/math], emit a photon towards the center and time how long it takes for the light to return to the ring, then you would measure the observed speed of light to be: [math]c_{obs}=\frac{c}{\sqrt{1-(\omega R)^2/c^2}}[/math] EDIT: I forgot to point out that the clock has to be moving with the ring. That's pretty much implied by "in a rotating frame of reference," but the distinction is important enough to be pointed out.