LHC, c, -273c and misunderstood science...
September 10, 2008 3:34 AM Subscribe
Two related questions about the LHC and science I don't really understand...
1) two beams of protons are running around at "99.999% the speed of light" (according to the scientist on Radio4). What's the collision speed? Are they colliding at 1c or 2c? I'm sure the answer is 2c, but then I get side tracked by high school talk about space ships with headlights and machine guns...
2) what would happen if I (or someone else) somehow managed to get a hand into the middle of the LHC so that it got hit my the particle beam? What about something relatively inert like gold or excitable like uranium?
[Well, obviously my hand would fall off at -272.8 celcius, and the magnetic integrity of things would go wrong. Imagine I could get my hand in there without knackering the rest of the machine...]
1) two beams of protons are running around at "99.999% the speed of light" (according to the scientist on Radio4). What's the collision speed? Are they colliding at 1c or 2c? I'm sure the answer is 2c, but then I get side tracked by high school talk about space ships with headlights and machine guns...
2) what would happen if I (or someone else) somehow managed to get a hand into the middle of the LHC so that it got hit my the particle beam? What about something relatively inert like gold or excitable like uranium?
[Well, obviously my hand would fall off at -272.8 celcius, and the magnetic integrity of things would go wrong. Imagine I could get my hand in there without knackering the rest of the machine...]
From the point of view of the experimental observers, it's more like 2c than 1c due to vector addition (I'm totally lying to you but it should satisfy your curiosity on a coarse level). More intuitively, it wouldn't have been worth it to design the LHC to have head-on collisions if you could get the same bang out of diverting the beam into a stationary target. But -- and this is an important point -- 1+1 does not equal 2 when you're talking about relativistic velocities. It's more like half and half makes three quarters, or really fast plus really fast makes really really fast, but still not quite as fast as the photons flying towards your face as you read this.
High energy particle physics doesn't really deal in multiples of the speed of light, though. Above a certain velocity, it's easier to think of the particles as having a certain energy relative to you, the observer, so you don't have to do lots and lots of math that looks like 99.999947% * c + 99.999921% * c. If you want to get really technical, you can look up things like Lorentz and gamma factor to learn more about it. Don't worry if it doesn't make intuitive sense; relativity is a real mindjob even for people who have been studying physics much longer than you have.
The energy is measured in electron volts, and google can convert them to units you might be more familiar with, like joules or gallons of gas.
If you got your hand in the beam you would A) have the worse case of frostbite ever because of the extreme cold, B) experience a Total Recall style decompression event because there's no air in the collider's main ring, C) have a glow-in-the-dark hand because high energy proton collisions tend to make ordinary matter radioactive, D) be so arrested because people aren't allowed underground now that the experiment is turned up, and E) get to foot the astronomically expensive bill for restarting the experiment since jamming your hand into the beam would require cracking the ring open and letting all the science out. You would also F) have no hand anymore because the power available in the beam is measured in terawatts. If you've seen Back to the Future ("1.21 gigawatts!"), a terawatt is like a thousand Mr. Fusions. The LHC's maximum power output is several terawatts.
posted by thalakan at 4:05 AM on September 10, 2008 [2 favorites]
High energy particle physics doesn't really deal in multiples of the speed of light, though. Above a certain velocity, it's easier to think of the particles as having a certain energy relative to you, the observer, so you don't have to do lots and lots of math that looks like 99.999947% * c + 99.999921% * c. If you want to get really technical, you can look up things like Lorentz and gamma factor to learn more about it. Don't worry if it doesn't make intuitive sense; relativity is a real mindjob even for people who have been studying physics much longer than you have.
The energy is measured in electron volts, and google can convert them to units you might be more familiar with, like joules or gallons of gas.
If you got your hand in the beam you would A) have the worse case of frostbite ever because of the extreme cold, B) experience a Total Recall style decompression event because there's no air in the collider's main ring, C) have a glow-in-the-dark hand because high energy proton collisions tend to make ordinary matter radioactive, D) be so arrested because people aren't allowed underground now that the experiment is turned up, and E) get to foot the astronomically expensive bill for restarting the experiment since jamming your hand into the beam would require cracking the ring open and letting all the science out. You would also F) have no hand anymore because the power available in the beam is measured in terawatts. If you've seen Back to the Future ("1.21 gigawatts!"), a terawatt is like a thousand Mr. Fusions. The LHC's maximum power output is several terawatts.
posted by thalakan at 4:05 AM on September 10, 2008 [2 favorites]
You can calculate the velocity yourself just by knowing that each proton beam is 7 Tev.
The rest mass/energy of a proton is 938 Mev. So, divide 7 Tev/938 Mev = 7460. That is, the protons are being accelerated to 7460 times their normal energy. But how fast is that?
From the relativity equation that OxFCAF posted above, 7460=1/sqrt(1-v^2/c^2). Solving for v/c we get 0.99999999!
Thats for each of the beams. Why bother with another beam just to get another fraction towards the speed of light? The velocities don't add but the energies do. So, two 7 Tev beams colliding can explore energies as high as 14 Tev. The higher we go, the more we might find. Supersymmetric particles, Higgs bosons, perhaps new stuff?
posted by vacapinta at 4:14 AM on September 10, 2008
The rest mass/energy of a proton is 938 Mev. So, divide 7 Tev/938 Mev = 7460. That is, the protons are being accelerated to 7460 times their normal energy. But how fast is that?
From the relativity equation that OxFCAF posted above, 7460=1/sqrt(1-v^2/c^2). Solving for v/c we get 0.99999999!
Thats for each of the beams. Why bother with another beam just to get another fraction towards the speed of light? The velocities don't add but the energies do. So, two 7 Tev beams colliding can explore energies as high as 14 Tev. The higher we go, the more we might find. Supersymmetric particles, Higgs bosons, perhaps new stuff?
posted by vacapinta at 4:14 AM on September 10, 2008
On #2, the LHC uses graphite blocks as beam dumps.
"Each beam dump absorber consists of a 7m long segmented carbon cylinder of 700mm diameter, contained in a steel cylinder, comprising the dump core (TDE). This is water cooled, and surrounded by about 750 tonnes of concrete and iron shielding."
The beam has to be played across the block to stop it vaporising the graphite - even dispersed, it raises the target area to 700degC. I think it's safe to say that the beam would take your hand off.
cool link
posted by Leon at 4:21 AM on September 10, 2008
"Each beam dump absorber consists of a 7m long segmented carbon cylinder of 700mm diameter, contained in a steel cylinder, comprising the dump core (TDE). This is water cooled, and surrounded by about 750 tonnes of concrete and iron shielding."
The beam has to be played across the block to stop it vaporising the graphite - even dispersed, it raises the target area to 700degC. I think it's safe to say that the beam would take your hand off.
cool link
posted by Leon at 4:21 AM on September 10, 2008
As to the question of what the beams might do to one's hand? You need to read about the LHC beam dumps. A properly dumped beam raises the temperature of a several-ton graphite slab several hundred degrees Celsius. In every way, the LHC is an engineering marvel.
posted by fatllama at 4:25 AM on September 10, 2008
posted by fatllama at 4:25 AM on September 10, 2008
From the point of view of the experimental observers, it's more like 2c than 1c due to vector addition (I'm totally lying to you but it should satisfy your curiosity on a coarse level).
The observational frame of reference doesn't matter, nothing is moving faster than 2c, ever. That's relativity in a nutshell.
posted by phrontist at 5:45 AM on September 10, 2008
The observational frame of reference doesn't matter, nothing is moving faster than 2c, ever. That's relativity in a nutshell.
posted by phrontist at 5:45 AM on September 10, 2008
"The observational frame of reference doesn't matter, nothing is moving faster than c, ever."
The question though is about a 'collision speed', which is not the speed that something is moving at. The collision is not an object. The collision speed is more like a quantity you define for convenience (admittedly because people tend to think more in terms of Galilean relativity where the collision speed is independent of your frame of reference), so there's nothing wrong with saying it is greater than c.
posted by edd at 6:07 AM on September 10, 2008
The question though is about a 'collision speed', which is not the speed that something is moving at. The collision is not an object. The collision speed is more like a quantity you define for convenience (admittedly because people tend to think more in terms of Galilean relativity where the collision speed is independent of your frame of reference), so there's nothing wrong with saying it is greater than c.
posted by edd at 6:07 AM on September 10, 2008
Do I mean Galilean relativity? Anyway, 'everyday' velocity additions. You know what I mean.
posted by edd at 6:08 AM on September 10, 2008
posted by edd at 6:08 AM on September 10, 2008
Do I mean Galilean relativity? Anyway, 'everyday' velocity additions. You know what I mean.
The use of the term "collision speed" indicated to me that twine42 may not have taken physics and so my response should err more towards being accessible than being technically correct.
posted by thalakan at 6:15 AM on September 10, 2008
The use of the term "collision speed" indicated to me that twine42 may not have taken physics and so my response should err more towards being accessible than being technically correct.
posted by thalakan at 6:15 AM on September 10, 2008
Response by poster: So... the collision speed is near 2c, but as far as each particle is concerned the other is only getting nearer at 1c? Or have I just confused things even more?
So, hands would go all 'splody and then vapourisey. Cool. Gold then, I assume, would vanish and then gold plate the graphite dump. Would uranium vapourise or start a nuclear reaction, or both? could you get a nuclear reaction in something that is travelling at near light speeds? presumably the neutrons couldn't move forwards to trigger the next split...
Just wait til next week when I get to ask a bastardised version of the aeroplane on a conveyor belt question involving infinite length conveyor belts, velcro tyres and planes powered by singularities... :)
posted by twine42 at 6:19 AM on September 10, 2008
So, hands would go all 'splody and then vapourisey. Cool. Gold then, I assume, would vanish and then gold plate the graphite dump. Would uranium vapourise or start a nuclear reaction, or both? could you get a nuclear reaction in something that is travelling at near light speeds? presumably the neutrons couldn't move forwards to trigger the next split...
Just wait til next week when I get to ask a bastardised version of the aeroplane on a conveyor belt question involving infinite length conveyor belts, velcro tyres and planes powered by singularities... :)
posted by twine42 at 6:19 AM on September 10, 2008
Response by poster: Physics? I took physics to GCSE level (age 16), Maths to A-Level (age 18) and then did two years of a three year engineering degree. This was, however, 10 years ago and my knowledge has wandered somewhat since.
My physics knowledge is now limited to wondering what happens when you wave at Schrodinger's pets...
posted by twine42 at 6:21 AM on September 10, 2008
My physics knowledge is now limited to wondering what happens when you wave at Schrodinger's pets...
posted by twine42 at 6:21 AM on September 10, 2008
Response by poster: ( but yeah... "speed" should have been "velocity" throughout... )
posted by twine42 at 6:23 AM on September 10, 2008
posted by twine42 at 6:23 AM on September 10, 2008
So... the collision speed is near 2c, but as far as each particle is concerned the other is only getting nearer at 1c? Or have I just confused things even more?
Yes, from the perspective of any observer in the system (the two particles, some experimental sensor, Colonel Sanders, whatever) all other observable objects are moving at or less than the speed of light. If that statement were ever untrue, reality might start to unravel because then things like causality stop working. The first rule of relativity club is that the speed of light is the same for all observers, full stop. The universe deals with this by screwing with the observers' passage of time so things like shining a flashlight off the bow of something going close to light speed itself don't violate that rule.
Now, collisions aren't really measured in terms of speed. If I have a trunk doing 60 and you have a trunk doing 60 and we play chicken and neither of us flinches, then 60 + 60 = zero miles per hour because the trucks have collided. (You may have seen the mythbusters episode where they did this.) At no point is anything involved doing 120 miles per hour. If you added a third truck and had a three-way collision, you don't get 180 miles per hour; you still get zero.
One way to measure how big a particle collision is is to see how much energy was used in creating the collision. That's how it's done in particle physics. X amount of juice is used to make the particles go Y fast, which results in a collision that dissipates Z joules of energy into making quantum piñatas out of two otherwise perfectly good particles. If you want a bigger collision, you don't try for three times the speed of light, you try for three times the energy.
So, hands would go all 'splody and then vapourisey. Cool. Gold then, I assume, would vanish and then gold plate the graphite dump. Would uranium vapourise or start a nuclear reaction, or both? could you get a nuclear reaction in something that is travelling at near light speeds? presumably the neutrons couldn't move forwards to trigger the next split...
Well, from the perspective of the uranium it's not travelling at light speed, you are. It's all relative in relativity. Nuclear reactions work just fine from the perspective of the uranium, but the universe fakes you out by adjusting how fast time passes for you to keep from violating the speed of light rule. This might alter your ability to observe certain aspects of the reaction but from the perspective of the particles everything's just fine.
Now if a fissionable target has atoms that have vaporized and are moving away from each other at relativistic velocities, you're probably right. You also have a lot of explaining to do because there's hot uranium gas in a multibillion dollar device that's very hard to clean!
posted by thalakan at 6:50 AM on September 10, 2008
Yes, from the perspective of any observer in the system (the two particles, some experimental sensor, Colonel Sanders, whatever) all other observable objects are moving at or less than the speed of light. If that statement were ever untrue, reality might start to unravel because then things like causality stop working. The first rule of relativity club is that the speed of light is the same for all observers, full stop. The universe deals with this by screwing with the observers' passage of time so things like shining a flashlight off the bow of something going close to light speed itself don't violate that rule.
Now, collisions aren't really measured in terms of speed. If I have a trunk doing 60 and you have a trunk doing 60 and we play chicken and neither of us flinches, then 60 + 60 = zero miles per hour because the trucks have collided. (You may have seen the mythbusters episode where they did this.) At no point is anything involved doing 120 miles per hour. If you added a third truck and had a three-way collision, you don't get 180 miles per hour; you still get zero.
One way to measure how big a particle collision is is to see how much energy was used in creating the collision. That's how it's done in particle physics. X amount of juice is used to make the particles go Y fast, which results in a collision that dissipates Z joules of energy into making quantum piñatas out of two otherwise perfectly good particles. If you want a bigger collision, you don't try for three times the speed of light, you try for three times the energy.
So, hands would go all 'splody and then vapourisey. Cool. Gold then, I assume, would vanish and then gold plate the graphite dump. Would uranium vapourise or start a nuclear reaction, or both? could you get a nuclear reaction in something that is travelling at near light speeds? presumably the neutrons couldn't move forwards to trigger the next split...
Well, from the perspective of the uranium it's not travelling at light speed, you are. It's all relative in relativity. Nuclear reactions work just fine from the perspective of the uranium, but the universe fakes you out by adjusting how fast time passes for you to keep from violating the speed of light rule. This might alter your ability to observe certain aspects of the reaction but from the perspective of the particles everything's just fine.
Now if a fissionable target has atoms that have vaporized and are moving away from each other at relativistic velocities, you're probably right. You also have a lot of explaining to do because there's hot uranium gas in a multibillion dollar device that's very hard to clean!
posted by thalakan at 6:50 AM on September 10, 2008
I mean trucks, not trunks, but now I'm thinking two people playing chicken with two trunks going at highway speeds would make for a good youtube video.
posted by thalakan at 6:53 AM on September 10, 2008
posted by thalakan at 6:53 AM on September 10, 2008
The best information I have suggest that the explosions shown in SciFi movies are unlikely. Your hands will swell and bruise but skin is pretty tough. At least one astronaut and one high-altitude skydiver have experienced exposure of a hand to near-vacuum conditions with no permanent ill effects. The primary risks of exposure to vacuum appear to be hypoxia and gas embolism if the whole body is exposed.
I'll agree with what thalakan that the speed of a collision is meaningless. When people say that a head-on collision of two cars at 60 is equivalent to hitting a concrete slab at 120, what they are really comparing is kinetic energy.
posted by KirkJobSluder at 7:19 AM on September 10, 2008
I'll agree with what thalakan that the speed of a collision is meaningless. When people say that a head-on collision of two cars at 60 is equivalent to hitting a concrete slab at 120, what they are really comparing is kinetic energy.
posted by KirkJobSluder at 7:19 AM on September 10, 2008
Actually, two cars colliding at 60 won't have the same kinetic energy as one hitting a slab at 120, because v is squared in the old E = 0.5*m*v^2 equation. Anyway, the point is, it's just not meaningful to talk about two things at c "hitting each other at 2c", any more than it is to talk about two things at 100kph "hitting each other at 200kph". That's just not how things work. A collision isn't a physical meaureable thing, it doesn't have a velocity, and at no point does anything involved in a collision increase its velocity.
posted by nicolas léonard sadi carnot at 8:40 AM on September 10, 2008
posted by nicolas léonard sadi carnot at 8:40 AM on September 10, 2008
What KirkJobSluder said about vacuum. Most of your body is surprisingly capable of standing up to vacuum with only minor injury.
I actually wonder if the beam would cause your hand to explode either. For one thing, it's narrow (16mm at the focus points, apparently?). For another thing, I'm guessing the protons will mostly go straight through your hand, dissipating a whack of their energy electromagnetically and a smaller whack in (rarer) nuclear interactions, but not dumping the whole beam energy into your hand— that's why the beam dump has to be 7m long, because the beam isn't absorbed instantly. Whether the amount of energy they leave behind is enough to blow a hole in your hand or just leave a really, really nasty pencil-thin heat- and radation-burned channel through your hand, I don't know.
Hmm, googling says the beam luminosity is 1034 protons per cm2 per second, but it's bursty, so only 3*1018 protons per second passing through your hand.
posted by hattifattener at 8:56 AM on September 10, 2008
I actually wonder if the beam would cause your hand to explode either. For one thing, it's narrow (16mm at the focus points, apparently?). For another thing, I'm guessing the protons will mostly go straight through your hand, dissipating a whack of their energy electromagnetically and a smaller whack in (rarer) nuclear interactions, but not dumping the whole beam energy into your hand— that's why the beam dump has to be 7m long, because the beam isn't absorbed instantly. Whether the amount of energy they leave behind is enough to blow a hole in your hand or just leave a really, really nasty pencil-thin heat- and radation-burned channel through your hand, I don't know.
Hmm, googling says the beam luminosity is 1034 protons per cm2 per second, but it's bursty, so only 3*1018 protons per second passing through your hand.
posted by hattifattener at 8:56 AM on September 10, 2008
Hrm, looking up proton therapy information, it does appear that protons zip through human tissue with a maximum ionizing effect at a depth proportional to their energy.
The other side of the story is that the beam has 725MJ of energy in it. Medical lasers operate on the mJ range. So even a small fraction of the beam's energy dumped into your hand would result in some nasty heating. My order-of-magnitude calculation is that the LHC has 10^14 times the energy of a medical laser. So the fact that only a small number of protons will be absorbed by human tissue is cold comfort if you find yourself standing in front of that beam.
posted by KirkJobSluder at 9:44 AM on September 10, 2008
The other side of the story is that the beam has 725MJ of energy in it. Medical lasers operate on the mJ range. So even a small fraction of the beam's energy dumped into your hand would result in some nasty heating. My order-of-magnitude calculation is that the LHC has 10^14 times the energy of a medical laser. So the fact that only a small number of protons will be absorbed by human tissue is cold comfort if you find yourself standing in front of that beam.
posted by KirkJobSluder at 9:44 AM on September 10, 2008
What about the near-miss situation. If I'm in a car going 60mph, and a car goes past me in the opposite direction at 60mph, doesn't it move past me faster than if I was standing still? Or is this one of those "airplane taking off from a runway that's a conveyor belt" sort of problems?
posted by rikschell at 9:54 AM on September 10, 2008
posted by rikschell at 9:54 AM on September 10, 2008
I don't see how driving cars is a good analogy for relativistic speeds.
In the near-miss scenario at 60mph, if you imagine that your reference frame is stationary, it does appear that the opposing car is coming at you at 120 miles an hour. And if you collide, it's a helluva lot more like hitting something immobile at 120 than it is hitting that same something at 60.
But, this doesn't apply at relativistic speeds, which is why we needed a particularly imaginative and smart fellow to tell us that shit at high speed doesn't act like shit at speeds we're used to. (Among other neat things he told us about, like space-time.)
As people have mentioned, it's the energy of the pair of smashing protons that you should evaluate to predict the results. For a massed particle, you spend asymptotic energy as you approach c. So, no matter how much energy you dump into it, you just get more 9's on your 99.9999% figures. But the lack of significantly increased velocity doesn't change the fact that you have added all that extra energy which will dissipate when it hits something suitably solid.
posted by Netzapper at 11:41 AM on September 10, 2008 [1 favorite]
In the near-miss scenario at 60mph, if you imagine that your reference frame is stationary, it does appear that the opposing car is coming at you at 120 miles an hour. And if you collide, it's a helluva lot more like hitting something immobile at 120 than it is hitting that same something at 60.
But, this doesn't apply at relativistic speeds, which is why we needed a particularly imaginative and smart fellow to tell us that shit at high speed doesn't act like shit at speeds we're used to. (Among other neat things he told us about, like space-time.)
As people have mentioned, it's the energy of the pair of smashing protons that you should evaluate to predict the results. For a massed particle, you spend asymptotic energy as you approach c. So, no matter how much energy you dump into it, you just get more 9's on your 99.9999% figures. But the lack of significantly increased velocity doesn't change the fact that you have added all that extra energy which will dissipate when it hits something suitably solid.
posted by Netzapper at 11:41 AM on September 10, 2008 [1 favorite]
Discussions about adding speeds in relativity seem to always get contorted trying to avoid doing any math. This is actually a problem that isn't too hard. All you need to know is the rule for adding velocities,
If I have a particle at 0.999 99c = c - 10-5c colliding with another going in the other direction with almost the same speed, I can add their velocities with the same rule:
If you can match the speeds/energies of the colliding beams to 1% (small=0.01) the "rest frame" of the collision is still moving at 10-7c, a respectable speed for a car. In practice the energies of the beams are mismatched by a few percent on purpose, so that most of the collision products go in one direction. Then you can have a "forward" detector and a "backward" detector that see different types of things: the forward detector sees everything, but only the really energetic detritus can go backwards.
To your other question: if the graphite beam dump is 7 m long and scatters every proton, and your (graphite) hand is 1 cm thick, it will scatter about one out of every thousand protons. The total power (energy * current) in the beam is a few megawatts; a thousandth of this is a few kilowatts. An electric stove eye is also a few kilowatts. Of course, an electric stove eye is a foot across, while the LHC beam is millimeters or microns across. The terribly exciting electric pickle demonstration involves about a half a kilowatt of power (I think the pickle draws four amps), but the entire volume of the pickle is involved.
Actually somewhat less heat would be involved, since it takes many scatters for the beam to stop completely; your hand would be a "thin" target. So most likely the beam would etch a nicely cauterized hole through your hand.
One of the processes in proton-nucleus collisions is for the nucleus to heat up and spit out neutrons, mesons, and other protons. This is called "spallation." There's a lot that can be done with the neutrons, in particular, and there are running spallation neutron sources in the US that stop protons on metals. The IPNS, with a uranium target, has recently shut down; the 800 MeV protons on tungsten neutron source in New Mexico has recently been supplanted in brightness by the 1 GeV proton on mercury neutron source in east Tennessee.
posted by fantabulous timewaster at 11:46 AM on September 10, 2008 [1 favorite]
v1+2 = (v1+v2) / (c2+v1v2).One mile per hour is 1.5×10-9c. So if your speeds are convenient to talk about in miles per hour, the approximation (c2+v1v2) = c2 is good to fifteen or eighteen decimal places. Usually we measure speeds to three decimal places or so, and what happens in the fifteenth decimal place isn't interesting. So your finely-honed intuition tells you that v1+2 = v1+v2 is mostly right.
If I have a particle at 0.999 99c = c - 10-5c colliding with another going in the other direction with almost the same speed, I can add their velocities with the same rule:
v1+2 = (small×10-5c) / c2 + (c2 + small ×2×10-5 + very small ×10-10).Now that we're talking about relativistic speeds, the pretty good approximation is (c2+v1v2) = 2c2. Adding c to c gives 2c/2c2=c.
If you can match the speeds/energies of the colliding beams to 1% (small=0.01) the "rest frame" of the collision is still moving at 10-7c, a respectable speed for a car. In practice the energies of the beams are mismatched by a few percent on purpose, so that most of the collision products go in one direction. Then you can have a "forward" detector and a "backward" detector that see different types of things: the forward detector sees everything, but only the really energetic detritus can go backwards.
To your other question: if the graphite beam dump is 7 m long and scatters every proton, and your (graphite) hand is 1 cm thick, it will scatter about one out of every thousand protons. The total power (energy * current) in the beam is a few megawatts; a thousandth of this is a few kilowatts. An electric stove eye is also a few kilowatts. Of course, an electric stove eye is a foot across, while the LHC beam is millimeters or microns across. The terribly exciting electric pickle demonstration involves about a half a kilowatt of power (I think the pickle draws four amps), but the entire volume of the pickle is involved.
Actually somewhat less heat would be involved, since it takes many scatters for the beam to stop completely; your hand would be a "thin" target. So most likely the beam would etch a nicely cauterized hole through your hand.
One of the processes in proton-nucleus collisions is for the nucleus to heat up and spit out neutrons, mesons, and other protons. This is called "spallation." There's a lot that can be done with the neutrons, in particular, and there are running spallation neutron sources in the US that stop protons on metals. The IPNS, with a uranium target, has recently shut down; the 800 MeV protons on tungsten neutron source in New Mexico has recently been supplanted in brightness by the 1 GeV proton on mercury neutron source in east Tennessee.
posted by fantabulous timewaster at 11:46 AM on September 10, 2008 [1 favorite]
Response by poster: Fantabulous - wow... a maths answer that made sense. Ta! That really is simple enough to be explained to 15 year olds.
So... if the LHC is 27km in circumference, the beams are travelling a loop in 27000/(c*0.999) = 9.01*10-5 seconds.
If a thousandth of the beam is scattered by my graphite hand, then it will take 1000 laps before the stream is trash *
Those 1000 laps will be up in less than a tenth of a second... 1/10 of a second to ruin a science experiment and take a couple of slices off one of my graphite limbs in the process...
Sorry... I'm not asking any more questions here, I'm just marvelling in the silly revelations I'm coming to. There's something incredibly cool about combining insanely large and insanely small numbers into something thats, well, real. :)
* I know this is wrong, but a) I can't be bothered to do the maths and b) my graphite fingers are crap at poking the calculator. I suppose they'd make good pencils though...
posted by twine42 at 12:10 PM on September 10, 2008
So... if the LHC is 27km in circumference, the beams are travelling a loop in 27000/(c*0.999) = 9.01*10-5 seconds.
If a thousandth of the beam is scattered by my graphite hand, then it will take 1000 laps before the stream is trash *
Those 1000 laps will be up in less than a tenth of a second... 1/10 of a second to ruin a science experiment and take a couple of slices off one of my graphite limbs in the process...
Sorry... I'm not asking any more questions here, I'm just marvelling in the silly revelations I'm coming to. There's something incredibly cool about combining insanely large and insanely small numbers into something thats, well, real. :)
* I know this is wrong, but a) I can't be bothered to do the maths and b) my graphite fingers are crap at poking the calculator. I suppose they'd make good pencils though...
posted by twine42 at 12:10 PM on September 10, 2008
twine42, that's high praise. Thanks.
My inner pedant just noticed I put dropped a c2. The rule is actually
posted by fantabulous timewaster at 2:07 PM on September 10, 2008
My inner pedant just noticed I put dropped a c2. The rule is actually
v1+2 = c2 × (v1+v2) / (c2+v1v2).This is why I usually use feet per nanosecond, so c = 1, argh.
posted by fantabulous timewaster at 2:07 PM on September 10, 2008
In 2003, two-thirds of the superconducting magnets in the Tevatron’s six-kilometer ring quenched at the same time. The beam drilled a hole in one collimator and created a 30-centimeter groove in another. That accident, while serious, was the only one in the accelerator’s 20-year history, and the machine was back up and running within two weeks. Could something similar happen on a larger scale at the LHC?Protecting the LHC from itself
posted by KirkJobSluder at 9:20 PM on September 10, 2008 [1 favorite]
Cosmic ray particles have been measured/estimated to have speeds of as high as 0.999... c, where the number of 9s is 22. (This is what my memory holds from having read Friedlander's book a year or so ago.)
posted by neuron at 10:05 AM on September 11, 2008
posted by neuron at 10:05 AM on September 11, 2008
neuron, that wasn't a direct speed measurement. You're probably thinking of the 1991 oh-my-god particle, which had a total reported energy 300,000,000 TeV (compared to 7 TeV at the LHC).
The oh-my-god particle didn't actually interact with any man-made detector. It interacted with the top of the atmosphere and make a whole shitload of protons, neutrons, electrons, positrons, muons, photons, pions, neutrinos, who knows what else. If you have a big group of charged-particle detectors on the ground, you see a whole bunch of muons and electrons in all of them at almost the same time. Newer arrays for hunting for cosmic rays also look at the fluorescence of the sky.
So the evidence isn't quite that a proton moving at (1 - 1022)c hit the top of the atmosphere. The evidence is that something made an awful lot of short-lived, relativistic subatomic particles in the upper atmosphere in a very short amount of time, and if those particles were distributed like a usual cosmic-ray shower then the total energy was fifty joules. It could have been
posted by fantabulous timewaster at 11:10 AM on September 11, 2008
The oh-my-god particle didn't actually interact with any man-made detector. It interacted with the top of the atmosphere and make a whole shitload of protons, neutrons, electrons, positrons, muons, photons, pions, neutrinos, who knows what else. If you have a big group of charged-particle detectors on the ground, you see a whole bunch of muons and electrons in all of them at almost the same time. Newer arrays for hunting for cosmic rays also look at the fluorescence of the sky.
So the evidence isn't quite that a proton moving at (1 - 1022)c hit the top of the atmosphere. The evidence is that something made an awful lot of short-lived, relativistic subatomic particles in the upper atmosphere in a very short amount of time, and if those particles were distributed like a usual cosmic-ray shower then the total energy was fifty joules. It could have been
- a single cosmic ray traveling at nearly the speed of light, given enormous energy by some unknown mechanism
- many billions of ordinary cosmic rays all arriving simultaneously, grouped by some unknown mechanism
- an ordinary cosmic ray shower with some unusual distribution in space, that tricked the detector
posted by fantabulous timewaster at 11:10 AM on September 11, 2008
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Using the formula from the Wikipedia article and simplifing 99.999% to just 99%, we can take w to be 0.99c, v to be -0.99c (it's moving in the opposite direction), and come up with w' = 1.98 / (1 - (-.9801) / (1^2)) = 99.9949% the speed of light.
For #2, the huge magnetic fields and freezing your hand off would probably be a much bigger problem than the kinetic energy of the beam itself. Each proton is accelerated to 7 TeV (see Wikipedia, which is actually an extremely small amount of energy - 1.12E-6 Joules.
posted by 0xFCAF at 3:54 AM on September 10, 2008