Tag Info

New answers tagged

0

Carlos, I’m going to answer your question from a completely different perspective than current physics. I have been working on a New Way of looking at physics from Inner Space through Outer Space for the past 7 years. Your question relates to Outer Space. I believe the COSMOS is a closed system and cycles along a Cosmological Entropy Wave. All Matter – ...


1

The luminiferous aether was postulated to explain the propagation of electromagnetic waves. The nineteenth century physics knew wave equations in a medium and could not think that a medium was not necessary to propagate the electromagnetic ones. That was the function of the luminiferous aether. The Michelson–Morley experiment was performed in 1887 by ...


0

There is the need of a new Einstein to say clearly that the speed of light c is the same as to have a aether. Because in the 20th last century it was not nice to talk about aether it comes out of use. But gravitation is some kind of aether and gravitation does influent light.


3

This is too long for a comment so I'll post it as an answer, even though this question is years old. If Alcubierre warp bubbles are physically possible, which is exceedingly unlikely, and if the equivalence principle is correct, you could definitely escape from a black hole in one, because there's nothing locally special about the event horizon. In a large ...


-2

The drive works by warping normal space, creating a bubble that kind of surfs through space time. I don't know what would determine the speed such a ship could achieve so not sure if a natural law would limit the ability to travel beyond visible space. Black holes exist because their extreme mass has warped space beyond to point where light can escape. It ...


3

The point is that during the ordinary phase of the Big Bang expansion, the difference $|\Omega-1|$ was rather dramatically increasing with time. Today, we know that $|\Omega-1|\lt 0.01$ or so. If we use the cosmological equations to reconstruct what $|\Omega-1|$ had to be when the Universe was a second old, or very young, we find out that the following ...


2

The special state of motion you're talking about is often called the Hubble flow. (edit: oops, Ben Crowell already mentioned this.) I think that in modern slow-roll inflation the source of this asymmetry is an asymmetry in the tiny (Planck-scale?) seed of inflation, whatever it was, inflated by a factor of $e^{60}$ or more. The inflaton potential has to be ...


0

The majority of the sources that LIGO (and other gravitational wave detectors) are aiming for are astrophysical (e.g. neutron stars, black holes, supernovae, pulsars). The expected cosmological gravitational radiation from standard inflationary models (see this recent article from P. Steinhardt) would be very weak in the LIGO band (10-1000 Hz). There are a ...


0

We can't peek deeper than a certain distance away from our current cosmic position, but we know for sure that the universe extends far beyond that cosmic horizon. In fact, for all we know, although our observable universe is finite, the full universe is infinite in size. If this is indeed true (and there is no single piece of evidence against an infinite ...


-1

There is no edge if you come along with the infinite concept. Update: Harvard complements my answer: http://www.cfa.harvard.edu/seuforum/faq.htm#s1 Galaxies extend as far as we can detect... with no sign of diminishing.There is no evidence that the universe has an edge. The part of the universe we can observe from Earth is filled more or less uniformly ...


1

The structure of general relativity does not allow the theory itself to be classified in terms of global, discrete symmetries such as time-reversal. The Einstein field equations don't refer to a time coordinate; they're expressed tensorially, which means that they are completely independent of what coordinates you choose. Since there is no guarantee that you ...


4

Nice question. First off, there's a definitional problem because we can't apply a Lorentz boost to the universe as a whole. Lorentz symmetry is a local thing. So when we talk about "Lorentz symmetry" for the universe as a whole, I think we have to keep in mind that we mean something a little different. Basically if the "Lorentz symmetry" has already been ...


8

The cosmic microwave background has a redshift of about $1100$, see here for instance. Keep in mind that the "surface of last scattering" that gives rise to the CMB in fact existed everywhere in space, it's just that the photons currently reaching us have $z\sim1100$.


0

Every single comment and answer was very useful. I think i have found an answer in an old paper by Clark Glymour (Minnesota studies in philosophy of science, volume III, pp.50-60): "It has recently been noted (Ellis, 1971; Dautcourt, 1971; Ellis and Sciama, 1972; Glymour, 1972; Trautman, 1965) that in some general relativistic cosmologies various global ...


0

As far as I have understood from this paper, they have given some observational limits to the value of $\Omega_{k,0}$, but this article concludes asserting that "there is no evidence from Planck for any departure from a spatially flat geometry". Taking $\Omega_{k,0}=0$ and the value for $\Omega_{r,0}$ given at this post, one can compute the above integral ...


2

As has been discussed in many questions around here (e.g. here), relativity tells us only about local properties and behavior of a space-time. There are some exceptions when we make global assumptions - if we have a space of globally and strictly constant positive curvature, non-trivial topology is imminent because the space has to be the 3-sphere ...


9

I'm not going to provide a full answer here, because I don't know the answer, but I want to give some statements that illustrate quite nicely the kind of problems one would face when determining topology of anything: We know spacetime is a manifold. That means, locally, it looks just like $\mathbb{R}^4$. That's already a bummer. We can't do jack at one ...


3

Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...


3

The spacetime outside a spinning mass is described by the Kerr metric. To explain how the Kerr metric produces frame dragging is hard, because it's not something for which there's an easy intuitive model. Frame dragging arises because the spacetime geometry links the angle measured around the spinning object to time, and this means the angle changes with ...


1

John Rennie's answer is good already, but I want to add a single point: These fluctuations are very very short. In quantum mechanics you've got Heisenbergs uncertainty principle, which is often stated as $$ \Delta x \cdot \Delta p \le \frac \hbar 2 $$ and which means, that for any quantum object (think of an electron or a positron created in such a vacuum ...


8

I think the key conceptual hurdle is that the vacuum state is not nothing. Quantum field theory describes matter as excitations in quantum fields. These quantum fields are very strange things, and I don't know of any easy way to explain to a non-physicist what a quantum field is. The key thing is that the quantum fields fill all of spacetime. So a vacuum is ...


1

You can get the Planck data from the LAMBDA Data products pages on the NASA web site.


8

Measuring $w$ is actually what I do for a living. The current best measurements put $w$ at $-1$ but with an uncertainty of $5\%$, so there's a little room for $w \ne -1$ models, but it's not big and getting smaller all the time. Indeed, we'd all be thrilled if, as measurements got more precise, $w \ne -1$ turns out to be the case because the $\Lambda$CDM ...


1

I like this video by MinutePhysics on this topic. It can clarify things as a primer. When you state that the universe expanded at a speed higher than the speed of light, you have to stop and ask what is actually meant by such a statement. What is moving with respect to what? In standard cosmology, we describe the universe expansion by the Hubble rate ...


0

Let us clearly draw the line between two things here, since the question can easily involve opinion based answers, which may also be dubbed non-mainstream (which isn't welcome on this site). 1) The existence of dark matter is generally believed by a majority of the Physics community, since astronomical observations, notably by the Planck space ...


6

Yes, there have been suggestions that such particles exist, and an example is the sterile neutrino. But your question is a little more involved than you might think at first sight. For example if the sterile neutrino only interacts through gravity what interaction caused it to be created in the first place? There is nothing in the Standard Model that could ...


2

This answer is within the current physics and theoretical understanding, which has developed a successful formalism that includes all the experimentally seen particles in the Standard Model. The model has been very successful in predicting several new particles using its symmetry and mathematics, the experimental observation of the Higgs boson serving as ...


6

Is it because the acceleration is too weak? It is too weak with respect to the four forces we measure. The fact that the four known forces are so much stronger means that agglomerates of particles, up to the scale of galaxies are not internally affected, they keep their structure intact, like the famous raisins in the rising bread. It is only at the ...


0

In "The Scalar-Tensor Theory of Gravitation", of Yasunori Fujii and Kei-ichi Maeda you can find explicitly the solution, in Appendix C (pag. 195). Personally, I really didn't like this book and even this demonstration it's very difficult to follow. So I did it in another way. Use the usual theory for the GR part, and isolate this term: $\int d^4 x ...


2

Your question can't be answered because the qualifier when it was only the size of our solar system is meaningless. The size of the universe is a rather vague concept. The universe may well be infinite (it's unlikely we'll ever know for sure) in which case it was always infinite and it doesn't have a size. You could ask about the size of the observable ...


0

One of the most troublesome of the anomalies with which the Big Bang is marred, is that it is not possible to physically describe the Big Bang process in an infinite universe. The reason for this impossibility is that it is equally impossible to speak of the concept of concurrency everywhere in such a universe. Yes, in such a universe even the concept of ...


3

There aren't E and B fields in the entire universe. For example, there are no electric fields inside a conductor. I'm sure there are quite a few other such examples. If you mean "why are there electromagnetic waves throughout the entire universe?", one answer is because the radiation field drops like $1/r$, so the field from a single source never ...


0

Dark energy due to a cosmological constant does not get diluted by metric expansion of space. However, this does not violate energy conservation as the increase in energy will be cancelled by gravitational potential energy. The problem with general relativity is (some would say arguably) not energy conservation, but energy localization: In a rotating frame ...


2

The cosmological constant is a constant energy density per unit volume of space, so as the universe expands this does indeed create energy as it creates new space. In this sense conservation of energy is violated. Actually this is less surprising than you might think. Conservation of energy is linked to a symmery called time shift symmetry by Noether's ...


1

Actually, energy is often not Conserved in general Relativity. For are more in deep explanation see: http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html But just notice that Dark Energy might not necessarily end up being the cosmological constant, but a new force field, so its behavior might differ from that of an actual cosmological ...


1

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$ Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$ If we observe blue light at 400 nm, we get a corresponding redshift ...


6

Actually the "last scattering surface" of the CMB corresponds to the transition of the interstellar/intergalactic medium from an ionized plasma to cooler neutral atoms, about 300 000 years after the big bang. Most atoms have excitation and ionization energies in the visible, so the CMB was probably visible when it formed. We can be a little more precise ...


1

Just to add what the other questions say, if the sizes of the atoms were changing, there would have to be some corresponding change in at least one of the fundamental constants. For instance, if the size of the Hydrogen atom changed, then the ground state of the hydrogen atom would no longer be governed by the Bohr radius: $$a_{0} = ...


4

You're actually pretty close to the correct method for estimating the age of the Universe, but $H$ is not constant with time, it is $H=H(t)$. One of the many ways of writing the equation to solve is: $$t(z) = \frac{1}{H_0}\int_z^\infty \frac{dz}{(1+z)E(z)}$$ Here $z$ is redshift; $z\rightarrow\infty$ at the Big Bang, and $z=0$ now, so if you integrate ...


1

The strictly true equation for $\epsilon$ is $\epsilon=\frac{|\dot H|}{H^2}$, which is never negative. This is often written as $\epsilon=-\frac{\dot H}{H^2}$ because $\dot H$ is usually a very small and negative number. As for $\eta$, if $\dot\epsilon$ is negative, $\eta$ will be as well. However, a negative $\dot\epsilon$ would tend to indicate that ...


3

The centre of mass of a system is simply the weighted average position of the mass distribution in that system. Since the universe is thought to be homogeneous and isotropic, any observer should roughly observe themselves as being at the centre of mass for their observable universe. However, I do not think that is quite the answer you were looking for. From ...


1

As I understand it, the solar system evolved from a massive molecular cloud. To me, this seems to break the second law of thermodynamics, as I think it suggests order from disorder. There are two problems here. One is the concept of entropy as disorder. A number of thermodynamics texts have now discarded this old concept. For one thing, it doesn't help ...


1

Other answers have made clear the 'flat' only implies infinite given additional assumptions around the topology. In short: A universe which is the same everywhere but not simply connected can be finite. It's worth mentioning that whilst the main working model assumes that the universe is simply connected, the actual topology is an open and serious ...


11

This claim is simply wrong. The flat hyperplane is of course infinite, but non trivial topologies can be flat and still finite. The simplest example is the 3-torus, but there are even the Klein bottle and the Hantzsche-Wendt manifold. See for example page 27 of Janna Levin - Topology and the Cosmic Microwave Background, which show you ten different closed ...


26

We need to be precise about the phrase the size of the universe. Specifically I'm going to take it to mean the maximum possible separation between any two points. In an infinite universe two points can be separated by an arbitrarily large distance, so if the maximum distance between two points is finite this means the universe must not be infinite. The ...


3

I think that it is important to note that (almost) everyone doing cosmology works within the framework of the FLRW universe. This implies that we assume that the universe is spatially homogeneous and isotropic, i.e. 'every place is the same (at least on large scales)'. Now, think of a flat, finite universe: Is it possible to maintain that all places are ...


3

In Minkowski spacetime the one way light travel time to a galaxy at proper distance $\chi$ is just: $$ t = \frac{\chi}{c} $$ so: $$ \chi = ct $$ As you say. However in an FRW universe the travel time is given by a different equation so the proper distance is not simply $ct$. Let's assume all motion is in the $x$ direction, so the metric simplifies to: ...


0

I give a general derivation even if it is an-isotropic. As it is your homework, I wont give all the detial derivations. Notation: $\eta_{\mu\nu}$ is mostly positive, let $K=g_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi $ $$ S_{\phi} =\int d^4x \sqrt{-g}{\cal L}(\phi,K) \\ T_{\phi}^{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta{\cal L}}{\delta g_{\mu\nu}} = ...


3

Forgive me in advance, this may get overcomplicated. I am going to give you the facts as scaled down as I can but still sufficiently detailed. I think providing you with what we have and allowing you to infer from it is the best way to avoid misrepresenting the answer. Here is the General Relativity equation that describes how gravity interacts with ...


4

I am going to answer the following questions: 1) Does dark energy have gravity? 2) Why does dark energy have constant density during the expansion of the universe? First about the term of dark energy - it is more of a label for many attempts of a solution to the problem of an accelerated expansion of the universe. One of the most promising (or say ...



Top 50 recent answers are included