11:00 PM
@heather with the specific one you're just not worrying about the effect of how dense your material is
@heather yeah, but that feels a bit weird
it can't lose 1600 MeV, can it?
so what if I phrase it as $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=1.6\:\mathrm{MeV/mm}$?

@EmilioPisanty ah, okay...that seems pretty important, because the denser the material, the more the particle bumps into other particles and so the more energy it loses.
@EmilioPisanty that makes a lot more sense

@heather exactly
@heather so there you have it: your electron will lose something like 1.6 MeV over the first millimeter
then after that you can't really use that rate anymore
remember how $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r$ increased with increasing $E$?
a.k.a.: remember how $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r$ decreased with decreasing $E$?

i think so...

$-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2$

it makes intuitive sense, anyway. the change in energy over distance will naturally decrease as energy decreases.

11:05 PM
so if $E$ changes, the rate will also change

@EmilioPisanty right

@heather not necessarily
depending on the mechanism, it's quite reasonable to have a material that's better at catching slow electrons
in this specific case, though, the rate does decrease with the energy

@EmilioPisanty hmm, okay

so, in this specific case, it's better to leave things at $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2$

okay, i've got a stupid question: how does this all connect back to the paragraph in my book? it said "For higher energy electrons the loss of energy as bremsstrahlung becomes increasingly important and the intensity of this varies as $Z^2$ where $Z$ is the atomic number of the medium. Thus, for example, 9 MeV electrons in lead lose as much energy due to bremsstrahlung as due to ionization." this gives the total energy loss...is kZ^2 a specific type of energy loss (due to bremsstrahlung)?

11:07 PM
and then use an explicit $Z$ and put in the $\rho$
$-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\rho\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(183\:\mathrm{m}^{-1}) E$
@heather yes

@EmilioPisanty and is k defined as we defined it, or as something else?

@heather $k$ is that whole thing before the $Z^2$

okay

though you need to keep in mind that depending on the material, $N$ and $\rho$ will also change

so in that case, what's the difference between the total energy loss and the energy loss due to bremsstrahlung?

11:11 PM
but saying "bremsstrahlung varies as $Z^2$" focuses on the effect of a single atom, which is what matters for the physics
@heather well, you'll have other mechanisms contributing to the energy loss

@EmilioPisanty how does it focus on only one atom?
i'm sorry, I know I'm being rather difficult

@heather as in, if you change the material you'll also change the density, but that's a "trivial" change in a way
the fact that the rate of energy loss is proportional to the number density $N$ essentially tells you that your electron is interacting with each atom individually, and then the rate gets added up

@BernardoMeurer new DJ K single with Jay and Beyonce

@EmilioPisanty okay, let me change what i'm asking, because i think it's part of the problem - what does it mean by the intensity of this varies? it's not directly the energy loss per unit distance, it's a multiplier by that energy loss per unit distance to "add in" bremsstrahlung?

diss track @ Drake

11:15 PM

@heather exactly. The contribution of bremsstrahlung to the total energy loss varies as $Z^2$.

Khaled is such a meme

@EmilioPisanty that makes so much more sense!

@0celo7 I got the detailed syllabus for Analysis II

11:17 PM
wow, that was like the best lightbulb moment ever
i think I actually get that paragraph now
thanks so so much for your patience and help @EmilioPisanty

@heather no worries
just keep your units straight from now on =P

Understanding pointers is better than sex

@EmilioPisanty no kidding, lol

@heather no, seriously, if you take one thing away from this talk, make it how to do units consistency checks at every point in your calculations

i honestly think that was one of the best things I learned today
i know now how to do a fairly simple check that makes sure i'm not doing anything especially stupid.
i will definitely use that in the future.

11:22 PM
@BernardoMeurer i'm scared

atta girl =)

@0celo7 Wanna see it?

unbelievable...i've moved on to page 4 =P

@BernardoMeurer no, busy

Alright

11:23 PM
I might actually survive section 1.1

@heather keep dmckee's comment in mind
the one on the star board
3 hours ago, by dmckee
@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.
an in-depth reading can be exhausting

and more importantly, it can keep you from actually getting the important higher-level points the text is trying to make

in which case, the in-depth-ness of your reading is actually detracting from what you might want to get out of the reading

11:25 PM
@skullpetrol what do you want, old woman

@0celo7 language

@EmilioPisanty What's offensive about an old woman?
What tells you @skullpetrol is not, in fact, an old woman?

It's an inside joke @EmilioPisanty
np

@skullpetrol what part of the chat moderation post wasn't clear? If it's unacceptable language, being an inside joke does not make it OK.
moving on, though

11:28 PM
@EmilioPisanty What in that sentence is "unacceptable language"?

@EmilioPisanty ??

Most importantly, it can keep you from actually getting the important higher-level points the text is trying to make
in which case, the in-depth-ness of your reading is actually detracting from what you might want to get out of the reading

@BernardoMeurer Nuclear Electronics, by P.W. Nicholson
re: in-depth reading, i have skipped a few small bits i didn't quite get, but a. I want to understand what's in this book really well, and b. this is sort of the introductory, very basics section, so I want to know it really well so I have a shot at the rest of the book.
of course, I could be taking it too far, but I did think that paragraph was important.

11:40 PM
@heather to be honest, I think you over-read that bit

Are there any books listed as recommended reading?

but ultimately, it's you that's in charge of how you read what you read ;-)
@skullpetrol what, Nuclear Physics for middle schoolers?

@heather have you thought any more about my question?

it's me who's in charge, but I'm a bad person to be in charge of anything, which is why I appreciate the advice @EmilioPisanty

@heather you're always in charge of your own learning

11:41 PM
@0celo7 to be absolutely honest, not much, i've been focusing more on your earlier proof question, with set theory

it's important to realize that

@skullpetrol for what?

@heather Hint: $E\cap F=(E\cup F)-(E\Delta F)$.

@EmilioPisanty i do realize that, i was merely stating I appreciate the advice.

@heather As I said, you need to know about Banach spaces before you can answer my integral question. But it's good to have in mind because the technique is very important.

11:43 PM
@heather happy to help

Are you a straight A student? @heather

@0celo7 Banach spaces =)
@skullpetrol yeah

cool

no
i am ahead in math though
i'm in geometry this year (8th grade)

That's a good start.

11:48 PM
@skullpetrol how old are you? (if you don't mind my asking)

@heather you only need the first 6 sections of Jost to answer my question

Jost?
"Postmodern Analysis"?

I'm just an "old woman" on the internet with no life.