Download Order-of-Magnitude Physics – Solution Set 4

Order-of-Magnitude Physics – Solution Set 4
Andrew Myers, Eugene Chiang, Ryan Olf, and Alwin Mao
1
Tabletop Dynamos
~ in conducting fluids are governed by the induction equation:
Magnetic fields B
~
∂B
~ + DM ∇2 B
~
= ∇ × (~v × B)
∂t
(1)
where the first term on the right-hand side governs how magnetic flux is carried along
by (“frozen into”) fluid parcels, while the second term is the Ohmic dissipation term
derived in class. The fluid velocity is ~v , time is t, and DM is the magnetic diffusivity.
Magnetic dynamos are fluid systems whose motions (~v ) can sustain magnetic fields.
(a) Were fluid motions in the Earth’s core to cease, how long would it take the Earth’s
magnetic field to decay?
If fluid motion were to cease, ~v = 0 and the only remaining term on the right-hand side would
be the Ohmic dissipation term. The equation reduces to a diffusion equation. Notice that
~
∂B
~
= DM ∇2 B
∂t
(2)
looks identical to the Heat Equation
∂u
= Dth ∇2 u
∂t
so by analogy to our method of cooking a turkey,
tcooked ≈
L2
Dthermal
⇒ tfield
decayed
(3)
≈
Rcore 2
DM
(4)
Note that the radius of the convective outer core (source of Earth’s magnetic field) is about 3000
km and the temperature of the outer core ranges from 3000-5700 K.1 For the brave estimator,
Rcore = 12 × R⊕ and Tcore 10 × Troom would have done quite nicely!
c2
For Rcore ≈ 20%×R⊕ and DM = 4πσ
, and taking σe of Earth’s Iron Core to be 1017 s−1 × TTroom
as
e
core
seen in class (or you can look it up), this is a time of about 200,000 years which is short compared
to the age of the Earth. The brave estimator would get 450,000 years with this approach. It’s
comparable to average geomagnetic reversal times!2
1
2
http://en.wikipedia.org/wiki/Outer_core
http://en.wikipedia.org/wiki/Geomagnetic_reversal
1
An alternative, more in depth exposition continues below for the intrepid explorer:
We saw in class that to estimate orders of magnitude, we can replace ∂B/∂t with B/t and
similarly replace DM ∇2 B with DM B/L2 , where L is the length scale over which B changes. To
see in more detail what this approximation means, consider the general solution to the diffusion
equation in one dimension:
B(x, t) ∝ exp −DM t/λ2 exp (ix/λ)
where λ is a constant with units of length. Upon adapting to initial conditions, the exp (ix/λ)
becomes a Fourier series expansion for B(x, 0). The Fourier component that decays slowest is the
one corresponding to the largest λ. The largest λ of importance (meaning, its Fourier component
has substantial amplitude) is the distance over which B is interesting, which we can take to be the
same L we cavalierly adopted above. If we then consider B “decayed” when it reaches 1/en , we’re
interested in the t for which DM t/λ2 ≈ n, or
t ≈ nλ2 /DM .
So the poor-man’s derivative chopping approximation is, in this case, akin to choosing a λ and
setting n = 1.
Take L ≈ 3000 km to be the radius of the Earth’s liquid outer core. From class, we know that
liquid iron has resistivity ρe ≈ 10−6 Ω m ≈ 10−16 s. Its magnetic diffusivity is then
DM ≈ c2 /4πσe ≈ c2 ρe /4π ≈ c2 × (10−16 s)/4π ≈ 7 · 103 cm2 s−1 .
Thus, t ≈ L2 /DM ≈ 1.3 · 1013 s ≈ 4 · 105 yr. Without fluid motions in the core, the Earth’s field
would quickly die out.
We might also expect t to be of order the duration of field transition, during which the field
direction flips (from N-to-S or S-to-N) and the field is weaker than on average. This is to be
compared with the field reversal period (time between flips), which is seen from the sea-floor record
to vary erratically with a period of 105 –107 years).
(b) Why is it so difficult to study dynamos using tabletop experiments?3
Dynamos require the
of magnitude of the flux
magnitude of the Ohmic
latter to define the figure
flux freezing term to win over the Ohmic dissipation term. The order
~ ∼ vB/L. This must exceed the order of
freezing term is ∇ × (~v × B)
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dissipation term, DM B/L (from part (a)). Divide the former by the
of merit:
(vB/L)/(DM B/L2 ) = vL/DM ≡ ReM ,
otherwise known as the “magnetic Reynolds number.” This number must greatly exceed unity for
dynamos to succeed. But for tabletop experiments involving, say, a rotating vat of liquid mercury
(which is conveniently liquid at room temperature), L ≈ 1 m, v ≈ 1 m s−1 (rotation every 6 seconds),
and DM ≈ 7 · 103 cm2 s−1 . So ReM ≈ 1, which makes it hard to do dynamo experiments because
the field is decaying as fast as it is being generated.
3
Despite the difficulty, the Princeton Plasma Physics Lab has performed laboratory experiments related to the
magneto-rotational instability, a dynamo mechanism in differentially rotating flows (http://mri.pppl.gov).
2
2
Fade to Black
White dwarfs represent collapsed end-states of solar-mass stars that have expired their
nuclear fuel. Typically they have masses M ∼ 1M and radii R ∼ 1R⊕ . Their interiors
are composed of a crystalline lattice of carbon and oxygen nuclei, bathed in a sea of free
degenerate electrons: in short, an ultra-dense metal made of carbon and oxygen.
Usually in stars, heat is transported by either convection or radiation. But in white dwarf
interiors, heat is transported by electron conduction. Overlying the metallic interior is a
thin shell of hydrogen-rich gas — the white dwarf atmosphere — that transports energy
by radiation.
Take the
(effective
(they are
this ratio
core temperature to be Tcore and the atmospheric temperature to equal Teff
temperature of a blackbody). When born, white dwarfs have Teff ≈ 3 × 104 K
UV-bright sources) and Tcore ≈ 1 × 107 K. Assume that Tcore /Teff maintains
of ∼300 over the entire cooling history of the white dwarf.
For this problem, neglect corrections to the specific heat that might arise from T <
TDebye .4
Shapiro & Teukolsky, in their classic textbook “Black Holes, Neutron Stars, and White
Dwarfs,” state that as white dwarfs cool, nearly all of their metallic interiors can be
modelled as isothermal. That is, the temperature equals Tcore nearly everywhere inside
the interior. However, they offer no quantitative justification. We try to provide one
here.
(a) Estimate the thermal conduction time across the white dwarf interior, as a function
of Tcore , assuming the interior is isothermal. Express in yr.
Easy but not quite correct way (will still get full credit): The easy way is to say the
conduction time is just like for turkeys, so tcond ∼ R2 /D where D ∼ λe ve /3 is the thermal diffusivity.
From class the mean free path of an electron in a sea of phonons is
λe ∼
2π 2 c
a.
3 kT
The confinement energy is roughly the electrostatic energy between nuclei, c ∼ Z 2 e2 /a. So we
have
2π 2 Z 2 e2
λe ∼
∼ 6 · 10−8 cm
3 kT
Now for the electron speed ve , we should use the speed at the top of the Fermi sea, because only
those electrons are free to change their energy states and thereby transport thermal energy from
place to place:
ve ∼ h/(me a)
4
These corrections turn out to be of order unity; if you look at Shapiro & Teukolsky, there are large numerical
coefficients that happen to cancel out any Debye corrections that we might naively make. In this regard, Kittel
(“Introduction to Solid-State Physics”) points out that the thermal conductivity of metals is roughly constant for
T > TDebye /10 ∼ 30 K but does not explain where the “10” comes from; see the Course Reader, page 28, for the
thermal conductivity curve of copper to verify that this is true.
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from the Heisenberg uncertainty relation (assuming non-relativistic — we will check this in just a
−1/3
second), where a ∼ ne
∼ (Zρ/AmH )−1/3 ∼ (ZM/R3 AmH )−1/3 and Z ∼ 7 and A ∼ 14 (average
between C and O). Then we get ve ∼ 8 × 1010 cm/ s which exceeds the speed of light c but is close
enough to it (factor of 3) that the actual speed must be about c — i.e., the electrons are mildly
relativistic. The astronomers reading this will recognize this to be true because our white dwarf of
mass 1M is precipitously close to the Chandrasekhar mass of 1.4M (above which the electrons
are too relativistic and can no longer support the star against gravity because their equation of
state is too soft). Upshot: ve ∼ c.
Putting it all together, we have
7 2 10 K
D ∼ 6 × 10
cm2 s−1
Tcore
and tcond ∼ 2 × 107 (Tcore /107 K) yr (easy way but not quite right: should get full credit anyway) .
More correct way (also gets full credit): So the easy way above misses the fact that the
electrons are degenerate and therefore most of them don’t participate in carrying heat. The only
heat-carrying electrons are at the top of the Fermi sea; these comprise only a fraction ∆n/n ∼
kT /F where F ∼ (1/2)me ve2 ∼ (1/2)me c2 ∼ 200 keV and kT ∼ 1 keV(Tcore /107 K) (remembering
that 1 eV = 1e4 K). So the true D is down by kT /F ∼ 1 keV/200 keV ∼ 1/200 which means
tcond ∼ 4 × 109 yr — and independent of Tcore (more correct way; also gets full credit) .
If just inserting the dilution factor of ∆n/n ∼ kT /F was too fast and cavalier for you, then we
can derive it a bit more carefully, using the full heat diffusion equation:
∂
(ρCV T ) ∼ ∇ · (−kc ∇T )
∂t
where the LHS is the time rate of change of the thermal energy density and the RHS is the
divergence of the heat flux, with the heat flux written in terms of the thermal conductivity kc . Now
the LHS is dominated by the ions which are not degenerate (the electrons are mostly degenerate
and so can’t contribute to the thermal energy density — they have no freedom to move around
in energy space). So on the LHS we must evaluate ρCV = ρion CV,ion ∼ 3knion . On the RHS,
the thermal conductivity is dominated by the free electrons (with their long mean free paths and
fast near-c speeds), and from lecture we know that for metals (read: degenerate free electron gas),
kc ∼ (λe ve /3)(3k/2)ne (kT /F ). So solve for t = tcond in the full diffusion equation (OOM style,
replacing ∂/∂t with 1/t and ∇ with 1/R):
tcond ∼
R2
R2
∼
kc /(ρion CV,ion )
(λe ve /3)(1/2)(ne /nion )(kT /F )
So indeed we see that the true conduction time is longer than R2 /(λe ve /3) by the Fermi-sea dilution
factor kT /F — with another correction factor (1/2)(ne /nion ) ∼ Z/2 ∼ 3. Including this factor of
3, we get tcond ∼ 1 × 109 yr—independent of Tcore (best way; also gets full credit) .
(b) Estimate the thermal cooling time of the white dwarf interior, as a function of
Tcore (= 300Teff ), assuming the interior is isothermal. Express in yr. The cooling time
is the thermal energy of the dwarf divided by the power (a.k.a. luminosity) it emits into
space.
4
The cooling time is the total thermal energy content of the dwarf, divided by the radiative
luminosity. The heat capacity of the metal is dominated by the ions, which each carry 3kTcore ’s
worth of energy per ion (neglecting Debye corrections). There are M /(AmH ) such ions. The
4 4πR2 .
radiative luminosity, if we approximate the atmosphere as a blackbody radiator, is σTeff
Then
3kTcore (M /AmH )
tcool ≈
.
4 4πR2
σTeff
⊕
The problem states that Tcore /Teff ≈ 300 for the entire evolution so we can write
tcool ≈
900k(M /AmH )
,
3 4πR2
σTeff
⊕
which evaluates to, for an initial Teff ≈ 3 · 104 K, tcool ≈ 4 · 107 yr 107 K/Tcore
3
.
(c) You should find from (a) and (b) that the conduction time should be shorter than the
cooling time, especially as Tcore drops. Given this inequality, show from the diffusion
equation that an isothermal interior is a good approximation.
Hint: white dwarfs are spheres.
If you did part (a) the “easy-but-not-quite-right” way, then initially, tcond /tcool ≈ 0.5, and this
ratio only gets smaller as the white dwarf cools5
If you did part (a) the “best” way, then initially, tcond /tcool ≈ 20 , and this ratio gets smaller
as the white dwarf cools. Sorry the ratio tcond /tcool didn’t work out so well doing it the correct
way — in fact, there are more complicated reasons why people believe white dwarfs are isothermal;
e.g., when the white dwarf is still a proto-dwarf, embedded in its host red giant, there are shell
sources of nuclear burning that can heat the dwarf from the outside in, and thereby flatten out
the temperature gradient (Brad Hansen, personal communication). In any case, even if the white
dwarf doesn’t start out with tcond /tcool < 1, it soon evolves to this condition: tcool scales steeply
−3 , which means that lowering T
with Tcore as Tcore
core by just a factor of 3 will bring tcond /tcool below
unity.
Assuming (as the problem asks us to do) that tcond tcool (which, as explained above, eventually becomes true even if it’s not true initially), then we conclude that the white dwarf doesn’t
actually cool — it keeps its Tcore constant over the conduction time tcond . Then in the heat diffusion
equation
∂
(ρCV T ) ∼ ∇ · (−kc ∇T )
∂t
we can drop the LHS ∝ ∂T /∂t compared to the RHS:
∇ · (kc ∇T ) = 0
5
Truth be told, our calculation of the electron mean free path is only good after the white dwarf’s interior has
crystallized, after Tcore < a few × 106 K. Initially, as Shapiro & Teukolsky state, the interior is a pure plasma. Then
the electron mean free path is set by the usual Coulomb cross-section in a plasma and the number density of ions.
This simple calculation is done by Padmanabhan (Theoretical Astrophysics, volume 2) — but he erroneously drops
the kT /F factor in estimating the true D.
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which assuming kc ∼ constant means that
∇2 T = 0
which in spherical coordinates reads
1 ∂
∇ T = 2
r ∂r
2
2 ∂T
r
=0
∂r
or
∂T
= constant.
∂r
Even our lecturer can solve this equation; the solution is
r2
T (r) =
A
+ B.
r
To avoid a singularity inside the sphere we have to set A = 0. Thus, for a sphere the only possible
distribution is T = constant . Voilà, the white dwarf is isothermal.
One may criticize the above approach for being too mathy and lacking physical insight (our
lecturer is no fan). It’s also, strictly speaking, unrealistic, because by saying T = constant, we are
forced to conclude that the conductive flux of energy throughout the star is identically zero, which
is not strictly true — there’s always a little bit of leakage — the white dwarf is shining after all.
In the following second approach, we try to be more realistic—without trying too hard. We try
to estimate the order of magnitude of the temperature gradient required for the conductive flux in
the stellar interior to match the radiative flux in the atmosphere, assuming that the lengthscale
over which the temperature changes in the interior of the star is of order the entire stellar radius.6
The statement that the conductive flux matches the radiative flux is, to order of magnitude,
4
kc ∇T ∼ σTeff
Now ∇T ∼ ∆T /R. What we want is ∆T /Tcore :
∆T
σ
Teff 3
∼ R⊕
T
Tcore
kc
Tcore eff
Plug in (from part a) kc ∼ (λe ve /3)(3k/2)ne (kT /F ) ∼ 1 · 1015 (cgs; and notice kc is independent
of temperature, because λe ∝ 1/T ) and use again Teff /Tcore = 1/300 to find
3
∆T
Teff
∼3
Tcore
3 × 104 K
For the white dwarf to be isothermal, we want ∆T /Tcore 1. The white dwarf may not be
isothermal when born (cf. the personal communication above from Brad Hansen), but since Teff
3 drops pretty quickly , and
drops as the white dwarf cools, the ratio ∆T /Tcore ∝ Teff
isothermality eventually becomes a good approximation . For example, when the white dwarf is
1010 yr old, Teff ≈ 5000 K (see below) and ∆T /Tcore ∼ 0.01.
6
In principle, it doesn’t have to be the entire radius. The conductive flux could well be limited to a thin layer just
beneath the radiative atmosphere—in which case the bulk of the star would indeed be isothermal.
6
(d) What are the luminosities of the faintest white dwarfs in the universe today?7 Express in solar luminosities L .
Set tcool in part (b) equal to the age of the universe, 1010 yr, and solve for Teff ≈ 5000 K. The
2 σT 4 ∼ 4 × 10−5 L , which is pretty close to the
faintest white dwarfs have luminosities of 4πR⊕
eff
actual luminosities of the faintest white dwarfs observed.
[For the sticklers in the audience (there aren’t any in this class, are there?) one can solve the
differential equation
4 4πR2
σTeff
dTcore
⊕
=
,
dt
3k(M /AmH )
which leads to approximately the same answer.]
7
Sometimes referred to as black dwarfs.
7
3
Hot Chips
One factor limiting computer chip speeds is the waste heat they produce. The more
computations per second a chip performs, the more heat it produces and the greater the
danger of electronic failure.
A chip is a collection of transistors and connecting wires. Functionally, a transistor is
an on/off switch. Turning the transistor “on” means charging it up — it takes energy
to charge it up, just as it takes energy to charge up a capacitor. The faster the CPU
speed, the more capacitors are being charged up per unit time (the faster the charges are
being shuttled around the chip).
Chip manufacturers are constantly finding new ways to shrink transistors to cram more
of them onto a given size chip (Moore’s Law). These days a transistor measures a ∼ 300
Angstroms across.
(a) Using the transistor size a, estimate the waste-heat power output of a modern chip
the size of your fingernail. Express in Watts.
Hint: in cgs, capacitance has very nice units.
As a naive astronomer who can barely fix his own toaster, I will say that functionally, a transistor
is a capacitor: both are devices that require energy to be charged up.
I learned in high school that the energy in a single capacitor is E = (1/2)CV 2 . Use the hint
that capacitance C in cgs units has units of length. Then C ∼ a ∼ 300 Angstroms. Figure the
voltage is about 1 V — this makes sense since energy levels for electrons in atoms are typically
measured in eV (one electron charge times one Volt). Semiconductor energy gaps are typically
measured in eV. Also remember that 1 V = 1/300 statvolt (statvolt is the cgs unit of voltage).
How many capacitors N can you fit on a chip of area A? Transistor area is a2 so figure
N ∼ A/2a2 , where the factor of 2 is for spaces between transistors and other kinds of electronic
devices. Take A ∼ 1 cm2 .
Finally, we want power P ∼ N × E × f × waste heat efficiency, where f ∼ 2 GHz ∼ 2 · 109 for
a modern chip; and the waste heat efficiency might be ∼50% (this is a common-sense guess).
Putting it all together, we get P ∼ (1/2)A/a2 × (1/2)aV 2 × f × 50% ∼ (1/8)(A/a)V 2 f ∼ 100
Watts.
A trip to http://en.wikipedia.org/wiki/List of CPU power dissipation figures reveals that our
estimate of 100 W is pretty much spot-on. I guess high school wasn’t all that bad.
For a fingernail-sized chip to produce 100 W of waste heat is huge — it’s about as much power
as a human being radiates. No wonder my lap gets really hot when I work with my laptop!
(b) The chip heats the air inside the computer casing. A cooling fan is used to flush the
hot air out. Use your answer in (a) to estimate how much hotter the air gets inside the
computer, in steady state. Express the change in air temperature, ∆T , in K.
This is the same as the “Bright Lights, Big City” problem.
Figure the computer casing is a tallish box of volume V ∼ 30 × 10 × 10 ∼ 3 · 103 cm3 . The fan
blows maybe 1 m/s, and so flushes out the air in the casing every ∼30 cm / (1 m/s) ∼ 0.3 s (this
is what Harte calls the “residence time” in his “Consider a Spherical Cow” book).
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So in 0.3 s, the chip releases 100 W × 0.3 s ∼ 30 J ∼ 3 · 108 erg which gets converted into the
thermal energy of the air in the casing: (3k∆T /2) × nair V , where nair ∼ 3 × 1019 cm−3 (a number
worth remembering).
Solve for ∆T ∼ 15 K which feels about right (it’s hot in there!), and is certainly right to OOM.
If you do this problem backwards, starting with the estimate that an order 30-100 Kelvin
temperature change will damage or at least accelerate the aging of a CPU, you can figure out how
much power a CPU can safely consume. And then with the same equations from part (a), you can
figure out why clock speeds are set to what they are and why only hobbyists with advanced cooling
techniques can and do “overclock” their CPUs.8
8
http://www.youtube.com/watch?v=UKN4VMOenNM
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