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I'm working with a plot whose backstory involves two alternate timelines colliding with one another. Our solar system is spared the worst of the side effects due to half of the mass of each version of the planet being "spun off" (ie. hand-waved away), but every other star in the universe minus the Sun is smooshed into its alternate self and promptly explodes.

What happens as the energy from these explosions reaches Earth? Do the inhabitants of Earth (with a 20-minutes in the future tech level, but also but dealing with the fallout of a world-altering cataclysm) have anything resembling long term prospects of survival, or are they doomed?

(Edit) Clarification: I don't envisage these explosions as full scale, outshine-the-galaxy supernova events: the two stars "colliding" with one another are already in the same location and moving with the same velocity. Rather, I envisage them as being more akin to ejecting the mass of the star into space, the way a red giant does at the end of its lifespan.

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    $\begingroup$ Our sun is unaffected. Re: the explosion scale, I'm adding a clarification $\endgroup$ – Quasar Mar 2 '18 at 4:59
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    $\begingroup$ Clearly I failed to make the question clear. Allow me to clarify again. This is not a question about every star in the sky going supernova at once from our perspective and scything our planet of life. This is a question about every star in the sky experiencing a classical nova, and becoming significantly brighter from our perspective as their light reaches us in order. $\endgroup$ – Quasar Mar 2 '18 at 9:38
  • $\begingroup$ What, exactly, do you mean by 'at once'? Since the effects would travel no faster than 'c', there would be some time lag between one star doing whatever you say it does and the nest furthest, and the furthest effects would reach us in several billion years. Or do you mean, they happen sequentially, over several billion years, from furthest to closest, so the effects all ARRIVE at earth at the same instant? $\endgroup$ – Justin Thyme Mar 2 '18 at 15:26
  • $\begingroup$ Since the earth has ALREADY been effected (the 'spin-off'), did this START at the earth, has it already happened elsewhere, is it still happening? Has it happened EVERYWHERE at exactly the same time? What exactly has happened to the Sun? Dark matter and dark energy? Are BOTH universes exactly identical? $\endgroup$ – Justin Thyme Mar 2 '18 at 16:53
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Given the edit to the OP, the destruction of the remaining stars would be spectacular to watch, but probably not fatal to humanity in the form described. The stars would certainly go through what could be millions of years of flashing out of existence like fireworks (given the relativistic distances, Proxima would flash and go dark in a little over 4 years, Barnard's Star just under 6, etc.) but life on Earth would still survive quite well. The night sky for many years would look relatively unchanged because the light of the exploding stars would take a long time at the speed of light to reach us.

The big issue is that you've just doubled the mass in the universe, without doubling the energy. That means that the super massive black hole at the centre of the Milky Way galaxy would still be attracting our solar system, but as the mass is now double, our current orbital speed no longer counterbalances the gravitational force and we would start falling into it.

Admittedly this may take longer than you'd think; the centre of the galaxy is still a long distance from us and the falling could bring us into an elliptical orbit instead of consuming us straight away. This orbit is unlikely to be stable for long, but we could potentially survive for as long as the Earth is meant to be habitable (another billion or so years) in such a galactic orbit.

All things considered, this is not a concern to us for some time but there is still a compelling argument to be made that it would affect us eventually. Even if the effects are not immediate, they would be permanent. If you actually have a way to make this happen, please don't try it at home.

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  • $\begingroup$ How long would it take for a double-mass Milky Way to collapse in on itself, though? If it takes longer than 5 billion years, we'll have other problems by then. $\endgroup$ – Quasar Mar 2 '18 at 5:04
  • $\begingroup$ A double mass Milky Way would not collapse in on itself. All the stellar orbits would change to more compact forms. But the distances between stars is vast and they would be unlikely to collide. $\endgroup$ – StephenG Mar 2 '18 at 15:08
  • $\begingroup$ "we would start falling into it." This suggests that you are engaging in the common error of treating the FoR of an orbiting body as an inertial frame of reference, and thinking that since the original gravitational force is balanced by centrifugal "force", additional force will simply pull the object straight down. "the falling could bring us into an elliptical orbit" There is no "could". The earth has tangential velocity, therefore WILL be in elliptical orbit. "This orbit is unlikely to be stable for long" There's no reason to think the orbit will be less stable than the current one. $\endgroup$ – Acccumulation Mar 2 '18 at 16:00
  • $\begingroup$ Doubling the 'mass' of the universe would probably trigger a phase transition, in which case the universe would probably instantaneously be reduced to a singularity, and every law of physics, and every physical constant, would become null and void. Or maybe not. $\endgroup$ – Justin Thyme Mar 3 '18 at 0:35
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    $\begingroup$ Actually, while I agree with Accumulation's answer, (and that's why I added my comments to the edit as shows) the trouble is we really don't know what would happen if we suddenly doubled the mass. Justin could equally be right about an instant singularity. All our science is currently based on realistic changes to mass so I'm comfortable being a little circumspect in this case. $\endgroup$ – Tim B II Mar 3 '18 at 2:27
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So apparently a purely mathematical explanation is not desired.

Here is the non-mathematical explanation.

See Collapse of the universe is closer than ever before

Sooner or later a radical shift in the forces of the universe will cause every little particle in it to become extremely heavy. Everything -- every grain of sand on Earth, every planet in the solar system and every galaxy -- will become millions of billions times heavier than it is now, and this will have disastrous consequences: The new weight will squeeze all material into a small, super hot and super heavy ball, and the universe as we know it will cease to exist.

This violent process is called a phase transition and is very similar to what happens when, for example water turns to steam or a magnet heats up and loses its magnetization. The phase transition in the universe will happen if a bubble is created where the Higgs-field associated with the Higgs-particle reaches a different value than the rest of the universe. If this new value results in lower energy and if the bubble is large enough, the bubble will expand at the speed of light in all directions. All elementary particles inside the bubble will reach a mass, that is much heavier than if they were outside the bubble, and thus they will be pulled together and form supermassive centers.

A phase transition means the collapse of the entire universe, back to the starting point of the 'big bang'.

It pretty much seems to me that the OP's scenario would surely probably almost certainly maybe perhaps undoubtedly trigger this bubble everywhere in the universe at once. No propagation. No traveling out at the speed of c. The number of Higgs bosons would double, in the same Higgs field. Apparently, you are doubling the 'mass' in the entire universe all at once. I can not see any result BUT that the existing physical constants would all be changed. But the math is purely hypothetical, as I elaborated in a previous answer, and is not certain.

From the same article:

Although the new calculations predict that a collapse is now more likely than ever before, it is actually also possible, that it will not happen at all. It is a prerequisite for the phase change that the universe consists of the elementary particles that we know today, including the Higgs particle. If the universe contains undiscovered particles, the whole basis for the prediction of phase change disappears.

"Then the collapse will be canceled," says Jens Frederik Colding Krog.

In these years the hunt for new particles is intense. Only a few years ago the Higgs-particle was discovered, and a whole field of research known as high-energy physics is engaged in looking for more new particles. At CP3 several physicists are convinced that the Higgs particle is not an elementary particle, but that it is made up of even smaller particles called techni-quarks. Also the theory of super symmetry predicts the existence of yet undiscovered particles, existing somewhere in the universe as partners for all existing particles. According to this theory there will be a selectron for the electron, a fotino for the photon, etc.

TL:DR

You can not look at this as simply merging universes, you are merging elementary particles from each universe. You are merging the basic physical constants. You are merging the physics and the rules of physics of two universes. But are you merging the fields? And do the elementary particles behave the same in each universe? Are the universal constants the same in each? Are the laws of physics exactly the same?

The question is purely hypothetical.

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All is totally explained in the following article.

Complete Hamiltonian analysis of cosmological perturbations at all orders II: Non-canonical scalar field

for example

In the context of cosmological perturbation theory, there are currently two approaches in the literature to evaluate the effective Hamiltonian — Lagrangian fomalism and Hamiltonian formalism. In case of Lagrangian formalism, the Lagrangian is expanded up to a particular order, i.e., if we are interested in obtaining third order interaction Hamiltonian, effective Lagrangian needs to be expanded up to third order and constraints are systematically removed from the system to obtain the effective perturbed Lagrangian. Then, the momentum π corresponding to ϕ is obtained as a polynomial of ˙ϕ and using order-by-order approximations, ϕ˙ is expressed as a polynomial of π. Next, using Legendre transformation, Hamiltonian is expressed in terms of π and ϕ. In order to express the Hamiltonian in terms of ˙ϕ and ϕ, only the leading order relation between π and ˙ϕ is used.

A very thorough mathematical hard science explanation.

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    $\begingroup$ @ F1Krazy I am glad you get the point. There ARE no 'layman's terms'. $\endgroup$ – Justin Thyme Mar 2 '18 at 15:34
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    $\begingroup$ This feels like it is a link only answer to me. You're not actually addressing the question in the OP. $\endgroup$ – James Mar 2 '18 at 15:42
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    $\begingroup$ Maybe add a TL;DR for both possibilities? Like, "If you use X, then bla bla bla, everyone dies. But if you use Y, then bla bla bla, and everyone dies instantly. I think I get what you're saying about non-propagating events happening everywhere all at once, even if I don't get the math, but a short explanation in everyday language would make things a lot clearer $\endgroup$ – AndyD273 Mar 2 '18 at 17:33
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    $\begingroup$ @AndyD273 This article does not have a TL:DR. It is ALL just more of the same. My TL:DR point would be that the OP question is so hypothetical that the math is indeterminate. 'For those kinds of models the approach is not applicable' pretty much sums up the TL:DR. But if you want a simpler explanation, see my new answer. $\endgroup$ – Justin Thyme Mar 2 '18 at 18:44
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    $\begingroup$ @Michael Kjörling♦ Physics modeling is ALL based on advanced mathematics today. The models just drop numbers out the end. It is up to physicists to interpret the numbers. The bottom line in this case is that there is no interpretation of the numbers except a mathematical one. I am not sure any physicist has even BOTHERED to interpret the numbers from such a scenario. But if the OP really wants the answer, perhaps the OP can do the math and interpret the numbers that drop out at the end. My money is on a phase transition of the universe, but I have no proof of it. $\endgroup$ – Justin Thyme Mar 3 '18 at 0:29

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