elfmotat

Some were figured out by analogy with classical physics, some by guesswork, and some by symmetries that the particular theory should have, etc. For example, in the Lorenz gauge the classical Maxwell's equations which describe the electromagnetic field are: [math]\square A^\mu=j^\mu[/math], where [math]A^\mu[/math] is the electromagnetic four-potential, and [math]j^\mu[/math] is the current density. When [math]A^\mu[/math] is quantized it becomes the photon field, i.e. the field whose excitations are photons. There's also a field which describes electrons, called the Dirac field [math]\psi[/math]. It turns out there's a conserved quantity made up from the Dirac field which acts like a current density: [math]j^\mu =-e\bar{\psi} \gamma^\mu \psi [/math]. If we make the assumption that the photon field couples to the Dirac field in exactly the same way that the Dirac field couples to the photon field, we can write, by the using the current-density relation above, the coupled photon-Dirac Lagrangian: [math]\mathcal{L}_{QED} = \bar{\psi} (i\gamma^\mu \partial_\mu -m)\psi -e\bar{\psi} \gamma^\mu \psi A_\mu -\frac{1}{4}F_{\mu \nu} F^{\mu \nu}[/math] I put the subscript "QED" there because this Lagrangian turns out to be exactly what describes Quantum Electrodynamics. Now, there are other ways of adding interaction terms, such as adding higher order powers of a field to its own Lagrangian. This describes a field which self-interacts. There are many things you can add, and some of them turn out to be interesting and physically meaningful. There are a lot of fields, one for each type of particle pretty much. I'm not sure what you're talking about with the temperature thing, but it's not clear whether or not there is some "unified field" from which everything else is an offspring at this point. The electromagnetic force and the weak force were successfully unified into the electroweak force, so that's a bit of success.

The usual definition of mass does not vary in different reference frame, and force is a covariant four-vector whose magnitude is the same in all frames.

As good as this is, crackpots don't tend to be very familiar with mathematical rigor. So while a consistency check is good for those of us who are familiar with relativity and the Poincare group, I doubt it will be very convincing to people who are "completely against relativity." I do appreciate input from competent people, so thank you.

Assuming there are no singularities, spacetime can always be considered locally Minkowskian, not just in a vacuum.

I did define it. No, that is not how I will define force, and that's not how Newton defined it either. Newton gave F=ma as an equation of motion, not a definition of force. He defined force as a quantifiable "push or pull," and he gave an example of a force: gravitation. The Lorentz force is another example. F=ma is not, nor has it ever been, a definition of force. Acceleration is absurdly simple to define: the time derivative of velocity. Before we continue, I think we deserve to know what point you're trying to make. Because you're coming across as an obnoxious troll who just wants to argue for the sake of arguing. It's clear you didn't come hear because you were confused and wanted clarification, you came here to be annoying.

My post above assumed a diagonal metric, which in general isn't true of rotating coordinate systems. The metric of a rotating frame will generally have terms like [math]dt d\theta[/math]. If you set a sphere into rotation then you will get a Kerr spacetime, which of course is not equivalent to Schwarzschild spacetime. You get new effects like frame-dragging and such. The example you asked me to do is far too complicated. The Einstein Field Equations can only be solved analytically in the most simple cases. Advanced numerical techniques are required for more complicated situations, like the one you describe.

Are you just trying to be obnoxious? In classical physics, mass is the constant of proportionality relating how much an object accelerates to how much net force was applied to it. On Earth, the easiest way to determine the mass of something would be to place it on a scale (measure its weight), then divide by the acceleration due to gravity at Earth's surface which is ~9.8 m/s2. If you don't have the luxury of being able to measure something's weight (if you're in orbit around a planet, in deep space, etc.), then I suppose the next easiest way to determine mass is to attach the object to the end of a spring with known spring constant, then set it into simple harmonic motion. If you measure its period to good accuracy, then its mass is [math]m=T^2 k /4 \pi^2[/math].

For the backward Euler's method, you could use something like the following: clc; clear; ti=0; tf=10; n=5000; h=(tf-ti)/n; t(1)=ti; y(1)=1; for i=1:n dydt=sin(t(i)); t(i+1)=t(i)+h; y(i+1)=y(i)+h*dydt; end; plot(t,y) I used dy/dt=sin(t) as an example function. To make it go backwards, all you have to do is make tf<ti, so that h becomes negative.

I'm not really sure what you're getting at. Do you accept that all data at this time suggests particles are point-objects, as predicted by the Standard Model? If so, what exactly is your point? Because point objects cannot, by definition, be rotating extended objects. If you don't accept this, did you read the wikipedia quote I provided above?

I'm confused now. Do you want to know about orbital angular momentum, or about spin? Because they're separate concepts. An electron has spin regardless of whether it's bound to an atom.

At a certain point trying to visualize what's going on isn't helpful, and can often be hurtful - a lot of misconceptions can spring up. As far as physics is currently concerned, particles are point-objects. They don't have any extended structure, so it doesn't make any sense to think of them as spinning like a top. I'm by no means familiar with the subject, but in String Theory spin is associated with a type of rotation motion of the string. The wikipedia had this to say: Spin: each particle in quantum field theory has a particular spin s, which is an internal angular momentum. Classically, the particle rotates in a fixed frequency, but this cannot be understood if particles are point-like. In string theory, spin is understood by the rotation of the string; For example, a photon with well-defined spin components (i.e. in circular polarization) looks like a tiny straight line revolving around its center.

Ya, momentum usually has a canceling affect to energy. I don't know whether or not you'll find it's a convincing argument, but a moving particle and a stationary particle have the same stress-energy trace [math]T[/math] (as they should, considering they only differ by a coordinate transformation). For example, consider a point particle stress-energy [math]T^{\mu \nu}=mu^\mu u^\nu \delta (x-x_0)[/math] where [math]x_0^\mu (s)[/math] is the path of the particle and [math]u^\mu=dx_0^\mu /ds[/math] is its four-velocity. First we can consider a particle at rest (assuming a diagonal metric), with [math]u^\mu = (a,0,0,0)[/math]. The stress-energy trace is then: [math]T=g_{00}T^{00}=g_{00} ma^2 \delta (x-x_0)[/math] We can then consider a moving particle, with [math]u^\mu = (d,b,0,0)[/math]. We know that [math]a^2 g_{00} = d^2 g_{00} + b^2 g_{11}[/math], so we can rewrite the four-velocity as: [math]u^\mu = (\sqrt{a^2 - b^2 \frac{g_{11}}{g_{00}}},b,0,0)[/math] The SET trace is then: [math]T=g_{00}T^{00}+g_{11}T^{11}=\left [g_{00} m \left ( a^2 - b^2 \frac{g_{11}}{g_{00}} \right ) +g_{11}mb^2 \right ] \delta (x-x_0) = g_{00} ma^2 \delta (x-x_0)[/math] The Ricci tensor is then, by the trace-reversed field equations, compensated for in the 11-component by what was added to the 00-component. This should correspond to a metric with identical geodesic equations to the first (though I don't know if an exact solution for these metrics is possible).

The time component of four-velocity increases when you move spacially. The length remains constant because the inner product isn't positive-definite.

Right. Well, depending on how far away you are from a massive object it may or may not offset the expansion. Two galaxies which are far apart will not be gravitationally bound, which is how we know the universe is expanding in the first place. No! I thought I did a decent job of explaining this in the last thread. You can approximate the predictions of General Relativity to obtain a "Newton-esque" force-law which includes the cosmological constant: [math]F=m\left (-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3}r \right )[/math] Notice how you don't see any explicit time-dependence in that equation. No constants are changing. No "empirical adjustments" are being made. 2 and 3 are extremely vague, and 1 contradicts observation.

There you go!

Your whole post is based on the assumption that the Cosmological Constant changes over time, despite the fact that there is no evidence to suggest this is the case. Hence the term Cosmological Constant. It is thought to represent the vacuum energy density, and there's no current reason to suspect it should change over time. In fact, if the CC were changing in time then it would have to be promoted to a scalar field, which would also (upon quantization) imply the existence of a spin-0 "Cosmological Particle." The actual name for such a particle is "dilaton." Most research on the possibility of a time-variable CC is in the context of Brans-Dicke gravity (a modification of General Relativity to include a scalar field). Dilatons aren't ruled out by experiment (though they certainly haven't been observed) and in fact they are typical in the various string theories. Also, I don't see your "hypothesis" anywhere. Just vague words slapped together.

What's the question mark for? You just proved it!

Neutrino masses are predicted to be of the order of about ~1 eV, with the lowest possible value being ~0.05 eV. The current upper limit on the photon mass is of the order ~10-17 eV.

I believe the derivation of that equation probably went something like this: start with the g-value on Earth's surface: [math]g=\frac{GM}{R^2}[/math] Now define g' as the g-value at a height h above the Earth's surface: [math]g'=\frac{GM}{(R+h)^2}=\frac{GM}{R^2} \left (1+\frac{h^2}{R^2}+\frac{2h}{R} \right )^{-1}=g \left (1+\frac{h^2}{R^2}+\frac{2h}{R} \right )^{-1}[/math] Now do a binomial expansion to get: [math]g'=g \left (1-\left (\frac{h^2}{R^2}+\frac{2h}{R} \right )+\left ( \frac{h^2}{R^2}+\frac{2h}{R} \right )^2 - \left ( \frac{h^2}{R^2}+\frac{2h}{R} \right )^3 + ... \right )[/math] Now if we make the assumption that h<<R, (h is very small compared to R) then terms with higher powers of (h/R) are essentially zero. This makes all terms in our expansion go to zero, except: [math]g'\approx g \left (1-\frac{2h}{R} \right )[/math] If you make h large compared to R, then the simplifying assumption we used no longer holds. In other words, the approximation fails when h becomes large. So h=infinity is definitely not going to work.

A better answer would be "because that's what we've observed."

Well, the first one is very easy. Take an object moving at a constant velocity [math]v[/math] in your coordinate system (can be a vector). Now boost to a new coordinate system moving at constant velocity [math]u[/math] with respect to the original (again, can be a vector). The acceleration of the object in this coordinate system is: [math]\frac{d}{dt}(v+u)=0[/math] So the object is still inertial in the new inertial coordinate system. The third law is more of a postulate that can't be "proven" one way or another, except that experiment supports it. The third law is also explicitly local, so I don't see how a boost would affect it.

The acceleration required to maintain constant "r" value, i.e. the equivalent of the Newtonian "g" value, is given by: [math]a=\frac{GM}{r^2\sqrt{1-2GM/rc^2}}[/math]

It's actually a spinor for fermions, not a vector. But the difference isn't really that important if you're just trying to get a feel for it.

Mike, may I ask why you don't just learn the physics for yourself? You certainly seem capable and genuinely interested. However, from a lot of the posts I've read, you appear to favor obscure philosophy and even sometimes blatant crackpottery. Why not just buy a few relativity and QM textbooks? There are good free online resources as well. If your goal is to understand how nature works, then you should make a real effort to understand how nature works.

Black holes don't have infinite mass. Infinite mass would imply infinite Schwarzschild radius, so the entire universe would be swallowed up.