# Can a galaxy be destroyed by a single but well-accelerated particle?

Can a galaxy be destroyed by a single but well-accelerated particle, like electron or neutrino?

For instance, one shoots a very high-energy electron in the direction of the galaxy. It collides with other interstellar particles, and due to very high energy a lot of new particles produced in such collisions, so a beam appears. As the beam approaches the galaxy in question, it grows and grows so to include now a lot of new particles, like an avalanche.

These particles are still so energetic that they crush all planets and stars on their way.

Is this possible?

• How would you define "destroyed"? – HDE 226868 Jan 10 '15 at 14:38
• @HDE 226868 planets and stars disintegrated, except black holes. – Anixx Jan 10 '15 at 14:47
• I have a hard time imaging anything with that kind of energy. You realize you would have to put the energy of millions of stars (maybe billions?) into this particle? – bowlturner Jan 10 '15 at 14:49
• @HDE226868 That is what I was thinking. 'No' isn't much of answer, but it is the right one... – bowlturner Jan 10 '15 at 14:55
• Anixx - A neutrino is worse, because it will interact even less with the objects around it. – HDE 226868 Jan 10 '15 at 14:56

What you're talking about here is basically the same thing as a cosmic ray air shower, except that it would have to take place in intergalactic space instead of in the atmosphere, and the amount of energy involved would have to be unimaginably higher. There are two factors that make this event rather different from an air shower:

1. There are fewer particles to hit in intergalactic space
2. This would require the energy released from the formation of an entire galaxy to somehow be concentrated in a single particle

Reason #2 is enough, on its own, that this would never happen in practice. But since the premise of the question appears to be that, somehow, that reason has been bypassed, let me go through the relevant calculations.

First of all, the amount of energy in the particle needs to be enough to cancel out the binding energy of the galaxy and all the stars and planets inside it. From this presentation, slide 10, suppose the galaxy's gravitational binding energy is $M(10^{-3}c)^2$, which works out to roughly $10^{53}\text{ J}$ assuming $M \approx 10^{12}M_\odot$. This would be the amount of energy required to separate the galaxy into individual stars. Then, let's approximate the amount of energy required to separate all the stars, planets, etc. into atoms as $10^{42}\text{ J}$ per solar mass, which gives another $10^{54}\text{ J}$ total. So the incoming particle will have to have $10^{54}\text{ J} \approx 10^{73}\text{ eV}$ in the galaxy's rest frame. (Actually a little more because it needs to transfer some energy to the remnants of the galaxy as kinetic energy, but this excess is something like a factor of $10^{-6}$ smaller and thus negligible.)

So suppose we have a particle of energy $10^{73}\text{ eV}$ somehow propagating through the universe. Now, we know nothing about how a particle with such a tremendous amount of energy would actually interact with ordinary matter. Such a high energy is firmly into the domain of (beyond-)nstandard-model physics. For purposes of a science fiction story, you could make it do all sorts of weird things.

But, sticking to the current science for the sake of argument, let's say you naively extrapolate the known behavior of high-energy scattering to this $10^{73}\text{ eV}$ cosmic ray. The next thing to figure out is the probability of the cosmic ray scattering off the particles it meets. And the relevant parameter to characterize this is the squared center-of-mass energy, $s$. For a collision between a massive particle in motion, with mass $m_1$ and energy $E_1 = \gamma_1 m_1 c^2$, and a massive particle at rest, with mass $m_2$ and energy $E_2 = m_2 c^2$, this is

$$s = m_1^2 c^4 + m_2^2 c^4 + 2E_1 E_2$$

Alternatively, for the same massive particle and a photon which has energy $E_2$ and is approaching the moving particle at angle $\theta$ (with $\theta = 0$ being a head-on collision), assuming $E_1 \gg E_2$, the CM energy is

$$s = m_1^2 c^4\biggl(1 - \frac{E_2}{E_1}\cos\theta\biggr) + 2(1 + \cos\theta)E_1 E_2 + \text{negligible terms}$$

So $s$ for an interaction between the cosmic ray and a massive particle is a fixed, very large value. Interactions of this sort generally get less likely as $s$ increases, so a particle with $10^{73}\text{ eV}$ is basically going to pass right through matter as if it doesn't exist. But for an interaction between the ray and a photon, $s$ varies depending on the angle. It goes as low as $s = (m_1 c^2)^2$, when $\theta = \pi$ (the photon and the cosmic ray are traveling in the same direction), and goes all the way up to more than $4E_1 E_2 \approx 10^{89}\,\mathrm{eV}^2$.

This is important because the interaction between two particles is most likely at a resonance, a center-of-mass energy which corresponds to the mass of some intermediate particle. For example, the delta baryon has a mass of $1232\,\mathrm{MeV}/c^2$, and therefore interactions between charged particles and photons are particularly likely when $s = (1232\,\mathrm{MeV})^2$. The cosmic microwave background (CMB) provides an ample supply of photons traveling in all directions, and thus any charged particle with enough energy to achieve $s \ge (1232\,\mathrm{MeV})^2$ in a collision with a CMB photon will be very likely to do so quickly. Particles with such high energy simply do not propagate very far through space. This effect is known as the GZK limit, and the associated energy cutoff is on the order of $10^{19}\text{ eV}$ (in the rest frame of the CMB). The exact order of magnitude varies by the type of particle involved, but regardless of what type of particle it is, anything that has $10^{73}\text{ eV}$ will be well over it.

In fact, a high-energy cosmic ray with enough energy to destroy a galaxy will hit not only the delta resonance, but the resonances of every particle in the standard model, and any unknown particles that may exist with higher masses up to a very high threshold. (Higher than the Planck mass, thus underscoring the need for a theory beyond the standard model to explain what happens.)

Anyway, the gist is that a particle with this huge amount of energy will pretty much immediately produce a shower of other particles of all sorts, with energies rapidly dropping as the shower progresses. This is actually perfect for your scenario, because it spreads out the immense amount of energy from one particle (which, as I mentioned, passes right through matter) to a broad swath of particles which is rather well distributed for galaxy destruction.

One might then consider the question of how close to the galaxy you have to produce the highly energetic particle in order to make this work. The answer to that depends on the characteristic length scale of the shower, which in turn depends on the scattering cross section and some complicated math that I don't want to get into now. If I figure it out later, I'll come back and add details, but for now, my conclusion is that if you ignore the main reason this could never happen, it actually seems quite plausible.

• Good answer, but would not the produced particles recombine so to make planets, stars etc inside this shower so the outcome would look more like a collision of 2 galaxies rather than evaporation of matter inside one? – Anixx Jan 11 '15 at 7:44
• You'd get a cloud of atoms propagating through space, and then after a few hundred thousand or million years, yes, some of them would probably recombine to make new stars and planets. – David Z Jan 11 '15 at 8:04
• Yes, a good answer. Although I am not sure this would count as shooting a galaxy with a single particle. It is more like shooting a galaxy with a huge blast of energy while using a single particle as a very brief, unnecessary, and increasingly inconvenient intermediate stage. Well, seeing as generating such a particle is implausible anyway, doesn't really matter. – Ville Niemi Jan 11 '15 at 13:27
• Assuming the produced particle shower is relatively uniform, I think this would require even more energy than you've estimated, since most of the particles would simply pass through the galaxy and not even hit any stars due their small cross section vs that of the galaxy. As an example, a very conservative estimate for Messier 5 gives seven orders of magnitude more energy needed if the shower is uniform. – cartographer Jan 11 '15 at 19:32
• @cartographer probably not, though it depends on what dark matter is. We know that it doesn't interact electromagnetically (hence "dark") but it does interact gravitationally. It almost certainly is some kind of massive particle, so the argument I made about how the cross section for massive particle interaction is low suggests that dark matter would not play a significant role in this. – David Z Jan 11 '15 at 21:24

No.

At these speeds, we have to take special relativity into account. The relativistic kinetic energy formula is $$KE_r=mc^2 \left(\frac{1}{\sqrt{1-v^2/c^2}}-1 \right)$$ Taking the limit as $v \to c$, we see that the kinetic energy becomes infinite.

So, actually, I was wrong before - there is no limit to how much kinetic energy a particle can have!

However, a particle would have to have a lot of energy to destroy even one star in the manner you described. It would need to have a kinetic energy upon impact of the star's gravitational binding energy: $$U=\frac{3GM^2}{5R^2}$$ where $M$ is the mass of the star and $R$ is the star's radius. For a star like our Sun, with $M=1.989 \times 10^{30}$ and $R \approx 696342000$, the gravitational binding energy would be $$U=\frac{3 \times 6.676 \times 10^{-11} \times (1.989 \times 10^{30})^2}{696342000} \approx 1.38 \times 10^{42} \text{ Joules}$$ Which means it would have a velocity of . . . something very large. The relativistic kinetic energy formula gives me something extremely close to $c$.

If you want to really shake things up, you have to rip the constituent quarks out of each proton and neutron - which is also impossible. The vast majority of the proton's rest mass is due not to the rest mass of the quarks, but to their energy. However, the strong nuclear force is a real - uh, tough opponent, so put it one way, and it's the strongest of all the fundamental forces. You can't really beat it, though you can try it.

Now do that for all of the hundreds of billions of stars in the Milky Way, as well as all the other bodies in the galaxy.

However, all of this is still possible - you'd just need a lot of energy.

• It is obvious that the speed should be close to c and u can totally omit any references to classical mechanic as being totally out of place here. – Anixx Jan 10 '15 at 15:51
• "At this speed, though, it would going too fast for it to successfully interact with any other objects." - prove. Photons go at speed of light and interact with everything. – Anixx Jan 10 '15 at 17:03
• @Anixx For the second-to-last comment, let me give an analogy. Let's say two planets move by each other. During that time period, they collide. During that collision, there will be a change in momentum. However, if the planets are moving faster, there will be a smaller impulse delivered, and so a weaker interaction. There's your proof. – HDE 226868 Jan 10 '15 at 17:27
• @Anixx Yes, the electron would interact, that is not the issue. The problem is that electrons mostly interact with electromagnetic fields and going fast would reduce the time for the fields to have effect, meaning it would really only interact on direct collision. Unfortunately normal matter is very sparse so odds of hitting anything are low, and even if you did you'd just have two particles moving too fast to really interact. Neutrinos are different because they mostly only interact by direct collision already, photons because they are bosons. – Ville Niemi Jan 10 '15 at 18:05
• @Anixx I wanted to show that the amount of energy that would be needed would be unrealistic. But just saying "You'd need an unrealistic amount of energy" doesn't have anything to back it up - and I'm guessing you'd agree with that. So I wanted to prove - if you can call it that - that the amount of energy needed would be incredibly large. – HDE 226868 Jan 10 '15 at 19:37

No, for all the reasons everyone else has said, plus you'd tear the particle apart and everything around it long before you poured the necessary energy into it. Here's some limits to the energy you can put into a particle.

2 x 10^12 K is the Hagedorn Temperature where hadrons (protons, neutrons, electrons) melt into quarks and gluons. So you can't use a proton, it has to be something smaller and more fundamental like a quark. And you have to hope that quarks don't break down into something else. The LHC regularly achieves these energies and no galaxies have yet to be blown apart.

1.41 x 10^32 K is the Planck Temperature above which our current physical theories break down. At TP the wavelength of the object is the Planck Length and we don't know what happens then. Since the energy needed to tear apart a single star is 1.38×10^42 Joules, and you're dealing with a miniscule amount of mass, and you want to tear apart billions of stars in a galaxy, it's fair to say you'll hit the Planck Temperature before you can get enough energy into your particle. Though it would be interesting to see someone do the math.

Since you'd need to accelerate your particle to very, very, very close to the speed of light, there's practical issues with how you'd accomplish that acceleration once it's in the 0.9999c region. Simply put, your particle will outrun your attempts to pour more energy into it. It has so much energy that curving its trajectory in a particle accelerator would prove unfeasible. You'd have to make a cosmic railgun on a galactic scale. Again, it would be interesting to see someone run the numbers on this.

And pumping too much energy into too small a space has an annoying tendency to form a black hole.

VSauce has a nice video about "absolute hot" called How Hot Can It Get with many sources in the description.

No. Any particle at such speed will very heavily interact with anything in the universe, from CMB photons and protons from distant stars to gravitons of gravitation waves, all these particles will be hugely spectrum-shifted for the propagating particle. Note that there is a lot of CMB photons around us (they are the most common particles in the universe), but we do not notice them because they have very small energy. The smaller energy, the more photons there is. So the cosmic vacuum as we see it will be a very dense medium from the particle's point of view at such speed.

Even passing any particle without direct collision will produce substantial gravitation waves.

This means the fired particle will be slowed down, lose its energy to radiation (both photonic and gravitational) and produce a lot of new particles.

The interaction will start shortly after the particle leaves the gun, may be some meters apart from it. This will produce a huge explosion with creation of a lot of matter-antimatter pairs and radiation. Some of it will be radiated away into every direction, not necessarily the same as the original one, the rest will produce a huge plasma fireball.

This fireball after travelling some thousands years towards the target galaxy will cool down and slow down. And inside it proto-stars, planetoids and may be, black holes will appear.

Upon reaching the target galaxy the result will be like a collision of a young galaxy with an older one.

Even if the total energy of this bullet would be enough to desintegrate the target galaxy, by the time it reaches the target the most energy will be either radiated away or in the form of stellar bodies, stars, planets and gas, moving at quite moderate speeds. They rather will merge with the target than disintegrate it.

• You mean gravity waves, not gravitational waves, right? Do you have anything to support the fireball idea? It seems highly unrealistic to me. – HDE 226868 Jan 10 '15 at 19:11
• Gravity waves are on the ocean, @HDE226868, and gravitational waves distort spacetime. I think the former could not be contrived to apply. – JDługosz Jan 10 '15 at 22:54
• @jdlugosz I meant the opposite of what I typed, sorry! – HDE 226868 Jan 10 '15 at 22:56

No, a galaxy is too big and too low-density

Consider what you're trying to do. You're trying to disrupt an entire galaxy! (As a note: I am assuming you will not consider "moved the galaxy 4in to the left" to be destroyed. You want to see structural changes)

You obviously need to impart kinetic energy into the galaxy. You can only really consider movement which is fast enough to cause serious damage. If your particles get too slow, they wont be disruptive enough to do any real effect to the structures held in place by gravity.

As a first attempt, lets try to play a game of billiards. Lets try to hit a planet, so we can send it careening into a sun, and try to make the sun a billiard ball. Right away, we see a huge problem: acceleration. Imparting all of your particle's energy into a planet is not easy. Consider a particle going just shy of the speed of light (rounded up to $c$ just to keep things simple). If that planet were Earth, the particle would go from one side of the earth to the other in 42 milliseconds. To do any real damage with our new billiard ball, it needs to be going pretty darn fast. Lets say we want to make the sun travel 1km/s in a direction of our choosing by hitting it with the Earth. For perspective, that's not even fast enough to escape the moon's gravitational pull, but we'll use these little numbers for now.

The sun is 330k times more massive than the earth, so the earth is going to have to be going roughly 330k times faster than our final speed of the sun. We need to impact the earth to make it go 330km/s. This should raise some eyebrows, because that's faster than the speed of light. The only way to add 1km/s to the velocity of the sun requires accelerating the Earth to relativistic speeds (where the kinetic energy effectively adds mass).

Now we see the real problem. Lets pretend we were okay with accelerating the earth to JUST 100km/s. No biggie. We only had 42 milliseconds between particle hitting the earth and passing through. That corresponds to 243 kilo-G's of acceleration on the earth. This is an unimaginably high number, and the very imaginable outcome will occur: the particle will break through the Earth, rather than taking the mass with it.

This problem occurs whenever we need to accelerate a mass to a very high speed (such as 20% the speed of light). Whenever we try to do this, we need to do it slow. Doing it fast will just cause the moon/planet/sun to shatter, distributing your energy in all directions, bleeding off your momentum. This will occur whether you hit your moon/planet/sun with a particle or another planet. Whenever you need to accelerate other particles to relativistic speeds, you'll find you just bust straight through them.

And this is what makes a galaxy invincible to such activity. Until you start sending high velocity black holes through, the galaxy wont even notice. And even they won't have the massive galaxy-shattering effects you are looking for. Consider we believe we have a black hole in the center of our galaxy, and its a reasonable sized one!

Now what if you really wanted to make this work. What if you were a god which knew where every particle was at all times, so you could cast your relativistic proton like a pool shark, and clear the table?

Your best bet would be to start millions of light years away, if not further. Line up your shot to collide with a bunch of intersteller hydrogen. But don't hit each one head on. Your goal should be to create a pressure wave of hydrogen moving forward. You want each collision to perfectly impart half of its momentum to the next particle to build the chain reaction. Take your time, and make sure to chalk the tip of your cue first.

Eventually you could create a wave which would act like a tsunami when it finally hit the galaxy, tossing it around like a crab ship in a stormy sea.

Of course, this was assuming you could predict everything. If you could do that, you could always just play the lottery. I hear the Betelgeuse pick-70 lottery has a really strong payout!

Yes. There is no limit to the amount of energy you can put into a mass-bearing particle by accelerating it. No matter how much energy you put in, the particle still won't reach the speed of light, so you can still add more energy. Assuming it will take a finite amount of energy to destroy your target galaxy, this means your idea is theoretically possible.

But practically, where would you get all the energy to put into your particle? Of course you can get energy from matter that exists in the universe. But I imagine you would need to consume the matter from very many galaxies in order to get the energy to destroy just one [citation needed].

In which case, it might just be easier to use your existing galaxy-consuming technology instead of the extravagant single-particle method.

• This is relevant: en.m.wikipedia.org/wiki/Oh-My-God_particle there is a limit as to how far such a particle will travel once created, so a practical limit on how energetic of a particle can be delivered to a given target. – JDługosz Jan 10 '15 at 22:46
• Thus the problem of the StarKiller in The Force Awakens. – Caleb Woodman Jan 29 '16 at 17:47

No. It might do some damage initially, but it can't keep doing damage indefinitely.

On the first collision, no matter how fast the particle is moving, it will impart energy to the other particles. To break the bonds holding planets and stars together, it will need to convert their potential energy (stored in the bonds) to kinetic energy. Keep in mind that total energy must stay the same, meaning that both particles will now be moving slower than the original particle's speed. This means that your "particle avalanche" will be continuously slowing down as more and more matter is added, and it must slow down rapidly to get other particles moving fast enough to escape a high amount of potential energy (in the bonds). This means that the destructive power of the particles will decrease accordingly, since individual particles will not have enough kinetic energy to continue destroying bonds. Additionally, such high energy collisions tend to do things other than just smash stuff apart. Consider fusion: high energy collisions of hydrogen atoms results in conversion of some matter to light (releasing photons on a wide range of the spectrum) and actually overcoming the necessary energy boundaries to form a nuclear bond. This actually reduces the overall kinetic energy of the matter by converting a lot of the energy into photons and potential energy. So the speed of the particles will not stay high enough to cause that level of destruction on a galactic scale.

Even if it were possible, it would be ridiculously impractical. Our galaxy is over 100,000 lightyears across. This means that it would take well over 100,000 years for the damage to propagate across, and probably longer for things to actually settle down. I imagine that natural processes would just begin anew in that time, with new nebulae and stars and planets forming eventually (though not within 100,000 years). And this is one of the smaller galaxies out there. By the time it was done, the inhabitants of the galaxy would have plenty of time to react or possibly retaliate (assuming practical space travel). They might even find a way to stop it. This also ignores the feat of imparting that much energy to a single particle in the first place without blowing yourself up first, which is far less practical than firing off a large number of very high energy particles.

• It will impact energy to the other particle - of course; that's part of @Anixx's assumption. What you're describing is similar to what was described in the question. – HDE 226868 Jan 10 '15 at 21:34
• @HDE226868 The point is that each time the particle collides, it's losing kinetic energy. The energy it imparts must be enough to break other particles' bonds, so the kinetic energy of individual particles will very rapidly decrease, reducing their destructive power significantly. I've edited to clarify. – jpmc26 Jan 10 '15 at 21:38

(Incorrect ideas from kinetic energy formulae edited out. Thanks to comments from HDE 226868 and BartekChom. Remainder applies, however.)

I would say that even if you had something with extremely high energy that would hit interstellar dust, it won't create a "beam", because the dust will 1) be destroyed and 2) still have its original velocity component, and go sideways, and so spread out. Moreover, by conservation of energy, I wouldn't think this would add any energy to your attack, though it would spread it out, making it more likely that some of it would hit something.

Which pertains to another difficulty: even if you did have a particle that somehow had crazy-high momentum, you might need to somehow aim it very carefully, or it might most likely just cruise through a galaxy without contacting any significant body.

Even if you do hit a large star with a ridiculous amount of focused kinetic energy, I know I don't know what that would do. Maybe a good theoretical question for a physicist. Even if you cause some sort of high-energy supernova of the star you hit, I don't know if that would result in a star's worth of also-super-powerful matter flying outwards, or what.

Finally, I think the main reason this probably would never happen would have to do with mechanics of how this much energy could be put into one particle, whether the particle wouldn't just change into something else during the process, or decay to something else along the way, and on top of that, how you would ever get that much energy in the first place.

• A few comments: 1) There's a factor of $\frac{1}{2}$ missing in the first equation. 2) I originally showed the classical calculations, but Anixx said I shouldn't put them in, as they confuse people. It mixes classical and relativistic physics. 3) Did you mean $mc^2$ (really $\frac{1}{2}mc^2$) at the end? :-) – HDE 226868 Jan 10 '15 at 20:25
• Thanks. Thanks for the 1/2 but it's fairly meaningless here except when going to units and calculating exact energy. As I wrote, I was taught the classical formula (in the 1980's), which from your answer would seem to be superceded. And yes, I meant mc^2 at the end. I'll edit my answer. – Dronz Jan 10 '15 at 22:39
• Why the downvote here? Is it the use of classical physics? – HDE 226868 Jan 11 '15 at 0:21
• Probably. In this case it does not make sense at all. Energy is not capped. Actually one can say that the speed of light cannot be exceded because the energy becomes infinite. Relativistic kinetic energy formula is in HDE 226868 answer. And annihilation gives rather $2mc^2$ - full energy from mass of both particles. – BartekChom Jan 11 '15 at 5:54
• Thanks @BartekChom. I'll edit to correct my answer even more. – Dronz Jan 11 '15 at 7:47