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0

That's a very nice answer by ACuriousMind. I would like to add something, though. GR is actually not like other gauge theories in some of its aspects (apart from having lots of similarities). For starters, it is background-independent and highly non-linear. In ordinary QFT we usually deal with perturbative expansions, which make sence only for weak-coupled ...


-1

The cosmic forces are changing the properties of elementary particles, gravity does not! See:"http://home.quicknet.nl/qn/prive/nj_dewit/quarks/index.htm"


0

It's probably best to think of the energy as living in the gravitational field, and not in the singularity itself. When you perturb the spacetime, then, you're perturbing the gravitational field, and the energy you get out comes from there. Causal information can't come from inside the horizon, after all.


1

The other forces are also just the result of "spacetime bending", just in a different way. There is no fundamental difference in the description of the other forces through gauge theories and gravity through relativity.1 The reason why it is often said that it is different is that our usual methods of quantizing a theory fail when applied to gravity. But to ...


0

Force is a classical concept that is useful in modeling the mesoscopic world, i.e the world of classical thermodynamics, mechanics and electrodynamics. Exchanged particles are quantum mechanical concepts which mainly work in small atomic size dimensions. There is continuity in physics going from mesoscopic to the microscopic frameworks, and continuity ...


5

To add to Hindsight's great answer: one of the reasons that the analogy fails is the same reason why Nordström's Scalar Theory of Gravitation fails: Waves on rubber sheets are described by linear wave equations; at least in the small amplitude limit. However, by analogy with Maxwell's equations, waves in gravitation should bear energy. But we are also ...


12

The rubber-sheet analogy is often used to "explain" the basics of GR to beginners, but actually it has nothing to do with real gravity. It acts much more like a scalar field (the up/down freedom degree) - and there were several attempts to build a scalar gravity. But the correct description turned out to be tensorial and purely geometrical. GR has 10 ...


1

Start with the lower expression: $$ (c/2)\eta^{bc}\eta^{ae}\partial_{a}\left(g_{be,c} + g_{ce,b} - g_{bc,e}\right) - (c/2)\eta^{ae}\eta^{bc}\partial_{c}\left(g_{be,a} + g_{ae,b} - g_{ba,e}\right).\\ = \frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{be,ca} + g_{ce,ba} - g_{bc,ea}-g_{be,ac} - g_{ae,bc} + g_{ba,ec}\right)\\ =\frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{ce,ba} - ...


2

The first reason is that your "distance" between geodesics is measured by a parallely propagated direction $\partial/\partial \phi$. If you take a look at the sphere, the difference $\Delta \phi$ does not correspond to the distance between the points on the geodesics. The distance between them would be measured by arc-lengths of great circles. But you are ...


9

The source of gravity is not mass, but stress-energy-momentum, so you are correct that the energy converted in this process already has gravity and that that gravity is only rearranged The change in the gravitational field needs time to propagate, though, and this does indeed happen at the speed of light.


0

$H$ tells us how fast the universe is expanding, relative to how much it has already expanded. It has units of inverse time. For example, if $H=0.1\ \mathrm{s}^{-1}$, then the universe is expanding by 10% every second. Suppose that the density of mass-energy in the universe was so small that deceleration was negligible, and suppose that at $t=1$ s, we have ...


0

The Weyl transformation (unlike diffeomorphisms) does not affect physical fields, only the geometry of space-time. Probably the best way to show that your equation is Weyl-invariant is to build an action which (when varied) yields the equation. In your example it would be $$ S[\Psi] = \int d^4 x \: \sqrt{-g} \: \bar{\Psi} \gamma^a e^{\mu}_a D_{\mu} \Psi. $$ ...


2

All the known laws of physics, including the exclusion principle, are believed to be valid at all times during the collapse, up until the matter that you're talking is just about to hit the singularity. ("Just about to hit" may mean when the density reaches the Planck density, so that quantum gravity effects become important, or it may be a little earlier, ...


0

I'll post the same answer here that I posted on Worldbuilding: Under the circumstances you describe, my immediate reaction is that it would not be possible. The issue here is that the cluster would be fairly unstable. The black holes would all be mutually attracted to each other, and would soon coalesce into one large black hole - taking the Imperial ...


2

I) In Palatini $f(R)$ gravity, the Lagrangian density is $$ {\cal L}~=~ \sqrt{-g} f(R), $$ with $$R~:=~ g^{\mu\nu} R_{\mu\nu}(\Gamma),$$ and where $\Gamma^{\lambda}_{\mu\nu}=\Gamma^{\lambda}_{\nu\mu}$ is an arbitrary torsionfree$^1$ connection. II) As OP mentions, the word Palatini refers to that the metric $g_{\mu\nu}$ and the connection ...


0

This article deals with passable wormhole http://scitation.aip.org/content/aapt/journal/ajp/56/5/10.1119/1.15620 Wormhole Trafficable properties As we have seen, there are several objections to the possibility of come true interstellar travel through holes black or wormholes from Schwarzschild. To make passable a wormhole should have the following ...


3

The total energy of the universe is a vexed issue since different commentators have different views about what the concept means. See the question Total energy of the Universe for a sampling of the various viewpoints. If you Google for zero energy universe you'll find several papers purporting to show that the total energy is zero. However since their ...


2

There are some qualifications on the equivalence principle that often aren't made clear in casual summaries. One is that the equivalence principle only works in the limit as the size of the spacetime region where measurements are performed approaches zero (or to use another term, in an "infinitesimal" region of spacetime)--in other words, you'd have to ...


4

They're not zero in general. For example, take flat Euclidean space in two dimensions, in polar coordinates: $$ ds^2=dr^2+r^2d\phi^2 $$ for which the nonzero Christoffel symbols are $$ \Gamma^r_{\phantom\phi\phi\phi}=-r,\quad \text{and} \quad \Gamma^\phi_{\phantom\phi \phi r}=\Gamma^\phi_{\phantom\phi r \phi}=\frac1r $$ Then $\Gamma^\lambda_{\phantom\phi ...


0

I'll try to write this up into a quick answer. Maybe it will help. What exactly is a? $a$ is the universal scale factor. It is also written as a function of time: $a=a(t)$. As you can see, it features prominently in the Friedmann equations (and is featured in the FLRW metric), and is important when studying the expansion of the universe. As is the ...


0

My suggestion would be to start with the ADM formalism. The details of the Kaluza-Klein seperation will be a little different, because they're foliating on a timelike fiber, and thus, your "special" dimension is spacelike, but the idea is essentially the same -- the lapse function is your dilaton field, and the shift is your vector potential. I'm sure that ...


9

Gravity definitely does exist. Einstein did not provide a model without gravity, he simply found a new way to think about it. Gravity, as we experience it, is a consequence of bodies moving along so-called geodesics, which can in simple terms described as "shortest paths through spacetime". The effect of gravity now arises if there is motion along geodesics ...


1

It's only a way of modeling motion as we see. You can think in that way about General Relativity, but physicists strongly believe that there exists theory of quantum gravity, which means that there exists a corresponding fundamental particle called the graviton. The most developed examples of quantum gravity theories are String Theory and Loop Quantum ...


33

Physics does not answer existential problems. It gathers data and observations and models them with mathematical equations and functions, and then can explain the data with the model and predict new observations. This has been going on for centuries, and what we see if we study the history of physics is that there are regions of validity for the ...


3

Here are couple of references that describe professional uses of a post-Newtonian formalism to model the planets and the Earth's Moon: Standish, et al. "Orbital ephemerides of the Sun, Moon, and planets," Explanatory Supplement to the Astronomical Almanac (1992): 279-323. The relevant equation is 8-1 on page 3. Petit and Luzum (eds.), "IERS Technical Note ...


0

Here is one way to derive the geodesic equations from the Euler-Lagrange equations. First consider a natural Lagrangian system $(M,L)$, where $L\in C^\infty(TM)$. Let $g$ be a Riemannian metric. Suppose in our mechanical system the net force is zero. That is, the Lagrangian is just equal to the kinetic energy, $$L(p,V_p)=\frac{1}{2}mg_p(V_p,V_p)$$In ...


3

To be exact Einstein made a claim that it is gravity that curves space-time. You can follow his reasoning in his "Relativity: The Special and General Theory." Einstein started off with comparing acceleration caused by gravity to acceleration in a lift (assuming it moves with accelerated motion) going up. He claimed that these two accelerations are ...


2

According to the excellent and very well researched scientific biography "Subtle is the Lord" by A. Pais, as late as 1912 Einstein was still assuming a flat Euclidean space (at that point he had been working on the general theory for 5 years). Then (in 1912) Some time between August 10 and August 16, it became clear to Einstein that Riemannian geometry ...


0

I think the answer to this is basically yes. You do have to be careful, because in general, you can't interpret inner products in relativity as measures of whether something is "orthogonal" to something else in the Euclidean sense. E.g., a lightlike vector has a zero inner product with itself. This is because the metric isn't the Euclidean metric. However, ...


1

To address the first question you can consider $$ \nabla_av^b := v^b ,_a$$ since $$ \nabla : \Gamma(E) \rightarrow \Gamma(E\otimes T^*M)$$ where E is any section(e.g. the vector field in question) and contracting it with the tangent vector field of the curve you get $$t^av^b,_a = 0 $$ this is similar to contracting a vector field with a dual vector $$ ...


2

Yes, you're exactly right, $\omega_b {C^b}_{ac} = {C^b}_{ac} \omega_b$. In general, the order of factors doesn't matter in a tensor expression like this. This is Wald, so technically you're supposed to think of those expressions as using abstract index notation instead of involving components in any particular basis, but it's also valid to think of them as ...


1

There is no definite SIZE to the universe as such. There is however a size to the OBSERVABLE universe. These are very different. And indeed the observable universe is defined by Einstein information caveat where information cannot propagate faster than light. Now, it is in this sense, that Newtonian mechanics fails us, as newtonian gravity (and grav. ...


2

Only in the case of a static spacetime is the metric derivable from a scalar potential. Cosmological spacetimes aren't static, so they can't be derived from a potential.


-2

Here is a short description of Big Bang: Over the years, proponents of the big bang have tried heroically to change the name. They are dissatisfied with the common, almost vulgar connotation of the name and the fact that it was coined by its greatest adversary. Purists are especially irked that it was also factually incorrect. First, the big ...


0

The Einstein-Hilbert action, from which the Einstein field equations are derived, is given by, $$S = \frac{1}{16\pi G_N} \int d^4x \, \sqrt{-g} \, \mathcal{R}$$ where $\mathcal{R}$ is the Ricci scalar, dependent on the metric $g_{\mu\nu}$ and its derivatives. The self-interaction and non-linearity of general relativity arises from this factor. To see this, ...


0

About.. locally measured your time never differ, aka, 'c' relative your wristwatch for example, although some might want to argue that a acceleration is different there, which I don't agree too. But we can keep it to uniform motions and then also conclude that this fact is what makes 'repeatable experiments' work, as well as constants. Assume this wrong, ...


1

To add to Qmechanic's Answer and TwoBs's Answer and answer "....what the heck is h then? Any arbitrary function?": $h$ is pretty much arbitrary. It is wontedly taken to be at least differentiability class $C^1$ (all first derivatives continuous) so that the Lie bracket of vector fields is defined as in Qmechanic's answer. You need to assume it is of class ...


3

Think of an infinitesimal Diff as of a translation where the the shift is space dependent, $x^\mu\rightarrow x^\mu+\epsilon^\mu(x)$. Now, you get that the generators are $L_\epsilon=\epsilon^\nu(x)\partial_\nu$ since $L_\epsilon x^\mu=\epsilon^\mu(x)$. They form an infinite space since $\epsilon^\mu(x)$ is a function that can be expanded in infinitely many ...


1

Formally speaking, given a (differentiable, finite dimensional) manifold $M$, then the (infinite dimensional) Lie group of (globally defined) diffeomorphisms (with composition $\circ$ as group structure) has the set $\Gamma(TM)$ of (globally defined, differentiable) vector fields as corresponding Lie algebra. This (infinite dimensional) Lie algebra ...


0

your Lagrangian is almost correct. But you also need to use a mass term that is conserved, which won't be the case of the mass term with your Lagrangian. If you use: $$ \mathcal{L}_m = \sum_p m_p \gamma_p^{-1}(-g)^{-1/2}\delta^{(3)}(x^j-x^j_p(\tau_p)),$$ where $\gamma_p=dx^0/cd\tau_p$ the Lorentz factor, $u^\mu_p=dx^\mu_p/cd\tau$ the 4-velocity of the ...


4

The criterion for gravitational radiation is (conjectured to be, pending direct evidence) a changing quadrupole moment in the mass distribution, so an accelerating mass distribution does not always radiate, but can do so if the acceleration changes the quadrupole moment. This is in contrast to electromagnetic radiation, which occurs when the charge ...


4

Yes. The electromagnetic field is nonzero, so the stress-energy tensor is equal to $\frac{1}{2}F_{ac}F^{c}{}_{b}- \frac{1}{8}g_{ab}F^{cd}F_{cd}$. And, of course, $8\pi G T_{ab} = R_{ab} -\frac{1}{2}Rg_{ab}$ EDIT: to clarify, in the riessner-Nordstrom spacetime, you have $g_{ab} = {\rm diag} (-\Delta, \frac{1}{\Delta}, {\rm sphere\; metric})$ and $F_{ab} = ...


4

Please let me first refer you to the following review by I. L. Shapiro, which contains a lot of theoretical and phenomenological information on spacetime torsion. The answer will be mainly based on this review. In the basic Einstein-Cartan theory, in which the antisymmetric part of the connection is taken as independent additional degrees of freedom, the ...


2

There seems to be some confusion regarding conservation laws vs. invariant quantities. Any (Lorentz) scalar, such as density, pressure, temperature, or charge, will be invariant as reference frames change. So too will any vector or higher other tensorial quantity. That is, even though different observers might assign different numbers to the components of ...


0

(This answer was written before this question was drastically rewritten, when the basic question was "Is it possible to define a conserved quantity in GR?".) Sure, there are derivative forms of conservation laws that hold in general relativity. For example, there's the conservation of the stress-energy tensor, $${T^{\mu\nu}}_{;\nu} =0\ .$$


0

I think you're misinterpreting what the no-hair theorems say. There's a no-hair theorem for stationary electrovac solutions. It applies to electrovac solutions, not to solutions containing any matter field you like; in fact, there are known counterexamples if you include certain types of matter fields. Also, it's a theorem about stationary solutions. A ...


0

The small angle approximation is very reasonable here. The angle of deflection predicted by the approximation is ${4GM \over rc^2}=8.49\times10^{-6}$ radians. Writing $\tan{\theta}$ instead of $\theta$ is just clutter. The difference between the two is $2.039\times10^{-16}$. In order to resolve this difference in angle in visible light, you would need a ...


1

The source of the potential energy is the fact that different observers have a different notion of "stationary", because the curvature of spacetime rotates the time of one observer into the space of another. We know from classical mechanics that energy is related to time translation symmetry, so it shouldn't be surprising that messing with the definition of ...


2

The F(L)RW metric comes with very few assumptions, though these are fairly strong: Spacetime is homogeneous. Spacetime is isotropic. Or, in other words, the cosmological principle is assumed. Philosophically this is very desirable, as the notion that there are preferred locations or directions in the Universe is, from a modern point of view, somewhat ...


3

We don't think the FLRW metric is valid throughout the entire history of the universe. If we take a metric of the form: $$ ds^2 = -dt^2 + a^2(t) d\Sigma^2 $$ then we expect this to be valid throughout the history of the universe as long as the universe is isotropic and homogenous. However we need to find the equation for the function $a(t)$, and this is ...



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