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In article <1992Apr14.213121.17723@lmpsbbs.mot.com> capurka@ssd.comm.mot.com (Tom Capurka) writes:
>Hello,
>I have a question about particle accelerators:  Why does there seem to be
>a terminal particle energy for a given particle, accelerator, related to
>the radius of the acclerator?  Is this because the radiation caused by
>the fact that the article is not in an inertial frame gets too large and
>approaches equilibrium state with the energy input?

Yes.

And this "bremsstrahlung" (literally 'braking-radiation') that arises from 
acceleration of the particle around a curve depends on mass.  Hence an 
electron synchrotron must have a larger radius than a proton synchrotron 
with the same energy.  

>Whatever the reason, is there any possibility that there exists
>a yet undiscovered trick around this so that one could have an SSC type
>acclerator (as far as particle energies are concerned) in a 1 meter box?

Highly unlikely.  What people are working on are compact linear 
accelerators (where this problem will not appear).  The effort is on  
the development of RF cavities that provide extremely high electric 
field gradients while maintaining good beam properties. 

>Sorry if this is too basic but a good library is far away and this may turn
>out to be more fun.

The best source is probably articles in Physics Today, where you can see 
nice color pictures of things like RF quadrupoles.  There was a cover story 
some years back and occaisional news stories about things like planned 
linear colliders.  Maybe one of the SLAC HEP-cats can add some to this. 
I think the former USSR was planning to build one at Serpukhov. 

--
J. A. Carr                                    |  "The New Frontier of which I  
jac@gw.scri.fsu.edu                           |  speak is not a set of promises
Florida State University  B-186               |  -- it is a set of challenges."
Supercomputer Computations Research Institute |   John F. Kennedy (15 July 60)

In article <1992Apr14.213121.17723@lmpsbbs.mot.com>, capurka@ssd.comm.mot.com (Tom Capurka) writes:
> Hello,
> I have a question about particle accelerators:  Why does there seem to be
> a terminal particle energy for a given particle, accelerator, related to
> the radius of the acclerator?  Is this because the radiation caused by
> the fact that the article is not in an inertial frame gets too large and
> approaches equilibrium state with the energy input?
> Whatever the reason, is there any possibility that there exists
> a yet undiscovered trick around this so that one could have an SSC type
> acclerator (as far as particle energies are concerned) in a 1 meter box?
> Or is this impossible in principle, given the state of physics today.
> 
> Sorry if this is too basic but a good library is far away and this may turn
> out to be more fun.
> 

I saw an answer posted about bremsstrahlung, which does pertain but isn't
exactly to the point. For circular accelerators, the bulk of the ring is not
devoted to accelerating elements (RF cavities), but to steering and focussing
magnets that keep the particles travelling around in a circle so that they can
return to the RF cavity for another kick in the butt. The analogy is that
schoolyard game tetherball, where a ball is tied to a pole by a rope, and
players slap the ball every time it comes around to get it really going. For
the Tevatron, for example, the protons circulate about a million times through
the RF cavity before they make it up to 900GeV.

The limitation, then, is the magnetic strength of the bending magnets. The
expression for this is
		p = q*B*r       (Elementary Physics II)
where p is the momentum of the particle, q its charge, B the magnetic field,
and r is the radius of curvature. For a synchrotron, B increases as p
increases, up to the limit the magnets can provide.

The research into "table-top" accelerators takes the approach of doing all the
acceleration in one linear shot, with some kick-ass accelerating fields, as
opposed to the circular technique, which uses the same modest accelerating
field over and over and over again.

The comment about bremsstrahlung does have some bearing on the radius in the
sense that at high enough momentum (or, more accurately, high enough Lorentz
factor gamma), trying to squeeze down the size of the circle by using stronger
magnets just doesn't pay off, because a smaller circle also means more energy
loss by radiation, to the point where the accelerating cavity produces no net
gain in energy. Case in point is HERA, which uses the same tunnel for
accelerating 820GeV protons and 30GeV electrons. The magnets for the protons
are high-field superconducting magnets, the magnets for the electrons are
lower-field conventional magnets. Why not use superconducting magnets for the
electrons to get a higher electron energy? Because bremsstrahlung for electrons
is so much worse than for protons. It doesn't pay to do it.

Paul Draper
University of Texas at Arlington
Draper@sscvx1.ssc.gov

In article <1992Apr15.114716.1@sscvx1.ssc.gov> draper@sscvx1.ssc.gov writes:
>
>I saw an answer posted about bremsstrahlung, which does pertain but isn't
>exactly to the point.    ....     

Mea culpa. 

Proton machines are never really bremsstrahlung limited, but electron 
machines are always limited by radiation.  The mass factor is huge.  
(It has been too long since I did that computation!)  That is why, as 
Draper points out, work on proton machines centers on magnet development 
while electron people are thinking about two linacs pointed at each other 
with novel acceleration technologies used also.  

So if the question is "why can't I do e+ e- and make toponium", the 
answer is bremsstrahlung, but if it is "why is the SSC so big", the 
answer is magnets.  And magnets tend to be limited by material properties, 
if that is your next question! 

--
J. A. Carr                                    |  "The New Frontier of which I  
jac@gw.scri.fsu.edu                           |  speak is not a set of promises
Florida State University  B-186               |  -- it is a set of challenges."
Supercomputer Computations Research Institute |   John F. Kennedy (15 July 60)

In article <1992Apr14.213121.17723@lmpsbbs.mot.com>,
capurka@ssd.comm.mot.com (Tom Capurka) writes:
|>Hello,
|>I have a question about particle accelerators:  Why does there seem to be
|>a terminal particle energy for a given particle, accelerator, related to
|>the radius of the acclerator?  Is this because the radiation caused by
|>the fact that the article is not in an inertial frame gets too large and
|>approaches equilibrium state with the energy input?

Well I'm not an accelerator physicist, but yes that's my understanding.
If you look in Jackson, you can see expressions for ultrarelativistic
synchotron radiation. The energy radiated goes down with increased
mass to charge ratio. Thus you can get higher energy protons in a ring
then you can electrons. This is why many of the electron expts. are linear
(SLAC). Note, you also get some radiation b/c of linear acceleration, but
it is usually a much smaler effect.

|>Whatever the reason, is there any possibility that there exists
|>a yet undiscovered trick around this so that one could have an SSC type
|>acclerator (as far as particle energies are concerned) in a 1 meter box?
|>Or is this impossible in principle, given the state of physics today.
Well, you coudn't put a colliding ring w/ SSC type energies into a box
with getting a great deal of radiation. So the competing effect you
describe above would be imensely increased. You might think of doing a
SLAC type accelerator in 1 meter. However, now you need to accelerate to
10 TeV in 1 meter. That means that the electric field gradient causing
the acceleration needs to be 10E13 V/m. This is a huge electric field
(it would probably be RF and not DC). BY comparison, the electric field
in an atom is on the order of 13eV/Angstrom or ~10e9 V/m. So if you try to
put a 10 TeV SLAC in a box your accelerationg field will suck electrons
off of your vacuum surfaces and you will get electrical breakdown. So
the trick you ask for seems impossible :<.

But wait, what if you use a plasma instead of a vacuum to guide your
beam. Then there is no problem with breakdown (you already are ionized).
If you send a strong Laser light pulse just ahead of your beam, it may cause
a Wake of extremely strong electric fields. Then you might be able to
get the SSC into a 1 meter box. These ideas have been around a while. It's
called the laser wakefield accelerator. If you interested, look in the
citation indices under John Dawson, UCLA; Toshi Tajima, Texas; and Chan
Joshi UCLA. My understanding is that both the theoretical and experimental
work is promising but that the problem is really difficult, so my opinion
as ot whether it will <ever> work is a definite maybe.

You might think of some other tricks also. The two big problems you need to
finesse are 1) a collider must reduce (or re-absorb?) its own radiation
so that you can tolerate a smaller radius of curvature (and thus decrease
the size);  or 2) you must increase the accelerating gradient in a linear
device to shorten the length (i.e. decrease the size).


|>Tom C.  <capurka@ssd.comm.mot.com>

John W. Cobb
jwc@fusion.ph.utexas.edu

In article <70425@ut-emx.uucp>, johncobb@ut-emx.cc.utexas.edu (John W. Cobb) writes:
> In article <1992Apr14.213121.17723@lmpsbbs.mot.com>,
> capurka@ssd.comm.mot.com (Tom Capurka) writes:
> |>Hello,
> |>I have a question about particle accelerators:  Why does there seem to be
> |>a terminal particle energy for a given particle, accelerator, related to
> |>the radius of the acclerator?  Is this because the radiation caused by...
> 
> Well, you coudn't put a colliding ring w/ SSC type energies into a box
> with getting a great deal of radiation. So the competing effect you
> describe above would be imensely increased...
> 
> You might think of some other tricks also. The two big problems you need to
> finesse are 1) a collider must reduce (or re-absorb?) its own radiation...

Just to reiterate a point I made earlier, the limitation for an SSC-type
machine is NOT the radiation, but the magnetic field used to constrain the beam
into a circle. The magnetic field required to collapse a 20TeV ring to the size
of a meter radius, whether it's electrons or protons or Buicks (the latter
suffering almost no synchrotron radiation at all), would be 60,000 Tesla!

Now in fact, for the SSC, the energies of the proton are high enough that
synchrotron energy losses are not completely ignorable (left as an exercise),
and this is why the magnets are not absolutely the strongest available...i.e.
the ring as designed was allowed to expand a little to lessen the losses due to
radiation. But believe me, the magnets will run out of steam long before you
ever worry about vaporizing the beam tube walls with gamma rays.

Paul Draper
University of Texas at Arlington
draper@sscvx1.ssc.gov
Mythink. Not UTAthink, not SSCthink, not GOVthink. Happy now?

In article <70425@ut-emx.uucp> johncobb@ut-emx.cc.utexas.edu (John W. Cobb) writes:
[question about syncrhotron radiation limits on particle accelerators]
>
>|>Whatever the reason, is there any possibility that there exists
>|>a yet undiscovered trick around this so that one could have an SSC type
>|>acclerator (as far as particle energies are concerned) in a 1 meter box?
>|>Or is this impossible in principle, given the state of physics today.
>Well, you coudn't put a colliding ring w/ SSC type energies into a box
>with getting a great deal of radiation. So the competing effect you
>describe above would be imensely increased. You might think of doing a

This is not practical, but just for fun, suppose that the walls of the
beamline were made of some material that would form an extremely high-Q
cavity for the synchrotron radiation. Then there would be a lot of
radiation, but it would be stored in the volume of the ring. If you
could engineer the phases correctly, you could get acceleration out of
it (I'm not talking perpetual motion here --- just a case where the
acceleration from the radiation field balances the loss to particle
kinetic energy due to emitting radiation a la thermal equilibrium within a
black body) your losses now are only the absorptive losses in the walls
of the ring, but I'm assuming (ridiculously, to be sure) very high Q.

Of course, the high-Q cavity would have a narrow bandwidth compared to
the bandwidth of the emitted synchrotron radiation, but this would just
mean that the particle could not emit in certain parts of the spectrum
We already know that atoms will not radiate in below-cutoff cavities.
Kleppner and Hinds & Haroche have demonstrated this.

The problem is making a cavity with high-Q across the bandwidth of the
synchrotron radiation. There's no such thing that I know of.

---Jon

Disclaimer: The government probably disagrees with my opinions.

In article <1992Apr16.155454.1@sscvx1.ssc.gov> draper@sscvx1.ssc.gov writes:
>Just to reiterate a point I made earlier, the limitation for an SSC-type
>machine is NOT the radiation, but the magnetic field used to constrain the beam
>into a circle. The magnetic field required to collapse a 20TeV ring to the size
>of a meter radius, whether it's electrons or protons or Buicks (the latter
>suffering almost no synchrotron radiation at all), would be 60,000 Tesla!

OK, maybe we can't get 20 TeV electrons into a meter even without
synchrotron radiation. However, although I'm a lowly atomic physicist
everyone I talk to in particle physics would prefer to look for new
particles with e+e- colliders, and there the practical limitation _is_
synchrotron radiation. The only reason to use protons, I'm told, is

1) to study proton structure

2) to avoid synchrotron losses

By the way, I can make my coffee cup move in a 1-meter circle at millions
of TeV with no difficulty (a million TeV = 0.1 Joule) so the only problem
with Buicks is making them small enough.

---Jon

In article <1992Apr16.155454.1@sscvx1.ssc.gov>, draper@sscvx1.ssc.gov writes...
> 
>Just to reiterate a point I made earlier, the limitation for an SSC-type
>machine is NOT the radiation, but the magnetic field used to constrain the beam
>into a circle. The magnetic field required to collapse a 20TeV ring to the size
>of a meter radius, whether it's electrons or protons or Buicks (the latter
>suffering almost no synchrotron radiation at all), would be 60,000 Tesla!
> 

Of course, if you had an accelerator that could yield 20 TeV/meter, then
there would be no need to make it a ring.  You would just make a 100 meter
long linear collider with 2000 TeV per beam.  Accelerators are only rings in
the first place because the feeble accelerating gradients must be applied over
and over again to get the high energies needed for useful physics.

-Scott

--------------------
Scott I. Chase
SICHASE@CSA2.LBL.GOV