# Tag Info

63

No, it's not. The radiation field in the interior of the Sun is very close to a blackbody spectrum. If you look in any particular direction the brightness (power per unit area) you see is $\sigma T^4$, where $\sigma$ is Stefan's constant. Even at any particular wavelength it is always the case that a blackbody of higher temperature is brighter than a ...

37

It depends on what object it's acting on. There are many objects, including stars, that have magnetic fields where Lorentz forces on charged particles like electrons and protons are stronger than the gravitational force on them. Also remember that the strength of the Lorentz force depends on the speed of the particle moving through it, so a fast enough ...

33

Human color vision is based on three types of "cones" in the eye that respond differently to different wavelengths of light. Thus, not counting overall brightness, the human color space has two degrees of freedom. In contrast, the spectra of stars are very close to a black body, which depends only on effective temperature. As one varies the temperature, the ...

22

Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less? Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction. Why does this not speed up, i....

17

Let's look at the proper magnetic force (as opposed to the Lorentz force on a moving, charged object described in @KenG's answer) on a specimen $S$ of magnetized material with mass $M_S$ as a way to try to compare. Let's arbitrarily assume it has a fixed, permanent magnetic moment $m_S$. We can't use iron because it will saturate too easily. Then let's ...

16

The total entropy actually increases, as the molecular cloud shrinks under gravity. It may seem that as the molecules are getting closer, they are more ordered, which means less entropy. That is however only one part of the process. The second (important) part is: when the molecules are closer, they also have higher kinetic energy (since they descended into ...

15

Coming from a different direction as @Rob's, Opacity and Thermal Radiation are orthogonal properties of a material. The photon flux at the interior of the sun is very high, so it is definitely not dark. However, it is opaque to virtually all light outside the sun. To provide an analogy, if you are in a sealed room with no windows, you cannot see anything ...

14

No. There is no consensus. The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows. If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of ...

14

Short answer: Without tunnelling, stars like the Sun would never reach nuclear fusion temperatures; stars less massive than around $5M_{\odot}$ would become "hydrogen white dwarfs" supported by electron degeneracy pressure. More massive objects would contract to around a tenth of a solar radius and commence nuclear fusion. They would be hotter than "normal" ...

14

Starting from a protostar, one would hope to be able to predict everything about its future development if we knew its initial mass, chemical composition and angular momentum. Mass is fundamental because it determines how much fuel the star will have and the pressure at its core. Composition is key because among other things it determines the opacity of the ...

13

No, the fusion rate of the Sun is not absolutely constant in time. The Sun is gradually becoming more luminous and that luminosity is provided for almost exclusively by fusion in the core. However, the rate of increase is not large, of order 10% per billion years. The fusion process is extremely slow (and inefficient in terms of energy release per unit ...

10

$$\frac{dP}{dr} = - \rho g,$$ is the equation of hydrostatic equilibrium, where $\rho$ and $g$ are the local density and gravity, $P$ is pressure and $r$ is the radial coordinate. This can be rewritten as $$\frac{d\rho}{dr} \frac{dP}{d\rho} = -\rho g.$$ Since $\rho$ and $g$ are positive numbers, the pressure gradient is negative. For all types of matter $P = ... 9 It isn't impossible, but the short answer is "no". A gravitational field will accelerate all matter and energy equally while a magnetic field will only accelerate moving electric charges (other magnets). The force due to gravity is proportional to the inverse square of the distance, and the force due to magnetism asymptotically approaches the inverse cube ... 8 Let me try to add some numbers to Steve's answer. The Sun's luminosity is about$L_{\odot}=4\times10^{26}\text{ J/s}$. Now, if we assume that the majority of that energy comes from nuclear fusion, we have $$L_{\odot}=\frac{dE}{dt}=\frac{d(mc^2)}{dt}=c^2\frac{dm}{dt}=c^2\dot{M}_{\text{nuc}}$$ Therefore, we can write the mass-loss due to nuclear fusion as$\...

7

The slowest reaction rate in the pp chain determines how quickly hydrogen can "burn" in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay. The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of ...

7

There is no specific answer to this -- anything from "just sits there" to flys away at high speed is possible. It all depends on the symmetry of the supernova (SN) explosion. Extensive modelling shows that the explosions can be quite asymmetrical, and if they are the gravitational waves created can give the new black hole (BH) quite a kick. If the ...

7

Neutron stars are one of the possible end products of the evolution of stars greater than around 8 solar masses. If you start out with a close binary pair of these fairly massive stars -- not common, but not rare, either -- the more massive star will evolve to a red giant and tides (or even friction) in the extended envelope will pull them closer together....

7

The Earth is a moving (actually, accelerating) platform from which we make our observations. If you want to describe the motion of a distant celestial body, then it does not make much sense to provide a geocentric velocity, because this will depend on exactly when the observations were taken (because the Earth orbits with a speed of about 30 km/s, but the ...

7

So if I read your question correctly, you're asking why as stars get dimmer, they are given higher magnitudes? The reason is purely historical. The ancient greeks assigned stars with 6 brightness levels or magnitudes. The brightest stars were of first magnitude and the least bright stars (to the naked eye) were of 6th magnitude. In the 1800's this system ...

7

Rob Jeffries has covered pretty much everything in his answer, but I'll add that this is a question with a long history, enough that there is a famous answer to it called the Russell-Vogt or Vogt-Russell Theorem. That states that composition and mass are the two key properties, assuming that the star is in hydrostatic equilibrium and derives its energy from ...

6

This is basic thermodynamics. When you compress a gas, you inject energy into it. Think of the pump you use to inflate the tires on your bike. It takes some force to move the piston, right? That effort is not wasted, but goes directly into the air in the pump. Now the air has more energy. But what happens to a gas when you put energy into it? It's ...

6

What is it? An IMF, $\Phi(m)$, is defined such as $\Phi(m){\rm d}m$ gives the fractions of stars with a mass between $m - {\rm d}m/2$ and $m + {\rm d}m/2$, and with a normalized distribution $$\int_{m_{\rm min}}^{m_{\rm max}}m\Phi(m){\rm d}m = 1\ M_{\odot}.$$ Note that these boundaries ($m_{\rm min}$ and $m_{\rm max}$) are ill-defined, but typically of the ...

6

Plasma consists of ionised particles. The electrons are not bound to their nuclei and are free to move around within the star. However, they are not free to leave the star. Without the electrons to maintain a neutral charge, there would be massive electrical repulsion from all the positive protons and helium nuclei, the result would probably be something ...

6

I think that there isn't a strict answer to this question. However, I believe the answer is that there's a difference between the core of a hydrogen-burning star and the core of a protostar or star-forming, gas cloud. For a hydrogen-burning star, the core, as you say, is the region of the star where fusion is taking place. This is surrounded by the ...

6

The energy derives from gravitational potential energy. The core of a bit more than a solar mass collapses from the size of the Earth to a 10km radius. Some of the gravitational energy (a tiny percentage) released is transferred to the overlying envelope and blasts it into space. Further energisation takes place due to radioactive decay. A ball of neutrons ...

6

We use the position of the sun (or more accurately the centre of mass of the solar system) as this gives us a very nearly inertial frame of reference. An inertial frame is one which is not changing its velocity. The surface of the Earth does not define an inertial frame. The Earth orbits the sun and so it is moving in opposite directions in summer and ...

6

In general, you can't. If obtaining spectra in regions where there is expected to be a spatially varying background then you either need to do long-slit spectroscopy so that you have a good measurement of the ISM contribution either side of your source, or you do integral field spectroscopy with the same idea. The problem is that the line strengths for the ...

6

TL;DR: the main sequence lifespan of the sun can be increased by a factor of 12.2. Perhaps the most complete astrophysical analysis of stellar engineering for extending Earth habitability is Martin Beech's book Rejuvenating the Sun and Avoiding Other Global Catastrophes (2008). In order to maintain the biosphere the sun's interior need to be mixed (in ...

6

The role of H$_2$ is to allow primordial gas to cool down sufficiently to allow the collapse to start and then to hold the gas as a relatively low temperature as it gets much more dense. The formation of H$_2$ is essential because atomic hydrogen simply has no way of cooling itself below temperatures that are capable of exciting the $n=2$ level and then ...

5

The wording provides a contrast with the Hayashi track phase that immediately precedes it, where the luminosity decreases by orders of magnitude with little change in effective temperature. You are correct though, the Henyey track for low-mass stars is not horizontal in the HR diagram, however it is a critical point that the luminosity does not change very ...

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