elfmotat

I believe it was Feynman who said that math is language + logic.

I believe you're thinking of divergence. Fields with no sources aren't necessarily conservative. For example, the E or B field of an electromagnetic wave: the curl of E is nonzero even in the absence of sources.

I'm going to change your variable O to an A, because I don't like it (it looks too much like zero). So we have: [math]U=\int_{0}^{\Delta l} F dl[/math] [math]\sigma =F/A[/math] [math]\epsilon =\Delta l/l[/math] We can first replace F with [math]F= \sigma A[/math]: [math]U=\int_{0}^{\Delta l} \sigma A dl[/math] Now we also have [math]\Delta l= l \epsilon [/math]. In the limit that [math]\Delta l[/math] becomes smaller and smaller it becomes [math]dl[/math], and [math]\epsilon[/math] becomes [math]d \epsilon[/math]. So our equation with differentials becomes: [math]d l= ld \epsilon [/math] Since we are changing variables we also need to change the limits of integration, which is where the strain in the upper bound comes from. (If change in length goes from 0 to Δl, then strain goes from 0 to ϵ.) So after all of our substitutions, the integral becomes: [math]U=\int_{0}^{\epsilon} \sigma A l d\epsilon[/math]

A field is defined as conservative if it can be written as the gradient of a potential function: [math]\mathbf{F}=\nabla f[/math] The definition is also equivalent to saying that its curl is zero, because: [math]\nabla \times \mathbf{F}=(\nabla \times \nabla) f =0[/math] The curl of a gradient is zero because partial derivatives commute. Physically, curl represents the amount of "circulation" in a field. Conservative fields are useful because the amount of work done in moving a point particle from one point in the field to another is independent of the path it takes. To see this: [math]W=\int_a^b \mathbf{F} \cdot d\mathbf{x}=\int_a^b \nabla f \cdot d\mathbf{x}= f(b)-f(a)[/math]

That's exactly what I'm saying. It is "wrong" because we start by asking the question, "what is the internal pressure some spherical mass distribution would need to resist collapse at R<9GM/4c2." The answer we get is infinity, so that means there was a flaw in our assumptions - namely that the sphere can resist collapse. In other words: the pressure required to resist collapse is infinite, therefore it does not resist collapse.

The original derivation is here: http://journals.aps.org/pr/pdf/10.1103/PhysRev.116.1027 .

The easiest way would be to use four-vectors. The four-momentum is: [math]p^\mu = (E/c, \vec{p})[/math] This can be found easily by simply defining four-momentum as mass times four-velocity, where four-velocity is the derivative of position [math]x^\mu =(ct,\vec{x})[/math] with respect to proper time. We know that the magnitude of the four-momentum must be constant for every observer: [math]p_\mu p^\mu = const.[/math] If we consider the frame where a particle is at rest, its momentum becomes: [math]p^\mu = (E/c,0)[/math] But when a particle is at rest its energy is just its rest energy, which is equal to its mass: [math]p^\mu = (mc,0)[/math] Its magnitude squared is: [math]p_\mu p^\mu = m^2 c^2[/math] In some other general reference frame we have: [math]p_\mu p^\mu = E^2/c^2 - |\vec{p}|^2[/math] But as I said these must be equal, so: [math]E^2/c^2 - |\vec{p}|^2= m^2 c^2[/math] You can do a little rearranging to get it into the more familiar form: [math]E^2 = m^2 c^4 + |\vec{p}|^2 c^2[/math]

Singularities exist within the context of General Relativity, which is a classical theory. It is suspected that a quantum theory of gravity would be able to describe the center of a black hole without singularities, but such a theory does not currently exist. It is generally believed that there's not "really" a singularity there (in the sense that there's actually a discontinuity in an otherwise smooth spacetime), but they are predicted by the math of GR. The reason we know, classically, that extremely dense objects must collapse to a singularity is due to a pressure analysis of spherical bodies. It is possible to show, with a few assumptions, that once a sphere is compressed to a radius of [math]R=9GM/4c^2[/math] that the amount of internal pressure required to resist gravitational collapse becomes infinite. (G= gravitational constant, M=mass of sphere, and c=speed of light.)

If you have time tar, you should watch this video. Hopefully it clears up some of your confusion. You seem to believe that uncertainty is just a matter of not being able to account for all of the variables. Feynman gives a good argument for why this isn't the case.

The following must be true for both SET's to represent the same physical observable (the energy-momentum): [math]\int d^3x T^{0 \mu}= \int d^3 x \Theta^{0 \mu}[/math] In my first post I said "spacetime" where I should have said "space," but what I meant was the above. Sorry for the confusion. Yes, div(B) must also be divergence free, as mandated by B's definition.

What I meant was, we have a mathematical model that agrees with experiment. QM isn't some untested hypothesis, and the math didn't come out of the sky. The math was tailor made to fit and predict experimental data, and it works.

Yes, B is antisymmetric in the first two indices. I should have mentioned that as part of its definition. If I were going to try to derive B, the way I would do it would be to show that if [math]\Theta^{\mu \nu}-\Theta^{\nu \mu}=-\partial_\sigma C^{\sigma \mu \nu}[/math], then the form of B must be: [math]B^{\sigma \mu \nu}=\frac{1}{2} (C^{\sigma \mu \nu}+C^{\mu \nu \sigma}-C^{\nu \sigma \mu })[/math] If you then take the angular momentum tensor (the kinetic part plus the spin) and impose Lorentz invariance as a symmetry of the field (i.e. you impose conservation of angular momentum), then you must have: [math]\Theta^{\mu \nu} - \Theta^{\nu \mu}=\partial_\sigma C^{\sigma \mu \nu}= -\partial_\sigma (\Pi^\sigma \Sigma^{\mu \nu} \phi)[/math]

It's not what anyone decided, it's what experiment tells us.

Okay TAR, I'll work out the math for you since you seem to be having trouble understanding that it's not just a matter of measurement, the momentum for a particle with known location is undefined, and vice versa. Say we measure an electron's momentum, and we have determined it to be exactly p0. That means the particle's wave function satisfies the following equation: [math]-i \frac{\partial \psi }{\partial x}=p_0 \psi[/math] (It must satisfy that equation because its momentum must be an eigenvalue of the momentum operator when its momentum is a single value.) The solution to this equation is: [math]\psi (x)=e^{ip_0 x}[/math] Its probability distribution is therefore: [math]P(x)= \psi^* \psi =e^{ip_0 x} e^{-ip_0 x}=1[/math] This distribution is not integrable over space. I.e. when you try to integrate it to see where the particle is likely to be you get nonsensical answers. (For a probability distribution to be valid, the integral of P(x) from -∞ to ∞ must equal one. If you try integrating the above, you get infinity.) Its position is undefined. You also can show that the opposite is true, i.e. if we have a single precise value for the position then the momentum is undefined.

What a constructive discussion this has been.

This isn't a theory, it's a bunch of word salad.

No problem. Good question.

There are really two types of stress-energy tensor: the Hilbert SET and the canonical SET. The Hilbert SET is defined by a functional derivative of the Lagrangian, and is always symmetric. This is the SET that appears in the Einstein field equations. The canonical SET is defined as the conserved Noether current that comes from spacetime translation symmetry. The canonical SET is not symmetric in general. Since Noether currents remain conserved with the introduction of a divergence term, both tensors can be related to each other by introducing another tensor (called the Belinfante tensor): [math]T^{\mu \nu}= \Theta^{\mu \nu} +\partial_\sigma B^{\sigma \mu \nu}[/math] where theta is the canonical tensor. Integrating T over spacetime is identical to integrating theta over spacetime, because the two only differ by a total divergence. So you can think of B as something that contains all of the information about the intrinsic angular momentum in the field. Explicitly we have: [math]T^{\mu \nu} \equiv \frac{2}{\sqrt{-g}} \frac{\delta ( \mathcal{L} \sqrt{-g})}{\delta g_{\mu \nu}}[/math] This is how T is defined (the definition comes from varying the Hilbert action to get the EFE's). If you work out the conserved current under [math]x^\mu \mapsto x^\mu +a^\mu[/math] ,[math]~~~[/math] [math]\phi (x) \mapsto \phi (x+a)=\phi(x) + a^\mu \partial_\mu \phi(x)[/math] (with infinitesimal a) for some field phi (could be scalar, vector, tensor, etc.), you get the quantity: [math]\Theta^{\mu \nu} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial^\nu \phi - \eta^{\mu \nu} \mathcal{L}[/math] Now we define the tensor B such that [math]T^{\mu \nu}= \Theta^{\mu \nu} +\partial_\sigma B^{\sigma \mu \nu}[/math]. If you go through the work, you get B to be of the form: [math]B^{ \alpha \mu \nu } = \frac{1}{2} \left (\Pi^{\alpha} \Sigma^{\mu \nu} + \Pi^{\mu} \Sigma^{\nu \alpha} - \Pi^{\nu} \Sigma^{\alpha \mu} \right ) \phi[/math] where [math]\Pi^\mu = \partial \mathcal{L}/\partial (\partial _\mu \phi )[/math] and the sigmas are the spin matrices. So, for example, with a spin-1 field like the EM field, we can plug in the EM Lagrangian to find the canonical SET, and we get: [math]\Theta^{\mu \nu} =F^\mu_{~\lambda} \partial^\nu A^\lambda-\frac{1}{4}\eta^{\mu \nu} F^{\alpha \beta } F_{\alpha \beta }[/math] Indeed this is not symmetric due to the presence of intrinsic spin in the field. If we use the vector spin matrices and plug everything into B, we get [math]B^{\alpha \mu \nu }=-F^{\alpha \mu } A^\nu[/math]. When we take its divergence and add it to the canonical SET, we get: [math]T^{\mu \nu}=F^{\mu}_{~ \lambda} F^{\lambda \nu}-\frac{1}{4} \eta^{\mu \nu} F^{\alpha \beta }F_{\alpha \beta }[/math] This tensor is symmetric, and it corresponds to what you would get with the above definition. This is the tensor that couples to curvature in the EFE's.

Remember that [math]\widehat{H} = (\widehat{A}^\dagger \widehat{A} +\frac{1}{2}) \hbar \omega_0 [/math]. What they're doing is rearranging the equation so that [math]\widehat{H}[/math] acts on the wavefunction before [math]\widehat{A}[/math] does. The reason they do this is so that you can replace the Hamiltonian with its eigenvalue energy, [math]E[/math]. To do this they used the commutator relation and factored out a lowering operator: [math]\widehat{H} [\widehat{A} \psi ]= (\widehat{A}^\dagger \widehat{A} +\frac{1}{2}) \hbar \omega_0 [\widehat{A} \psi ]= \hbar \omega_0( \widehat{A} \widehat{A}^\dagger -1+ \frac{1}{2}) [\widehat{A} \psi][/math] [math]=\hbar \omega_0(\widehat{A} \widehat{A}^\dagger \widehat{A}-\widehat{A}+\frac{1}{2} \widehat{A})\psi = \hbar \omega_0 \widehat{A}( \widehat{A}^\dagger \widehat{A}+ \frac{1}{2} -1) \psi =\widehat{A}(\widehat{H}-\hbar \omega_0)\psi [/math] Now, remember that the Hamiltonian acting on a state is its eigenvalue, which is the energy of the state: [math]\widehat{H}\psi =E \psi [/math]. So we have: [math]\widehat{H} [\widehat{A} \psi ]=\widehat{A}(E-\hbar \omega_0)\psi =(E- \hbar \omega_0) [\widehat{A} \psi ][/math] So we see that the lowering operator has the effect of lowering the energy by an amount [math]\hbar \omega_0[/math].

It's not endless energy. The energy stops when the water runs out, and to get more you need to bring more water up the hill, i.e. expend mechanical energy to increase the water's potential. Or you can evaporate the water by supplying heat energy, and have it condense and precipitate at the top of the hill. (The sun can do this for you.) Either way you need to expend energy to get water to the top of the hill for the generator to be of any use.

The color of an object is defined as the particular way in which our brain interprets the frequency of light that reflects off it and reaches our eye. A blue object is defined as blue because the light reflecting from it is blue. Why would we define its color to mean the frequencies of light that it does absorb, i.e. the ones we can't see? That makes no sense to me.

Any software that can graph vector fields would work. Matlab, Mathematica, Maple, etc. For example, this plot took me about 30 seconds in matlab. It corresponds to the electric field of a positive point charge. EDIT: Well, swansont's suggestion puts mine to shame.

What do you mean by "what does the 1 represent?" It doesn't represent anything - it's just a number. If you know that something has mass m and is moving at velocity v, then you plug m and v into the equation to get p.

Only approximately. The form of relativistic 3-momentum is different: [math]p=\frac{mv}{\sqrt{1-v^2/c^2}}[/math]

Sure, a strong enough electromagnetic field could certainly disrupt brain function, though I have no idea how strong it would have to be for the changes to be noticeable.