0
$\begingroup$

An electroweak star is a theoretical exotic star in which the radiation pressure from its core keeps it from collapsing into a black hole. Because of its theoretical nature, it's hard to find a lot of information about this concept, so I was hoping for some more intelligent and knowledgeable people to explain something about them to me.

First of all, would an electroweak star radiate in a similar way to other stars? And if that is the case, would its radiation create a "goldilocks' zone" in which a habitable planet could conserve a biosphere? Let's assume that this habitable planet was an extrasolar capture and as such, it managed to not get obliterated by the supernova that created the electroweak star in the first place.

And lastly, how would this radiation compare to other star types? Would it be on the lower side, like other dwarf stars, or would its insanely hot core allow it to shine stronger than most other stars?

$\endgroup$
4
  • 1
    $\begingroup$ Too many questions. $\endgroup$ – puppetsock Oct 21 '20 at 18:27
  • $\begingroup$ Hi, Erik - this is a really interesting question, but as puppetsock said, it's got about two or three questions in one. They're related, but I'd recommend starting by just asking one, and then asking the others in separate questions after you get answers to the first. Hopefully, that will help you get answers that are more focused. $\endgroup$ – HDE 226868 Oct 21 '20 at 18:33
  • $\begingroup$ Electroweak stars are expected to be very compact, more compact than neutron stars. This means that goldilock zone, if exist, should be narrow, and very close to the star. Second, habitability would depend on the star's surface temperature (which is difficult to estimate). Too low or too hot would mean that the system it's unwelcoming to the life as we know it. $\endgroup$ – Alexander Oct 21 '20 at 18:41
  • $\begingroup$ There is also the matter of life expectancy of such an object. You put a hard-science tag on there. I should think it would require quite a bit of research to determine the characteristic time for such a thing. If it has a "Goldilocks" zone, but only for a couple million years, it's not really a good candidate for life to evolve. $\endgroup$ – puppetsock Oct 22 '20 at 0:45
1
$\begingroup$

The electroweak star is a very brief (on cosmological scale) transient state. It results from a supernova explosion, for a very limited range of star sizes.

It lasts only about 10 million years.

It is about 8km radius. Core temperature is 10^15K, but only for a very small region of the core that actually produces energy.

The extreme vast bulk of this energy is in the form of energetic neutrinos, and thus virtually irrelevant. We don't actually know if such a flood of neutrinos has significant effect on matter. Neutrinos are very non-interacting, but the flux rate near the star will be of ludicrous magnitudes! Remember that it is actual pressure from this neutrino flux that is preventing the star from collapsing into itself as a black hole.

So you have a 2-3 Sol mass star, smaller even than a Neutron star, with a surface radiating high in the x-ray or soft gamma regime. And absolutely flooding the space around it with neutrinos.

There will be no goldilocks zone.

$\endgroup$
1
$\begingroup$

According to the electroweak star wikipedia article, the star has a temperature of $10^{15}$ degrees Kelvin. This article provides a little more information, that the star is about 8.2 kilometers in diameter.

The power being radiated from the electroweak star is, then, $P = A \epsilon \sigma T^4$ watts per meter squared. $\epsilon$ can be simplified as 1.0 and dropped. Therefore ${P \over A} = \sigma T^4$.

Once you leave the surface of the star, the "surface area" increases by $4 \pi r^2$, but to save you some trouble you can just use ${{A_n} \over {A_0}} = ({r_n \over r_0})^2$. The power (P) doesn't change, so the radiant power at any distance from the star is ${P \over A_n} = {P \over A_0} {{A_0} \over {A_n}} = {P \over A_0} ({r_0 \over r_n})^2$

enter image description here

The Circumstellar Habitable Zone (goldilocks zone) is defined as a range from the star where liquid water can exist ($0^o C$ to $100^o C$ / $274^o K$ to $374^o K$). Using the same radiant power equation, this is where $\sigma T^4$ is between $(274)^4$ to $(374)^4$ times $\sigma$

$({P \over A})_{goldilocks} = \sigma (274)^4 to \sigma (374)^4$

Plugging in the values from the star, you can find the Goldilocks zone and skip a bunch of math $\sigma T_{star}^4 ({r_0 \over r_n})^2 = \sigma (274)^4$. You can now drop the $\sigma$ and solve for $({r_0 \over r_n})$, which is equal to $({r_0 \over r_n}) = \sqrt{(274)^4 \over (10^{15})^4} \rightarrow \sqrt{(274)^4 \over (10^{60})}$. I'm going to flip that, because what I want to know is $r_n$ which is $r_n = r_0 \sqrt{(10^{60}) \over (274)^4}$. Which I'm calculating as between 7.1 x $10^{24}$ and 1.3 x $10^{25}$ times $r_0$, which is half 8.2 kilometers (4.1 km). The would put the final CHT for this star to be several thousand light years away from it.

At only 1.3 times the mass of our Sun, the CHT is way too far away to be effected by the electroweak star's gravity. Safe to say there isn't a habitable zone, unless a debris field provides some sort of shading.

Per Wein's displacement, the wavelength of the star would be about $\lambda_{peak} = {b \over T}$ would be in the range $10^{-12}$ m, which is between x-rays and gamma rays.

I wonder if the temperature cited in the Wikipedia article is too high. It's nearly one million times the temperature of the hottest known star. Intuitively, a proto-blackhole (which is what an electroweak star will eventually become) isn't generating scalding amounts of heat out to near-galactic radii.

I did a little more digging, and found the paper that describes electroweak stars. It mentions that although the core is a few GeV, a lot of this energy is left in the gravity well, and only a few hundred MeV escape.

Using that, and converting from MeV to Kelvin, the surface temperature of the star is in the range of $10^{12}$ degrees Kelvin.

Using the new surface temperature, the Goldilocks zone for this star is still unreasonably far away. Maybe some other mechanism is in play further cooling an electroweak star.

$\endgroup$
1
  • $\begingroup$ This is a good writeup, but the wikipedia article states that the 1E15K burning is in an area the size of an apple (the core of the star) rather than the entire star. Using that value, I still get a value which is ludicrously high (10^49 watts). For reference that's 10^5 supernovae per second. I think the clue is in the gravity losses - as you mentioned, that brings you down to ~10^12K, at which point you have a more reasonable 10^37 watts (which is still insane). $\endgroup$ – Nikhil Murali Oct 22 '20 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.