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Home , Lambda baryon

Ask a scientist answers. GN

I quote the beginning of the question, then my answer:

Q11: “Since beta decay is when …”

A11: Good question. The weak force is responsible for changing one into another, in this case a d-quark into a u-quark. Because the electric must conserved one needs to emit a negative charge (the is neutral, the is +1, so you need a -1- type ). In principle there could be other -1 emitted (say a mu-) but nature is lazy and will go with the easiest to come by (as in the “lowest possible energy” needed to do the job) particle, an . Of course, because we have an electron in the final state now we also need a matching anti-, as the number (quantum number that counts the , , and their kind – remember the slide with the felines in my presentation) needs to be conserved as well.

Now, going back to your original question “…where they come from?” the energy of the original nucleus. Remember E=m * c^2. That is a recipe for creating particles (mass) if one has enough initial energy. In this case in order for the beta decay to occur (naturally) you need the starting nucleus to have a higher sum of the energies of the daughter nucleus one gets after decay and the energies of the electron and antineutrino combined. That energy difference can then be “spent” in creating the mass of the electron and antineutrino (and giving them, and the daughter nucleus some kinetic energy, assuming there are energy leftovers).

Q12: “Also, how does one of the down…”

A12: Nice follow-up. Remember that we said the quantum picture for the fundamental forces is that there are “force carriers” (gauge ) that mediate the interaction (charged particles exchange , strong nuclear force works via the exchange of , etc.). The weak force has both neutral force carriers (Z-zero) as well as charge gauge bosons (W+ and W-). When one of the latter is exchanged the net result is changing the charge of the initial particle by +/-1; in this case we go from -1/3 to 2/3, which is a +1 leap in charge. You tell the audience which gauge must have been exchanged.

Q13: “What happens if a quark and anti-quark of a different type collide?...”

A13: Very clever of you to specify “of a diff. type”. Particle- collisions result in annihilation (as in your example) which produces energy (i.e. we’re reading E=mc^2 from right to left). That energy can be (but is not limited to) in the form of photons; incidentally for e+ e- annihilation you will produce not one but two photons, otherwise you will find it difficult to conserve momentum (yes, conservation laws are very important in physics!).

Going back to your question: if the quark and antiquark cannot annihilate (because they are of different type) then they just scatter of one another. If the energies of the two are low, then this scattering is elastic (think of two bowling balls, of different masses, colliding). If there is more energy in the collision then the interaction can be inelastic, and the energy lost by the quark-antiquark colliding can be used to produce other particles and/or photons (providing that conservation laws are observed, of course: so you will not be able to produce a proton in the interaction unless you have enough energy and produce an anti-proton as well).

The same goes for electron – (anti)quark: you get either elastic or inelastic scattering depending on how much energy is involved. This process is the backbone of almost all experiments carried out at Jefferson Lab.

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Q21: “If electrons are prevented …”

A21: The generally accepted view is that electrons orbit around the nucleus so that the attractive electric force between an electron and the positive in the nucleus acts as a centripetal force. So something like kZe^2/r^2 = m v^2/r

Depending on the nucleus the RHS of the above will sprout a gamma factor if the electron is moving at relativistic speeds.

Because only certain orbits are permitted one gets the energy levels that you mentioned.

Assuming that an electron is already on the lowest energy level of a given it cannot go any lower than that. If a sufficiently low energy FREE (as in not bound in an atom!!!) electron is available, some nuclei can capture them, effectively turning the electron plus a proton from the nucleus into a neutron (a neutrino is also emitted in this process). This reaction mechanism is called “electron capture” or “inverse beta decay”. Dr. Banu might be able to tell you additional details about this.

Q22: “…Bose-Einstein condensate?”

A22: B-E condensates apply to integer spin particles. Neither the proton, nor the electron match that description, as they are spin ½ entities.

Q23: “From what I heard; supporters of M theory…”

A23: Modern science works on the interplay between theory and experiment and one needs experimental evidence in order to give credibility to ones theories. One can have isolated +ve and –ve charges (think protons and electrons) but we haven’t (yet) found an isolated North (or South) magnetic pole. That would be the definition of a – an isolated N/S pole. We haven’t found one as of now but people are indeed searching as finding such a thing will require major reworking of Maxwell’s equations (i.e. the equation that describe electromagnetic interactions).

One last point here: field lines are not “emitted” nor are they absorbed by poles. Field lines are the trajectory that a positive electrical charge will follow is brought in that region of space and left free.

Q3: “Is the high cost of the CLAS…”

A3: Compared with other detectors the CLAS spectrometer is actually on the lower end of the $$ scale.

In order to achieve the kind of precision needed for current (and near future) nuclear and particle physics detectors need to operate with very high precision and reliability. Typically one needs: - timing resolution better than 1 billionth of a second, - ability to operate at high rates (the TOF counter currently under construction in our JMU lab should be able to provide timing measurements with a ~150ps resolution for an incident flux of up to 1MHz particles per scintillator bar) - very good energy resolution (parts per million if not parts per billion) - detector efficiency (at least) in the higher 90%.

To achieve these goals one must design and operate nuclear/particle physics detectors (and the accelerators used to produce the primary beams these detectors need) under extreme conditions (very high, or very low pressures, extreme temperatures, very high voltages, high magnetic fields, radiation, etc.). This puts significant constraints on the design and materials that need to be used to build these machines.

So the quick answer to your question is that the cost is driven both by the specific design and materials needed to build these detectors as approved by DOE/NSF.

Incidentally, the process of getting one of these detectors built is very deliberate as one first needs to establish the need for such device, then that need must be identified as one of the priorities of the nuclear/particle physics community at the present time. Subsequently one generates a conceptual design of the detector. Once approved the conceptual design needs to evolve into an actual design (here is where engineers of various specialties (civil, electrical, mechanical, etc.) get heavily involved). As the final design is approved one then sends the various major subassemblies (support structure, cryogenic magnets, various electronic interfaces, etc.) out for bid to industry. Finally, after the whole detector is built and assembled it needs to pass the final inspection and we are able to operate it.

In the process of costing a particular detector or piece thereof the funding agencies require a seven-level breakdown of expenses down to things like “how much does it cost for a technician to glue a one inch strip of material to another”. One also needs to provide a list of “value engineering” solutions (cost saving measures involved in the design and/or building of the detector) used in this project.

Q4: “You tell us that a proton p (uud) = 1 charge…”

A4: “…what do these values represent” The proton charge +1 and the u (2/3) and down (-1/3) are all in terms of the “”, i.e. the magnitude of the smallest possible free charge. Because the Coulomb, which is the SI unit for measuring electric charge, is such a large unit (or the elementary charge is so much smaller than a Coulomb), the elementary charge is: e = 1.6 x 10^-19 C

So the absolute charge of the proton is q_p = 1.6 x 10^-19 C or q_p = 1 x 1.6 x 10^-19 C in units of 1.6x10^-19C the charge of the proton is q_p = 1 !!!

Objects that contain quarks fall into two categories: (made up of quark-antiquark pairs), and (made up of three quarks). Exotics for both baryons and mesons (things like a five quark or a two quark – two anti-quark mesons) are predicted though not observed as of now.

Example of “things” that contain quarks other than ups and downs: K+ (u anti-s), Lambda baryon (uds), J/Psi (c anti-c), Omega- baryon (sss) and the list goes on (on several hundred pages in the last Parrticle Data Book.

Q51: “As I understand gravity…”

A5: For a physicist “understanding” a phenomenon is being able to uncover and (attempt to) solve the differential equations that govern that particular aspect of nature. At that level image of gravity as a deformation of space-time and the exchange of “” is exactly the same.

Q52: “Your example of the string…”

A52: You are absolutely right, string theory, or the newer “brane” (as in “membrane”) theories are very intriguing beasts indeed. While I cannot claim that everything makes so much more sense in a n-dimensional world (put 12, 34, etc for “n”) as some of my theorist friends would say I am, like most everyone in the physics community, (still) waiting for a few definite experiment-testable predictions that string theory will put forth. After all I hope that we established that the only way in which physics makes progress is through the interplay between theory and experiment.

Thank you. GN