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Magnetic work
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: September 25, 2017)
When we discuss the magnetic work in the thermodynamics, we realize that the internal
energy is expressed in terms of different forms,
MdBTdSdUR , BdMTdSdUK
where the suffixes R and K denote the names of Reif and Kittel, for convenience. There are two
gedanken experiments on the magnetic work, (1) Kittel and (ii) Reif. The expression of the
internal energy is different, depending on the experiments. First we show such gedanken
experiments.
In the quantum mechanics, the mutual energy should be equal to MdB as is expected from
the Zeeman energy. This means that Reif’s expression is close to the prediction from the
quantum mechanics. Note that the origin of the magnetic energy lies in the quantum mechanics,
but not in the classical physics since no work can be done for the Lorentz force.
Here we discuss the validity of the two expressions based on the quantum statistical
mechanics. To this end, we use the simple model of spin 1/2 paramagnet. We show that the
Reis’s expression can be derived from this method.
It is important to stress that the results of the thermodynamic properties based on the Reif
notation are the same as those based on the Kittel’s notation. The thermodynamics of
superconductivity can be discussed mainly in terms of the Kittel’s notation.
1. Gedanken experiment on the magnetic work by Kittel (classical method)
Fig. Magnetic material (cylinder) inside the solenoid with area A1 and distance l.
The magnetic material is put inside the coil. The magnetic field is produced by current
through the coil. The magnetic material can be magnetized.
For a coil having 1n turns per unit length, we apply the Ampere’s theorem.
JHc
4 ,
lHlInc
dc
dd 111
44)(
aJlHaH ,
or
lHlInc
111
4
or
111
4In
cH
Fig. Ampere’s law for the magnetic field of the solenoid
The magnetic flux density B inside the magnetic material is
01 4 MHB
where 0M is the magnetic moment per unit volume. The magnetic flux 0 is given by
BAHAA 21210 )(
When 1I is increased, 1H and 0M cahnges. Then B also changes. Thus the voltage V is
generated inside the coil,
])[(11
21
2110
1dt
dBA
dt
dHAAln
cdt
dln
cV
The work done by the power supply during the time dt,
dBH
VdHH
VV
ldBAH
ldHAAH
dt
dBA
dt
dHAAdt
lH
dt
dBA
dt
dHAAln
cdtI
VdtIdWK
44)(
]4
)(4
])[(4
]))[(1
(
)(
121
121
21
1211
21
211
21
2111
1
where
111
4In
cH
, lAV 11 , lAV 22
Noting that
01 4 MHB
we have
012
2
11
11
211
1
121
121
8
)(44
44)(
dMHVHV
d
dHdBH
VdHH
V
dBH
VdHH
VVdWK
This is the work done on the system consisting of the sample plus field. We use M which is the
total magnetic moment of the sample, the magnetic field H, and volume V as
MMV 02 , HH 1 , VV 1
So we have
HdMVH
ddWK
8
2
(1)
The internal energy is given by
8
2VHdHdMTdSdWTdSdU KK (2)
When we use B instead of H in Eq.(2), we have the final form,
8
2VBdBdMTdSdUK (2’)
2. Gedanken experiment on the magnetic work by Reif
We consider the force exerted by a magnetic field H on a magnetic moment represented by a
small rectangular current loop. There is a net x component of force shown in Fig. below, where I
is the current and M is the magnetic moment of the loop. The force on a large sample can be
regarded as due to the superposition of forces on many such infinitesimal moments.
Fig. ])(
)([1
2 dxx
xHxHIdy
cF
, )(1
1 xIdyHc
F
The magnetic moment is defined by
Idxdyc
IAc
M11
Net x-component for the force
x
HM
dxdyx
H
c
I
dyHc
Idx
x
HHdy
c
IFx
)(
Then the work done by the system is
MdHdxx
HMdxFW xout
The work done on the system (Reif) is
MdHdWdW outR (3)
dx
dy
I
F2F1H x
This is the work done on the system in some given magnetic field. It is the work done on the
system consisting of the sample alone. The internal energy is given by
MdHTdSdUR (4)
When we use B instead of H in Eq.(4), we have the final form
MdBTdSdUR (4’)
or
BdMMBdTdSdUR )(
3. Relation between KU and RU
We now consider the relation between two expressions KU and RU . We introduce *U as
MBUU R * .
Thus we have
BdMTdSdU *
The internal energy KdU can be rewritten in the exact way, as
)8
(
8
2*
2
VBUd
VBdBdMTdSdUK
which leads to
8
2* VB
UUK , MBUUR *
For now we omit the term 8
2VB. Then we have
BdMTdSdUdUK * ,
MdBTdSMBUddUR )( *
4. Thermodynamic relations by Reif
MdBTdSPdVTdSdUR
leading to the corresponding relation
MP , BV
Fig. Born diagram using the definition by Reif; MP , BV
where
P, B (intensive variables)
M, V (extensive variables)
S
B
T
M
U
FG
H
Helmholtz free energy:
STUF RR , MBFG RR
MdBSdT
TdSSdTMdBTdS
STddUdF RR
)(
B
R
T
FS
, T
R
B
FM
Gibbs free energy:
MBFG RR
BdMSdT
BdMMdBMdBSdT
MBddFdG RR
)(
RG is a function of T and M.
((Thermodynamic analysis))
BT T
M
B
S
(Maxwell’s relation)
and
B
T
S
T
S
B
S
BT
BS
TB
TS
BT
SB
BT
ST
SB
ST
B
T
),(
),(
),(
),(
),(
),(
),(
),(
),(
),(
or
BBTB
B
T
S T
M
C
T
B
S
C
T
T
ST
B
ST
B
T
where BC is the heat capacity at fixed B.
((Example))
For the paramagnet, M is given by
BTM )(
with
T
CT )(
where C is the Curie constant. Using the expression of M, we have
2
)(
T
CB
T
TB
B
S
T
<0 (Isothermal process)
This means that in the isothermal process, the entropy decreases with increasing magnetic field.
We also mote that
0
SB
T (Isentropic cooling)
which means that the temperature decreases with decreasing B during the adiabatic process.
5. Thermodynamic properties by Kittel (Stanley)
BdMTdSPdVTdSdUK
leading to the corresponding relation
BP , MV
Fig. Born diagram using the definition by Kittel; BP , MV
S
M
T
B
U
FG
H
where
P, B (intensive)
M, V (extensive)
Helmholtz free energy:
STUF KK ,
BdMSdT
TdSSdTBdMTdS
STddUdF KK
)(
M
K
T
FS
, T
K
M
FH
KF is a function of T and M.
Gibbs free energy:
MBFG KK
MdBSdT
BdMMdBBdMSdT
MBddFdG KK
)(
KG is a function of T and B.
B
K
T
GS
, T
K
B
GM
((Maxwell relation))
BT T
M
B
S
, MT T
B
M
S
MS S
B
M
T
, HS S
M
B
T
(a) Isothermal process
Maxwell’s relation:
BT T
M
B
S
(2)
For the paramagnet, M is given by
BTM )(
with
T
CT )(
where C is the Curie constant. Using the expression of M, we have
2
)(
T
CB
T
TB
B
S
T
<0
This means that in the isothermal process, the entropy decreases with increasing magnetic field.
(b) Isentropic process
TB
T
B
T
B
S
B
S
C
T
B
S
T
ST
T
B
S
T
S
BT
ST
BT
BS
SB
ST
B
T
1
),(
),(
),(
),(
1
,(
),(
where CB is the heat capacity at constant B and is positive and 0
TB
S. Thus we have
SB
T
which means that the temperature decreases with decreasing B during the adiabatic process.
So we note that the Maxwell’s relation (1) for the Reif type is the same as that (2) for the Kittel
type.
6 Magnetic work derived from quantum statistics (MacDonald)
The magnetic work is equal to zero in classical physics since the scalar product of the
Lorentz force and the displacement vector is always equal to zero. On the other hand, in quantum
mechanics, the magnetic work is equal to a Zeeman energy given by a form BM . In this
sense, the magnetic work should be treated in terms of the quantum mechanics. To this end we
consider a simple case such as a paramagnet where the electron with the spin (S = 1/2) is under
the presence of an external magnetic field B.
There are two states (up state and down state) of the magnetic moment B ). Both states are
assumed to be nondegenerate. We apply an external magnetic field B in the up direction. So that
a positive magnetic moment has an negative energy BB , while a nagative magnetic moment
has a positive magnetic energy BB . The energy difference is BB2 .
The probability for finding the system with the positive magnetic moment is
)exp()exp(
)exp(
BB
BP
BB
B
The probability for finding the system with the negative magnetic moment is
)exp()exp(
)exp(
BB
BP
BB
B
The internal energy U is given by
)tanh(
)exp()exp(
)]exp()[exp(
)(
BB
BB
BBB
PBBPU
BB
BB
BBB
BB
The magnetization M:
)tanh(
)exp()exp(
)]exp()[exp(
)(
B
BB
BB
PPM
BB
BB
BBB
BB
The entropy S:
)tanh()]cosh(2ln[
)lnln(
BBkBk
PPPPkS
BBBBB
B
Then we have
)]}cosh(2ln[(
)]}cosh(2ln[()tanh({)tanh(
BTk
BBBTkBB
STUF
BB
BBBBBB
In another approach, using the partition function Z for the canonical ensemble defined by
)cosh(2
)exp()exp(
B
BBZ
B
BB
we can also calculate the Helmholtz energy F, the entropy S, and the magnetization M as follows.
The Helmholtz free energy:
)]cosh(2ln[ln BTkZTkF BBB
The entropy S:
)}tanh()]cosh(2{ln[ BBBkT
FS BBBB
B
The magnetization M:
)tanh( B
B
FM
BB
T
The expressions of S and M thus obtained are the same as those derived above. Since F is a
function of B and T, dF can be expressed by
dTT
FdB
B
FdF
BT
The derivatives TB
F
and BT
F
can be evaluated as
MBB
FBB
T
)tanh(
S
BBkBkT
FBBBBB
B
)tanh()]cosh(2ln(
Thus we have
SdTMdBdF
The internal energy U is related to the Helmholtz free energy F as
STUF
We now consider the Gibbs free energy defined by
MBFG
leading to the relation
BdMSdT
BdMMdBMdBSdT
MBddF
MBFddG
)(
)(
So the Gibbs free energy G depends on T and M,
dMM
GdT
T
GdG
TM
We now derive the expression of magnetic field B in terms of M and T. To this end, we start with
1)2exp(
1)2exp()tanh(
B
BB
M
B
BB
B
The magnetic field B can be uniquely obtained as
)]1ln()1[ln(2
1
1
1
ln2
1
ln2
1
BBB
B
B
B
B
B
B
MM
M
M
M
MB
Since
M
MB
B
BB
)exp(
F can be also evaluated as
)]1)(1ln[(2
12ln
1
]
)1)(1(
2ln[
1
]))((
2ln[
1
]ln[1
)]exp()ln[exp(1
BB
BB
BB
B
B
B
B
B
BB
MM
MM
MM
M
M
M
M
BBF
The Gibbs free energy is obtained as
)]1ln()1()1ln()1[(2
12ln
1
1
1
ln2
1)]1)(1ln[(
2
12ln
1
BBBB
B
B
BBB
MMMM
M
M
MMM
MBFG
or
)]1ln()1()1ln()1[(2
12ln
1
BBBB
MMMMG
Thus we have
B
MM
M
G
BBBT
)]1ln()1[ln(2
1
and
S
MMMMkk
Gk
T
G
BBBB
BB
M
B
M
)]1ln()1()1ln()1[(2
12ln
2
Then dG can be expressed by
BdMSdT
dMM
GdT
T
GdG
TM
Here we note that
B
B
B
BB
B
B
B
B
B
B
B
B
B
BB
B
B
B
B
BBB
BBB
B
M
M
M
MM
M
M
M
M
M
M
M
M
MM
M
M
M
M
MBBB
BBBk
S
1
1
ln]
11
2ln[
1
1
ln]
1
1
1
1
ln[
ln]ln[
})]exp()ln[exp(
)tanh()]cosh(2ln[
or
)]1ln()1()1ln()1[(2
12ln
BBBB
BB
MMMMkkS
Using the relations
BdMSdTdG
SdTMdBdF
MBSTUMBFG
we get
MBSTGU
and
MdBTdS
BdMMdBTdSSdTBdMSdT
MBdSTddGdU
)()(
In summary, it is reasonable to conclude that from a view point of quantum statistical physics,
MdBTdSdU and BdMSdTdG
The expression for dU is the same as that derived by Reif and the expression for dG is the same
as that derived by Kittel. We agree with the comment by Robertson that the magnetic work done
on the system by the mutual energy must be MdB
7. Problem and solution (1)
K. Huang
Introduction to Statistical mechanics, second edition (CRC Press, 2010)
((Problem 3-8))
Thermodynamic variables for magnetic system are B, M, T, where B is the magnetic field, M is
the magnetization, and T the absolute temperature. The magnetic work is
BdMdW ,
and the first law states BdMTdSdU . The equation of state is given by Curie’s law
T
BCM ,
where C (>0) is the Curie constant. This is valid only at small B and high T. The heat capacity at
constant B is denoted by BC , and that at constant M is MC . Many thermodynamics relations can
be obtained from those of a PVT system by the correspondence PB , MV
(a) Show
M
MT
UC
, BB
BT
MB
T
UC
(b) From the analog of the energy equation and Curie’s law, show that
0
TM
U
This is the analog of the statement that the internal energy of the ideal gas is independent of the
volume.
(c) Show
C
M
T
CBCC MB
2
2
2
____________________________________________________________________________
Here we use the correspondence relations such that
BP (intensive quantity)
MV (extensive quantity)
PdVTdSdU HdMTdSdU
____________________________________________________________________________
((Solution))
(a)
HdMTdSdU
The heat capacity:
M
MT
STC
,
HHH
HT
MH
T
U
T
STC
(b)
MTT T
HTH
M
STH
M
U
where we use the Maxwell’s relation:
MT T
H
M
S
M is given by the Curie law as
S
M
T
B
U
FG
H
BT
CM
Noting that
T
B
C
M
T
B
M
we have
0
MT T
BTB
M
U
(c)
T
M
M
B
M
BT
MS
T
BT
MT
BT
MS
TMT
MST
T
STC
),(
),(
),(
),(
),(
),(
),(
),(
or
T
BT
B
BTTB
T
TB
TB
T
M
B
M
T
M
B
ST
T
ST
T
M
B
S
B
M
T
S
B
M
T
B
M
T
M
B
S
T
S
B
M
TC
][
Using the Maxwell’s relation
BT T
M
B
S
Then we get
T
BBM
B
M
T
MT
CC
2
For the paramagnetic system,
BT
CM
2
2
T
CBCC BM
or
C
M
T
CBCC MB
2
2
2
8. Problem and solution (2)
K. Huang
Introduction to Statistical mechanics, second edition (CRC Press, 2010)
((Problem 3-9))
Define the free energy ),( TMF and Gibbs energy ),( TBG of a magnetic system by analogy
with the PVT system.
(a) Show
BdMSdTdF , MdBSdTdG
(b) Show
BT T
M
B
S
hence
2T
CB
B
S
T
.
(c) With the help of the chain rule, show
TC
CB
B
T
BS
((Solution))
Helmholtz free energy F
Gibbs free energy G
HdMTdSPdVTdSdU
STUF , HMFPVFG
where BP , MV (Kittel)
(a)
BdMSdT
TdSSdTBdMTdS
TdSSdTdU
STUddF
)(
S
M
T
B
U
FG
H
MdBSdT
MdBBdMBdMSdT
BMFddG
)(
(b)
BT
GS
, TB
GM
B
TB
BTT
T
M
B
G
T
BT
G
TB
G
T
G
BB
S
2
2
Since
BT
CM ,
2T
CB
T
M
B
2T
CB
B
S
T
(b)
Noting that
1),(
),(
),(
),(
),(
),(
TS
TB
BT
BS
SB
ST
we get the relation
1
TBS S
B
T
S
B
T
Since
CB
T
B
SS
B
T
T
21
B
BB
CT
Q
T
ST
we have
TC
CB
T
CT
CB
T
S
B
S
B
S
S
T
B
T
BB
B
T
TBS
2
___________________________________________________________________________
REFERENCES
C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, 1980).
F. Reif, Fundamentals of statistical and thermal physics (McGraw-Hill, 1965).
H.E. Stanley, Introduction to Phase Transition and Critical Phenomena (Oxford, 1971).
H.S. Robertson, Statistical Thermodynamics (PTR Prentice Hall, 1993).
D.K.C. MacDonald, Introductory Statistical Mechanics for Physicists (Dover, 2006).
H.B. Callen, Thermodynamics (John Wiley, 1960).
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