@Kaumudi.H It would be more than my life's worth to not be using it when my niece comes to visit, but in fact it is the perfect size for my coffee mug.
@Slereah Voyager was a mix. Some sciency "oh no the organic black hole has got us" episodes, some "the prime directive blah blah" episodes, some "does the doctor have a conscience" episodes...
As male individuals were not biologically meant to nurse their offsprings, they sported much smaller breasts and produced no milk. Mandalore the Lesser (then a gladiator),[9] Aron Peacebringer (a planetary leader)[10], and Anakin Skywalker (in certain circumstances, such as on Nelvaan) would freely exhibit them.
physics.stackexchange.com/questions/313117/… - with the section on "For higher energy electrons the loss of energy as bremsstrahlung becomes increasingly important" would it make sense to think of it as objects with greater energies have a harder time braking, so they lose more energy?
@heather The decay energy is distributed between the nucleus, the beta particle (i.e. electron), and the antineutrino. Therefore, beta particles show a continous spectrum from about zero to a characteristic maximum beta energy.
A neutron is a neutral particle which is merely some times massive than an electron
what makes it so unstable outside the nucleus that it has a half life only of about 12min?
How much does thermal expansion affect neutron stars? Would the loss of temperature cause a neutron star to be more densely packed and thus collapse into a black hole?
@Loong i'm sorry, i don't understand your explanation. what do you mean by "beta particles show a continuous spectrum"? Is this the spectrum of all possible energies for the beta particle?
@heather If the decay was to proton (or other heavy remnant) + electron there would only be one way to share the energy between the two particles that conserved momentum.
So a measurement of the electron energy spectrum would show a single isolated peak (modulo instrumental blurring and scattering tails).
But with three particles there are many ways to share out the energy depending on the relative angles of the particle emissions. The result is a broad electron energy spectrum.
You should be able to convince yourself of this using a classical model of an object blown apart into either two or three discrete pieces of prearranged masses by an explosive that delivers fixed energy to the parts. No need to bother with relativistic energy and momentum to get the idea.
@BernardoMeurer I have a decided loathing for Halliday's books. Have since the 90's. Unfortunately for you there are some very good scientists who consider them very good. So you're facing mixed recommendations.
@0celo7 It is the need to conserve both momentum and energy that forces your hand, because the energy available comes from the mass difference and so is fixed for any given decay. Work the problem in the frame of the parent particle for simplicity.
@heather Well, we only measure the electron energy (neutrino escapes undetected and the recoiling proton or nucleus is much lower energy and therefore hard to detect).
@BernardoMeurer I used Giancolli in college and didn't hate it. I have a desk copy of the newest edition and find it no worse than the competition.
There is something to be said for Shankar's books: amazon.com/… REeatively readable, so you might like them.
@0celo7 If you want the exact shape of the distribution, you could use the Fermi theory of beta decay. However, you should ask a physicist about it. ;-)
"For energies up to 10 MeV, electrons lose their energy to the detection medium mainly by excitation and ionization of the electrons of the medium, as in the case of heavily charged particles." @Loong that was the quote involving the 10 MeV threshold
@heather The energy of this approximate threshold depends on the absorbing material (e.g. air, water, lead). Do they mention the material in your text?
It does mention that the "loss of energy as bremsstrahlung becomes increasingly important and the intesity of this varies as $Z^2$ where $Z$ is the atomic number of the medium."
otherwise, that's literally the beginning of the section.
::chuckles:: We have the same problem in trying to decide what books to adopt for courses. Unfortunately, the school makes us stick with anything we choose for at least three years, so the process is fraught.
@heather Depending on your goals in this reading it might (or might not) be worthwhile to just take a few things on faith at this point and come back to the subject when you have accumulated a little more knowledge of how these ideas are used.
@heather @heather My above-mentioned plot is for water. In that plot, the threshold looks more like about 100 MeV. But for a high-Z element like lead, a value of 10 MeV could be correct.
Just judging from the way the plot is present he may be using ROOT, which has some surprisingly good facilities for that kind of thing which makes it relatively easiy to generate plots.
okay, another question: when it says that "the intensity of this varies as $Z^2$ where $Z$ is the atomic number of the medium" what does it mean? is $Z^2$ literally the point at which the energy loss due to bremsstrahlung equals the energy loss due to ionization?
that makes the most sense intuitively, but it doesn't make much sense mathematically - lead's atomic number is 82, and $82^2$ is 6724 - not sure how that relates to 10 MeV.
@Slereah These "theorems" often restrict very strongly the class of theories they are talking about. I'd bet that e.g. gauge theories coupled to matter are not in the class of theories forbidden here, otherwise the Yang-Mills mass gap question wouldn't be so hard.
@heather No. The statement says that you have some quantity of interest $I(Z)$ that depends on $Z$ and they're telling you that it goes like $Z^2$, which means that there is some constant $k$ such that $I(Z) = k Z^2$.
This is only useful if you're willing to consider different media with different $Z$
So, if you have some medium with $Z=Z_1$, and you want to compare the quantity of interest $I(Z_1)$ in that medium with, say, some other medium with $Z=Z_2=\frac12 Z_1$ (such as lead as medium 1 with $Z_1=82$ and niobium as medium 2 with $Z_2=41$), then you would have $$I(Z_2)=kZ_2^2 = k(\tfrac12 Z_1)^2 = \frac14 kZ_1^2 = \frac14 I(Z_1).$$
i.e. "if you halve the $Z$ then you get a quarter of the response".
so using a different $Z$'s equation to solve for another $Z$?
sort of?
so $k$ changes from element to element?
it says 9 MeV electrons in lead lose as much energy due to bremsstrahlung as ionization...but wait, Loong said it was 10 MeV and that's what it implied earlier....
and how does the energy of the electron come into play in that equation? it has too..
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@heather my Knoll says $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{NEZ(Z+1)e^4}{137m_0^2c^4}\left(4\ln\frac{2E}{m_0c^2}-\frac43\right)$, but I haven't checked it.
How much does thermal expansion affect neutron stars? Would the loss of temperature cause a neutron star to be more densely packed and thus collapse into a black hole?
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