What-If 0073: "Lethal Neutrinos"

[davidstarlingm]

Randall did say "energy delivered to your retina"

Ooh, good point. I missed that.

Though there's still a chance he's talking about electromagnetic radiation alone, rather than some of the other types of energy released in both types of explosions.

[brenok]

But I would also like people not to encourage stars to explode.

Why not? As long as it is not our star...

A supernova can kill us from thousands of lightyears away

[Ehsanit]

But I would also like people not to encourage stars to explode.

Why not? As long as it is not our star...

A supernova can kill us from thousands of lightyears away

Meh, not for thousands of years.

[phlip]

I came here hoping someone would explain why ln(x)^e approaches x. As a computer scientist, I haven't done derivatives in years and I'm predisposed to keep it that way while the streak is hot.

(I adore math and study it on my own, but I prefer mathematical theories which have virtually no applications. The calculus of limits is far too useful.)

It's not so much that it "approaches" x, or even is a particularly relevant number... rather, it's easily confused with e^ln(x), which is equal to x by definition (for real x greater than 0, at least... let's leave complex multivalue logarithms for another time). So, d(e^ln(x))/dx is 1, but d(ln(x)^e)/dx is very much not.

The point being that ln(x)^e is a very weird construction. It's not something that's likely to ever come up in an actual mathematical application. So when you see it, you have to double-take, hence the analogy to "lethal dose of neutrinos". cf the mouseover text for Approximations.

[Klear]

cf the mouseover text for Approximations.

I read the mouseover text. Then I googled "Twin prime" and read the wikipedia article. This part caught my eye:

On April 17, 2013, Yitang Zhang announced a proof that for some integer N that is at most 70 million, there are infinitely many pairs of primes that differ by N.[1][2] Zhang's paper was accepted by Annals of Mathematics in early May 2013.[3] Terence Tao subsequently proposed a Polymath project collaborative effort to optimize Zhang’s bound;[4] as of November 22, 2013, James Maynard claims to have reduced the bound to N = 600.[5]

Holy shit, it's happening right now!

[cjmcjmcjmcjm]

A black hole would make a good shield for neutrinos. A neutron star might work too.

Now I'm wondering how much of a neutron star we would need between the sun and Earth to absorb all those neutrinos.

[cjmcjmcjmcjm]

But I would also like people not to encourage stars to explode.

Why not? As long as it is not our star...

A supernova can kill us from thousands of lightyears away

Meh, not for thousands of years.

Unless the star in question went supernova thousands of years in the past and we only fAAAAAAAAAAAAAAAA!~!<sup>‡

‡</sup>and the room was slowly become vaporise

[Aiwendil]

This also reminded me of a quotation from a physicist who introduced the neutrino but now I can't find it anymore. Maybe it was Fermi but I'n not quite sure. The quote something like "I've committed the ultimate sin in physics - I've introduced an undetectable particle".

That was Wolfgang Pauli.

[Mousepup]

Re: lethal dose, I'd like some help getting a sense of perspective here.

For stuff like gamma rays, relativistic baryons and leptons and whatever other things we think of as hazardous radiation, a Sievert for a 50-ish kilogram human is a quantity which it makes sense to measure in Joules, right? For thermal radiation or functional equivalent, which I'm guessing isn't measured in Sieverts, a lethal dose is maybe kJ or MJ neighborhood, I'd reckon. (by burning and/or other heat damage)

So, the "hazardous" in "hazardous radiation" means it's more damaging to us than stovetop or microwave radiation per unit energy by hundreds of thousands of times, or millions of times. That's the sort of terms I'm looking for.

For Neutrino radiation, is a lethal short-term dose in the neighborhood of grams, kilograms or megagrams? I'm looking for a unit of mass, so that I can compare quantitatively to those J and MJ, to help get my head around the stupendosity of "lethal neutrinos."

Sadly, I'm "pretty sure" the required dose is still low enough that the mechanism of organic damage is ionising radiation, as opposed to gravitational interaction. (Gravity for the star itself is enough to be hazardous, but not to such extent that I expect the portion converted to neutrinos to be ...uh, grave.)

[feedayeen]

A black hole would make a good shield for neutrinos. A neutron star might work too.

Now I'm wondering how much of a neutron star we would need between the sun and Earth to absorb all those neutrinos.

Given that a neutron star is only a few miles in diameter, I'm afraid it won't make much of a difference to the net neutrino flux of the Earth. Perhaps we could create a safe zone for the best (richest) of us to avoid this hazard.

Alternatively we could use these ultra dense objects as a gravitational lens to deflect neutrinos from the Earth.

[january1may]

For Neutrino radiation, is a lethal short-term dose in the neighborhood of grams, kilograms or megagrams? I'm looking for a unit of mass, so that I can compare quantitatively to those J and MJ, to help get my head around the stupendosity of "lethal neutrinos."

Sadly, I'm "pretty sure" the required dose is still low enough that the mechanism of organic damage is ionising radiation, as opposed to gravitational interaction. (Gravity for the star itself is enough to be hazardous, but not to such extent that I expect the portion converted to neutrinos to be ...uh, grave.)

A quick WolframAlpha count seems to say that, at a distance of several AU, the rest mass of all passing neutrinos (not only radiation-providing ones, but I suppose they all provide gravity) would be somewhere between kilograms and megagrams per square meter (the uncertainty being due to unclear neutrino mass).
Of course, when it comes to objects travelling close to the speed of light, I'm not sure whether "rest mass" is even a meaningful way to calculate gravitational interaction. And of course any gravitational interaction in this case would be mostly cancelled off by the simple fact of neutrinos going in all directions roughly equally (assuming, of course, that they are - I'm not entirely sure).

[Showsni]

I believe the derivative is eln(x)^e-1/x, correct?

That's what I get...

If h(x) = f(g(x)) then h'(x) = f'(g(x))g'(x)

Here, h(x) = ln(x)^e

So f(x) = x^e

g(x) = ln(x)

f'(x) = ex^e-1

g'(x) = 1/x

h'(x) = f'(g(x))g'(x) = eln(x)^e-1/x

Although, "ln(x)^e" seems a little confusing. Is it (ln(x))^e or ln(x^e)? What's the derivative of the second? e/x?

[airdrik]

Although, "ln(x)^e" seems a little confusing. Is it (ln(x))^e or ln(x^e)? What's the derivative of the second? e/x?

Function calls are higher precedence (to just about anything), so ln(x)^e is equivalent to (ln(x))^e.
for ln(x^e), you just swap your f and g:

So f(x) = ln(x)

g(x) = x^e

f'(x) = 1/x

g'(x) = ex^e-1

h'(x) = f'(g(x))g'(x) = 1/(x^e)(ex^e-1) = e/x

[brenok]

Isn't it much more simple to use the property that ln(a^b) = b ln(a)?

[GreenTom]

Anyone else remember "Anvil of Stars." Only SF book I can think of where lethal neutrinos were used, and I think roughly correctly.

Test at (spoiler) http://books.google.com/books?id=kFAhNMnnNuUC&pg=PA165&lpg=PA165&dq=anvil+of+stars+neutrinos&source=bl&ots=cOutJWxOfa&sig=Cy_TmYlyc0bZng-RlNYDnsZZy_U&hl=en&sa=X&ei=VUWeUsbXLfjNsQTm2YLYBA&ved=0CC4Q6AEwAA#v=onepage&q=anvil%20of%20stars%20neutrinos&f=false

[airdrik]

Isn't it much more simple to use the property that ln(a^b) = b ln(a)?

Yeah, you certainly could do that (honestly, I haven't done this kind of math in quite a while, and it was a bit of a stretch for me to remember things like 1/(x^y) = x^-y and x^ax^b = x^a+b)

[Mousepup]

Ooooooh, dang. I assumed that gravity must act on all mass rather than just rest mass, since it acts on light. (and my guess was still low!) Well, I'll add finding a list of what doesn't interact to my to-dos.

I was also planning on using a "spherical chicken" approximation treating the whole blast as instantaneous and the star as a point source, totally didn't even think about what relativistic effects would do to the pull of the passing wave of Neutrinos.

[Diadem]

Ooooooh, dang. I assumed that gravity must act on all mass rather than just rest mass, since it acts on light. (and my guess was still low!) Well, I'll add finding a list of what doesn't interact to my to-dos.

It does actually. Gravity acts on energy. All kinds of energy, whether it is rest mass or kinetic energy, or electromagnetic energy, or even potential energy. Under earth-like conditions everything but the rest mass is pretty negligible though.

If you choose to use the concept of relativistic mass, then gravity definitely acts on this mass.

[davidstarlingm]

Gravity is simply reference-frame dependent. The relativistic mass gained by approaching the speed of light only exists in the reference frame of the very-fast-moving object, and thus gravity does not operate on it directly.

[Diadem]

Eh? In the frame of the moving object, the moving object is standing still, there is no kinetic energy and gravity thus doesn't act on it.

The gravitational interaction between two objects does not depend on the absolute velocity of the objects. Of course, that's the principle of relativity. But it does depend on their relative velocity.

Stop thinking of gravity as acting on mass. Think of it as acting on energy. Energy bends spacetime, gravity is this bending of spacetime.

[davidstarlingm]

Eh? In the frame of the moving object, the moving object is standing still, there is no kinetic energy and gravity thus doesn't act on it.

The gravitational interaction between two objects does not depend on the absolute velocity of the objects. Of course, that's the principle of relativity. But it does depend on their relative velocity.

Yeah, sorry, got those things flipped. Gravitational interaction does depend on relative velocity, but not on absolute velocity. I knew there was some significance there, but I must have gotten them mixed up.

If gravity had any interaction with absolute velocity, then an object with a sufficient absolute velocity would collapse to a black hole under its own weight, and this would allow us to measure absolute velocity and thus determine the rest reference frame of spacetime.

[gmalivuk]

Isn't it much more simple to use the property that ln(a^b) = b ln(a)?

Not for the actual equation under discussion, which is ln(a)^b rather than ln(a^b).

[brenok]

Isn't it much more simple to use the property that ln(a^b) = b ln(a)?

Not for the actual equation under discussion, which is ln(a)^b rather than ln(a^b).

I was talking about the post above me, but maybe I should have quoted it for unambiguity.

[Trickster]

I came here hoping someone would explain why ln(x)^e approaches x. As a computer scientist, I haven't done derivatives in years and I'm predisposed to keep it that way while the streak is hot.

(I adore math and study it on my own, but I prefer mathematical theories which have virtually no applications. The calculus of limits is far too useful.)

It's not so much that it "approaches" x, or even is a particularly relevant number... rather, it's easily confused with e^ln(x), which is equal to x by definition (for real x greater than 0, at least... let's leave complex multivalue logarithms for another time). So, d(e^ln(x))/dx is 1, but d(ln(x)^e)/dx is very much not.

The point being that ln(x)^e is a very weird construction. It's not something that's likely to ever come up in an actual mathematical application. So when you see it, you have to double-take, hence the analogy to "lethal dose of neutrinos". cf the mouseover text for Approximations.

I do get the idea, but either Google is wrong, or you guys are wrong (so far I think three forum users have insisted it doesn't approach x).

I fear that I am predisposed to trust my Google overlords. ^@.@^ <-- (hypnotized wuffy)

[Trickster]

A black hole would make a good shield for neutrinos. A neutron star might work too.

No. Placing a black hole between you and a supernova would actually make it worse. Think about how the area outside of the photon sphere will bend radiation and hot particles.

[Diadem]

I do get the idea, but either Google is wrong, or you guys are wrong (so far I think three forum users have insisted it doesn't approach x).

I fear that I am predisposed to trust my Google overlords. ^@.@^ <-- (hypnotized wuffy)

Google is right. But that graph isn't a graph of y=x. The starting point is wrong, and the graph is not entirely straight. It looks straight, but that's because you're not taking into account the big picture.

Plotted from 0 to 10
Plotted from 0 to 100
Plotted from 0 to 1000
Plotted from 0 to 10,000

As you can see, it's not straight.

[davidstarlingm]

A black hole would make a good shield for neutrinos. A neutron star might work too.

No. Placing a black hole between you and a supernova would actually make it worse. Think about how the area outside of the photon sphere will bend radiation and hot particles.

**immediately begins fantasizing about a day when warfare will be conducted on a galactic scale using black-hole gravitational lenses to turn supernovae into neutrino lasers**

[rmsgrey]

A black hole would make a good shield for neutrinos. A neutron star might work too.

No. Placing a black hole between you and a supernova would actually make it worse. Think about how the area outside of the photon sphere will bend radiation and hot particles.

**immediately begins fantasizing about a day when warfare will be conducted on a galactic scale using black-hole gravitational lenses to turn supernovae into neutrino lasers**

Isn't that book 8 of the Lensman series?

[davidstarlingm]

A black hole would make a good shield for neutrinos. A neutron star might work too.

No. Placing a black hole between you and a supernova would actually make it worse. Think about how the area outside of the photon sphere will bend radiation and hot particles.

**immediately begins fantasizing about a day when warfare will be conducted on a galactic scale using black-hole gravitational lenses to turn supernovae into neutrino lasers**

Isn't that book 8 of the Lensman series?

Haven't read it, but great minds and all that.

Sudden inspired and probably totally inane thought: you can create an artificial black hole if you cram enough bosons into a small enough space. Black holes can also be used for gravitational lensing. And sufficiently small ones can also act as extremely hot, bright sources of energy thanks to Hawking radiation's inverse relationship to mass.

Is there thus any arrangement of micro-black-holes which would be able to produce a self-sustaining, controllable chain reaction wherein Hawking radiation from one set of micro black holes was gravitationally focused by another set of micro black holes to create a third set of black holes which then started the cycle over? Like some sort of gravitational/radiation dynamo. Could either be contained (perhaps by some sort of orbital confinement) or directed along some axis.

Other than the obvious weaponization, it could potentially be used for peaceful power generation or gravitational manipulation, the production of discrete gravity waves, and so on. Maybe even for some sort of gravity propulsion.

[january1may]

I do get the idea, but either Google is wrong, or you guys are wrong (so far I think three forum users have insisted it doesn't approach x).

I fear that I am predisposed to trust my Google overlords. ^@.@^ <-- (hypnotized wuffy)

As far as I can tell, ln(x)^e<=x for all x>1, and the equality only happens for x=e^e. (If you substitute x=e^t, you get t^e<=e^t, which is, I believe, a fairly well-known inequality.)
Unfortunately, e^e is just over 15, and the Google graph cuts off at just under 15. So it does indeed look like something approaching x, because the actual line goes tangent to y=x just barely outside the visible graph (and then, of course, it goes below y=x again; ultimately, as x approaches infinity, the slope approaches zero).

[Cynical Idealist]

The fraction stopped by your shield is negligible (unless your shield is like a billion miles of solid lead), so you still have to deal with all the neutrinos, but now you also have to deal with the radiation the shield is giving off due to neutrinos hitting it.

Surely the shield isn't going to get that radioactive. And the radiation will be generated evenly throughout the shielding, so most of it should be blocked by the rest of the shielding (if your have shielding thick enough to block any noticeable amount of neutrinos).

Also the mean free path of a neutrino in solid lead is 22 light years, so I think that the fraction stopped by a billion miles (~1.4 light hours) of lead is going to be negligible.

[gmalivuk]

I do get the idea, but either Google is wrong, or you guys are wrong

Or you're wrong.

Classic first-graphing-calculator mistake, not trying different scales to see if the graph keeps looking like you think it does right near the origin.

A black hole would make a good shield for neutrinos. A neutron star might work too.

No. Placing a black hole between you and a supernova would actually make it worse. Think about how the area outside of the photon sphere will bend radiation and hot particles.

Like any kind of lens, a gravitational one doesn't magically focus everything right where you happen to be sitting. And even if you do happen to be sitting where it focuses most of the radiation, it will bend the paths of all the massive particles being thrown out more than it bent the light, so you won't be at the focal point of any of that.

[Trickster]

Or you're wrong.

Cheerfully admitted, but see the point previous about hoping I could avoid doing Actual Math this time around. (Although I failed--I wandered into the Hockey Puck comic thread and had to calculate the number of nucleons in a regulation-size puck made of neutronium. Ah well.)

A black hole would make a good shield for neutrinos. A neutron star might work too.

No. Placing a black hole between you and a supernova would actually make it worse. Think about how the area outside of the photon sphere will bend radiation and hot particles.

Like any kind of lens, a gravitational one doesn't magically focus everything right where you happen to be sitting. And even if you do happen to be sitting where it focuses most of the radiation, it will bend the paths of all the massive particles being thrown out more than it bent the light, so you won't be at the focal point of any of that.

The focal point doesn't matter. I'm assuming the person is smaller than any reasonable black hole, and the supernova is considerably larger and far enough away for the black hole to block it if it did not warp spacetime.

Let me provide an analogy. If a black hole lies between you and a star, you can see the same star twice in the night sky. Removing the black hole would reduce it to a single star, but would not double the amount of light you see vs. when it was blocked. You're getting radiation from more angles.

[gmalivuk]

Radiation from more angles doesn't mean more radiation, though.

[davidstarlingm]

Radiation from more angles doesn't mean more radiation, though.

I'm sure it wouldn't be too hard to do some ray-tracing and come up with an equation to predict the amount of incident radiation for any spacing/BH size.

[Mousepup]

I'm under the impression that "mass" and "energy" are by default taken literally, so they mean the same thing as each other in absence of qualifiers like "rest mass" and "kinetic energy."

Back to the present subtopic, If the source and the lens are of just the right sizes and distance to produce a cylindrical beam, it's trivial to find a place to take shelter that's outside that beam but still inside the occlusion cone created by the lens' capture area. If the beam's converging instead, the area to avoid is also limited in range because it's gonna start diverging after the focus; outside this concentration area, the lens is providing shelter.

Ah, derp. Thinking just a wee, teeny bit beyond the maximally simple thought experiment, there's a nice range of configurations which produce HOLLOW cylindrical beams, so at least out beyond the foci for all the points in the source' volume, you can actually hide directly behind the shield in relative shade while being completely surrounded by that beam.

Either way, for me the takeaway is simply that attenuation distance for any kind of absorptive shielding is so huge that it's effectively indistinguishable from just using good old-fashioned distance from the source, so gravitational lensing is the most likely use of matter for reducing exposure/safe distance.

[Trickster]

Radiation from more angles doesn't mean more radiation, though.

This is true. I'm basing my opining here on intuition. You're going from laminar flow of radiation to cross-flow, but the radiation is being stretched in multiple directions. I'd have to do math. Ugh. I'm going to have to do this, I just know it. Can't get it out of me noggin.

[Trickster]

I'm under the impression that "mass" and "energy" are by default taken literally, so they mean the same thing as each other in absence of qualifiers like "rest mass" and "kinetic energy."

Back to the present subtopic, If the source and the lens are of just the right sizes and distance to produce a cylindrical beam, it's trivial to find a place to take shelter that's outside that beam but still inside the occlusion cone created by the lens' capture area. If the beam's converging instead, the area to avoid is also limited in range because it's gonna start diverging after the focus; outside this concentration area, the lens is providing shelter.

Ah, derp. Thinking just a wee, teeny bit beyond the maximally simple thought experiment, there's a nice range of configurations which produce HOLLOW cylindrical beams, so at least out beyond the foci for all the points in the source' volume, you can actually hide directly behind the shield in relative shade while being completely surrounded by that beam.

Either way, for me the takeaway is simply that attenuation distance for any kind of absorptive shielding is so huge that it's effectively indistinguishable from just using good old-fashioned distance from the source, so gravitational lensing is the most likely use of matter for reducing exposure/safe distance.

I'm not so sure. There is zero perfect shade, because even just outside the photon sphere you'll be hit by radiation pulled toward the hole. Even within the sphere but outside of the event horizon you get hit. It's a question of quantity.

[davidstarlingm]

Wouldn't there be a gradient behind the black hole, where one region had much-more-concentrated radiation and the region behind it had much-less-concentrated radiation? Something like this, perhaps:

[Mousepup]

I don't know how "so sure" is, but I did not mean to suggest that there's perfect shade anywhere, just that it has to create relative shade for some volumes somewhere in order to concentrate radiation anywhere else- which sounds kinda dumb when I put it that way, but... uh, yeah.

I was also picturing things kinda backwards. Relative to David's plot there, I was imagining putting the lens closer to the source, so the beam-side focal point is farther away instead: the red region is about where I was imagining the source, so the beam on the far side is much more elongated. If it helps, I only pictured it that wacky way because I was kinda starting from the mental image of a cylindrical beam setup as described upthread, not for any sane reason.