The h Bar

@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.

Emilio Pisanty

or do that ↑

way easier

heather

9:11 PM

you can just ignore something in an equation like that?

oh.

Emilio Pisanty

@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

heather

oh, ok

Emilio Pisanty

hence the change in units

heather

rest mass of the electron, let me see...wikipedia says 9.109×$10^{−31}$ kilograms

Emilio Pisanty

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

Loong

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 $Z$.

heather

9MeV*82(82+1)

9 MeV*82*83

61254 MeV??? that can't be right.

Emilio Pisanty

@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

heather

what units do i have besides MeV?

Emilio Pisanty

in particular, you're missing an $e^4$ and an $(m_0c^2)^2$

Loong

@EmilioPisanty I haven't checked. It's not clear in this paragraph of the book.

heather

9:24 PM

@EmilioPisanty but Loong said i could just focus on the part I used...

Loong

@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 $I(Z)=kZ^2$

Emilio Pisanty

@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 $\sim1$

heather

@Loong well.. it multiples out to $EZ^2+EZ$ right, if you distribute?

Emilio Pisanty

@heather don't distribute

heather

@EmilioPisanty why not?

Emilio Pisanty

9:27 PM

for $Z\gg1$, you can mostly assume $Z(Z+1)\approx Z^2$

heather

@EmilioPisanty ok

so...then E=k, basically

Emilio Pisanty

@heather no

heather

?

Emilio Pisanty

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 $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{NEZ^2e^4}{137m_0^2c^4}$

0celo7

@EmilioPisanty is the term in parentheses $\sim 1$ for the usual energies?

Emilio Pisanty

9:30 PM

@0celo7 not really, but it will be relatively tame

it's relatively easy to calculate if you want to

$m_0c^2\approx 0.5\:\mathrm{MeV}$

0celo7

rest energy is that small?

Emilio Pisanty

for an electron? yes

0celo7

With E-31 I guess it is

Emilio Pisanty

so at $E=9\:\mathrm{eV}$ you have $2E/m_0c^2\approx 36$

Loong

@EmilioPisanty 0.5 MeV

Emilio Pisanty

9:32 PM

@Loong d'ough

and therefore $\ln(2E/m_0c^2)\approx3.5$

0celo7

so, like .36?

Emilio Pisanty

so the factor in brackets is ~14

not quite unity, but close enough

0celo7

@EmilioPisanty Uh, 2*9/500 is not 36

Emilio Pisanty

dammit

that's $E=9\:\mathrm{MeV}$, I believe

@0celo7 that's 2*9/0.5

0celo7

@EmilioPisanty yes

Emilio Pisanty

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 $k\approx\frac{e^4}{10m_0^2c^4}E$

heather

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?

loltospoon

I heard that computer bits are a quantum mechanical effect (unrelated to quantum computing/Qbits). Is this true?

0celo7

jesus, what unit is psi*lbm

er

psi*ft^3?

@loltospoon all of matter is a quantum effect, you have to be more precise

heather

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

loltospoon

@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.

0celo7

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

Emilio Pisanty

@heather $e$ is the electron charge, probably in gaussian units

$c$ is the speed of light

heather

1.6021765 × 10 −19 coulomb for $e$ is what comes up

Emilio Pisanty

@heather no, there's a units issue

that value is in SI units

Loong's formula is most likely in gaussian units

heather

::sighs:: okay, googling

Emilio Pisanty

i.e. if you want it in SI, you probably have to substitute the $e^2$ for $\frac{e^2}{4\pi\epsilon_0}$.

heather

9:49 PM

what's $\epsilon_0$

Emilio Pisanty

@heather usually known as "the permittivity of the vacuum"

heather

i think i'll stick with gaussian units

0celo7

@heather Why are you trying to do this?

heather

they sound much simpler right now

@0celo7 i'm trying to understand this paragraph in my book

and miserably failing

0celo7

why are you reading a book like this without knowing basic electrostatics

loltospoon

9:51 PM

SI units are way simpler

0celo7

@loltospoon $c=\hbar=G=k=1$

the simplest units

Emilio Pisanty

@heather long story short, $e_\mathrm{gaussian}^2=\frac{e_\mathrm{SI}^2}{4\pi\epsilon_0} = 1.4\times10^{-9}\mathrm{m\:eV}$.

loltospoon

You'll use them in high school and in the first 2 years of undergrad if you are in the U.S.

Emilio Pisanty

@0celo7 leave her alone, she's struggling enough as it is

@loltospoon let's not discuss the SI system

heather

@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

Emilio Pisanty

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

heather

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.

Emilio Pisanty

@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

heather

okay

Loong

@heather 0.5109989461(31) MeV

Emilio Pisanty

heather

9:56 PM

@EmilioPisanty good to know.

Emilio Pisanty

@heather yeah. This goes into the calculations of $\ln(2E/m_0c^2)$ above too.

heather

so $\frac{e^2}{0.5}9\text{MeV}$ is what we're at now?

@EmilioPisanty oh, yeah, right!

Emilio Pisanty

@heather never drop the units

heather

ack, right

$\frac{e^2}{0.5\text{ MeV}}9\text{ MeV}$

Emilio Pisanty

OK, so you're at $k\approx\frac{e^4}{10m_0^2c^4}E$

note that that's $k\approx\frac{(e^2)^2}{10(m_0c^2)^2}E$

heather

9:58 PM

oh so you multiply the 10 by 0.5?

so then you'd get 5 MeV in the denominator

no...

Emilio Pisanty

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

heather

oh, it's MeV squared...i didn't even notice that...

Emilio Pisanty

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, $k\approx\frac{(e^2)^2}{10(m_0c^2)^2}E\approx (8.3\times 10^{-31}\:\mathrm m^2)E$

heather

and E here is 9 MeV

Emilio Pisanty

with me so far?

@heather yes

heather

10:05 PM

@EmilioPisanty I think so

Emilio Pisanty

@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?

heather

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

Emilio Pisanty

@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 $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$

heather

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?

Emilio Pisanty

does that make sense units-wise?

heather

10:08 PM

or, wait, no, there's the units of N...

Emilio Pisanty

@heather precisely

so what are those?

heather

N is atomic density, right?

Emilio Pisanty

yup

0celo7

@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

heather

so...number per units cubed?

0celo7

10:09 PM

Do you have any idea?

heather

so maybe meters cubed @EmilioPisanty?

meters is a bit big though...

Emilio Pisanty

$N$ is in number of atoms per unit volume

@heather it's going to be huge anyway

heather

@EmilioPisanty true

Emilio Pisanty

i.e. Avogadro's-number-big

0celo7

not if you measure in planck units

Emilio Pisanty

10:10 PM

@0celo7 ::rolls eyes::

heather

@0celo7 ::groans::

rob

@0celo7 Heat capacity per number of atoms, since there aren't atomic degrees of freedom?

Emilio Pisanty

@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?

heather

@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?

Emilio Pisanty

10:11 PM

@heather nope

heather

but i thought we were dividing out N?

Emilio Pisanty

$N$ is in $1/\mathrm m^3$

@heather we will

for now just focus on the equation I just put up

0celo7

@rob hmmm

Emilio Pisanty

$-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$

0celo7

is that why $\gamma$ seems to be the same for them?

heather

10:13 PM

1/m^3 *m^2 *MeV= 1/m * MeV?

0celo7

I recall statistical mechanics very vaguely

heather

or am i doing it wrong again?

Emilio Pisanty

@heather yeah, that's correct

the right-hand side is in MeV/m

rob

@0celo7 I always forget what γ is in gas mechanics

heather

okay.

Emilio Pisanty

10:14 PM

is that what you'd expect for the left-hand side as well?

0celo7

@rob $C_v/C_p$

heather

what's that $r$ subscript?

Emilio Pisanty

@heather not sure, but it doesn't affect the units

heather

oh, okay.

so...E would be MeV again

Loong

@heather "radiative"

heather

10:15 PM

and x would probably be meters

and you're dividing again

so MeV/m

@EmilioPisanty yes, i think

Emilio Pisanty

@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?

heather

the negative of the change in energy over distance

0celo7

@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.

Emilio Pisanty

@heather precisely

0celo7

10:16 PM

but its worded like a trick question

heather

in which case the units do indeed make sense

Emilio Pisanty

@heather excellent

what we just did is called a dimensional-analysis consistency check

rob

@0celo7 What process?

Emilio Pisanty

it's really important to do this at pretty much every step in any calculation

heather

is that basically a fancy term for a common sense check =P @EmilioPisanty

Emilio Pisanty

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

heather

@EmilioPisanty that makes sense.

0celo7

@rob straight old heating

Emilio Pisanty

@heather in particular, it stops you from making some of the mistakes you were doing earlier

0celo7

it's a one-line problem, so I'm suspicious

heather

that's always a good thing =)

Emilio Pisanty

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?

Emilio Pisanty

(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)

0celo7

@rob constant pressure

actually, no qualifiers

just heated

the internal energy of an ideal gas depends only on the temperature

Emilio Pisanty

@heather OK, so we have $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$

0celo7

so the other variables are not relevant anyway

Emilio Pisanty

10:21 PM

let's make that do something useful, shall we?

rob

@0celo7 There you go.

0celo7

@rob ?

heather

@EmilioPisanty sounds good =)

Emilio Pisanty

@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

heather

okay

rob

10:23 PM

gtg, cheers all

heather

have a good day @rob

Emilio Pisanty

so, can you: (i) get the units of that quantity, and (ii) relate $N$ and $\rho$?

heather

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 $\rho$?

0celo7

@rob @EmilioPisanty So in an isothermal expansion of argon, I will not see any change in internal energy, right?

Emilio Pisanty

@heather the units of $\frac1\rho \frac{\mathrm dE}{\mathrm dx}$

@0celo7 no idea

heather

10:30 PM

@EmilioPisanty well rho is the mass density, not atomic density...so 1/(grams/m^3)*MeV/m

Emilio Pisanty

@heather yes

and from there?

heather

or m^3/grams*MeV/m

or MeV m^2/grams?

Emilio Pisanty

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

heather

=D

Emilio Pisanty

OK, so, next: how do you relate $\rho$ and $N$?

heather

10:33 PM

well, the units of N are 1/m^3

Emilio Pisanty

@heather not just the units

an actual equation

heather

oh

Emilio Pisanty

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 $A$

heather

right...

Emilio Pisanty

the question is what is $A$

heather

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

Emilio Pisanty

yeah

heather

oh...duh. the mass of each atom?

Emilio Pisanty

@heather exactly

call that $m_\mathrm{Pb}$

how much is that?

heather

3.4406366 × 10^-22 g ± 1.6605389 × 10^-25 g

according to the google

the second part is just an accuracy thing

Emilio Pisanty

sounds about right

@heather yeah

heather

10:37 PM

so really we have 3.4406366 * 10^-22 grams

Emilio Pisanty

you can forget about the uncertainty bit

OK, so we have $\rho=m_\mathrm{Pb}N$ and $\mathrm{Pb}=3.44\times10^{-25}\:\mathrm{kg}$

heather

right

Emilio Pisanty

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

heather

@EmilioPisanty shouldn't that be rho/m_Pb not N/rho?

Emilio Pisanty

$-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=\frac{8.3\times 10^{-31}\:\mathrm m^2}{m_\mathrm{Pb}} E\:Z^2$

heather

10:40 PM

nvm you're doing it to both sides

Emilio Pisanty

@heather exactly

and then I'm using $\rho/N=m_\mathrm{Pb}$ in the denominator on that last bit

heather

right

so why is that manipulation useful?

Emilio Pisanty

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

heather

oh, right.

@EmilioPisanty 2.4127907 * 10^-6 m^2/grams

Emilio Pisanty

correct

no, wait, not quite

mind your units

heather

10:44 PM

m^2 is on the top divided by grams for the mass on the bottom

Emilio Pisanty

@heather but you divided by the mass in kilograms ;-)

heather

oh duh =P

m^2/*kilo*grams would be the units then

Emilio Pisanty

exactly

0celo7

@heather kilo*gram?

Emilio Pisanty

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?

heather

10:46 PM

@0celo7 there's an asterisk on the other side - it's meant to italicize kilo because i made a mistake

0celo7

ah

I should write italics in mathfrak

heather

@EmilioPisanty to multiply it all out?

Emilio Pisanty

@heather pretty much

0celo7

Wow, $\mathfrak{Maroon}$ $\mathfrak 5$ and $\mathfrak{Future}$ collab

heather

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

Emilio Pisanty

10:49 PM

@heather see? now that factor of $82^2=6724$ doesn't seem so huge, does it?

heather

so 0.16621524 m^2 MeV/kg

@EmilioPisanty not at all

oh, 82^2 not 87^2

Emilio Pisanty

OK, so that's the specific stopping power

heather

whoops

Emilio Pisanty

@heather ;-)

what about the actual stopping power, in MeV/m?

heather

0.14765904 m^2 MeV/kg is the specific stopping power

@EmilioPisanty which equation was that again?

Emilio Pisanty

10:52 PM

$\frac{\mathrm dE}{\mathrm dx}$

instead of $\frac1\rho\frac{\mathrm dE}{\mathrm dx}$

heather

oh, right

$-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$

Emilio Pisanty

@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?

heather

according to the google, 11340

(kg/m^3)

Emilio Pisanty

yup

heather

1674.45351 MeV/m^3 then?

Emilio Pisanty

10:55 PM

and we had $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2=0.14\:\mathrm{MeV\:m^2/kg}$

heather

ack, darn the m^2

Emilio Pisanty

i.e. $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=0.14\:\mathrm{MeV\:m^2/kg}$

heather

let me do that again =)

1587.6 MeV m^2 kg/ kg m^3

or 1587.6 MeV/m

Emilio Pisanty

I get 1654 MeV/m, but that's about right

now, is that a useful unit?

heather