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Cold Fusion Experiment

  1. Nov 28, 2011 #1
    Hi, I have a question, why wouldn't this work:

    Imagine you have a two large parallel plates working as a capacitor with small holes in the center of the two plates. You put a potential over the plates. The electric field is then zero outside of the capacitor. You have Deuterium on one side and send in ionized hydrogen trough the hole of the other plate. The hydrogen will accelerate towards the opposite hole and smash into the deuterium on the other side making a nuclear reaction.

    No energy is put into the system, right? Where is then the kinetic energy given to the electron coming from? And why would not the fusion work?
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  3. Nov 28, 2011 #2


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    implies a potential

    Umm - does one see an inconsistency here? Accelerate (a mass) - over distance = ?

    A p + d reaction is possible, and if successful, produces 5.5 MeV, but I believe the cross-section is relatively small.

    One is describing more or less the principle behind a neutron source, or so-called neutron howitzer, in which a tritiated target is bombarded by deuterons (the d+t reaction produces alpha + n). The electrons have to stripped from the deuterium in order to accelerate the deuterons (the nucleus of a deuterium atom) to the target. The deuterons (those scattered without producing a fusion reaction) and fusion products recombine with the electrons in the target.

    If one is accelerating protons or deuterons, usually in the keV range, that's not exactly cold fusion.

    With regard to practicality, one has to look at the energy invested which is subsequently lost to scattering in the system without causing fusion reactions. If the energy from the fusion reactions > energy input, then one would have a net production of energy, and a potentially viable energy source.
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  4. Nov 28, 2011 #3
    Hi Astronuc, thanks for your reply.

    If you first charge up the capacitor and then disconnect the power source. Wouldn't it be possible to accelerate infinitely many protons trough the device? Where is their kinetic energy coming from?

    I see, but what if you use a huge transformer to charge up the capacitor? Or put a large number of these devices in a line? Will you still get scattering?

    But other than the initial power needed to charge up the capacitor, what else energy is needed to be put into the system?
  5. Nov 28, 2011 #4


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    One could produce a static potential, but one still has to ionize the hydrogen, and separate the electrons and protons. That requires an input of energy. The accelerating potential can be static - but it's difficult to keep it truly isolated - there is charge leakage.

    A static potential accelerates a charge, but it's not that simple.

    The scattering occurs when the projectile, let's say protons in this case, hit the target material and immediately interact with the atoms through ionization and scattering of the nuclei.

    One may find that more energy is lost to scattering than produced from fusion.

    With respect to cold fusion, how would one propose to produce useful (electrical or thermal) energy?
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  6. Nov 28, 2011 #5
    Yes, I wasn't thinking of that. All that potential energy did we give to the proton in the first place by removing the electron.

    If you confine the experiment in vacuum with no photons, why will there still be charge leakage?

    But if you put many of these devices in a line the protons will have enough energy to break the coulomb barrier and few of them will be scattered due to quantum dynamics, right?

    If there is a nuclear reaction there will always be thermal energy released, right? Ex. Translation, vibration or angular momentum of the products. The energy needed to ionize the hydrogen is far less than the energy received from the fusion, the magnitude of 10^3. I would guess that a Stirling engine connected to a generator has a efficiency of more than 0.001?
  7. Nov 28, 2011 #6


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    Of course one would accelerate charges in a vacuum, otherwise the charges would collide with gas atoms and lose energy. There is no perfect insulator, and there is some high voltage potential with which one is accelerating protons (or some positively charge nuclei).

    The purely static potential is problematic. One has to move the electrons to the negatively charged target. If one is thinking of cascading accelerating potentials, that's not really how one would do it for a low (few kV) potential. The static (relatively constant) potential is just a DC potential. The scattering occurs because the target is comprised of atoms, and the projectile particles travel through the electrons surrounding the atoms, which are ionized, and scatter off the nuclei. The coulomb interaction range is much greater than the range of nuclear interactions.

    If there is a nuclear reaction, there will be some energy released, e.g., kinetic (thermal) energy of the products, or in the case of p + d, the kinetic energy of an He3 nucleus + a gamma ray, and the gamma ray would then scatter off atomic electrons (Compton effect) or be absorbed (photoelectric effect). That energy heats the solid target, which could in theory be used to heat a working fluid.

    The 13.6 ev ionization energy of hydrogen is certainly much less than the MeV levels of energy liberated in fusion reactions, but one has to consider the reaction rate for fusion vs scattering, which is a function of the cross-section of the reaction, and densities of the target and projectile particles. If one obtains one fusion reaction for every 100,000 projectile particles, then there isn't much net energy produced, especially if one uses a thermodynamic cycle with a conversion efficiency less than 30-40%.

    I haven't looked at the p+d or p+t cross-sections for fusion reactions, so I really don't know how low they are. Most fusion systems are considering d+d or d+t fusion, and some p + B11.
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  8. Nov 28, 2011 #7
    We only have two metal plates that are charged, no insulator in between. Isn't the vaccum then a perfect insulator? I meant that we have no photons that may knock out any electrons due to the photoelectric effect so the charge should stay the same?

    Ok, I didn't think of that the protons have to pass trough the electron cloud. I have to read more about scattering. But lets say we ionize the target? I ionizing of the target would only require another ~10 eV, correct?

    Or maybe we make a cavity of the target, just like the case of black body radiation from a hole. The accelerated particle will enter trough the hole, be scattered around until it does fusion in the wall within the cavity. But then we would need another reaction because one of the reactants have to be made solid so we can make the cavity. Would that work?
  9. Nov 29, 2011 #8


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    Also consider that the plates have to be held up with something. There could be some charge leakage there, not just between the plates. Also, many of the ions themselves will impact the negatively charged plate after scattering, reducing it's charge.

    The deuterium? Then you have a plasma that is attracted to the negative electrode.

    It will not scatter until it fuses. I haven't done the math, but after just 1 scatter from the fuel or the container and I would expect enough energy to be lost to almost guarantee that it will not fuse. Then you have to worry about particles from the container getting thrown into your vacuum and spoiling future reactions by intercepting the protons before they reach the fuel. PLUS these reactions create lots of heat which will have to be dealt with.

    Unless your fuel is in a plasma state, there are simply too few fusion events compared to non fusion events to ever do any meaningful amount of fusion. In a plasma state the particles can retain much of their kinetic energy after collisions, allowing them to have many events before fusing or being lost. This increases the fusion rate per energy cost.
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  10. Nov 30, 2011 #9
    If the plates are large, the field outside is zero. You might aswell put the whole device inside a faradays cage. Do the experiment in space or use magnetic felds at the edges to keep the plates from touching eachother. Or use a very good insulator and the leakage will be minimal.

    Outside of the plates the field is zero.

    Ok, but lets say that we ionize the fuel and remove the electrons. Then there will be no scattering right?
  11. Nov 30, 2011 #10


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    We can reduce the leakage to small amounts sure.

    Only at large distances from both plates. When the ions exit the negative plate they will be attracted right back to it because they are very very close to it. Look up a Fusor. The inner grid can be negatively charged. The ions are attracted to it and accelerate until they are inside the sphereical grid. Once inside they cease accelerating and coast through until the pass to the outside of the inner grid. Once beyond the grid again, they decelerate and are then attracted back through the middle again. A similar effect would happen here.

    Scattering mostly occurs through collisions of ions on ions. Due to the extremely low mass of electrons they don't really affect the KE of the ions. They do some, but not much compared to an ion vs ion collision. Plus removing the electrons makes it extremely difficult to contain the plasma now that all those protons will repel each other and have no electrons to counteract it.
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  12. Nov 30, 2011 #11
    The field outside two infinately large plates is zero. Their fields cancel out. We will have a small field, but its not enough to deaccelerate the protons that much.

    If the speed is way over the required to break the coloumb barrier, then why will so much scattering occur?
  13. Nov 30, 2011 #12


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    Alright. I'm not that familiar with capacitors and such. I was not aware of this.

    Because protons and other nuclei aren't hard little balls. The have a tremendously tiny size and the repulsion from one proton on another is enormous even before you reach the point where they would fuse. If your ions don't approach each other exactly right they simply don't fuse. (Ignoring Quantum effects) And once you pass a certain velocity the ions have too much energy and start to have a reduced chance to fuse.
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  14. Nov 30, 2011 #13
    Oh really.. I am just thinking the analogy of a plasma. You have protons bouncing around and colliding untill they fuses. In our case we have protons accelerated in a direction. They enter a target where the nucleis almost standing still, they should also start bounce around and fuse. If we do the experiment where the fuel is at 0 K shouldn't we be able to match the right speed?
  15. Dec 1, 2011 #14


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    No. When an ion impacts an atomic nucleus and doesn't fuse it imparts a large amount of energy into that atom. That nucleus then transfers that energy to the surrounding medium. The net result is a fuel ion that bounces around until it loses enough energy to capture an electron from the material and is now either trapped inside or is a free gas particle. Either way the fuel ion has been lost with all it's original energy.

    In a plasma state the goal is to get the majority of the ions to a high enough velocity to fuse. Then when they impact each other and don't fuse they will both be at nearly the same velocities as they were before the collision. A small amount of energy may be lost as radiation, but not much. In this way a single ion can have thousands of collisions and still retain enough energy to continue until it fuses. The key is to be able to keep the ions at this level of energy and to compress them enough so that they do experience enough collisions to fuse in a short amount of time, as confining them takes both energy and space. The higher the density of the plasma the greater the reactions per unit of time and per unit of volume of the plasma.
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