Back to normal service now. I still don't understand circular motion (and in particular I don't understand angular momentum), but I've managed to muddle my way through the equations.
To business. Essentially a sci-fi circular accelerator is like the LHC on steroids. Projectile goes round and round the loop, and then it's let to fly off at a tangent towards its target. The assumptions are much the same as those for the linear case covered
- The accelerator fires discrete shots.
- The yield is limited by material strength.
- The accelerator structure is a rectangular torus. This is a sensible shape that will make the maths easier than for a circular torus. The major radius is also significantly greater than the minor width, again for mathematical simplicity.
- The projectile is fired at ultrarelativistic speeds. Again this lets us simplify the equations.
- Recoil can be neglected.
Unlike in the linear case, a circular accelerator does not have to bring its projectile up to speed within a limited distance; rather the projectile can "spin up" over many laps. The accelerator structure is thus required to exert a centripetal force needed to keep the projectile going round in circles, as well as a linear accelerating force which can be arbitrarily small and shall thus be neglected. The equation for centripetal force in Newtonian mechanics, F = mv2/r, is of course not valid in special relativity, so we will derive the ultrarelativistic equation.
Those who last did vector calculus more recently than several years ago will have to forgive my doubtless slightly sloppy notation.
The diagram above shows the projectile with an initial momentum p. After a short time interval dt, its momentum is now p'. During that time interval it travels through an arc with length vdt and angle dθ. Its momentum keeps the same magnitude - we're considering uniform circular motion - but rotates through that same angle.
s/r = dp/p
If dθ is small, then s ≈ vdt, ie the arc and chord are nearly the same length (and tend towards the same length as dθ shrinks to zero). As such
vdt/r = dp/p
Rearranging, the equation for force can be found.
F = dp/dt = pv/r
The above equation is valid equally in Newtonian mechanics and special relativity, since it's just derived from geometry. The expression for momentum, however, varies. In the ultrarelativistic limit,
v = c
E = pc
F = E/r
Thus we have the equation linking centripetal force, energy, and radius of curvature for an ultrarelativistic projectile.
In the linear case, we were able to consider the forces as being borne simply in compression. For a circular accelerator, however, the stresses are more complex. A key simplifying assumption I feel justified in making is that the force is spread essentially equally about the ring, because the projectile and thus the centripetal force are moving much faster than the speed of sound in the accelerator structure. With that assumption, the situation can be considered as analogous to a pressurised cylinder.
The above shows a cross-section through a cylinder with an outward force acting on it. In our accelerator, the outwards force is the reaction to the centripetal force on the projectile. In the analogous pressurised cylinder, it's fluid pressure on the inner walls. Either way, the outwards force results in a circumferential, or 'hoop', tension in the accelerator structure.
Assuming the projectile travels centrally around the accelerator, the radius of curvature r corresponds to the mean radius of the cylinder. If the cylinder then has length l and wall thickness w, and is thin-walled (wall thickness small compared to radius), the pressure on the inside walls is given by the force divided by the surface area of the inside
σp = F/2πlr
The hoop stress is then give by the following equation, which can be derived by considering half of the cylinder as a free body
σθ = σpr/w
In both the above equations, the radius should correctly be the inside radius of the cylinder. However using the mean radius does not in practice result in a large error for the peak hoop stress; even for w = 0.2r, normally regarded as thick-walled, the value is just 11% too low.
As in the linear case, this stress is limited by material strength. The hoop stress cannot be greater than the ultimate tensile strength σt of the material of the accelerator. Equating σθ and σt, and substituting in expressions for σp, we can get an equation relating projectile energy, material strength, and accelerator geometry.
σt = F/2πlw
σt = E/2πlwr
The volume of the accelerator structure can be approximated as
V = 2πrlw
And thus the equation for energy simplifies down to
E = σtV
Apart from being limited by tensile not compressive strength, it's exactly the same as for the linear accelerator! As a result, I shan't be doing a worked example this time round.
Power we cannot determine from the geometry for a circular accelerator in the way we did for the linear case. Ignoring losses, it could be computed simply from the desired firing rate and could be arbitrarily low. In practice, however, losses may exist, such as synchrotron radiation.
The key consequence is thus that a circular accelerator design will have mass similar to a linear one, the exact ratio depending on the available materials' performances in compression compared to in tension. The circular design though will require less power, probably orders of magnitude less, to run. Within a certain range, the yield depends just on the overall mass of the accelerator structure, and not on the exact shape.
If the first assumption is violated, it actually won't make a difference, due to the assumption made during the working out that the centripetal force is essentially equally spread around the accelerator anyway.
If the second assumption is violated then an entirely different approach will be required to determine the yield.
If the third assumption is violated, a more detailed mathematical treatment will be required. In particular, the equations given here will tend to underestimate the stresses, and thus overestimate the yield, for 'thick walled' designs, especially for the extreme case where the structure is a solid disc which might be practical for a compact accelerator. That said, I believe the error does not increase without limit, so the calculations here will still be within the right order of magnitude.
If the fourth assumption is violated, likewise a more detailed treatment will be required. The mass and velocity of the projectile will have to be considered separately, instead of just its total energy. Unlike the linear case though, we won't have to deal with a varying speed.
The violation of the fifth assumption has two implications. Firstly, while the projectile is circulating, the whole accelerator structure ought to also circle due to conservation of momentum. This could simply be mitigated by having multiple smaller projectiles to balance things though. Secondly there is the linear recoil upon releasing the projectiles, which like in the linear case will sap some energy and so manifest as an inefficiency.
I plan on in future returning to issues of accelerator design, for both the linear and circular cases, and considering arguments based on magnetic field strength for when that is the specific accelerating force. For now, I feel confident in the basic conclusion that circular designs are better than linear ones - a conclusion backed up by most real-world particle colliders using circular designs.
PS: Suggestions for what to tackle next are welcome!
ReplyDeleteFirst, of course, thank you - you've clearly put a lot of work into this article. But... (and there aways is a 'but')
ReplyDeleteI would contend that while a circular accelerator is more efficient in terms of space and equipment (as the same track is used repeatedly), it uses more energy than a linear one.
This is because travelling in a circle requires a constant adjustment of the tangental vector and thus constant acceleration to maintain the circular path, whereas a linear path only requires acceleration to increase the projectile's speed.
Thanks for the feedback!
DeleteFor an object in circular motion, the centripetal force does no work. This is because work is the force acting on an object times the distance moved in the direction of the force, and the centripetal force is always at right angles to the motion. So if everything was 100% efficient it wouldn't take any energy to supply the force to maintain the circular path. Only the force to speed up the projectile, which will be tangential, will require power input to maintain.
In reality of course things are *not* 100% efficient, so some energy will be required for the centripetal force. Resistive magnets consume lots of juice for example, and even superconducting magnets can need a power input to overcome flux-flow resistance, as well as cooling. This may or may not be small compared to the energy that goes into the projectile.
If the projectile is charged and the centripetal force provided by a magnetic field - which is the one proven method to make high-energy particles go in circles - then there will be losses due to synchrotron radiation. These may be considerable, and I plan to address just how big they are when I do a follow-up to this piece.
And thank you for the reply.
DeleteHowever, I have thought of another issue - firing patterns. In a linear accelerator, you can put delays between projectiles as you wish, e.g. projectile, projectile, PAUSE, projectile, and a set period of time later these will leave the barrel. However, a circular accelerator can only 'spin up' a certain number of projectiles at a time, and more projectiles cannot enter while it is 'spinning up'. This means that circular accelerators will always have a period where they aren't firing anything, whereas a linear accelerator can have several projectiles at different stages of acceleration at once and can thus fire continuously.
Or, to put it another way, a circular accelerator with a circumference a quarter the length of a linear accelerator can fire projectiles with the same acceleration time, exit velocity, and energy consumed at a quarter of the size and equipment required.
ReplyDeleteHowever, it can only fire a quarter of the number of projectiles over the same period of time as a linear accelerator, and with an unavoidable delay between firings.
This can be used by a sufficiently intelligent enemy to predict when a certain gun will be unable to fire,m and exploit that.
There are a few ways to get around this requirement. The key lies in a term used by actual particle accelerator designers: "storage ring." The giant circular structures at the LHC and other, smaller installations definitely do *accelerate* the particle beam, but they also serve to *store it.* The beam does not initiate within the storage ring; there is an entirely separate set of smaller linear and circular accelerators designed to do that.
ReplyDeleteBy attaching a linear-accelerator 'gun' to your storage ring and using *it* to get projectiles up to speed, they can be introduced to the storage ring without posing a collision hazard once they arrive in the ring. One might also make an analogy to merge lanes on a freeway on-ramp, which permit cars to join the freeway's traffic safely, without requiring the freeway drivers to slow down.
On another note, firing solid projectiles at ultrarelativistic speeds is... prohibitively difficult in most settings. The extreme forces acting on both the projectile and the accelerator's bracing mounts are only part of the problem. Another part is that any energy leakage that heats the projectile will probably be enough to vaporize it thousands of times over. A reasonable rule of thumb for speeds that can be measured as fractions of c *well below* c itself is that the kinetic energy of a projectile traveling at 0.1c is approximately one kiloton TNT-equivalency per kilogram of mass.
ReplyDeleteAt 0.5c this becomes 25 kt per kilogram, times a relativistic Lorentz factor that is starting to get big enough to be noticed even for purposes of our rough first-order approximations.
At "ultrarelativistic" speeds (where the relativistic kinetic energy of the particle greatly exceeds its rest mass-energy), by definition the kinetic energy of the projectile is well in excess of (mc^2): in other words, over twenty megatons per kilogram!
If even a small fraction of that energy is dumped into heating the projectile, the energies are sufficient to volatilize any imaginable physical projectile millions of times over.
It is probably more practical to model circular accelerators for relativistic particle beams, where the beam can be contained more safely and more easily. We already have the technology to contain a particle beam with a total kinetic energy of 350 megajoules; the LHC does exactly that.