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9:01 PM
@EmilioPisanty, okay, so let's say I'm given 9 MeV electrons in lead. How can I find I(Z)?
 
@heather no idea
 
Z is 82 in this case
 
not sure what it is you're actually doing
but I can help with the "varies as $Z^2$" bit
 
I(Z) is the loss of energy as bremsstrahlung (related to the loss of energy as ionization, i think)
 
@heather your 9 MeV would be the $E$ in my equation
 
9:04 PM
@Loong what is N?
 
$N$ is atomic density
 
oh, and what's $m$ and $c$?
atomic density of the medium...?
(in case you couldn't tell already, i've got no idea what I'm doing =P)
 
$m_0$ is the rest mass of the charged particle (in this case, your electron)
@heather yes, atomic density of the medium, e.g. number of lead atoms per cubic centimetre
 
@Loong is there a "standard" number for lead that the book might be using?
 
@heather probably
solid metallic lead will have a pretty fixed density
 
9:09 PM
let me go google that
 
> Density near r.t. 11.34 g/cm3
(from Wikipedia)
 
but that isn't atomic density - is it?
 
@heather that's the mass density
you want the number density
 
@heather This factor is usually ignored first, and the resulting stopping powers are given in MeV cm^2/g. You can multiply the results with the actual density later.
 
or do that ↑
way easier
 
9:11 PM
you can just ignore something in an equation like that?
oh.
 
@heather you're not ignoring it
you've got an equation of the form $a=b\times c$, but the effect of $b$ is boring
so instead you calculate $c$ and you report it as $a/b$ instead of giving $a$.
if you're sloppy, you might even give it the same name, but by rights you shouldn't
 
oh, ok
 
hence the change in units
 
rest mass of the electron, let me see...wikipedia says 9.109×$10^{−31}$ kilograms
 
Also, note that anything with an actual number density in it is bound to have some ridiculously big numbers in it (of the order of Avogadro's number), which makes it a lot less helpful
having something in terms of the mass density is a lot more helpful
 
9:18 PM
@heather You can just focus on the $EZ(Z+1)$ part of the equation. So the energy loss is approximately $kEZ^2$ for large $Z$.
 
9MeV*82(82+1)
9 MeV*82*83
61254 MeV??? that can't be right.
 
@Loong what units is that in?
I'd assume gaussian units for the $e^2$ term, right?
@heather mind your units
that can't be right because the units are not right, which means that you're missing other factors that can completely change the order of magnitude
 
what units do i have besides MeV?
 
in particular, you're missing an $e^4$ and an $(m_0c^2)^2$
 
@EmilioPisanty I haven't checked. It's not clear in this paragraph of the book.
 
9:24 PM
@EmilioPisanty but Loong said i could just focus on the part I used...
 
@heather focus on this part to answer your question:
37 mins ago, by heather
hmm, i think i can move on from this section once i understand one thing: how does the electron energy play a role in the equation $I(Z)=kZ^2$
 
@heather as in: the bit in brackets will be a comparatively tame dimensionless factor of order unity
so you can just assume that the stuff in brackets is $\sim1$
 
@Loong well.. it multiples out to $EZ^2+EZ$ right, if you distribute?
 
@heather don't distribute
 
@EmilioPisanty why not?
 
9:27 PM
for $Z\gg1$, you can mostly assume $Z(Z+1)\approx Z^2$
 
@EmilioPisanty ok
so...then E=k, basically
 
@heather no
 
?
 
If you have $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{NEZ(Z+1)e^4}{137m_0^2c^4}\left(4\ln\frac{2E}{m_0c^2}-\frac43\right)$
which you appoximate to $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{NEZ(Z+1)e^4}{137m_0^2c^4}$
and from there to $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{NEZ^2e^4}{137m_0^2c^4}$
 
@EmilioPisanty is the term in parentheses $\sim 1$ for the usual energies?
 
9:30 PM
@0celo7 not really, but it will be relatively tame
it's relatively easy to calculate if you want to
$m_0c^2\approx 0.5\:\mathrm{MeV}$
 
rest energy is that small?
 
for an electron? yes
 
With E-31 I guess it is
 
so at $E=9\:\mathrm{eV}$ you have $2E/m_0c^2\approx 36$
 
@EmilioPisanty 0.5 MeV
 
9:32 PM
@Loong d'ough
and therefore $\ln(2E/m_0c^2)\approx3.5$
 
so, like .36?
 
so the factor in brackets is ~14
not quite unity, but close enough
 
@EmilioPisanty Uh, 2*9/500 is not 36
 
dammit
that's $E=9\:\mathrm{MeV}$, I believe
@0celo7 that's 2*9/0.5
 
@EmilioPisanty yes
 
9:35 PM
anyways, @heather
the factor in brackets is $\sim14$
so if you want to be a bit more accurate you can cancel it out with the $1/137$ to leave $1/10$
that leaves you with $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r\approx\frac{NEZ^2e^4}{10m_0^2c^4}$
and you want to express that as $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r/N=kZ^2$
then you have $k\approx\frac{e^4}{10m_0^2c^4}E$
 
whoa, let me catch up...i follow up until your approximation of $k$
actually, no, I don't follow to how you get your equation after the equation where you cancel to leave 1/10
oh, wait...nevermind, I follow
what is $e$ and $c$ here? the constants or are they something else?
 
I heard that computer bits are a quantum mechanical effect (unrelated to quantum computing/Qbits). Is this true?
 
jesus, what unit is psi*lbm
er
psi*ft^3?
@loltospoon all of matter is a quantum effect, you have to be more precise
 
so for 9 MeV electrons in lead, k*6724 is what I need to solve
$\frac{e^4}{10m_0^2c^4}E$
so...10*9.109×$10^{−31}$ kilograms, or 9.109$\times 10^{-30}$ kilograms
 
@0celo7some guy told my friend that he didn't believe that the ideas from QM were correct. The guy was a computer engineer. So my friend told him that without the theory of QM we would not have been able to create a classical computer bit.
 
9:47 PM
@loltospoon He is talking about transistors
But you don't need computer bits to see QM at work.
The sun is a quantum mechanical effect
 
@heather $e$ is the electron charge, probably in gaussian units
$c$ is the speed of light
 
1.6021765 × 10 −19 coulomb for $e$ is what comes up
 
@heather no, there's a units issue
that value is in SI units
Loong's formula is most likely in gaussian units
 
::sighs:: okay, googling
 
i.e. if you want it in SI, you probably have to substitute the $e^2$ for $\frac{e^2}{4\pi\epsilon_0}$.
 
9:49 PM
what's $\epsilon_0$
 
@heather usually known as "the permittivity of the vacuum"
 
i think i'll stick with gaussian units
 
@heather Why are you trying to do this?
 
they sound much simpler right now
@0celo7 i'm trying to understand this paragraph in my book
and miserably failing
 
why are you reading a book like this without knowing basic electrostatics
 
9:51 PM
SI units are way simpler
 
@loltospoon $c=\hbar=G=k=1$
the simplest units
 
@heather long story short, $e_\mathrm{gaussian}^2=\frac{e_\mathrm{SI}^2}{4\pi\epsilon_0} = 1.4\times10^{-9}\mathrm{m\:eV}$.
 
You'll use them in high school and in the first 2 years of undergrad if you are in the U.S.
 
@0celo7 leave her alone, she's struggling enough as it is
@loltospoon let's not discuss the SI system
 
@0celo7 a. because i'm interested b. because my dad recommended this book and wants me to learn it c. because i can be idiotic like that =P
 
9:53 PM
@heather OK, next factor: $m_0c^2$, where $m_0$ is the rest mass of the electron and $c$ is the speed of light
don't bother putting in $m_0$ in kilograms
if you're doing particle physics, the only worthwhile figure is $m_0c^2=0.5\:\mathrm{MeV}$ itself
 
that's just how it was defined on the wiki article @EmilioPisanty (in kilograms) what units does it need to be in?
@EmilioPisanty oh, that just comes out to 0.5 MeV?
that's nice.
 
@heather you don't normally bother with $m_0$ itself, you mostly just want $m_0c^2$ as is
@heather have a look at Wikipedia
 
okay
 
@heather 0.5109989461(31) MeV
 
9:56 PM
@EmilioPisanty good to know.
 
@heather yeah. This goes into the calculations of $\ln(2E/m_0c^2)$ above too.
 
so $\frac{e^2}{0.5}9\text{MeV}$ is what we're at now?
@EmilioPisanty oh, yeah, right!
 
@heather never drop the units
 
ack, right
$\frac{e^2}{0.5\text{ MeV}}9\text{ MeV}$
 
OK, so you're at $k\approx\frac{e^4}{10m_0^2c^4}E$
note that that's $k\approx\frac{(e^2)^2}{10(m_0c^2)^2}E$
 
9:58 PM
oh so you multiply the 10 by 0.5?
so then you'd get 5 MeV in the denominator
no...
 
nope
the denominator is $10(m_0c^2)^2=10(0.5\:\mathrm{MeV})^2=10\times0.25\:\mathrm{MeV}^2$.
see why the units are important?
on the numerator too
 
oh, it's MeV squared...i didn't even notice that...
 
you had $e^2=1.4\times10^{-9}\mathrm{m\:eV}$
so $e^4\approx 2\times10^{-18}\mathrm{m^2\:eV^2}$
all told, $k\approx\frac{(e^2)^2}{10(m_0c^2)^2}E\approx (8.3\times 10^{-31}\:\mathrm m^2)E$
 
and E here is 9 MeV
 
with me so far?
@heather yes
 
10:05 PM
@EmilioPisanty I think so
 
@heather OK, so what do you reckon is up with that meter-squared inside the brackets there?
does that square with the units that you'd expect?
 
well you had e^2 in the units m MeV so that would square obviously to m^2 MeV^2 and then the MeV^2 cancels with the one on the denominator
so it makes sense mathematically
 
@heather cool
and in terms of the bigger expression?
i.e. in terms of what we're building towards
just as a reminder, we wanted $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r/N=kZ^2$.
i.e. we have $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
 
well, it'll end up being m^2*MeV which I suppose might make sense - I guess it represents the distance the electron travels at a certain energy before energy lost due to bremsstrahlung radiation becomes comparable to energy lost due to ionization, maybe?
 
does that make sense units-wise?
 
10:08 PM
or, wait, no, there's the units of N...
 
@heather precisely
so what are those?
 
N is atomic density, right?
 
@rob My thermo prof said that some property of noble gasses was the same but I didn't write it down. I thought it was $C_v$ but that's not true
 
so...number per units cubed?
 
10:09 PM
Do you have any idea?
 
so maybe meters cubed @EmilioPisanty?
meters is a bit big though...
 
$N$ is in number of atoms per unit volume
@heather it's going to be huge anyway
 
@EmilioPisanty true
 
i.e. Avogadro's-number-big
 
not if you measure in planck units
 
10:10 PM
@0celo7 ::rolls eyes::
 
@0celo7 ::groans::
 
rob
@0celo7 Heat capacity per number of atoms, since there aren't atomic degrees of freedom?
 
@heather the important thing here is checking that our equation still makes sense dimensionally
what dimensions does $(8.31\times10^{-31}\:\mathrm m^2)N$ have?
 
@EmilioPisanty well I suppose...wait! it's going to end up being m*MeV right? and that makes perfect sense!
 
rob
@0celo7 Compare to N_2 and O_2, which have rotational excited states at room temperature, but different rotational energy spectra because the rotor masses are different?
 
10:11 PM
@heather nope
 
but i thought we were dividing out N?
 
$N$ is in $1/\mathrm m^3$
@heather we will
for now just focus on the equation I just put up
 
@rob hmmm
 
$-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
 
is that why $\gamma$ seems to be the same for them?
 
10:13 PM
1/m^3 *m^2 *MeV= 1/m * MeV?
 
I recall statistical mechanics very vaguely
 
or am i doing it wrong again?
 
@heather yeah, that's correct
the right-hand side is in MeV/m
 
rob
@0celo7 I always forget what γ is in gas mechanics
 
okay.
 
10:14 PM
is that what you'd expect for the left-hand side as well?
 
@rob $C_v/C_p$
 
what's that $r$ subscript?
 
@heather not sure, but it doesn't affect the units
 
oh, okay.
so...E would be MeV again
 
@heather "radiative"
 
10:15 PM
and x would probably be meters
and you're dividing again
so MeV/m
@EmilioPisanty yes, i think
 
@heather precisely
does that make sense as a unit for the quantity?
we want $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r$
what does that actually mean?
 
the negative of the change in energy over distance
 
@rob My homework wants to know if the $\Delta U$ in some process changes if I replace neon with argon. I think it's a "yes" because they have different heat capacities
 
rob
@0celo7 Right. Because in that ratio the mass of the atom cancels out, and the ideal gas equation $pV = NkT$ treats pressure and volume the same way as far as storing energy in the gas.
 
@heather precisely
 
10:16 PM
but its worded like a trick question
 
in which case the units do indeed make sense
 
@heather excellent
what we just did is called a dimensional-analysis consistency check
 
rob
@0celo7 What process?
 
it's really important to do this at pretty much every step in any calculation
 
is that basically a fancy term for a common sense check =P @EmilioPisanty
 
10:17 PM
i.e. making sure that the units on both sides still coincide
@heather a specific sort of common-sense check, if you will
 
@EmilioPisanty that makes sense.
 
@rob straight old heating
 
@heather in particular, it stops you from making some of the mistakes you were doing earlier
 
it's a one-line problem, so I'm suspicious
 
that's always a good thing =)
 
10:18 PM
like forgetting that the $e^4$ was an $e^4$ instead of an $e^2$
ditto for the $m_0^2c^4$
if you keep the units and you work everything through correctly, the units should come out correct
if they don't, you know you've made a mistake and you need to double-check everything
 
rob
@0celo7 Constant volume? constant pressure?
 
(of course, just because the units come out right doesn't mean you haven't made a mistake, but it's a good tool to catch the obvious ones)
 
@rob constant pressure
actually, no qualifiers
just heated
the internal energy of an ideal gas depends only on the temperature
 
@heather OK, so we have $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
 
so the other variables are not relevant anyway
 
10:21 PM
let's make that do something useful, shall we?
 
rob
@0celo7 There you go.
 
@rob ?
 
@EmilioPisanty sounds good =)
 
@heather now, $N$ is some unmanageable number
which is good, because that area there is pretty tiny
but what Loong said is the standard is the specific stopping power: the energy loss per unit distance per unit density
i.e. $\frac1\rho\frac{\mathrm dE}{\mathrm dx}$ where $\rho$ is the mass density
 
okay
 
rob
10:23 PM
gtg, cheers all
 
have a good day @rob
 
so, can you: (i) get the units of that quantity, and (ii) relate $N$ and $\rho$?
 
so the units of density are 1/m^3 and the units of the other part are MeV/m, so multiply it out and you get MeV/m^4, which doesn't seem right
wait, are you looking for the units of N or the units of $\rho$?
 
@rob @EmilioPisanty So in an isothermal expansion of argon, I will not see any change in internal energy, right?
 
@heather the units of $\frac1\rho \frac{\mathrm dE}{\mathrm dx}$
@0celo7 no idea
 
10:30 PM
@EmilioPisanty well rho is the mass density, not atomic density...so 1/(grams/m^3)*MeV/m
 
@heather yes
and from there?
 
or m^3/grams*MeV/m
or MeV m^2/grams?
 
1 hour ago, by Loong
@heather This factor is usually ignored first, and the resulting stopping powers are given in MeV cm^2/g. You can multiply the results with the actual density later.
i.e. yes
 
=D
 
OK, so, next: how do you relate $\rho$ and $N$?
 
10:33 PM
well, the units of N are 1/m^3
 
@heather not just the units
an actual equation
 
oh
 
they need to be directly proportional
i.e. if you cram twice as many atoms into a given unit volume then it will have twice as much mass
so you need to have $\rho=AN$ for some $A$
 
right...
 
the question is what is $A$
 
10:35 PM
i mean thinking about it, the difference is $\rho$ uses a mass per unit volume
A has to have the units grams
 
oh...duh. the mass of each atom?
 
@heather exactly
call that $m_\mathrm{Pb}$
how much is that?
 
3.4406366 × 10^-22 g ± 1.6605389 × 10^-25 g
according to the google
the second part is just an accuracy thing
 
sounds about right
@heather yeah
 
10:37 PM
so really we have 3.4406366 * 10^-22 grams
 
you can forget about the uncertainty bit
OK, so we have $\rho=m_\mathrm{Pb}N$ and $\mathrm{Pb}=3.44\times10^{-25}\:\mathrm{kg}$
 
right
 
let's put that into $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
it gives us
$-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)\frac N\rho E\:Z^2$
i.e. in other words
 
@EmilioPisanty shouldn't that be rho/m_Pb not N/rho?
 
$-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=\frac{8.3\times 10^{-31}\:\mathrm m^2}{m_\mathrm{Pb}} E\:Z^2$
 
10:40 PM
nvm you're doing it to both sides
 
@heather exactly
and then I'm using $\rho/N=m_\mathrm{Pb}$ in the denominator on that last bit
 
right
so why is that manipulation useful?
 
So, can you calculate that $\frac{8.3\times 10^{-31}\:\mathrm m^2}{m_\mathrm{Pb}}$?
@heather because we explicitly care about that $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r$, remember?
that's the stopping power per unit density
 
oh, right.
@EmilioPisanty 2.4127907 * 10^-6 m^2/grams
 
correct
no, wait, not quite
mind your units
 
10:44 PM
m^2 is on the top divided by grams for the mass on the bottom
 
@heather but you divided by the mass in kilograms ;-)
 
oh duh =P
m^2/*kilo*grams would be the units then
 
exactly
 
@heather kilo*gram?
 
so you have $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2$, then
what else do you need?
 
10:46 PM
@0celo7 there's an asterisk on the other side - it's meant to italicize kilo because i made a mistake
 
ah
I should write italics in mathfrak
 
@EmilioPisanty to multiply it all out?
 
@heather pretty much
 
Wow, $\mathfrak{Maroon}$ $\mathfrak 5$ and $\mathfrak{Future}$ collab
 
2.196*10^-5 m^2 MeV/kg Z^2 i think
and then in this case we're using lead
Z^2 = 87^2 if i remember right
 
10:49 PM
@heather see? now that factor of $82^2=6724$ doesn't seem so huge, does it?
 
so 0.16621524 m^2 MeV/kg
@EmilioPisanty not at all
oh, 82^2 not 87^2
 
OK, so that's the specific stopping power
 
whoops
 
@heather ;-)
what about the actual stopping power, in MeV/m?
 
0.14765904 m^2 MeV/kg is the specific stopping power
@EmilioPisanty which equation was that again?
 
10:52 PM
$\frac{\mathrm dE}{\mathrm dx}$
instead of $\frac1\rho\frac{\mathrm dE}{\mathrm dx}$
 
oh, right
$-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
 
@heather no, don't go that far back
we had $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2$
so you just need to multiply both sides by $\rho$
what's $\rho$ in kg/m^3?
 
according to the google, 11340
(kg/m^3)
 
1674.45351 MeV/m^3 then?
 
10:55 PM
and we had $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2=0.14\:\mathrm{MeV\:m^2/kg}$
 
ack, darn the m^2
 
i.e. $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=0.14\:\mathrm{MeV\:m^2/kg}$
 
let me do that again =)
1587.6 MeV m^2 kg/ kg m^3
or 1587.6 MeV/m
 
I get 1654 MeV/m, but that's about right
now, is that a useful unit?
 
@EmilioPisanty, did you multiply by 0.15 instead of 0.14?
 
10:58 PM
that factor of ~1600 feels pretty big to me
@heather doesn't really matter, remember all the approximations we did?
 
i've practically forgotten what these numbers mean
@EmilioPisanty true.
 
make it 1600 MeV/m and it's in the right range
OK, so you have $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=1600\:\mathrm{MeV/m}$
 
what's the difference between specific and actual stopping power (intuitively)?
@EmilioPisanty or the negative of the change in energy over distance is 1600 MeV/m
 

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