Which way does antimatter fall?

Antihydrogen

The force of gravity on antimatter has never been directly measured but a growing number of physicists believe that such an experiment is within their grasp. Today, a group attempting to design an experiment called AEGIS (Antimatter Experiment: Gravity, Interferometry, Spectroscopy) outline their plans to measure this force.

In some ways it’s an ambitious plan. The team wants to build AEGIS at CERN, the European particle physics laboratory near Geneva, where the building blocks of antihydrogen, low energy antiprotons and positrons, are in relatively good supply.

The idea is to fire a beam of antihydogen atoms at a target and see how much they are deflected by gravity.

That’s easier said than done. Creating a beam of this stuff turns out to be remarkably tricky. The problem is that it’s easy enough to trap antiprotons and positrons in electromagnetic fields. It’s even fairly straightforwad to put them together so that they form antihydrogen. The problem is that antihydrogen is neutral and simply falls out of the trap. So some way has to be found to collect and trap these antiatoms.

I know what you’re thinking: why not do the experiment with antiprotons or positrons instead.

People have tried but it’s been impossible to completely remove any residual electromagnetic fields from such experiments. These are many orders of magnitude stronger than gravity and so even the smallest trace of them deflects charged particles by an amount that overwhelms the effect of gravity. That’s why neutral antihydrogen is so important.

Why bother? There are several flavours of general relativity that allow antimatter to experience an opposite gravitational force compared to ordinary matter. Finding evidence for this (or ruling it out) will have important consequences for some serious cosmological conundrums such as why we see so little antimatter around and the value of the cosmological constant.

If these guys get the go-ahead, it’ll be a few years before we hear back from them, but it’ll be worth the wait.

Ref: arxiv.org/abs/0805.4727: Formation Of A Cold Antihydrogen Beam in AEGIS For Gravity Measurements

26 Responses to “Which way does antimatter fall?”

  1. [...] 2008, 1:47 pm Filed under: Uncategorized This is a very interesting experiment proposal.

  2. Kent says:

    Directly observing the effects of gravity on anti-protons which, as the article explains, is very difficult because of the weakness of the gravitational force compared to the EM. I wonder if it is possible to see compare gravity induced phase-shifts in the interference patterns between protons & anti-protons?

    You split one beam into two beams, one higher & the other lower, and then recombine them to observer the interference. If anti-matter is “repelled” by gravity then the anti-proton pattern should have a negative phase shift (since the potential is positive) compared to the proton pattern. In fact, any the magnitude of the phase shift might give some idea on the strength of the gravitational interaction… Anyone want to calculate what height difference & length of proton/anti-proton guides are needed?

  3. Bent Schmidt-Nielsen says:

    Dicke claimed that his Eötvös equivalence experiments could be used to rule out and opposite gravitational sign for anti-mater due to the differing virtual anti-mater content of the different materials in his experiment.

    http://www.kfki.hu/~tudtor/eotvos1/onehund.html
    http://www.mazepath.com/uncleal/eotvos.htm

  4. R.Mirman says:

    The cosmological constant is trivially 0. With it in Einstein’s equation one side transforms as a function of a massless representation (which presumably everyone agrees with) while the other is a momentum-0, representation. This is like equating a vector and a scalar, mathematically inconistent so impossible. It is trivially 0 unfortunately, else gravitation would have a fascinating property: a wave would be detected an infinitely long time before emission. See the MRPG book. Or use the example in MTW.

    MRPG;
    Massless Representations of the Poincaré Group
    Massless Representations of the Poincaré Group
    electromagnetism, gravitation, quantum mechanics, geometry
    (Commack, NY: Nova Science Publishers, Inc., 1995; republished by Backinprint.com)
    Geometry requires general relativity, which is thus the quantum theory of gravity. Trivially the cosmological constant is 0 as are the reasons for gauge transformations and CPT.

  5. [...] ¿En qué sentido cae la antimateria? [ENG]arxivblog.com/?p=450 por Kartoffel hace pocos segundos [...]

  6. Rob says:

    OK, maybe a naive question here: but if when you combine the antiprotons and positrons, the “antihydrogen … simply falls out of the trap”, then why couldn’t you put a target below the trap, and another above, and see which one is hit?

  7. Justin says:

    @ Rob on Jun 4, 2008:

    There’s no way to ensure the particles are created with zero momentum I suppose. But you could do it in a big, big space to ensure momentum was overcome by gravity. Sadly then you have the problem of creating the antimatter in the first place back again.

  8. ko says:

    I culd be wrong here, but:
    “The problem is that antihydrogen is neutral and simply falls out of the trap.”

    isn’t this what they are looking for??

  9. Jeff says:

    It seems to me that this effect would be detected next to black holes very clearly. I think they are trying too hard. Black holes do spew x-rays. Perhaps the x-rays are the result of annihilations with normal matter as they are “ejected” from the gravity of the singularity. Have we thought of that yet?

  10. Andrzej says:

    @Jeff:

    No, because X-rays are not nearly energetic enough. Unless we are talking about very light particles like neutrinos (about which we know so little), even annihilation of electron-positron pairs would produce gamma rays with energy of approximately 1 MeV.

    I suppose one cannot rule out these gamma rays being red-shifted (to practically anything) by the gravity of black hole, but then the question remains: how did the antimatter get there in the first place?

    As far as X-rays from black holes go, we have a much simpler and plausible explanation now (Bremsstrahlung radiation, EM radiation by accelerating charged matter). As for other, more complicated explanations, extraordinary claims require extraordinary evidences.

  11. Hamish says:

    @ko:

    No, the word fall implies an effect of gravity but, as I understand it, the antiatoms are very energetic so the direction in which they leave the penning trap(?) is barely effected by gravity at all. And if they were not very energetic they would still have momentum.
    On the page linked to in the article (http://arxiv.org/abs/0805.4727) it says an effect of the electric fields that would accelerate the beam is extreme cooling.. obviously useful as the effect of gravity would be more apparent on a slower moving object.

  12. Cliff says:

    This will measure the gravitational effect that
    matter has on anti-matter. If this is different
    than matter on matter then the question becomes
    what about anti-matter on anti-matter?

    In a practical sense it is impossible to measure
    anti-matter on anti-matter gravitational
    effects, but if there is a sign change like
    for electric charges, then one would expect
    that gravitational repulsion between matter
    and anti-matter would go along with attraction
    between anti-matter and anti-matter.

  13. I thought that this had already been examined by looking at the effects of the tides on the beam position at Fermilab and CERN. Is that just an urban legend?

  14. Mel Landin says:

    Perhaps a dumb question, but if they fall out of the trap, which way do they fall?

  15. Mel Landin says:

    Wouldn’t they register an anti-gravity deflection by a mini-black hole? Couldn’t the velocity be increased to measure predictable relativistic effects on mass? Wouldn’t a bose condensate slow it down long enough to truly “fall” out of a quantum doughnut, such as an Abrikosov vortex? Or stabilize a beam within EM forces between two superconducting blocks of similar EM charge but differing densities and count the matter-antimatter explosions on one side vs the other? It sounds like a scale issue which could be solved by redirecting beams at two theoretical positions in orbit with detector satellites, or two blimps with detector mesh for skin and reactive gases.

  16. [...] I guess it’s never been empirically tested. (See Baez’ Physics FAQ) Anywayway, CERN is doing an experiment on it. It’ll almost certainly be one of those experiments confirming what we already know. [...]

  17. Zephir says:

    Naivelly speaking, if the matter and antimatter are recombining into pair of photons, which are both attracted by gravitational field being indistinguishable each other, it’s improbable to expect, the antimatter would behave differently with respect to gravitational field.

    The Dirac’s concept of matter has considered, the antiparticles are hollow “bubbles” of the Aether, so they would exhibit an antigravity behavior due their negative curvature.

    By Aether Wave Theory the explanation of antimatter is different. The vacuum has character of nested foam here and every observable matter is formed by dense blobs of foam undulations, exhibiting surface parity (the foam gets more dense under shaking). The particles corresponds the undulations on the inner side of foam membranes, which are having slightly lower radius, being slightly more dense, stable and massive by such way, while the outer surface corresponds the antiparticles.

    http://superstruny.aspweb.cz/images/fyzika/aether/cpinavriance.gif

    Nevertheless, both ways of Aether foam undulation are increasing the vacuum density by the same way, therefore I don’t expect different behavior of matter and antimatter in gravitational field. If it would be different, it would mean, the AWT concept of matter particle is wrong on behalf of Dirac’s model.

    On the other hand, the AWT models explains the CP symmetry violation and the disappearance of antimatter during Universe formation easily. The tenuity of foam bubbles corresponds the small CP symmetry violation, which is quite low at the present state of vacuum, because both inner, both outer surfaces of foam are the nearly same curvature. During inflation the Aether foam has condensed and the thickness has made the CP violation a much more pronounced phenomena.

    http://superstruny.aspweb.cz/images/fyzika/aether/aether_density.gif

    By AWT the Universe matter has passed by the same evolution, like the matter escaping from quasar. The fast cooling of vacuum can lead to separation of thermodynamically metastable mixture of particles and antiparticle, for example in the gravitational gradient near black holes, where the antiparticles can be still observed.

  18. [...] Which way does antimatter fall??? [...]

  19. Jim Pivarski says:

    > I thought that this had already been examined by
    > looking at the effects of the tides on the beam
    > position at Fermilab and CERN. Is that just an
    > urban legend?

    The tides misalign the magnets, causing the beam to be deflected simply because they’re being pointed the wrong way.

    > Perhaps a dumb question, but if they fall out of
    > the trap, which way do they fall?

    Perhaps a bad choice of words. I think KFC meant they fly out of the trap— their direction depends much more on their initial velocity than any acceleration they feel.

    A question: a few years back, I saw a great talk on the effect of gravity on slow neutrons from a reactor. They were made so slow that their deBroglie wavefunctions were macroscopic, and solved Schrodiner’s equation for slow particles falling and bouncing on a table. Couldn’t a similar thing be done for antineutrons? Aren’t there any nuclear reactions that would produce slow antineutrons? Building antihydrogen seems like a lot of work to answer the same question.

  20. Brian H says:

    Jim;
    Yes, there are such nuclear reactions. First, start with 1 kg. of anti-uranium235 …

  21. Kent says:

    @Jim Pivarski: The experiment you mention is called GRANIT. It exists at the ILL research facility in France where I work. In fact, I work on the future source of neutrons that they will be using.

    Anyway, as you mentioned, gravity on neutrons is well understood. Anti-neutrons however are a whole different ball game. I don’t know of any way to produce them. There have been many searches for neutron to anti-neutron oscillations (much like the neutrino oscillations) however none have been observed. At a recent conference a next generation experiment was proposed. Keep an eye out on this if we can find the support!

  22. Codifex Maximus says:

    I read a few years back that some researchers had been able to take hydrogen atoms in a near vacuum and converge them using laser beams. They supposedly had gotten multiple atoms to nearly stop moving altogether and occupy the same point in space and time.

    If I’m not too far off the mark, could not this be used to collect and then accelerate the anti-hydrogen atoms at the target? I guess it would depend on how the anti-hydrogen atoms reacted to the lasers…

    Codifex

  23. TF L says:

    A variation of GR that I am working on predicts zero attraction between equal masses of matter and antimatter. However, both would be attracted to a large neighboring mass, but at different rates. If the large mass is normal matter, the smaller normal matter mass would accelerate more rapidly than the small antimatter mass. Vice versa if the large mass is antimatter.

  24. Joe nahhas says:

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    The problem that the 100,000 PHD Physicists could not solve

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    All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
    r = r (x, y, z). The state of any object in the Universe can be expressed as the product

    S = m r; State = mass x location

    P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

    = change of location + change of mass

    = m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

    F = d P/d t = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r

    = m γ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

    In polar coordinates system

    r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

    F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m’[r'r(1) + rθ'θ(1)] + (m”r) r(1)

    F = [d²(m r)/dt² - (m r)θ'²]r(1) + (1/mr)[d(m²r²θ')/d t]θ(1) = [-GmM/r²]r(1)

    d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

    Let m =constant: M=constant

    d²r/dt² – r θ’²=-GM/r² —— I

    d(r²θ’)/d t = 0 —————–II

    r²θ’=h = constant ————– II
    r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ’(d u/d θ) = -h (d u/d θ)
    d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ’(d²u/d θ²) = – (h²/r²)(d²u/dθ²)
    [- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²
    2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²
    d²u/dθ² + u = GM/h²
    r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)
    r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]
    r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

    r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

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    If λ(r) ≈ 0; then:

    r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

    θ’(r, t) = θ’[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}

    h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
    h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

    θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²
    θ’ (0,t) = θ’(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} Exp (-2iwt)

    θ’(0,t) = θ’(0,0) [cosine 2(wt) - ỉ sine 2(wt)] = θ’(0,0) [1- 2sine² (wt) - ỉ sin 2(wt)]
    θ’(0,t) = θ’(0,t)(x) + θ’(0,t)(y); θ’(0,t)(x) = θ’(0,0)[ 1- 2sine² (wt)]
    θ’(0,t)(x) – θ’(0,0) = – 2θ’(0,0)sine²(wt) = – 2θ’(0,0)(v/c)² v/c=sine wt; c=light speed

    Δ θ’ = [θ'(0, t) - θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second
    {(180/π=degrees) x (36526=century)

    Δ θ’ = [-720x36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

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    The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²—) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
    v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

    v = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her
    Let m = mass of primary; M = mass of secondary

    v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]
    v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com

  25. Hasanuddin says:

    What is the gravitational nature between matter and antimatter? Boy, I’m glad people are finally truly discussing this. Back in 2007 I began a deductive analysis that cracks this very question. The result is a new cosmologic model that adheres to all tested and verifiable evidence (though does overturn a few theoretical and unverified popular assumptions.) If you’re interested in this topic, you’ll be happy to know that the new model is currently being slowly unfolded at: http://www.scientificconcerns.com/Forums/viewtopic.php?f=32&t=776