New answers tagged

-1

Set up a strong temporary field and then hit the planet a couple of times with a big rock to magnetize the core.


0

Hopefully an observer can chime in with better advice... but: Generally, people calculate the (average) brightness in rings around the center, then plot that brightness as a function of the projected radius (i.e. angle, e.g. in arcseconds). You can check to make sure your solid angle is correct (for example you say $\delta$ is the complementary angle to ...


0

The scale-factor of the universe changes significantly over that period of time, so you can't calculate distance as simply $d = v \cdot t$. You have to actually integrate over the expanding spacetime metric, i.e. $$s = \int_{z_1}^{z_2} \frac{c_s}{H(z)} dz$$ Where $c_s$ is the speed of sound (and I think that should be $c/\sqrt{3}$ instead of $c/2$, ...


6

We want the Newtonian limit of the Einstein Field equations for nonzero vacuum energy(=cosmological constant). As $\rho_\mathrm{vac}=\Lambda/4\pi G$ is a mass(=energy) density, Poisson equation is $$ \Delta\Phi=4\pi G\rho(\boldsymbol r)-\Lambda \tag{1} $$ If we assume spherical symmetry, and point-like source $\rho\sim\delta(\boldsymbol r)$, the ...


3

You get an extra term that increases with r: $$a = -\frac{G\cdot M}{r^2} + j\cdot r$$ with j as the repulsive component.


8

Background Stars are composed of plasmas, which are an ionized gas that exhibit a collective behavior much like a fluid. There are two important aspects of plasmas to keep in mind. The first is that they act like very highly conductive metals in that the electrons can move very freely in order to cancel out any charge imbalance. The consequence is that ...


9

Overall, a star stays more or less neutral. This is true for all stellar objects beside black holes. I am using a simple calculation that can be found in a footnote of https://arxiv.org/abs/1001.3294 on p. 11 chap. 2. Suppose the star has an overall charge of Z times the elementary charge, $Ze$, and we consider the Coulomb repulsion of a test particle, say ...


0

In the case of the second order acceleration, two clouds are approaching, therefore the energy the charged particle gains comes from the energy of the clouds. In the case of the first order acceleration, the charged particle gains energy as it moves repeatedly through the shock front. The region before the shock front (upstream) moves at higher speed than ...


1

Note You should clarify your statement from "...a charged particle cannot gain energy from a magnetic field..." to "...a charged particle cannot gain energy from a static magnetic field..." There is nothing wrong with energy transfer from time-varying magnetic fields. Background If the spatial gradient in the magnetic field is slow enough such that the ...


8

The Big Bang Theory is a much more general and less specific description of our theory about the origin of the Universe than the $\Lambda{\rm CDM}$ model (by the way, I don't think that the hyphen is written in that acronym). The Big Bang Theory says that the Universe was expanding and the distances between two places where galaxies sit today used to be ...


2

The important point here is that there is no thermodynamic limit for gravitating systems, and thus there is no well-defined temperature. This is, perhaps, not a completely intuitive result, but it comes from work on the stability of matter. This is not as glamorous as it sounds, but revolves around the need to show that the energy of matter is an extensive ...


1

Mike's answer is good but there Is more. You can actually see it in the papaer he referred you to, it is an excellent paper. It not only describes the possible sources, but it points to the new physics that it might see. eLisa, also called NGO, will be sensitive to gravitational waves from the early universe (and after), down to 10 to the minus 18 sec after ...


6

A typical velocity dispersion in a globular cluster is 10 km/s. For a typical 1 solar mass subgiant in an old globular, then equating the kinetic energy to $3kT/2$, we get $T = 5\times 10^{60}$ K. Doesn't seem that helpful really... The concept of temperature is only ever applied in a relative sense - i.e. some component is hotter than another. Can't say ...


2

Temperature is not useful concept for describing clusters of stars or other gravitational systems, because such systems are not in the realm described by thermodynamics. There is no way to set up thermodynamic equilibrium - globular clusters partly evaporate and core implodes. Also the velocity distribution can't be Maxwell-Boltzmannian, because very fast ...


4

Space is, as you say, good for removing a lot of the background noise that spoils LIGO's data — like seismic noise, disturbances from traffic and logging activity, people shooting at the beam tubes, etc. But another important reason to go into space is so that you can basically make a much larger version of LIGO. LIGO's arms are 4km long; eLISA's arms will ...


1

Firstly, you must know that there are many models for inflation which give different results to your a) and b) questions, and we still don't know which is the right one. I'll try to answer regarding the most accepted and simple models. a) During the period of inflation the distance between two separated points in the Universe increased at least ...


-2

Utilising Black Holes as a potential energy source Have I got a surprise for you! I'm aware of the Penrose process and the basic physics behind that. Be wary of Penrose. He has a habit of appealing to Einstein's authority then flatly contradicting the guy. And then he'll tell you about the parallel antiverse and other fairy tales: Also, I ...


2

Great question. Black holes are some of the brightest objects in the universe. While we think they require the Blandford-Znajek (BZ) mechanisms to produce things like Relativistic Jets, the bulk of the light (emission) they produce is just the efficient thermalization of gravitational energy when material falls into (`accretes' onto) them. The simplest way ...


2

The vast majority of the star like objects we see in the sky are stars in our own galaxy. Assuming the accelerated expansion is due to a cosmological constant, and assuming the value of the cosmological constant does change (it's currently of order $10^{-52}\,\text{m}^2$) the expansion will never be strong enough to disrupt the Milky Way. So our night sky is ...


8

There are a few big differences: Tidal effects change the rate at which the orbit decays The neutron stars touch before the black holes would have merged Ejected matter can contribute to the gravitational-wave signal The merged neutron star can have "mountains" that keep radiating Probably the most important effect: the matter in NSs can emit photons and ...


6

There are several things going on and you seem to have a good grasp of the factors involved. White dwarfs are produced from progenitors with main sequence masses between around 8 solar masses (any more massive and it leads to a supernova and a neutron star) at the top end and about 1 solar mass at the bottom end. This lower limit is nothing fundamental, it ...


1

The fact that the gyroradius is small compared to the Galaxy size leads to a multitude of collisions between the CR and the galactic magnetic field (compare ~pc CR vs ~kpc CR gyroradii). Each collision helps diffuse the particle, disassociating it from its original direction (i.e., makes isotropic).


4

The galactic magnetic field is fairly irregular on distance scales that are small compared to the size of the galaxy (although there does appear to be structure to the magnetic field associated with the spiral arms). In a uniform magnetic field, a charged particle would follow a nice spiral trajectory. In an uneven and varying magnetic field, charged ...


3

A massive object (such as a galaxy) along the line of sight to a distant bright source (such as a quasar) bends the light along its path. If the "lensing" object is massive enough and the geometry is right, the background object can be seen as multiple sources. For instance, here is a galaxy (central point) and four images of a single quasar: For a ...


3

This is an active (hot?) topic of research, in fact I attended a workshop on the subject just last week. In brief, no one has found a dark matter (DM) halo yet that does not host a galaxy, though we would very very much like to! The first reason it's so difficult to find a DM halo that does not have a galaxy is that a common working definition of a galaxy ...


18

Interstellar space is an excellent vacuum, but it's not a perfect vacuum. For example Earth is constantly bombarded with protons from the solar wind, which stream outward uninterrupted until the heliopause when matter from other stars becomes more dominant. If there were, say, an antimatter star nearby, the place where its stellar wind of antiparticles met ...


0

Without having any experimental data at hand, I guess that most planets formed from a uniformly rotating dust disc, and thus their rotational and orbital momentum have the same sign. However, upon random tangential impacts, some of them (Venus, Uranus..) could change their original axis of rotation, and most probably it happened so early we will not find ...


3

The correct formula is actually $$M = \frac{4\pi^2 a^3}{GP^2}$$ and is a form of Kepler's third law. $M$ in this formula is the central mass which must be much larger than the mass of the orbiting body in order to apply the law. In reality the formula that should be used is $$M_1 + M_2 = \frac{4\pi^2 a^3}{GP^2},$$ where $M_1+ M_2$ is the sum of the masses ...


0

Your intuition that that this equation actually gives the mass of the body being orbited is exactly right. In general, there is no way to infer the mass of a body using any measurements of its response to gravitational forces, because in its equation of motion, $$m\mathbf a=m\mathbf g,$$ where $\mathbf a$ is the body's acceleration and $\mathbf g$ is ...


0

If black holes can swallow any object, Black holes have a very strong gravitational field, this field from a distance attracts any other matter and the Newtonian approximation when there is a distance is adequate. When two strong gravitational fields get close enough General Relativity equations have to be used, but the "attraction" is there and very ...



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