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damage? What changes if the parked car has its brakes on? Motion Mountain The Adventure
of Physics copyright Christoph Schiller June 1990October 2016 free pdf file available at
www.motionmountain.net 4 from objects and images to conservation 111 TA B L E
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tutors comments on an assignment or with the grading of an assignment. You should try your
best to attend the tutorials. This is the only chance to have face to face contact with your tutor
and to ask questions which are answered instantly. You can raise
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is given by Fdrag/A = V/d, where V/d = Vx /z is the velocity gradient and is the shear
viscosity. In steady state, the applied force balances the drag force, i.e. F + Fdrag = 0. Clearly in
the steady state the net momentum density of the fluid does not ch
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(nV ) = 0 . (8.78) The second bracketed term on the RHS vanishes because of particle continuity,
leaving us with s + V s s = 0 (since V = 0 on average, and any gradient is first order in
smallness). Now thermodynamics says ds = s T p dT + s p T dp = cp T
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opposite sign on the northern and southern hemispheres and thus prove the rotation of the
Earth. (In the First World War, many naval guns missed their targets in the southern hemisphere
because the engineers had compensated them for the Coriolis effect in
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rad can be measured, a limit given by the fluctuations of the air in the atmosphere. In space,
such as for the Hubble telescope orbiting the Earth, the angular limit is due to the diameter of
the telescope and is of the order of 10 nrad. Practically all s
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called staircase formula. Why does it hold? Challenge 304 s Most animals have an even number
of legs. Do you know an exception? Why not? In fact, one can argue that no animal has less
than four legs. Why is this the case? On the other hand, all animals wi
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detailed balance is f 0 ( ) f 0 ( 1 ) f 0() f 0(1 ) = w 1  1 w 1  1 . (8.54) Under
this condition, the collision term vanishes for f = f 0 , which is the equilibrium distribution. 376
CHAPTER 8. NONEQUILIBRIUM PHENOMENA 8.3.6 Kinematics and cross sectio
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() 2 ! . (8.348) Then Fourier transforming the source function J(t), it is easy to see that Z[J] =
Z[0] exp 2 Z d 2
J() 2 1 + 2 2 ! . (8.349) Note that with (t) R and J(t) R we have () = () and J
() = J(). Transforming back to real time, 18A discussion o
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Suppose we have some averaged intensive quantity which is spatially dependent through T (r)
or (r) or V (r). For simplicity we will write = (z). We wish to compute the current of across
some surface whose equation is dz = 0. If the mean free path is , the
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since the cubic term prefers positive m. 330 CHAPTER 7. MEAN FIELD THEORY OF PHASE
TRANSITIONS Figure 7.15: Behavior of the quartic free energy f(m) = 1 2 am2 1 3 ym3 + 1 4
bm4 . A: y 2 < 4ab ; B: 4ab < y2 < 9 2 ab ; C and D: y 2 > 9 2 ab. The thick black
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a > 0 is small, we have m2 0 = a/3b and m (q) = h(q) 2a + q 2 , (7.247) 7.9.2 Domain wall
profile A particularly interesting application of GinzburgLandau theory is its application toward
modeling the spatial profile of defects such as vortices and domai
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nonequilibrium distribution to converge to the Boltzmann distribution. 372 CHAPTER 8.
NONEQUILIBRIUM PHENOMENA Figure 8.1: Level sets for a sample f(x, p, t) = Ae 1 2
(xpt) 2 e 1 2 p 2 , for values f = Ae 1 2 2 with in equally spaced intervals from = 0.2
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energy consumption per mass has been achieved. This is indeed the case; death occurs for most
animals when they have conRef. 98 sumed around 1 GJ/kg. (But quite a bit later for humans.)
This surprisingly simple result is valid, on average, for all known a
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antiferromagnet in an external field. The phase diagram is symmetric under reflection in the h =
0 axis. However, we can see from the figure that there will be three solutions to the mean field
equations provided that mA mB < 1 at the point of the solutio
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> 0 we can ignore the sextic term for sufficiently small u, and we recover the quartic free energy
studied earlier. There is then a second order transition at r = 0. . 334 CHAPTER 7. MEAN FIELD
THEORY OF PHASE TRANSITIONS Figure 7.18: Free energy (u) = 1
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performed with spheres.) What happens to the energy? Is the structure shown in Figure 87
possible? Does a wall get a stronger jolt when it is hit by a ball rebounding from it or when
it is hit Challenge 230 s by a ball that remains stuck to it? Motion Mou
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amplitude of 19.2 . Nutation occurs because the plane of the Moons orbit around the Earth is
not exactly the same as the plane of the Earths orbit around the Sun. Are you able to confirm
that Challenge 267 e this situation produces nutation? Astronomers a
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(0)
+ Xn =1 R,i Zt 0 ds e(ts) L (s) . (8.391) Note that the first term on the RHS is the solution
to the homogeneous equation, as must be the case when (s) = 0. The solution in eqn. 8.391
holds for general Q and (s). For the particular form of Q and (s) i
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variations of a fraction of a millisecond of arc due to the tides (from hpiers.obspm.fr/ eoppc).
Even though at first derided across the world, Wegeners discoveries were correct. Modern
satellite measurements, shown in Figure 109, confirm this model. For
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temperature, the thermodynamics scale. This will be discussed later. 3.4 Specification of
Interpolation The way we establish the temperature of a body on either the Celsius scale or the
Fahrenheit scale is what we refer to as the specification of interpol
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1 , corresponding to the eigenvalue 1 + . Clearly when < 1 the free energy is unbounded from
below, which is unphysical. We now set f m = 0 , f = 0 , (7.363) and identify four possible
phases: Phase I : m = 0, = 0. The free energy is fI = 0. Phase II : m
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container, first shoot up in the air. See the video at www. youtube.com/embed/_dQJBBklpQQ
and the story of the discovery at stevemould.com. Challenge 249 ny Can you explain the effect
to your grandmother? Motion Mountain The Adventure of Physics copyright
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length 8 are a large square, a rectangle and a corner. The large square gives a contribution =
N(tanh J) 8 There are two orientations for the rectangle. Including these gives a factor of 2, =
2N(tanh J) 8 Finally, the corner graph has four orientations, g
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but is flattened at the poles, confirmed the rotation of the Earth. Figure 95 illustrates the
situation. Again, however, this eighteenth century measurement by Maupertuis* is not
accessible to everyday observation. Then, in the years 1790 to 1792 in Bolog
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+ e t 1 e (tt ) . (8.413) We then fnd, for t > t , x(t) x(t ) x(t) x(t ) = 2kBT M t +
kBT 2M 2e t + 2e t 2 e (tt ) e (t+t ) . (8.414) 424 CHAPTER 8.
NONEQUILIBRIUM PHENOMENA In particular, the equal time autocorrelator is x 2 (t) x(t)
2 = 2kBT M t + kBT 2
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www.motionmountain.net 144 5 from the rotation of the earth (The centre of the Earths core is
solid; this was discovered in 1936 by the Danish seismologist Inge Lehmann (18881993); her
discovery was confirmed most impressively by two British seismologists
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" E N i=1 p 2 i 2m + m 2 x 2 i 2 # . Here (x) = 1, x > 0 0, x < 0 is the Heaviside unit
step function. We have d(x)/dx = (x), thus the structure function (E) = (E)/E is given by
the equation (E) = Z V dx1 Z V dx2 .Z V dxN Z d p1 Z d p2 .Z d pN " E N i=1
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Consequently, hWi = hV 2 i/R = (v 2 0/R) 3p(1 p) 2 +12p 2 (1 p) +9p 3 = (3v 2 0/R)
p(1+2p). Now let us turn to parallel connection. Let us for a while introduce conductance Gi =
1/Ri of the batteries, as well as conductance G = 1/R of the resistor. The vo
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want to look in another, the eye can accomplish this change in different ways. The option chosen
by the human eye had already been studied by medical scientists in the eighteenth century. It is
called Listings law.* It states that all Challenge 301 s axes