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The Decay Rate of a Hypothetical Antimatter Container

  1. Dec 29, 2016 #1
    Hi All,

    First time poster here, and I've got a couple questions. Straight up, I'm writing a sci-fi story where antimatter is utilised as fuel (feel free to laugh at the cliche), and I want to portray it as realistically as possible. To that end, I've come up with a storage method and I want to see if the math holds up.

    So, here are the parameters:
    1. assume you have exactly one kilogram of antimatter
    2. assume the entire mass uniformly consists of anti-Alpha particles (nuclei of 2 antiprotons and 2 antineutrons)
    3. assume this mass is contained in a vessel with an electromagnetic field powerful enough to suspend the total mass from coming into contact with any other matter
    4. assume that the vessel is equipped with an internal reaction chamber that has a source of normal Alpha particles, and that the vessel is continuously drawing Alphas and anti-Alphas into this chamber at a consistent rate
    5. assume the annihilation energy released in this reaction chamber is fully converted to electrical energy and used to power the electromagnetic field to contain the remaining mass
    So, with all that out of the way, two questions:
    1. how much energy in joules per second would be needed to maintain a containment field strong enough to maintain the repulsive forces necessary to keep the antimatter contained?
    2. based on that, what would the rate of consumption of stored antimatter in metric mass units per hour have to be to make that much energy?
    I'm hoping it will last years, preferably decades, ideally centuries before being depleted. But if it only lasts days or hours, then it's not a practical solution (even in fiction) and I'll have to have a rethink.
     
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  3. Dec 30, 2016 #2

    mfb

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    Antihydrogen would make the storage much easier: you can freeze it to a solid, electrically charged block, and levitate that. Spin it up to stabilize it in a suitable outer field, and cooling the whole setup should be the only power consumption you have. In space far away from stars, passive cooling could be sufficient and your storage can be stable as long as the hull doesn't get damaged from impacting dust particles.

    Order of magnitude estimate for anti-alpha particles: 1 kg of those have a charge of 48 MC (Megacoulomb). You cannot trap it with a purely electrical potential, because you cannot create a minimum of a potential in free space. You'll need a magnetic field. Just plugging in numbers leads to a magnetic flux of 1010 T m2. If your trap has a diameter of 1 kilometer, it probably needs a magnetic field strength of the order of 18,000 T. As comparison: the largest magnets we have can produce a field of 12 Tesla with a side-length of the order of 10 meters. And there is no realistic way to scale up the field strength, as the materials used don't work with fields beyond ~20 T.
     
  4. Dec 30, 2016 #3
    The maximum energy you can get from 1kg of antimatter should be given by the formula ##E=mc^2## so it should be for 1Kg, ##E=9x10^{16}joule\approx 2.5x10^{10}KWh##. Even if the whole system that does the containment has the power requirements of 1MW(MegaWatt)=1000KW , the energy could last for ##2.5x10^7## hours. That's around 3000 years.

    It would last for around 3 years though if the containment system requirements are 1GW (GigaWatt) (but I suppose that's unrealistically high, however I cant give details for the proper containment system of antimatter because I don't know much in this area of science).

    As to the rate of consumption of antimatter, again if we suppose a containment system of 1MW, the rate should be (if my calculations are correct) around ##10^{-8}\frac{gr}{sec}##
     
    Last edited: Dec 30, 2016
  5. Dec 30, 2016 #4

    mfb

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    The electromagnet of the LHCb experiment at CERN, with a diameter of just a few meters and a magnetic field of about 1 T, needs 4 MW to operate. A kilometer-sized 100+ Tesla magnet wouldn't work with 1 GW, and with superconductors you need even larger magnets which lead to all sorts of other problems.
     
  6. Dec 30, 2016 #5
    I have to admit I don't know much about the magnetic containment of antimatter, neither about superconductors.

    What's the stronger electromagnet made from superconductor up to date?

    Don't forget we are in the realm of sci-fi here, maybe in future some superconducting material is discovered that can support million of amperes (and hence million of Tesla).
     
  7. Dec 30, 2016 #6

    mfb

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    Superconductivity breaks down beyond some field strength, the value depends on the material, the temperature and the current density. We have superconductors that are easy to handle that can handle ~10 T. Coils made out of high-temperature superconductors are harder to manufacture, but they can support ~20 T fields.

    Sure, that is not a fundamental limit, new materials might support higher fields, but (a) kilotesla look unlikely and (b) we can't make interesting calculations if we don't even know the order of magnitude of science-fiction materials.
     
  8. Dec 30, 2016 #7

    Vanadium 50

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    Be careful when chipping off a piece!
     
  9. Dec 31, 2016 #8
    Ok one last question from me, why the current and field strength in superconductor electromagnets are not of the same order of magnitude or about the same (for example tens of KAmperes with a few KiloTesla). From Wikipedia :
    So 46KA current but only 13.5T magnetic field, why is that? Is it that materials with high magnetic permeability that could be used as cores break down under strong mangetic fields?
     
    Last edited: Dec 31, 2016
  10. Dec 31, 2016 #9

    Vanadium 50

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    Answer #1: Why would you expect the numbers to be the same when they measure different things? Compare the earth's mass with the number of seconds in a day. Very different - and an equally meaningless comparison.

    Answer #2: it's closer if you measure the magnetic field in Gauss instead of Tesla.

    Answer #3: in MKS units, μ0 = 4π x 10-7 N/A2. This is ultimately what causes the "mismatch".
     
  11. Dec 31, 2016 #10

    mfb

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    Ferromagnetic materials make great coils for low fields (~2T), but due to saturation they don't work beyond that.

    μ0 = 4π x 10-7 T/(A/m) makes the connection between Tesla, Ampere (and meters) easier to see. You need about 1 MA per meter (of coil length) to get a field of 1 Tesla.
     
  12. Dec 31, 2016 #11
    So we should suggest the OP to introduce at least two new materials that are discovered in the future,

    one for a super-superconductor that can support million of Amperes without overheating,

    and one for a super-ferromagnetic material that suffers from saturation at the million Tesla range.
     
  13. Dec 31, 2016 #12
    I have to admit, I wasn't sure this thread would get a great deal of traffic after being moved to this forum. Clearly I was mistaken. The responses have been incredibly insightful, and I thank you all for the valuable insights. I will definitely take them into account while writing.

    In the meantime have some more questions that aren't related to my original questions, and the one I'm stuck on at the moment is terminal velocity.

    I have an object (we'll call the object "Iggy") falling from orbit and colliding with the Earth's surface. I'd like to know
    1. how fast Iggy would be falling by the time it hit the surface
    2. how long Iggy would take to get there
    3. how much G-force Iggy would experience on impact
    So, with that in mind: v = the square root of ((2*m*g)/(ρ*A*C))

    g = standard Earth gravity, so 9.8 m/s^2
    m = Iggy's mass, which is give or take 400 metric tons
    rho = fluid density, which would be standard Earth atmosphere (I don't know the actual number), assume surface temperature of 27 degrees C and clear skies at roughly equatorial latitude
    A = Iggy's area, which I haven't quite figured out yet because I haven't decided what the shape should be, but for the sake of filling in for this variable, assume a rectangular shaped object with an area of 3 * 30m^3
    C = drag coefficient, which is where I'm having trouble, but assume Iggy is falling on a flat 90m^2 face OR a flat 30m^2 face, based on it's area

    Hopefully that should be enough information. If it isn't, let me know what other numbers I need to provide.
     
    Last edited: Dec 31, 2016
  14. Dec 31, 2016 #13

    PeterDonis

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    Are you asking if this is the correct formula? It looks OK to me.
     
  15. Dec 31, 2016 #14
    Nah, I know it is correct, that's not what I was asking. My questions are listed above the equation
     
  16. Dec 31, 2016 #15

    PeterDonis

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    The equation already answers question #1 (assuming that the fall is from high enough that terminal velocity is reached--since you said the fall is from orbit that shouldn't be a problem). For question #2 you need to know the altitude from which Iggy's fall begins (actually that will give you an approximate answer, a more accurate answer gets complicated). For question #3 you need to know how quickly Iggy stops on impact.
     
  17. Dec 31, 2016 #16
    Starting altitude: 350km

    Iggy's composition (for density calculation): varying degrees of mostly steel, aluminium, titanium and ceramics.
    Composition of Iggy's impact sight: H20 with varying levels of NaCl, MgCl, CaCl and other salts somewhere along latitude 0.
     
  18. Dec 31, 2016 #17

    PeterDonis

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    Then the simple approximate answer to question #2 is to compute how long it takes to go that distance at the velocity ##v## that comes out of your equation.

    The simple answer ignores the fact that Iggy has to accelerate for some period of time before reaching terminal velocity; but the time it takes to accelerate to terminal velocity, as given by your equation, is probably pretty short compared to the total time to fall, so ignoring the initial acceleration won't matter much.

    However, given a fall from 350 km, there is a much bigger complication: most of the fall will be through vacuum, not atmosphere, at least to a very good approximation. So what will actually happen is that Iggy will accelerate downward for most of the fall, reaching a speed that's probably much faster than the terminal velocity given by your equation; then, during the last, say, 50 km of the fall, Iggy will start to be slowed by air resistance (and also heat up a lot, hopefully Iggy has a good heat shield). I'm not sure whether air resistance will actually be able to slow Iggy, by the time it reaches the ground, to the terminal velocity computed by your equation or not; that would take a detailed calculation taking into account the variation in air density with altitude.
     
  19. Dec 31, 2016 #18

    PeterDonis

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    If Iggy is falling into the ocean, you can get at least a rough estimate of the deceleration rate by computing the drag force due to seawater using a similar formula for the drag in air, just with ##\rho## being the density of seawater rather than air.
     
  20. Dec 31, 2016 #19

    mfb

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    If Iggy is falling from an actual orbit (low Earth orbit?), it will enter the atmosphere at a speed of at least ~7.5 km/s. Everything after that depends on the angle of impact, its shape, mass and so on.
     
  21. Dec 31, 2016 #20
    Not that any of this matters from a story perspective; the only thing that's narratively significant is "big thing hits surface hard" - but being able to answer these questions with mathematically accurate answers would help with credibility.

    To that end, a third question I have which is related to both the previous questions is this:
    1. Should a release of energy equivalent to 860 gigatons of TNT be set off all at once at a single point on the Earth's surface, would that amount of energy be enough to render the entire surface uninhabitable?
    2. If not, how much energy would be needed to do so? Is it even at all possible?
    3. If so, would the image of a "wave of fire" enveloping the Earth from the epicentre be scientifically accurate?
    I got a figure of 860 gigatons based on the figure given by Delta^2 for 1kg of antimatter having an energy density of 9*10^16 joules, multiplied by 40,000 (40 metric tonnes of stored antimatter). Please advise if my math is wrong.

    edit:
    I also just realised that the 860 gigatons accounts for only half the energy released, since 40 tonnes of antimatter accounts for only half the total mass being annihilated. So assume I meant 1720 gigatons.
     
    Last edited: Dec 31, 2016
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