New answers tagged general-relativity
1
Joshphysics has already given a nice answer showing that in 2+1 dimensional Einstein gravity any metric is locally equivalent to a metric of constant curvature. As dilaton mentioned in a comment this in particular means that there are no local excitations.
The updated question also asks about 1+1 dimensions. In this case the answer is even simpler: the 1+1 ...
4
There is nothing "wrong" with the Einstein field equations in $2+1$ as indicated by the comments, but they do have interesting, significantly restricted behavior in $2+1$ dimensions.
For example, the Wikipedia page referred to by Olof in the comments says that in $2+1$, every vacuum solution is locally either $\mathbb R^{3,1}$, $\mathrm{AdS_3}$, or ...
0
The Schwarzschild spacetime has Killing vector fields $\partial_t$ and $\partial_\phi$ that give conservation of specific energy and angular momentum, respectively:
$$\begin{eqnarray*}e = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\lambda}\text{,} &\quad\quad& l= ...
0
The formula for the Hubble Parameter is
$$ H_{a} = \text{H0}\sqrt{\text{$\Omega $R} \cdot a^{-4}+\text{$\Omega $M} \cdot a^{-3}+\text{$\Omega $K} \cdot a^{-2}+\Omega \Lambda } $$
where only
$$ \Omega \Lambda $$
is for dark Energy. The other Omegas are for radiation, matter and curvature. Reducing to dark energy reduces the formula to
$$ H_{vac} = ...
1
If you regard curvature as a form of energy density (see Chris White's post), then yes, you should set $k=0$. If not, then $H$ will only approach a constant value as $a\rightarrow \infty$:
$$
H^2 \rightarrow \frac{8\pi G}{3}\!\rho_\Lambda \qquad \text{for $a\rightarrow \infty$.}
$$
You can check that the general solutions of
$$
\dot{a}^2 - \frac{8\pi ...
0
My knowledge of physics does not really extend to these realms, and I apologize if I am wrong or off topic, or talking to far above my head. It is often funny how the mathematization of things can force some level of belief or unification,
even though I am just trusting the mathematician.
I understand that the Kaluza-Klein theory from the twenties provides ...
1
Yes, you should probably assume $k = 0$, since otherwise the statement is not true, as you have shown.
The nature of the Friedmann equations allows us to rewrite the $k/a^2$ term as though it were itself a source of energy density $\rho_k$ varying as $a^{-2}$, complete with its own fraction $\Omega_k$ of the critical density. This is done by writing
$$ ...
0
Yes. Since the Einstein field equations consider the stress energy momentum tensor, it includes momentum density too, with energy density. You may want to learn about the "Kerr-Metric" and more generally, the "Kerr-Newmann metric" which are more general than the Schwarzschild metric". Yes, they can sometimes be measured from the Earth, but I think that is ...
-2
Firstly, gravity affecting time is completely different from gravity affecting mass. Do a search on "gravitational time dilation". Secondly, gravity affects light because it has energy. This is governed by the Einstein Field Equation. Thirdly, the black-holes tag is irrelevant. Fourthly, the visible-light tag doesn't apply here. It applies to optics only, ...
-2
All this time, the field of physics did not seriously consider the possibility of existence of negative mass (energy) in a general state. The standard explanation of negative mass is that the state of low energy is stable when a negative energy level exists and that the lowest state of energy is minus infinity. Thus, this means that all positive mass emits ...
1
The TL;DR version: even if we could form a synthetic event horizon, it wouldn't help us learn about the black hole interior.
The long version:
The phenomena you describe where light essentially orbits a black hole is called the "photon sphere" and it doesn't happen at the event horizon. The radius of a black hole, $R$ is where the event horizon is and the ...
2
A more recent alternative to Deser's work is that of Gull, Doran, and Lasenby. Framed in the mathematical of geometric (real Clifford) algebra and its associated calculus, this framework presents gravity as a gauge field on a Minkowski spacetime. The formulation is clearly inspired by relativistic quantum mechanics and tetrad approaches, but it has some ...
1
First, drop $\lambda$, because it is useless (can be absorbed into $L$). Second, try contracting in $a$ and $c$ only. This contraction should also be zero.
4
To rephrase the question slightly, you are asking for one of the Betti numbers of the (3+1)-dimensional manifold corresponding to one of the solutions of the Einstein field equations that corresponds to charged or rotating black hole.
The Betti numbers of a manifold are topological invariants that intuitively represent the number of non-contractible ...
1
I suppose $f$ is just an arbitrary scalar function on the manifold. I'm not well-versed with the concept of Ricci flow, so I'll try to give a simple operational answer. I also don't understand what exactly you're looking for.
The Ricci scalar $R$ roughly represents the amount of energy stored in spacetime (as curvature). The dilaton is a scalar field which ...
-1
For above question, my answer is yes! Actually, for this question, a number of authors, including Rosen, Kraichnan, Gupta, Weinberg, Feynman, Deser, Grishchuk, T. M. Nieuwenhuizen, and Logunov, etc., have discussed the utility of introducing the metric of flat background space-time into GR. Those theories called the bi-metric gravitation or the ...
1
There's a great discussion of this sort of thing in the first few pages of a paper by Penrose. Basically, to get an integral conservation law, you need the divergence of a vector to be zero. The energy-momentum tensor satisfies a differential conservation law, of course. But there's no associated quantity that you can generally integrate over a volume on ...
0
Let us denote
\begin{align}
\xi_1 = (0,1), \qquad \xi_2 = (-e^x, e^x/t), \qquad \xi_3 = (e^{-x}, e^{-x}/t)
\end{align}
Each of these killing vectors leads to a conserved quantity
\begin{align}
c_1 &= \dot x_\mu\cdot (\xi_1)^\mu = \dot x t^2 \\
c_2 &= \dot x_\mu\cdot (\xi_2)^\mu = \dot t e^x +\dot x te^x \\
c_3 &= \dot x_\mu\cdot ...
0
The main question has already been answered by joshphysics. For the remaining, rescale the variables as
$$ U~:=~cu, \qquad V~:=~cv. $$
The two equations becomes
$$ \dot{U}^2 ~=~U^4+U^2, \qquad \dot{V}~=~U^2 ,$$
with full solution
$$U(t)~=~\pm {\rm csch}(t-t_0), \qquad V(t)~=~\coth(t-t_0)+V_0. $$
OP's sought-for equation now follows from
$$(V-V_0)^2 ...
1
The FLRW metric can be static, this is the solution that Einstein concocted before Hubble observed the expansion of the universe. The only way that Einstein could make his equations static was by introducing the infamous cosmological constant $\Lambda$. The general FLRW metric has the form
$$
\text{d}s^2 = -c^2\text{d}t^2 + a(t)\left[\frac{\text{d}r^2}{1 - ...
3
As noted above in comments, I'm not competely sure I understand the question. But anyway, I'll give it a shot.
The answer is model-dependent. The standard cosmological model at the moment is the Lambda-CDM model. This model has various parameters. Depending on these parameters, the spatial curvature can be positive, negative, or zero. Observation puts ...
4
I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations.
Notice that the metric is invariant under translations of $v$. The associated killing vector is $\partial_v$ which in turn gives the following conserved ...
3
After the question was originally asked, the OP changed it to exclude the Big Bang. I don't understand the motivation for imagining that there would be a boundary anywhere else. The following answer addresses the question as originally asked.
First, we should recognize that any answer to this question is going to be model-dependent. The Big Bang is the only ...
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If I'm understanding the question correctly, it's referring to a universe that (1) has a spatial topology that wraps around, and (2) has cosmological conditions such that a timelike curve can circumnavigate the universe (in the sense of reuniting with a geodesic that has been at rest relative to the CMB). I assume that "looped" doesn't refer to closed ...
3
The gravitational field can indeed be assigned an energy. Unfortunately though whereas for, say, the EM field you can define an energy density at a point ($\bf{E}^2+\bf{B}^2$), for the gravitational field you can't do this. - Whichever way you define the energy in terms of the Christoffel symbols, you run into the problem that you can make them, and hence ...
3
Spacetime (probably) does indeed have at least one boundary. Crazy Buddy mentioned three related questions in his comment, and reading these will help you understand why spacetime has a boundary in the past i.e. the Big Bang. This is a singularity and it is a boundary because you cannot follow geodesics back through it to earlier times.
If the universe were ...
1
I'm not sure what OP exactly is requesting, but OP's equation follows e.g. from the general fact that for an arbitrary 2D surface, the Ricci tensor
$$ R_{\mu\nu} ~\propto~g_{\mu\nu} $$
is always proportional to the metric tensor $g_{\mu\nu}$. This is basically a consequence of that in 2D the Riemann curvature tensor is complete determined by the scalar ...
0
There are many different types of orbits according to General Relativity - more that that according to Newtonian mechanics. For example:
http://arxiv.org/abs/1207.7041 Characterization of all possible orbits in the Schwarzschild metric revisited
(skip right to the figures in the end of the article). Elliptical path is one particular case of these.
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General Relativity deals with curvature of Spacetime, not just curvature of Space. You can't ignore time because clocks are affected throughout the universe. Spacetime events are what we measure and are independent of observers.
Now, let's come to point: You're asking why we're able to measure effects of Spacetime curvature with classical way when reference ...
6
I don't understand why we are able to see and measure curvature /
warping of space at all.
The Earth's surface is curved and this can be observed via the vast number of pictures of the Earth from space that now exist.
However, the surface curvature can also be "seen" via measurements on the surface itself.
For example, if one were start at the North ...
5
Curvature affects how objects in the universe move and interact with one another, and these effects can be measured.
Take, for example, the phenomenon of gravitational lensing. Because spacetime curvature can deflect the path of light, we can potentially observe light coming from objects that are directly behind other objects. Here's a nice picture.
As ...
1
I like previous answer but:
1) I believe that in the provided formula the mass of the electron should have a power of one (not two)
2) It is valid for electrons only, because it uses their Compton wavelength.
By the way, there is such a thing as "Caianiello’s maximal acceleration". In his 1985 paper Caianiello demonstrated the existence of a maximal ...
2
The Kepler problem: the motion of a probe mass around a massive (spherical non-moving) body.
Newtonial mechanics gives Keplerian orbits (ellipses, parabolas and hyperbolas).
General Relativity modifies these orbits (some acquire perihelion precession, some change the period, and some become infalling spirals).
And Quantum Mechanics states that a position ...
1
There is a relativistic effect, but it's very tiny at that speed (0.24c). In fact, the effect is even smaller, since only the radially most distant particles from the rotation center are traveling with 0.24c. (The speed decreases with 1/r)
See for example this nice plot of relative mass vs. velocity, taken from gutenberg.org
P.S.: I calculated the mass ...
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I'm not very keen on GR, but there are lots of problems that can be solved both in QM and NP; there are, for example, the free particle, the harmonic oscillator, the box potential and the infinite potential well. Solutions are very different; in fact, they are described by totally different mathematical tools (points on a phase space in Classical mechanics, ...
3
You are talking about relativity and gravity together so the question can only be answered in the context of general relativity, but concepts like gravitational potential energy and gravitational force acting over a distance are Newtonian and do not really carry over to general relativity.
However, the gravitational field does contribute to total energy and ...
3
Depends on what you're doing. General relativity handles it for you, in the sense that the Einstein field equation links geometry to the non-gravitational stress-energy tensor. That general relativity is non-linear can be interpreted in part as gravity itself contributing to gravity, but it's generally not even possible to localize gravitational energy in a ...
4
You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then ...
1
The Schwarzschild metric is, in $-+++$ sign convention and units of $c = 1$ is
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1-\frac{2M}{r}} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\text{.}$$
We can index the coordinates arbitrarily, but let's take them in the typical order: $(U^0,U^1,U^2,U^3) ...
4
A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise.
What you probably find surprising is that $\delta g_{\mu\nu}$ and $\delta g^{\rho\sigma}$ are not related to each other by simply raising the indices $\mu,\nu$ or lowering the indices $\rho,\sigma$. Indeed, they're not related in this way. In this case, ...
1
Young neutron stars and the winds they energize, lay cause to some of the most extreme
physical environments in the universe. The exact plasma and wind production
mechanism are not well understood, but the basic picture is as follows.
At the stellar surface, the pulsar’s huge magnetic fields and rapid rotation induce
enormous electric fields within the ...
1
For the answer by PMay:
However, both the observer at the center axis and the observer at the perimeter would agree that the circumferance of the perimeter is $2\pi R$.
That is not true for the observer at the perimeter. He/she is moving with an acceleration, and from his/her point of view, space would be distorted, gravitational force, time dilation ...
0
Researching the Sagnac effect led me to Born coordinates for analysis if rigid rotation. I am still working through it, but it got me thinking about an intuitive way to understand the problem. Imagine the x' axis of the rotating frame of reference as being wrapped around the perimeter of the space station. The observer at the center axis of the space ...
3
Gravity is not special at all. It seemed to be special at dawn of the 20th century but now the picture is different.
Fields are more than just forces. Fields can have their intrinsic dynamics, solitons, topological features, nontrivial vacuum.
As of force aspect, electromagnetic field makes a 4-force $qF^{\mu\nu}u_{\nu}$, and gravitational field makes a ...
1
The answer is the energy goes into the gravitational field.
If you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is
$E = Mc^2 + \frac{P}{a} - \frac{3a}{\kappa} (\frac{da}{dt})^2 = 0$
$M$ is the fixed mass of cold matter in the volume, ...
2
These are not two different effects. They are the same effect as viewed in two different frames of reference. They shouldn't be added. If they were both calculated correctly, they'd be equal to each other.
They are not equal to each other, and that's because the calculation in the rotating frame is effectively assuming the existence of a gravitational ...
0
When a particle is deflected by gravity the gravitational field will also be modified by the particle. To form a conservation law for momentum you need to take into account the momentum in the gravitational field as well as the particle. This can be done e.g. using pseudo-tensor methods.
This works but remember that momentum is a relative concept. Even in ...
0
The short answer is "yes". The energy lost from the photons is taken up by the energy in the gravitational field. Of course energy is a relative concept but if you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is
$E = Mc^2 + \frac{P}{a} - ...
0
The only thing that prevents us defining a total conserved energy for the entire universe is that if the universe is infinite then the total energy could be infinite or indeterminate.
The statements that say energy is not conserved in general relativity are wrong, irrespective of who says them. You can define energy over any finite volume of space and you ...
0
The problem with this question is that gravitational potential energy between massive objects is a Newtonian concept but the question of energy conservation in cosmology can only be discussed properly in terms in general relativity.
The general answer is that energy is always conserved if you take into account the energy in the gravitational field as well ...
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