Stars generate their energy by fusing lighter elements into heavier elements. The most common reaction in Sun-like stars is the conversion of hydrogen to helium via the proton-proton chain, but heavier elements can also be synthesized, typically in more massive stars.

Before nuclear fusion was proposed as the source of energy for stars, there were other ideas put forth, notably the release of energy via gravitational contraction (i.e. the Kelvin-Helmholtz mechanism). Some stars release energy in this way while on the Hayashi track, but this is only for a short part of their lives.

Could star-like objects exist that produce energy and support themselves against the force of gravity by means other than nuclear fusion? I'm also interested in ways that a civilization could make one of these star-like objects.

Note, and a reminder to folks writing up new answers: This is a question. An answer needs to prove that the mechanism given will be a stable source of energy that will last on timescales similar to those of typical stars, without any catastrophic events. A helpful text can be found here.

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

• Too tired to make this a proper answer, but: Thorne–Żytkow object – Ilmari Karonen Aug 17 '15 at 23:39
• Comments are not for extended discussion; this conversation has been moved to chat. – Tim B Aug 18 '15 at 12:47
• "without any catastrophic events"? Most regular stars don't manage that either :) – Michael Borgwardt Aug 19 '15 at 9:35
• Dark stars fueled by dark matter annihilation are not excluded. – BartekChom Dec 31 '15 at 18:48
• How hot and how bright does it have to be? Jupiter radiates more heat than it receives from the sun, so is technically an energy source that's not powered by fusion, but it's only infra-red with no visible light. – Mike Scott Feb 22 '16 at 16:29

So, I'm going to alter my answer somewhat to address whether there are other means by which a star-like object may exist, besides the one we already know about. In that regard, I'm afraid my ultimate answer to this is going to be 'No' or at least 'Not with our current laws and/or understanding of physics.'

However, lets start by defining additional parameters that are necessary

Size

To be 'Star Like' we need to be about the same size as a star. Something that has a notable gravity well. In your comments, you also nixxed black holes, so those are out as they are not as stable, and to 'produce' light, would need a constant feed of matter supplied to their accretion disk. And even then, that light is produced in part by fusion.

Self-sustaining and Regulating

We need a reaction that can keep itself going, but not go too fast. Fusion is great for this...the primary requirement for Fusion is 'high density, high temperature.' Gravity creates that environment all on its own, and the Fusion reaction in a star stabilizes it. Gravity smashes all the Hydrogen together, increasing pressure and temperature until fusion begins. If the rate of fusion gets too low, the pressure from gravity speeds up the fusion. If the rate of fusion gets too high, the outward pressure of the reaction reduces the density, and ultimately slows the fusion back down. It's a reaction that is perfect for running an extremely long time...particularly since most of the star's supply of hydrogen is NOT held at fusile pressures at any given time.

Must produce enough energy to prevent Fusion

With enough temperature and pressure, fusion will begin. With massive enough elements (iron or heavier), this is a endothermic reaction which results in the death of the star. Our 'pseudo-star' needs to produce enough energy to resist gravity crushing it down and igniting fusion, or it will either turn into a star, or destroy itself as the high-mass fusion sucks all the energy out of the pseudo-star (depending on the mass of the materials it is made of). To produce that much energy, but still live on...we need an extremely high-energy reaction, comparable to Fusion, but efficient enough to maintain pace. So, really, we need something with a similar energy density to Hydrogen, when used for Fusion.

So, with these extra constraints in place, lets look at some options for producing this sort of energy.

1: Chemical Reactions: These are out. Completely. Fusion and Fission produce millions of times more energy per kg of reactant than any chemical reaction. In order to maintain a reaction that could keep up with Fusion or Fission, we'd burn through our fuel supply very quickly ref ref2 ref3

2: Fission. The reverse of fusion seems like a good idea at first, but there are a few problems. First off, Fission is a runaway reaction. You need a critical mass of the compound and then some neutrons need to be fired at the compound at a very high speed. From there on, unless controlled by external sources, Fission will spread exponentially through the fuel, and continue spreading and increasing in rate until the entire amount of fuel has been consumed. A star-sized fission reaction would be more like a supernova than a star. The second problem is that fission needs an ignition source, so it likely cannot occur in nature. ref ref2

3: Antimatter: The only reaction we are aware of that is more powerful than Fusion. But, again, anti-matter is a runaway reaction. If matter and anti-matter are mixed, they will annihilate any time the two pieces come into contact. There is only one means to regulate an antimatter reaction, and that is to intentionally keep the matter and anti-matter separate until you want them to react. This is not something that would happen in nature. ref

We do not currently know how Anti-matter interacts with gravity, but the two main possibilities are that it either obeys gravity normally, or is repulsed by gravity.

In the former case, the antimatter and matter would be pulled together by gravity and either annihilate entirely, or if enough came together at once (even by coincidence of convection) would unleash enough energy to shred our 'pseudo-star' to pieces. Even if it could pull itself back together somehow, we do NOT have a stable system here.

In the latter case, matter and anti-matter would repulse each other on a gravitational level, and without someone intentionally throwing them at each other, they would not naturally interact, and would certainly not form a star-like object

4: White Hole. A fun piece of theoretical physics that has largely been debunked. The idea was originally that black holes were holes in spacetime, and white holes were where everything they sucked up was expelled. All that light it spewed out would look an awful lot like a star. However, this only fits with the 'eternal blackhole' model for black holes, and a black hole created by gravitational collapse, and balanced by hawking radiation (read: every black hole we have ever found) do not allow for the existence of a White Hole. Furthermore, White Holes violate the Second Law of Thermodynamics, in that they actually would decrease entropy. The only remaining theory supporting White Holes is that they are actually a 'Big Bang,' not a Star-like object. ref

Conclusion

Unless I skipped some form of energetic but self-regulating reaction, it doesn't look like this is feasible, even from a purely logical standpoint. With our current understanding of the universe, it does not appear that a star-like object that fits all the necessary longevity requirements is possible.

• @Aron Not quite. The neutrons exert pressure preventing complete collapse. – PyRulez Aug 18 '15 at 2:23
• Boson degeneracy pressure dominates in a neutron star. – Aron Aug 18 '15 at 2:25
• Fission isn't always runaway, even in nature. True, it doesn't have the quality of fusion which is extremely hard to start and keep running, but there's somedecent evidence that there were natural fission reactors on Earth, running for hundreds of thousands of years (en.wikipedia.org/wiki/Natural_nuclear_fission_reactor). Of course, it would be quite a different beast to make this work as a star (this particular reactor depended on liquid water, so it needs pressure and low temperature). But the biggest problem is getting enough uranium in the first place - there's too little of it. – Luaan Aug 18 '15 at 7:42
• It's also not true that fission needs an ignition source. You just need enough fissionable material in a given volume. – trichoplax Aug 18 '15 at 13:57
• Fission goes on in every piece of fissionable material, as neutrons (from outside or natural decay of the material) hit other nuclei. Between natural decay and this natural fission, RTGs put out quite some energy, albeit only for a short time (on the stellar timescale). What you mean is a chain reaction, which is indeed tricky (but possible) to balance on the point where it is self-sustaining but not runaway. But since the question is hard science, finding that much fissionable material in one "star" is unlikely. – DevSolar Aug 20 '15 at 8:55

The closest mechanism I can find would be a pulsar star.

I'm not an astrophysicist so my explanations may not be the best, but:

1. A pulsar is a neutron star, which holds itself together without using fusion1

2. They can produce equal to/more energy2 than the sun

3. They actually do exist (it isn't science fiction)

# Notes

1

Neutron stars are very hot and are supported against further collapse by quantum degeneracy pressure due to the phenomenon described by the Pauli exclusion principle, which states that no two neutrons (or any other fermionic particles) can occupy the same place and quantum state simultaneously.

Source: Wikipedia: Neutron Star

2 A calculation (Page 118) of the Crab Pulsar states that it has an energy loss of 4.5*10^31 J/s. But it only radiates 1% (4.5*10^29 J/s) through X-rays and Gamma rays. The sun's total energy output per second is estimated to be 3.8*10^26 J/s. To put these calculations in perspective:

The tiny Crab Pulsar, which is not much more than 10 kilometers in diameter, powers the enormous energy output of the Crab Nebula, which is 10 light years in diameter. To set things in perspective in terms of relative sizes, this is as if a 1 kilometer wide volume of space were radiating strongly at various wavelengths and most of the power were being supplied by a single hydrogen atom at the center of that volume!

• Unfortunately I think what would happen here is you surround your pulsar with gas, that gas then gets compressed by the pulsar's gravity and fusion starts, at which point you're in violation of the answer. Alternatively it will get turned into more neutronium, and before long you wont' have your star anymore. Spinning the gas doesn't help because you can't keep it in a sphere shape. – Dan Smolinske Aug 17 '15 at 18:59
• The thing is, pulsars radiate in jets that can be incredibly thin. Radio-quiet neutron stars are thought to be pulsars, but their beams simply aren't oriented towards us - which shows that pulsars only radiate in a small range of directions. – HDE 226868 Aug 17 '15 at 20:28
• @HDE226868 That's true, but the question you posed was "Could star-like objects exist that produce energy and support themselves against the force of gravity by means other than nuclear fusion?". The beams are thin, but they still do produce energy. And yeah, wasn't that the point? Neutron stars aren't actual stars [any more] and fit the question. – Thatguypat Aug 17 '15 at 20:38
• Pulsars do radiate energy, as does any object that is hotter than the cosmic background radiation. The only energy they produce, though, comes from cooling and physical contraction, so in a relatively short time, cosmically speaking, they run down. – Oldcat Aug 17 '15 at 21:25
• @Aron Black holes lost the battle against gravity (or GR). Neutron stars did not, otherwise they would be black holes. – hyde Aug 18 '15 at 14:54

# Use a Quasi-star.

The solution I think will finally work is to use a quasi-star, a theoretical object from the early universe consisting of a black hole of perhaps $10M_{\odot}\text{-}100M_{\odot}$ surrounded by a gas envelope of up to $1000\text{-}10000M_{\odot}$. These objects generated energy from gravitational potential energy as matter from the inner boundary of the envelope fell into the central black hole. Fusion did not take place in the envelope, meaning that young, small quasi-stars could have appeared, to the naive observer, to be simple very massive stars.

Basically, a quasi-star is a black hole surrounded by a large cloud of gas around a black hole. It's extraordinarily massive, and looks a lot like a giant star. The big difference, though, is that a quasi-star produces energy from changes in potential energy caused by the black hole sucking in gas - there isn't any significant fusion happening. (Summary suggestion courtesy of AndyD273.)

The goal of this answer is to determine some properties of a quasi-star that could fit our specifications. Most of the answer is math, graphs, and code; the above summary is probably the most qualitative explanation I have. I’ll create an approximate polytropic model via numerical integration after determining some of the thermodynamic quantities in the object’s core. Polytropes are generally very good approximations to stars and star-like objects at most places inside them, and I’ve found that my results appear to match more detailed models.

My primary references here are Ball et al. (2011) and Fiacconi & Rossi (2016). There are some differences in equations, which I’ll point out, but it turns out that they’re actually negligible for the right parameters.

# Polytropes

I'm going to start this answer with a review of polytropes and some simple methods used to create reasonable models of quasi-stars. Fiacconi & Rossi justify the choice of a polytropic model (with $n=3$) by writing

the envelope represents the majority of the mass and volume of a quasi-star and convective regions can be described accurately by an adiabatic temperature gradient

In short, the conditions in most parts of the envelope are non-relativistic and are similar to those inside a large star. Polytropic models for stars are quite well represented using $n=3$.

A polytrope is an object that obeys the equation of state $$P=K\rho^{(n+1)/n}\tag{1}$$ where $P$ and $\rho$ are density and pressure, $K$ is a constant, and $n$ is the polytropic index. The quasi-star can be assumed to be in hydrostatic equilibrium, meaning that pressure (dominated by radiation flowing outward) balances the force of gravity: $$\frac{dP}{dr}=\frac{\rho GM}{r^2}$$ where $G$ is the gravitational constant, $M$ is the mass contained within $r$, and $r$ is the radial coordinate.

By inserting the polytropic equation of state in the equation of hydrostatic equilibrium, we eventually arrive at the Lane-Emden equation: $$\frac{1}{\xi^2}\frac{d}{d\xi}\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n\tag{2}$$ where $\theta$ is a specific function relating to the main thermodynamic variables (density, pressure, and temperature) and $\xi$ is a dimensionless radius. Analytical solutions only exist for three values of $n$: $n=0$, $n=1$, and $n=5$. Unfortunately, the case we’re interested in is for $n=3$, applicable to most main sequence stars as well as quasi-star envelopes. Therefore, we have to use numerical methods.

We can make the Lane-Emden equation easier to solve by casting it in the form of a pair of coupled differential equations: $$\frac{d\theta}{d\xi}=\phi,\quad\frac{d\phi}{d\xi}=-\theta^n-\frac{2}{\xi}\phi\tag{3}$$ The normal practice is to solve these via a Runge-Kutta method, typically of fourth order (denoted RK4). In general, RK4 is superior to most lower-order methods. However, for some cases, it’s not necessary. I’ve found that for small enough step sizes for each method, the Euler method works nearly as well, and is simpler to write - and computationally a bit cheaper. I’ll end up using it here. To hopefully convince you of this, I’ve implemented both methods in Python for an $n=3$ polytrope. I used a step size of $h=\Delta\xi=10^{-4}$, and I got excellent results:

The top graph plots $\theta_{\text{RK}4}(\xi)$ and $\theta_{\text{E}}(\xi)$ for $0\leq\xi\leq10$, where $\theta_{\text{RK}4}$ and $\theta_{\text{E}}$ are the solutions of the Lane-Emden equation using the RK4 and Euler methods, respectively. Some of the values (where $\theta<0$ are unphysical, but I’ve plotted them anyway to show the long-term behavior. The bottom graph plots $\theta_{\text{RK}4}(\xi)-\theta_{\text{E}}(\xi)$. The values for this are quite small, less than $\sim10^{-5}$ for most $\xi$.

# Key properties of quasi-stars

Most treatments of quasi-stars use slightly different forms of the Lane-Emden equation, with solutions called loaded polytropes which have cusps near the center. All have different boundary conditions than than ours. Our conditions were $$\theta(\xi_0)=1,\quad\phi(\xi_0)=0,\quad\xi_0=0\tag{Ordinary b.c.’s}$$ When modeling a quasi-star, however, we do not integrate from $\xi_0=0$, but from a radius $r_0$ related to the Bondi radius $r_B$ of the central object. In terms of unscaled distances, this is given by Fiacconi and Rossi as $$r_0=br_B=b\frac{GM_{\bullet}}{c_{s,0}^2}\tag{4a}$$ where $M_{\bullet}$ is the mass of the black hole and $c_{s,0}$ is the speed of sound in that region. Their substitution for $r_B$ appears to be smaller by a factor of four; however, this discrepancy goes away for the right choice of $b$. The authors use several other important quantities and relationships: $$\xi_0=\frac{3b}{2}\phi_0\tag{4b}$$ $$\phi_0\approx2q,\quad q\equiv M_{\bullet}/M_*\tag{4c}$$ $$\rho_0=\left[\frac{(n+1)^3}{4\pi G^3}\right]^{\frac{1}{4}}\frac{\phi_0^{1/2}P_0^{3/4}}{M_{\bullet}^{1/2}}\tag{4d}$$ where $M_*$ is the envelope mass. It should be noted that the $\phi_0$ in these equations is not quite the same as the $\phi_0$ used in the classical Lane-Emden equation; I’ll get back to this later. Ball et al. give us another relevant relationship between $\xi_0$ and $\phi_0$: $$\phi_0=\frac{1}{2n}\xi_0+\frac{2}{3}\xi_0^3\tag{4e}$$ This would seem to not be compatible with $(\text{4a})$ for most $\phi_0$ and $\xi_0$. However, it seems that this all works out. First, Fiacconi and Rossi describe $b$ as “of the order of few”. That should indicate that $1\leq b\leq10$, give or take. If we choose $b=4$, then their $\xi_0$-$\phi_0$ equations gives $\xi_0=6\phi_0$. Now, we also know that $q\simeq10^{-4}$ to $10^{-2}$. If we take $q=10^{-3}$, and use $\phi_0\approx2q$, we get $\phi_0\simeq2\times10^{-3}$. Plugging this into $(\text{4e})$ gives us $$\phi_0=\frac{1}{2n}\xi_0+\frac{2}{3}\xi_0^3=2\times10^{-3}$$ via Wolfram Alpha, or $\xi\simeq0.012=6\phi_0$. Both equations are nearly in agreement.

We want our quasi-star to be relatively tiny, as quasi-stars go, so let’s say that $M_{\bullet}=1M_{\odot}$. Since $q=10^{-3}$, that means that $M_*=100M_{\odot}$, giving us a total mass of $M_{\text{tot}}=M_{\bullet}+M_*=101M_{\odot}$. That’s reasonable - much more massive than the Sun, but still able to pass for a normal star. We should also choose a suitable central pressure - perhaps $P_0\simeq5*10^{10}\text{ erg cm}^{-3}=5\times10^9\text{ J/m}$. Plugging this into $(\text{4d})$ gives us $\rho_0\simeq5.426\times10^{-5}\text{ g cm}^{-3}$. This matches the progression of densities of the models of Ball et al. (Figure 1 and Table 1; their lowest $M_{\bullet}$ is $5M_{\odot}$, with a central density of $\sim8.71\times10^{-5}\text{ g cm}^{-3}$). Both results are much lower than the central density and pressure of the Sun.

In a polytrope, the speed of sound is given by $$c_{s,0}^2=\gamma\frac{P_0}{\rho_0}\tag{5}$$ where $\gamma=(n+1)/n$, and so in our case $\gamma=4/3$. Therefore, we find that $c_{s,0}^2=1.473\times10^11\text{ (m/s)}^2$. Plugging this into $(\text{4a})$ gives us a radius $r_0$ of $1.802\times10^{9}\text{ m}\simeq2.6R_{\odot}$. The Bondi radius is then $r_B=r_0/4\simeq0.65R_{\odot}$. This again matches up; the Ball models had $r_B\simeq1.66R_{\odot}$ for $M_{\bullet}=5M_{\odot}$. The progression seems to make sense.

# Boundary conditions

We’re now ready to integrate the Lane-Emden equation for a quasi-star. First, we set it up as a different pair of coupled differential equations: $$\frac{d\theta}{d\xi}=-\frac{1}{\xi^2}\phi,\quad\frac{d\phi}{d\xi}=\xi^2\theta^n\tag{6}$$ The boundary conditions here are $$\theta(\xi_0)=1,\quad\frac{d\theta}{d\xi}|_{\xi_0}=-\frac{\beta}{\xi_0^2},\quad\xi_0=r_0/\alpha\tag{Quasi-star b.c.’s}$$ where $$\beta\equiv\frac{M_{\bullet}}{4\pi\rho_0\alpha^3}\tag{7}$$ The scaling factor $\alpha$ can be determined from its definition. Plugging in our values for $r_0$ and $\xi_0$, we get: $$\alpha=r_0/\xi_0=\frac{1.802\times10^9\text{ m}}{0.012}=1.502\times10^{11}\text{ m}$$ Therefore, we get $\beta=8.59\times10^{-4}$. Our boundary conditions can now be rewritten as $$\theta(\xi_0)=1,\quad\frac{d\theta}{d\xi}|_{\xi_0}=-\frac{8.59\times10^{-4}}{\xi_0^2},\quad\xi_0=.012\tag{Quasi-star b.c.’s}$$

# The Euler method

I’d like to first review the Euler method. Let’s say we have an ordinary first-order differential equation of the form $$\frac{dy}{dx}=g(y,x)$$ with appropriate boundary conditions; that is, we know $x_0$ and $y_0=f(x_0)$. We want to find approximate values for the function $y=f(x)$ over some interval of $x$ starting at $x=x_0$. We use the approximation $$\frac{dy}{dx}\approx\frac{\Delta y}{\Delta x}$$ and choose some small $\Delta x$. We then use the first equation to find $$\Delta y\approx g(y,x)\Delta x$$ and iterate along the interval: $$x_{n+1}=x_n+\Delta x,\quad y_{n+1}=y_n+\Delta y_n=y_n+g(y_n,x_n)\Delta x$$ This is the type of method I implemented along with RK4 to produce the first graphs, of an ordinary $n=3$ polytrope. For a system of differential equations like these, the extension is simple; we just have more functions like $g(y,x)$.

# Results

Now we can finally create our models. I used the same step size as in my original example - $\Delta \xi=10^{-4}$ - and plotted $\theta(\xi)$ on both normal and logarithmic $\xi$-axes, to show both the dramatic central cusp and the fact that for a loaded polytrope, $\xi_0\neq0$. I wrote the code in Python 3:

import numpy as np
import matplotlib.pyplot as plt

n = 3
dxi = 10**(-4)
xi0 = .012
phi0 = 8.59*10**(-4)

def dtheta(phi,xi):
return (-phi/(xi**2))*dxi
def dphi(theta,xi):
return xi**2*(theta**n)*dxi

Xi = [xi0]
Theta = [1]
Phi = [phi0]

while Theta[len(Theta)-1] > 0:
Xi.append(Xi[len(Xi)-1] + dxi)
Theta.append(Theta[len(Theta)-1] + dtheta(Phi[len(Phi)-1],Xi[len(Xi)-1]))
Phi.append(Phi[len(Phi)-1] + dphi(Theta[len(Theta)-1],Xi[len(Xi)-1]))

plt.figure(1)
plt.subplot(211)
plt.plot(Xi,Theta)
plt.title('Quasi-star solution to the Lane-Emden equation for $n=3$')
plt.ylabel('Solution')
plt.subplot(212)
plt.title('Quasi-star solution with logarithmic scale')
plt.ylabel('Solution')
plt.semilogx(Xi,Theta)
plt.show()


That’s pretty painless, and quick to write. Here’s the output:

I also did a comparison between a loaded polytrope and a normal polytrope for $n=3$, again to emphasize the cusp:

Finally, here’s a set of graphs I made of normalized temperature, density and pressure for an $n=3$ loaded polytrope and an $n=3$ normal polytrope. While both temperature profiles are quite similar, there is a sharp difference in density and pressure near the core.

Now, remember that these are merely normalized values, and should be multiplied by the central parameters, but the point remains: Quasi-stars are much different than normal stars.

# Evolution

The one remaining question is whether or not our quasi-star will remain stable for any significant amount of time. It’s certain that the mass of the central black hole will change, as the object is powered by accretion from the inner edge of the envelope. Eventually, the quasi-star will be pretty much a black hole with a little bit of gas around it. In the shorter term, the stability of the envelope, for instance, poses a potential problem. It will also be losing mass, as well as accreting it from a disk that may form, surrounding the entire object.

Ball et al. found that $$\dot{M_{\text{BH}}}\propto M_{\text{BH}}^2\rho^{(3-\gamma)/2}\tag{8}$$ To within an order of magnitude, this remains around $\sim10^{-4}M_{\odot}$ per year for many quasi-stars at various stages of evolution. Assuming that the envelope mass of our quasi-star is $\sim100M_{\odot}$, then the envelope should be entirely accreted on a timescale of one million to ten million years, up to a factor of a few. That’s quite reasonable; massive stars typically travel through the main sequence on the order of one million to ten million years, and so while the quasi-star may live for only a short amount of time in comparison to the Sun, its lifetime is reasonable when compared to massive stars.

# Will fusion be possible?

One key assumption of quasi-star models is that any fusion is completely negligible. This particular quasi-star is certainly not normal, so I'd like to double-check and see if there will in fact be little or no fusion. We can do this by calculating the reaction rates of the quasi-star compared to those of the Sun. (I'd like to assume that any fusion occurs via the p-p chain. The CNO cycle is not possible in a star without carbon, nitrogen or oxygen!)

The rate of energy generation $q{ij}$ from a reaction of particles $_i$ and $_j$ is $$q_{ij}=\overbrace{C_1\left(\frac{1}{1+\delta_{ij}}\right)\frac{1}{A_iA_j}\frac{1}{AZ_iZ_j}}^{S_{ij}}X_iX_j\rho\tau^2e^{-\tau}Q\tag{9}$$ where $C_1$ is some collection of constants not unique to the stellar environment, $\rho$ is the density, $X_I$ and $A_i$ represent mass fraction and mass number (while $A$ is reduced (atomic!) mass), $Z_i$ is atomic number, $Q$ is the energy released per reaction, $\delta_{ij}$ is the Kronecker delta, and $\tau$ is a peculiar function of temperature: $$\tau=C_2\left(Z_i^2Z_j^2AT^{-1}\right)^{1/3}=D_{ij}T^{-1/3}\tag{10}$$ where $C_2$ is another constant. I've scrunched together a bunch of terms together as $S_{ij}$; you'll see why later.

Let's find the ratio of $q_{ij,\odot}$ (the Sun) to $q_{ij,*}$ (the quasistar): $$\frac{q_{ij,\odot}}{q_{ij,*}}=\frac{S_{ij}X_{i,\odot}X_{j,\odot}\rho_{\odot}\tau_{\odot}^2e^{-\tau_{\odot}}Q}{S_{ij}X_{i,*}X_{j,*}\rho_{*}\tau_*^2e^{-\tau_*}Q}\tag{11}$$ Both factors of $S_{ij}$ cancel out (as do the $Q$s), leaving us with something much simpler. For the quasi-star, we can look at the highest values of temperature and pressure, the central values we picked earlier. I'll take $\rho_{c,*}=5.426\times10^{-5}\text{ g cm}^{-3}$ and $T_{c,*}\simeq3.5\times10^{5}\text{ K}$, as per Table 1 of Ball et al. I already assumed that the quasi-star envelope is pure hydrogen, so $X_i=X_j=1$.

For the Sun, I'll use the BS05(AGS, OP) solar model by John Bahcall. This gives $X_i=X_j=0.36462$, $\rho=1.505\times10^2\text{ g cm}^{-3}$, and $T_{c,\odot}=1.548\times10^{7}\text{ K}$. Substituting some of this gives $$\frac{q_{ij,\odot}}{q_{ij,*}}=\frac{(0.36462)^21.505\times10^{2}\text{ g cm}^{-3}\tau_{\odot}^2e^{-\tau_{\odot}}}{(1)^25.462\times10^{-5}\text{ g cm}^{-3}\tau_*^2e^{-\tau_*}}=3.66\times10^5\frac{D_{ij}^2T_{\odot}^{-2/3}e^{-\tau_{\odot}}}{D_{ij}^2T_*^{-2/3}e^{-\tau_*}}$$ More substitutions give $$\frac{q_{ij,\odot}}{q_{ij,*}}=2.9\times10^4e^{\tau_*-\tau_{\odot}}$$ In our case, $D_{ij}$ is $42.46\mu^{1/3}$, with $\mu$ being the reduced mass (note that $\mu\neq A$), if $T$ is expressed in mega-Kelvin (i.e. millions of Kelvin). We can figure out that $\tau_*>\tau_{\odot}$ because $T_{c,*}<T_{c,\odot}$, and therefore $$e^{\tau_*-\tau_{\odot}}>e^0=1$$ Therefore, $\frac{q_{ij,\odot}}{q_{ij,*}}\gg1$, and it appears that any fusion will be insignificant.

# Conclusion

I proposed that a low-mass quasi-star - a black hole surrounded by a large star-like gaseous envelope - could have similar properties to a massive star of $\sim100M_{\odot}$. I took central pressure and density values of $5\times10^9\text{ J/m}$ and $5.426\times10^{-5}\text{ g cm}^{-3}$ for a quasi-star of envelope mass $100M_{\odot}$ around a black hole of mass $1M_{\odot}$. Temperature, density and pressure profiles using a polytropic approximation show that fusion is unlikely, even at the center, and therefore the sole source of energy from the quasi-star should be from accretion by the black hole. The object should remain stable for one million to ten million years, which is a reasonable lifetime.

I'll begin my response with an acknowledgement to those answers that came before mine. Each of them has, in one way or another, informed and inspired the avenues of my mind to construct a device which I hope fulfils the requirements of the question.

The first and most prevalent restriction everyone appears to have assumed (and I work under no assumption) is that this object must be star-sized. Given the restriction that the object must not be fusion powered, it must therefore also not reach 'critical mass', so to speak, which is roughly 80 times the mass of Jupiter.

These restrictions seem to be a more reasonable requirement:

• $0.087 \text{ M}_{☉}$ OR about $1.73 \times 10^{29}$ kg in mass
• $4.85 \times 10^{23}$ Watts in power output
• 14.35% luminous efficiency OR 98 lumens/Watt

For these last two requirements I've referenced the least luminous normal star from Wikipedia and the efficiency of our own star, as we likely wish this light-emitting object to be of some use to an orbiting body.

So, now to the 'idea' without the maths, as this idea requires exotic materials that have not yet been discovered.

Having done more reading, I've found several possible variations and limitations.

For example, if I stick my first idea and use a thixotropic fluid and a rheopectic fluid with unique properties (change of density when shifting from liquid to solid and being brittle when solid) causing them to exchange places in the 'mantle' via cracks, fissures, pieces moving around and liquidating again, generating frictive and compressive forces on each other, exhibiting one or a few luminescent properties, such as piezoluminescence, cryoluminescence etc.

I'm hoping the stellar sized masses and movements of these liquids will generate enough energy via several Luminescent types that it will generate light via a combination of these means.

The reliance on these luminescent properties means relying on a poly-cyclical production of energy, hopefully not all at the same time, but unfortunately on geological scales. This creates a need for some kind of structure to store the energy and release it continuously.

This structure will be some kind of stellar sized geodesic structure, the purpose of which will be to sit at the equilibrium point of the two fluids and can perform a variety of functions:

I originally suggested some further improvements, notably the use of a crystalline core that is piezoluminescent/electric and/or the suspension of tiny crystals with similar properties within the two fluids.

I believe now that using a crystalline core could be very problematic. Given the pressure it would have to withstand the core would likely crack if not overheat and become molten, losing it's function as another source of luminescence or electrical current. An interesting way around this would be to make several smaller stellar bodies that orbit each other... but this might 'diffuse' the light too much.

However, using the small crystals (and 'small' here could mean quite massive) in suspension could prove quite fruitful. 1 cm3 of quartz, for example, produces 12,500 Volts when placed under 2 kN of correctly applied force. Much more exotic crystalline structures may exist in the future. A quick glance online reveals great candidates, such as lead magnesium niobate–lead titanate.

I'm a philosophy major, so the maths is really difficult for me (read: impossible without much more time and practice).

However, a cursory glance around Google shows that even at perfect energy reclamation levels only 10% of the gravitational force will be converted into useful energy. But this is likely to be nearer to 1%. This is purely based on the piezoelectric effect.

Essentially, what I'm making is a giant gravity fed electrical device that could be used to power some kind of super large LED? It's a space-lightbulb!

It couldn't be perpetual, but I have no way of measuring how long it would last. Perhaps the crystals will eventually dissolve or be crushed into dust and float away into space? Perhaps the two liquids will eventually reach some sort of equilibrium? Perhaps these materials can never exist? (Although I'd never say never!)

I do not believe it could lead to any catastrophic events, though.

I can now give more than two references :)

• This is one of those really creative answers that I've been looing for, so +1. I hope this will let you add in more references to support your points. Welcome to Worldbuilding! – HDE 226868 Sep 1 '15 at 11:49
• @HDE226868 Thank you for your support! I've made some edits to the post but, on the whole, I am thoroughly under-educated to make this a hard-science answer. My only hope is a loophole caused by the conditions that warrant science fiction type answers :) – equilibriator Sep 1 '15 at 19:57
• Okay, so it's clear that this answer isn't yet hard-science. It relies on some speculative thinking, and it could be completely wrong, as far as I know. But you use some out-of-the-box thinking, and there's potential in this for a great idea - not just an answer, but a great idea. So I'm going to award you the bounty on the condition that you develop it a bit more, to the best of your ability. I can help out with anything, if need be. But I think I see the beginnings of something really cool here. – HDE 226868 Sep 1 '15 at 22:50

Note: This answer is not even close to being finished. I’m putting it out there as a sort of sanity-check, so I can get some input as to whether or not my idea is totally crazy or not. Links and more numbers will be coming.

Introduction

When I wrote this question, I thought that the Kelvin-Helmholtz mechanism would not be a good solution to the problem. It’s simple to see that if a Sun-like body were to produce energy at the same rate as the Sun in this way, it would run out of energy after ~107 years. This is something that has caused me to discard other ideas as well, something I’ll call the timescale problem.

I looked at using some form of accretion in various ways. I was already familiar with Thorne-Żytkow Objects (TŻOs), which had been suggested by a couple people. A TŻO is an M-type red giant or red supergiant that has had its core replaced by a neutron star. Nuclear fusion continues in the upper layers of the star, while the inner envelope is accreted by the new core, producing energy. Demetri’s answer talked about several pros and cons that are quite important. Unfortunately, the disadvantages (which I have added to) outweigh the advantages (see Thorne & Żytkow (1977)):

• Nuclear fusion still happens in the upper layers.
• The envelope will last for a time on the order of ~107 or ~108 years, which is too short.
• There is the potential for instabilities in various layers of the envelope.

Another possibility that crossed my mind was to use a quasistar, essentially an extremely massive protostar whose core collapses into a black hole. The disadvantages are that the lifetime of the envelope would be about the same as that of a TŻO, and the protostar would have to be at least 1,000 times the mass of the Sun (see Begelman et al. (2008)).

One final speculative option I came up with was also mentioned: a dark star. This would be a mixture of dark matter and normal matter that generates energy via annihilation between neutralinos. The downsides are twofold: The “star” would have a diameter between 4 AU and 2,000 AU, and would not emit light in the visible portion of the spectrum.

These are the most well-studied types of exotic stars. It should be apparent that these could not be good substitutes for a star and conform to the time and luminosity requirements I set down. The solution I present here is far more mundane, at least in terms of the star’s composition.

I propose using the Kelvin-Helmholtz mechanism to power a star like a T Tauri star. The timescale problem can be solved by periodic mass loss and replenishment recurring every Kelvin-Helmholtz time, by way of a disk in and out of which the tar oscillates. Nuclear fusion will not happen because temperatures in the core of the star will not have reached high enough levels.

1. The star

The Kelvin-Helmholtz mechanism transfers gravitational potential energy into radiated energy. The derivation is simple. The total radiated gravitational potential energy is $$U_r=\frac{3M^2G}{10R}$$ or, setting $C=\frac{3}{10}$, $$U_r=\frac{CM^2G}{R}$$ I use $C$1 [Footnote: Some authors use $\eta$.] here because this is not quite correct. The proportionality is correct, but there needs to be an indicator of how well the object compresses. This can vary greatly; for example, for Jupiter, $C\approx0.03$. In the present case, however, we will take $C=\frac{3}{10}$.

Given that luminosity is energy over time, we can write $$\frac{U_r}{t}=L\to t=\frac{3M^2G}{10RL}$$ We might naively substitute in $L=L_{\odot}$, the luminosity of the Sun, and do the same for the mass and radius, and calculate the Kelvin-Helmholtz timescale. But this will not give the correct time for a star of such mass and radius, for several reasons:

• There is no reason for the given luminosity to be the luminosity produced by such a star. It would only tell us how long a body acting like the Sun but producing energy via the Kelvin-Helmholtz mechanism would last.
• The radius of such a star will change over time as contraction goes on. The same should hold true for luminosity, in certain cases.

To accurately come up with a model, we must look to some real cases of stars contracting like this. Such stars are pre-main-sequence stars, living on either the Hayashi track (for lower-mass stars) or the Henyey track (for higher mass stars). Stars on the Hayashi track decrease in luminosity over time while retaining a constant temperature; stars on the Henyey track increase in temperature over time while retaining a constant luminosity. After a certain amount of time, they join the main sequence, as nuclear fusion sets in.

Kumar (1962) provides an alternate expression for the energy released by contraction (we keep an extra term, assuming non-zero initial and final radii, the importance of which will be explained later): $$t=\frac{GM^2}{28\pi\sigma T_{\text{eff}}^4}\left(\frac{1}{R_2^3}-\frac{1}{R_1^3}\right)$$ Note that there is an extra term for the initial radius. This is in part because of a different derivation and in part because we need to assume a finite initial radius, unlike most models.

The effective temperature of a star on the Hayashi track can easily be calculated: $$T_{\text{eff}}=(2600\text{ K})\mu^{13/51}\left(\frac{M}{M_{\odot}}\right)^{7/51}\left(\frac{L}{L_{\odot}}\right)^{1/102}$$ where $\mu$ is the mean molecular weight of the gas particles. This last factor of luminosity shows that the temperature of a star on the Hayashi track is only very slightly luminosity-dependent. In this case, I’ll choose to drop that term entirely, setting it equal to 1. This means that if we set the final and initial radii constant, the time spent on the Hayashi track is entirely mass-dependent, with the exception of composition. We can say $$T_{\text{eff}}^{-4}\approx(2600^{-4})\mu^{-52/51}\left(\frac{M}{M_{\odot}}\right)^{-28/51}$$ and then $$t\approx\frac{(2600)^4G}{28\pi\sigma\mu^{52/51}}M^2\left(\frac{M}{M_{\odot}}\right)^{-28/51}\left(\frac{1}{R_2^3}-\frac{1}{R_1^3}\right)$$ Now, I could simply set $t$ to ~109 years, then use that to find the mass of the star by picking two radii as guesses. But the big issue there is that low-mass, low-luminosity stars stay on the Hayashi track for longer. Therefore, any object that stayed on the Hayashi track would be pretty dim for much of that time. So this is pretty pointless.

This is why it’s necessary for the star to go through cycles of contraction. In order for a star on the Hayashi track to have a high enough luminosity, it must have a certain mass. However, it’s time on the Hayashi track will not be long enough for my needs. Therefore, it must continue on this track in a circular evolutionary track.

Each cycle begins with the gaining of a large circumstellar envelope of radius $R_1$. Gravity forces the star to contract to a final radius of $R_2$. At the end of this contraction, some mechanism must cause mass loss, so that succeeding envelopes do not become excessively big and cause the star to eventually begin hydrogen fusion. This mass loss will take away all but a small core. The star will then gain a new envelope, and the cycle repeats itself.

The mass gain mechanism will be discussed further later on, but I will discuss the mass loss problem now. The most tempting option is to have a strong stellar wind blow away excess material. In fact, T Tauri stars often have strong stellar winds, sometimes called T Tauri winds, or bipolar outflows related to astrophysical jets. The problem here is that these winds only set in after nuclear fusion has begun.

Another issue with that is that stellar winds are normally pretty regular. The type of mass loss I’m looking for would be sudden, violent, and short-lived. So at the moment, I’m in a bit of a bind as to what to do about that. I suspect disk-star interactions could end up stripping away the envelope and replacing it with a less dense one, but I’d need simulations to prove that.

2. The Disk

For the disk, I’m picturing something in the vein of an accretion disk. It will have to be dozens of solar masses in mass, and it will need to be quite wide. A better unit of measurement might be light-years, not AU. It will also have to be thick. For the density profile, I'm thinking of using a Plummer-Kuzmin model profile: $$\Phi(r,z)=-\frac{G\mathcal{M}_{\text{disk}}}{\sqrt{r^2+(a+\sqrt{z^2+b^2})^2}}$$ where $\Phi(r,z)$ is gravitational potential and $a$ and $b$ are constants.

The disk’s composition will be mostly dust and gas, in the form of molecular hydrogen (possibly non-ionized). It shouldn’t be too hot or dense - again, I need to prevent nuclear reactions from happening in the disk or during accretion.

To analyze the motion of the star, in its Sitnikov-style orbit, I’ll use a Lagrangian: $$\mathcal{L}=\frac{1}{2}M\dot{z}^2-M\Phi$$ I’ve restricted the motion to be linear in the $z$-axis, so our only relevant Euler-Lagrange equation is $$\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{z}}\right)=M\ddot{z}=\frac{\partial\mathcal{L}}{\partial z}$$ This then becomes $$\ddot{z}=\frac{G\mathcal{M}_{\text{disk}}z\left(a+\sqrt{b^2+z^2}\right)}{\sqrt{b^2+z^2}\left(\left(a+\sqrt{b^2+z^2}\right)^2+r^2\right)^{3/2}}$$ The star will begin at $r=0$, and, given that there will be radial symmetry and no radial forces, it will stay there. Therefore, $$\ddot{z}=\frac{G\mathcal{M}_{\text{disk}}z\left(a+\sqrt{b^2+z^2}\right)}{\sqrt{b^2+z^2}\left(a+\sqrt{b^2+z^2}\right)^3}=\frac{GMz}{\sqrt{b^2+z^2}\left(a+\sqrt{b^2+z^2}\right)^2}$$ This is a second order nonlinear differential equation of the form $$y''=f(y)$$ where $y=g(x)$. Here, $y=z$ and $x=t$. We can solve for $t$ as a function of $z$. The solution is $$t=\pm\left(-C_2+\int\left[C_1+2\int f(z)dz\right]^{-1/2}dz\right)$$ Integrating $f(z)$ doesn't appear to be possible analytically, though I'm trying numerically. One (inelegant) way you could do it would be to approximate the Taylor series of $f(z)$ up to some $O(z^N)$ for sufficiently large $N$. Then you could integrate that, then maybe take the expression inside the brackets and create a Taylor series for that, and integrate.

The upside of all this is that if you know the velocity of the star at $z=0$, you can find its maximum height (use conservation of energy), and from that you can find its period, $P$, and $P/2$.

My main worry with this setup is stability. A Sitnikov planet is unstable against radial perturbations. Depending on the density profile of the disk, this may or may not be the case. Now, the case of a ring-like object providing the potential for the body to oscillate in may not have such instabilities. This page explores some of the properties of a toroidal planet, including possible orbits for its moons.

Believe it or not, there are stable (at least in the short-term) orbits that run through the center of the torus!

I'd say it's possible for the same kind of stability to be possible here, even if the hyperboloid is "straighter". We can decompose the density profile of the disk into such a torus, at some stable distance from the center, and a less dense region involved in the active accretion. This could lead to orbits similar to those computer for the moon and the toroidal planet.

• Mathematica chokes on the integral as well. I think you'd be way better off doing numerical simulation. – 2012rcampion Feb 22 '16 at 1:07
• Actually, it looks like you might have a mistake: due to symmetry, z'' should be antisymmetric with respect to z, but the expression you give is actually symmetric. (Also, shouldn't you be able to set r=0 if the star is oscillating through the center of the disk?) – 2012rcampion Feb 22 '16 at 1:57
• @2012rcampion It looks like I plugged something in wrong when differentiating. Thank you. And yes, I'll set $R=0$, which should be helpful, at least in the new D.E. – HDE 226868 Feb 22 '16 at 16:16

I'll be editing, cleaning this up, and fleshing it out more later (and replacing the wikipedia links) but wanted to put down what I have for now before I forget or lose all the pages.

We can have a Artificial Star with a diamond core, heated by an antimatter reaction engine to 3,000 C. That's not quite enough for a star like our sun, but it can heat the surrounding hydrogen enough to get into the red/orange range.

Based on Samuel's excellent answer here, we can have a stellar-sized diamond in the 253,000km and 573,000km range. That's larger than a solar core, which means that we can surround it with hydrogen and keep it from triggering fusion.

Diamond conducts heat well, and can be heated safely up to ~3,000 C when under pressure. Like when it's in the center of a star.

3,000 C is right around the surface temperature of a red giant, so we can get that kind of light.

TODOs:

• Determine if the mass of the stellar diamond is greater than the mass of the core, and if so if it's enough more to trigger fusion.
• Figure out how to keep the antimatter engine from being destroyed by pressure, ruining the entire thing.
• What supplies the antimatter engine with either antimatter and matter, or else the energy to produce them? – trichoplax Aug 18 '15 at 13:47
• @trichoplax: I'm thinking a ball of antimatter stored in the center, the size to be determined based on the stellar energy required. – Dan Smolinske Aug 18 '15 at 14:29
• Ah I see. So the "engine" is some technological means of containing a huge ball of antimatter and releasing it slowly into the surrounding matter of the star? – trichoplax Aug 18 '15 at 14:31
• @trichoplax: exactly. I'm assuming the antimatter will need to be at least planetary sized, I need to lookup some ballpark figures on the energy output of a star and see how big it would need to be. It might be that the antimatter required is too large, I'm not sure. – Dan Smolinske Aug 18 '15 at 15:43
• Why not concerve power and only beam light directly at the planet, following it as it orbits. It will use far less fuel than a star. – JDługosz Aug 19 '15 at 8:19

## Nope.

You'll get a star sized object releasing energy via fusion unless it's made of elements heavier than iron. So you can make a star from iron, but it won't release much energy.

Unfortunately getting too many fission products together without a neutron moderator will release more energy than the gravitational binding energy of the star and it will blow itself apart. All the good neutron moderators are lighter than iron, so getting a star's mass of them together will simply cause them to undergo fusion. There are some self-limiting nuclear fuels, like uranium zirconium hydride, which has a negative fuel temperature coefficient of reactivity, but I doubt you could create a star made of it and maintain those properties.

Any other reaction will not prevent fusion or will not last long enough to be near the same lifetime as a traditional star.

Thorne–Żytkow objects are an example, if they exist (and there is evidence that they do).

These stars can have two different power sources:

• Nuclear fusion, albeit under rather unusual conditions.
• Accretion onto the NS core.

The second is much more efficient (energy output ~$0.12mc^2$ vs ~$0.007mc^2$). The primary source of power for an actual TZO may be either, however. This is because the convective envelope can bring down new fusion fuel and sweep away the fusion ash.

Since accretion onto an NS releases much more energy per unit mass than fusion does, it is fully capable of supporting the TZO against gravity for as long as the TZO exists, even if fusion is not the primary source of energy. Lifetime is likely to be rather short (< 1 million years). However, this is because stellar winds eject the outer envelope, not because accretion cannot provide enough power.

While fusion will occur in any TZO, in an accretion-driven TZO the amount of energy from fusion is much lower.

It's hard to conceive of a star that doesn't supply itself by nuclear fusion, except those that already exist.

# Some background

Nuclear fusion is a process by which two nuclei of two atoms fuse together (hence nuclear fusion). It can only happen under immense temperature and pressure.

A star is a huge object. At its core, the pressure and temperature are immense, and nuclear fusion will start happening because of this.

# Some numbers

$$$$\tag{1} T_{\text{eff}} \propto \sqrt{M}$$$$

Using $T_{\text{eff ☉}} = 5780\text{K}$ and $M_☉ = 1.988 \times 10^{30} \text{kg}$, from (1) we get: $$T_{\text{eff}} = k\sqrt{M}$$ $$k = \frac{T_{\text{eff}}}{\sqrt{M}}$$ $$$$\tag{2} k = 4.1 \times 10^{-12}$$$$

We know that objects as "small" as 3 Jupiter masses are stars, though fusion only occurs from 13 Jupiter masses (that'll be relevant in a moment).

Thus, of the vast uncountable trillions of stars, anything over $2.47 \times 10^{28} \text{kg}$ will undergo nuclear fusion. Using our numbers from (2), that's:

$$T_{\text{eff}} = 4.1 \times 10^{-12} \times \sqrt{2.47 \times 10^{28}}$$ $$\approx 644 \text{K} = 371 \text{°C}$$

which is a teeny-tiny temperature, in stellar terms. Of course, their cores are hotter, and that's why the fusion occurs, but their effective temperature is tiny. Thus, the vast majority of stars are hotter.

# Conclusion

The majority of stars will power themselves by nuclear fusion. Whether other methods are possible is slightly irrelevant, because those methods' existence will not deny the fact that due to the immense amount of energy released in fusion, nuclear fusion happening will be the primary power source of a star. Excluding, perhaps, inherent black holes.

The minority of stars - and these stars already exist - will exist without being powered by nuclear fusion. Neutron stars, while still categorised as stars, don't undergo fusion.

• Brown dwarfs may undergo fusion of deuterium, though, which provides some power. Also, they're often categorized as sub-stellar objects, not stars. – HDE 226868 Aug 17 '15 at 20:31
• @HDE226868 Let me just change that to read "neutron stars"... – ArtOfCode Aug 17 '15 at 20:33
• I suppose that's okay, though I'm still tempted to be a stickler and classify them as "stellar remnants", as I wrote on another answer. – HDE 226868 Aug 17 '15 at 20:35

In Palimpsest, Stross uses a "necrostar" to keep the earth habitable for orders of magnitude longer than the sun would last.

This was a black hole surrounded by gas, gravity being the ultimate energy source, giving nearly 100% mass to energy.

The whole reason for replacing the sun was to make it last (much) longer.

## Theoretically possible, I think. Using our friend thorium.

Here on earth, Th is essentially 100% Th-232 so let's consider it pure.

Th is not fissile (good for our purposes), and it does have a nice long-lasting 14.05 billion year half-life.

If you consider the entire thorium series, you have 42.6 MeV of energy output (including neutrinos) which is also 6.83e-12 joules. Of course, we lose the nearly all of the neutrino heat (just as we do for our sun).

One mole of Th thus eventually yields 4.11e12 J, or one kg yields 1.77e13 J. Half is released in the first 14 billion years.

One kg of Th-232 thus releases about 20 mWatts, so you need 5 tonnes to power a 100 watt bulb. Congratulations, you now have a very low power, but very long lasting RTG.

Our sun has a mass of about 2e30 kg. So let's replace it with 2e30 kg of thorium. We now have a 4E28 W RTG. Our sun actually produces only about 4E26 watts, so this is 100 times as powerful as our sun, and it should be considerably smaller too considering the relative density of the materials.

The actual amount of Th-232 you need to make your own fusionless sun will depend upon the exact conditions you want, but as a first approximation, a mere 1e28 kg of Th should be close enough as the total power output will be about the same as our sun.

Making this star is easy. Step 1: Start with 1e28 kg of Th, Step 2: throw it all into a big pile and stand way back.

Finding that much thorium will not be easy. Mass of universe is about 3e52 kg, Th is perhaps 1 part in 1e13, so rounding to nearest power of 10, 1e39 kg of Th in the universe. This is enough to build 1e11 thorium stars. Hmmm, maybe not so hard after all if you are sufficiently motivated to scavenge a galaxy for all of its thorium.

If you find that as time passes the star is no longer as shiny as you like, toss a little more thorium on the fire every 10 million years or so.

Are we done? Not fusion powered. Power output similar to a star. Lifespan similar to a star. No disasters in the next few billion years.

An accretion disk, fed by a brown dwarf or gas giant planet.

• Not a star (although the central body would usually be a stellar remnant; a white dwarf, neutron star or small black hole)
• Supported against gravity by angular momentum
• Produces energy by converting the gravitational potential energy of infalling matter into radiation.

For the white dwarf and neutron star central body, radiation is emitted when the infalling matter collides with the surface; there is additional energy release from fusion at this stage. If the central body is a black hole, there is no surface to collide with but the event horizon has such a small circumference (Schwarzschild radius of a Sol-mass black hole is 2953m) that frictional heating of infalling matter causes it to radiate away almost all of its gravitational potential energy.

If this is an engineered structure, multiple gas giants could be kept in reserve; such a structure could last for trillions of years.

• Would the energy be radiated spherically symmetrically? In most cases, two polar jets would form. Also, if the fuel is a brown dwarf or gas giant, the whole thing won't last for long. – HDE 226868 Aug 18 '15 at 15:53
• @HDE226868 thermal energy radiated by the disk will be emitted isotropically. Polar jets are (we think) powered by magnetic processes and by the ergosphere, so they shouldn't steal a significant amount of energy. If the fuel store is Jupiter, and the structure has the same power output as the Sun, the timescale for total conversion at 10% efficiency is 1.4 billion years. If the fuel store is a maximal brown dwarf of 75 Jupiter masses, it'll last 105 billion years. – ecatmur Aug 18 '15 at 16:41
• Can you show calculations for those timescales? – HDE 226868 Aug 18 '15 at 16:58
• @HDE226868 that's a model-free calculation; I just put "((Jupiter mass) * (speed of light in vacuum squared) * 10%) / (Sun luminosity)" into Wolfram Alpha to get 1.41 Gy. The difficulty would be managing accretion flow to ensure constant solar-equivalent luminosity. – ecatmur Aug 18 '15 at 17:17
• Maybe a quibble, but wouldn't a massive output of hard radiation be a problem for those wanting to use this as a star? – Gary Walker Dec 31 '15 at 19:05

A massive object doesn't necessarily collapse due to gravity. Gravity will certainly pull the component part(icle)s together, but it can't hold them there. With gravity alone, each part would be in orbit; converting between kinetic energy and gravitational potential energy forever.

If we introduce another force, like electromagnetism, then two things happen. Firstly, energy can be converted in new ways; for example, particles can emit electromagnetic radiation, and hence decrease their kinetic + potential energy, causing their orbits to decay. Secondly, new interactions between particles can occur, ie. collisions, which disrupt their orbits. That's why Hydrogen clouds collapse into a disc.

What if we don't have a force like electromagnetism acting on our particles? This is the case in some models of dark matter. In these models, dark matter interacts via gravity but not via electromagnetism, so it would not collapse in the way that baryonic matter does. A collection of perpetually-orbiting dark matter particles is called a dark matter halo. Whilst normally discussed on galactic scales, in principle such a halo could have the size and mass of a star; especially if they're engineered that way.

So, dark matter halos don't need an energy source to avoid gravitational collapse; but they're not very "star-like". Is there a way they could emit energy like a star? WIMP models allow dark matter particles to interact via a/the weak force; in this scenario, collisions can occur, just very rarely. Many experiments are attempting to detect such collisions. These rare collisions provide a mechanism for regulating some other reaction, eg. matter/antimatter annihilation, to prevent the normal run-away chain-reactions mentioned in other answers

Brown dwarfs are star-like objects, which get a lot of their energy from gravitational collapse. If you imagine a brown dwarf without any Deuterium (also without any Lithium if the brown dwarf massive enough to fuse Lithium), then the entire energy output comes from gravitational collapse, plus radioactive decay of whatever radioactive elements. This would be dim and quite quickly cooling object, but still star-like without any fusion. To increase lifetime, add radioactive elements with long half life, for example Thorium-232 has half-life of 14.05 billion years, so enough of it should keep things warm for quite a while, without causing chain reaction. If you want natural mechanism to produce a "star" like this, then you probably need to get inventive, but it could be "easily" constructed by any Kardashev Type 3 civilization.

Another, more exotic object would be a neutron star or a white dwarf surrounded by stable accretion disk, which would lose matter in with a stable rate. To prevent fusion of the infalling matter, it would have to be composed of heavy enough elements, so that they will not undergo fusion as they collide with the neutron star. This kind of object would probably need to be made, at least if it needs to be long-lived and have a relatively stable energy output. The upper limit for lifetime is probably until the neutron star collapses into a black hole, so depends on desired energy output, determined by mass flow of the infalling matter. You might have to add some extra objects to destabilize the disk of material just right to keep things falling in at a desired rate.

You could also imagine a black hole with such an accretion disk, but then there will be no actual collision with the central object, so you would have to have so much infalling matter, that it will collide with itself and heat up that way. Without doing any maths, I think this kind of object would be quite short lived, before it would run out of infalling matter. Another exotic options might be black hole and/or neutron star in a binary system, where you might be able to have controlled rate of matter falling in and generating energy for longer, than a single neutron star could do (before collapsing into a black hole).

Can a question also be an answer? Because I don't have the physics to answer this and am not sure anyone does.

A magnetic field has an energy density. One of the most exotic known astrophysical objects actually observed is the magnetar. It's a neutron star that formed with a magnetic field of up to 10E11 Tesla. Its energy output for tens of thousands of years is released by the decay of that magnetic field.

For comparison the greatest magnetic field created on Earth is around a hundred Tesla. The field around a magnetar will distort atomic nuclei into threadlike formations. It will cause a vacuum to become birefringent. Some observations of magnetars ( glitches and anti-glitches) are not well understood.

The energy stored by a given volume of magnetic field goes as the square of that field.

My question. Other than the gravitational collapse of a high enough density of energy into a black hole, is there any upper limit on a magnetic field's strength?

If not, then a "super-magnetar" provides a physical alternative to nuclear fusion for a source of stellar quantities of energy over stellar amounts of time. What form it might take and how it might arise, I cannot begin to imagine.

If you want to stray from hard science to handwavium then is there a new form of matter to be found closer to the black hole limit? And might it be capable of technological manipulation?

No idea on the natural one. But I am assuming there is some value in providing an artificial solution. And that the hard science tag applies to the main question. (ie, I won't do the numbers for a side question.)

Assume you have a starless Earth size and composition planet you wish to provide with an artificial sun that would look convincingly sun-like for a pre-industrial civilization you wish to plant on the planet. Further assume the planet also has a moon identical to ours. Or at least reasonably similar.

By coincidence Sun and Moon are roughly the same size from Earth. Thus if you heat the surface of the moon to same temperature as the photosphere of the Sun. It will be reasonably "Sun-like" and provide the same amount of light and energy. The spectrum, among other things, will be different, so a civilization with industrial technology and any astronomical knowledge will not confuse it with a real star.

The important thing is that Moon is much smaller than any kind of stellar object. And it can be smaller than Moon if you make the orbit smaller. This means that much less energy is needed and much less of the power will be wasted on irradiating empty space. For example, if the moon is tidally locked, only the surface of the planet facing side needs to be heated to full temperature.

These factors lower the energy and power needed enough for radioactive decay to be a viable method for providing it. IF we also assume that your artificial sun only needs to work a million years or so. I doubt that the several billion years needed for natural evolution are possible. For an ark type scheme you might be able to get away with a lifetime of a hundred or even ten thousand years. Lifetime here means that the power hasn't been reduced noticeably. Only heating the surface helps here a lot since the rest of the moon acts as a heat sink reducing output at the beginning.

Only real issue is gathering the required amount of radioactive material. I can't really think a solution other than using nuclear explosions to irradiate (and pre-heat) the surface. And fusion devices would probably be needed to avoid a shortage of fissile materials. A civilization that actually does artificial suns for starless planets might have a better solution though. For example, a fission powered particle accelerator might be able to irradiate the surface since less of energy would be wasted on light and kinetics (explosions). Although with underground explosions of proper yield the waste should be minimal.

## protected by guildsbountySep 2 '15 at 15:24

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