Christensen’s Physics of Diagnostic Radiology

NUCLEAR MAGNETIC RESONANCE 439

are related in a manner analogous to inches can note that if the nucleus has no spin (I
and feet):
= 0; i.e., it has no angular momentum), it
1. The Bohr magneton is used to ex­
press the MDM of electrons. will have no MDM. Nuclei with no MDM
will not be detectable by NMR. Therefore,
2. The nuclear magneton is used to ex- all nuclei whose mass number A (protons
plus neutrons) and atomic number Z (pro­
press the MDM of nuclei. tons) are both even cannot be used in NMR

The MDM of nuclei are not calculated, but studies. Table 23-l presents a number of
instead are measured for each individual
nucleus. These values are then expressed elements that are of interest in medical
as a certain number of nuclear magnetons. NMR.
Similarly, the MDM of the proton or neu­
tron is measured in the laboratory and Alignment of Nuclear MDM in a
found to be different from the nuclear Magnetic Field
magneton. Even more strange, the MDM
of the nuclei cannot be calculated by add­ We have tried to show you and to explain
ing up the MDM of all the protons and why some nuclei have spin angular mo­
neutrons in the nucleus. What all this mentum and a magnetic dipole moment.
teaches us is that nuclei, protons, and neu­ The term MDM can be translated, if you
trons are not the simple structures we once are so inclined, to "tiny magnet." MDM
thought them to be. To be complete, we might be considered as that property of a
will list the values of the nuclear magneton nucleus that causes it to behave like a tiny
magnet (if the tiny magnet is spinning
(f.LN), the MDM for the proton (f.Lp), and the around its north pole-south pole axis, it will
MDM for the neutron (f.Ln): possess spin angular momentum). When
placed in a magnetic field, the tiny magnets
f.LN = 5.05 x 1 D-27 J/tesla will try to align along the field. Unfortu­
flop = 2.7928f.LN nately, things in the tiny world don't always
f.Ln = -1.9128f.LN behave as we anticipate. In the present ex­
ample, the tiny magnet is allowed only a
where 1-LN is the nuclear magneton, f.Lp is the limited amount of alignment in the field.
Here we have a quantum rule very similar
proton MDM, and 1-Ln is the neutron MDM. to the electronic condition in atomic struc­
For spinning particles, the MDM is al­ ture that electrons can only be in certain
energy shells.
ways along the spin axis. (We described the
spin axis as the pencil passing through the We must digress a moment and mention
quantum physics. Quantum physics dic­
center of a ball, Figure 23- 1.) A positive tates certain rules that govern the behavior
of objects the size of electrons and nuclei.
value indicates that the MDM and the an­ Large objects, such as our examples of
gular momentum (remember the ball, pen­ balls, bar magnets, and wire loops, are not
cil, and right-hand rule) are in the same limited by these quantum rules, which is
direction. A negative value indicates that why the analogy between magnets and
the MDM points in the opposite direction spinning nuclei is not always precise. For
from the angular momentum. example, magnets are able to align them­
selves exactly with a magnetic field, but
Magnetic Dipole Moments of Nuclei. spinning nuclei are limited in their align­
The MDM of nuclei depend on the number
and arrangement of the protons and neu­ ment (Fig. 23-5). The possible alignments
trons. We have already seen, though, that
the protons and neutrons produce spin­ (orientations) are shown for nuclei with
up-spin-down pairs in the nucleus. The nu­
clear MDM is not just a simple addition of positive MDM and I = l,.,f. (Fig. 23-5A) and
the MDM of the nucleons. In fact, the I = 1 (Fig. 23-5B). (We really don't think
MDM of nuclei have been measured. We

440 NUCLEAR MAGNETIC RESONANCE

Table 23-1. Table of Nuclear Properties

ISOTOPE MDM SPIN, I GYROMAGNETIC LARMOR FREQUENCY
x NUCLEAR RATIO
MAGNETON FOR H = 1 T
(x 107 HzfT) MHz/T

n -1.91 '/, -18.3 29.16
'/, 42.58
'H(p) 2.79 26.8
6.53
2H 0.85 1 4.1 10.70

13c 0.70 '/, 6.7 3.08
,.N 40.05
'•F 0.40 1 1.9 11.26
11.09
2.63 '/, 25.3 17.24

23Na 2.21 % 7.1

27AI 3.64 % 7.0
31p
1.13 '/, 10.8

Example: MDM of 'H = 2.79 x 5.05 x 1D-27 joules/tesla
Gyromagnetic Ratio of 1H = 26.8 x 107 Hz/T

that medicalNMR will be applicable to nu­ would move into the two orientations, with
clei with higher values of I, with the pos­ some in one and some in the other (Fig.
sible exception ofNa.) Refer to Table 23-1 23-5A). As might be expected, the two ori­
for some nuclei that might have medical
applications. entations have two different energy values.
The up orientation has the lower energy,
The nucleus with I = �can spin in either and the down orientation a slightly higher
of the two orientations shown in Figure one. The number of nuclei that move into
23-5A, but no others. The nucleus cannot the lower energy state (up orientation) is
come into exact alignment with the mag­ only a little greater than those that move
netic field. The bar magnet and wire loop into the higher energy state (down orien­
would sooner or later come into exact
alignment, because they are too large to be tation). The ratio of the numbers is deter­
dominated by the quantum rules. Figure
23-5B shows that there are three orienta­ mined by the difference in energy between
tions for the I = 1 nuclei. the two states, the magnetic field, and the
temperature. For instance, in a population
Suppose we have a large number of nu­ of 2,000,000 nuclei, 1,000,000 + 1 might
clei of spin I = �. With no applied mag­ go to the lower energy state, while
netic field H, the nuclei (and MDM) would 1,000,000 - 1 might go to the higher state.
be in random orientation in all directions. This isn't very much difference. For a sam­
By applying a magnetic field, all the MDM ple with about 1023 nuclei, the difference
in the population of the two states would
AB be about 1017 nuclei. The NMR signal is
Figure 23-5 Nuclear alignment in a magnetic produced only by the 1017 excess nuclei.
field Therefore, of all the nuclei, very few par­
ticipate in the NMR signal.

When the little magnets reach the ori­
entation allowed by the quantum rule, the
magnetic field is still trying to move them

into exact alignment. This means that there
is still some torque on each of the little mag­
nets. In the next section, we will see that
this remaining torque is responsible for the
MDMs precessing about the magnetic field.

Why do we include a discussion of mag­
netic dipole moment? First, it gets us think­
ing of certain nuclei as tiny (nuclear-sized)

NUCLEAR MAGNETIC RESONANCE 441

magnets. Second, we can see how these tiny the side of the top touches the floor. (That's
magnets align themselves in an external when it goes rolling across the floor.) The
magnetic field. Now that we understand slowing down is a result of frictional forces
this, we have the M part of NMR. at the point of rotation (and even frictional
forces between the top and air molecules).
Angular Momentum and Precession Two assumptions have been made that are

We have just described a nucleus with rather subtle: 1) there is sufficient frictional
spin angular momentum and a MDM in a
magnetic field. The nucleus still has some force to hold the tip of the top in place;
torque on it. In this section we need to de­
scribe the effect of this torque on the mo­ and 2) the top is symmetric about the spin
tion of the nucleus. We feel, and most other
experts agree, that the best way to visualize axis. If the first were not true, we would
the motion of the nucleus is to describe the see something other than a top spinning
motion of a spinning top in a gravitational on a fixed tip. More than likely, the tip
field. Once again we are trying to use a would be moving in a circle on the floor. If
large object to illustrate a quantum-sized the second assumption were not true, there
object. Fortunately, in this example, we will would at least be wobble in the precessional
not find a discrepancy in the analogy be­ motion and, very likely, no spinning about
cause of the size of the two objects. One of a single axis. We would really hate to try
the most interesting motions in the world to spin a top with chewing gum on its side.
is the motion of a tilted spinning top. (We
should really be talking about spinning We are tempted to explain why a spin­
jacks because one of us QED) could never ning top precesses but a nonspinning top
wrap a string around a top, throw it down, falls off its tip. As a matter of fact we tried,
and get it to spin. His always just bounced but got bogged down in such matters as
off the floor. But he could really spin those perpendicular vectors called torque,
jacks with his thumb and finger.) A top that weight, angular momentum, and the
is spinning at some tilted angle (we nor­ change in angular momentum. Therefore,
mally define tilted as being at some angle we are going to assume that you believe
to up and down that we automatically de­ that a spinning top will precess when in a
fine as the direction of gravity) will con­ gravitational field, and that a nonspinning
tinue to spin with its axis at the tilted angle, top will fall over onto its side.
but the axis of rotation will move in a cir­
cular path about the direction of gravity. Larmor Frequency
The motion of the axis about the direction
When we wrote this section we used H
of gravity is called precession. instead of B to indicate the magnetic field.
Please feel free to substitute B for H if you
The precessional path of the center of wish. Most other MRI texts use B, while
many physics texts use H.
the top is shown in Figure 23-6A as the
All we need for precessional motion is a
circle at the apex of the top. Of course, this
point (apex) is just exactly where the spin symmetric body with spin angular mo­
mentum and some torque that is perpen­
axis (L) emerges from the spinning top. (In dicular to the angular momentum. We
physics texts the letter L is used to repre­
have just that situation with a spinning nu. ­
sent angular momentum, so we use it here.)
Remember, the spin angular momentum is cleus in a magnetic field (Fig. 23-6B). Fur­
also along the spin axis and, in this case,
thermore, there is no frictional force, so
will point up from the top (along L). We the nucleus will not slow down and fall
sideways . The frequency of precession,
know that a toy top will slow down and that
the precessional circle will get larger until called the Larmor frequency, is of funda­

mental importance in NMR. You will recall
from Chapter l that the Greek letter v (nu)
was used as the symbol for frequency. Sim-

442 NUCLEAR MAGNETIC RESONANCE
A.
B. Ho

--'"'��if ....

L L

Figure 23-6 Precession

ilarly, vL is the symbol for the Larmor fre­ MDM to the nuclear spin angular mo­
quency (sometimes the letter f is used for mentum Ih. Therefore,
frequency). The Larmor frequency de­
pends on the magnetic field and the gy­ 'Y f.l
romagnetic ratio, 'Y (gamma). This rela­
tionship is expressed by =ih

VL = -yH/2'7T 'Y = gyromagnetic ratio
vL = Larmer frequency
'Y = gyromagnetic ratio f.l = MDM
H = magnetic field
I = nuclear spin
(We sometimes see the Larmor frequency
n = h/2'7T, h is Planck's constant
written as w = -yB. Here omega, w, is really
the angular velocity, and w = 21Tv.) We h = 6.6 x 1� joules· sec

assume that this nucleus is in the presence The values of 'Yare included in Table 23-1.
of other nuclei so that we can use the mag­
The gyromagnetic ratio is a unique value
netic induction, B, rather than the mag­ for each type of nucleus. For example, the
'Y values of the hydrogen isotopes (IH, 2H,
netic field, H. If the nucleus were in the 3H) are all different from each other, and
region by itself, we would properly use the also different from the 'Y values of helium
isotopes.
magnetic field (H), since there would be no
Now we have gotten where we promised
induction. And, as we say in the addendum we were going. The mysterious spin an­
gular momentum and magnetic dipole mo­
at the end of this chapter, in the body B ment have formed a ratio called the gy­
romagnetic ratio. The gyromagnetic ratio
and H are so nearly the same that the terms and magnetic induction (from our imaging
can be used as synonyms (even though such magnet) produce the Larmor frequency of
use is not correct in terms of the physics of precession. Hydrogen nuclei (protons) pre­
the situation). cessing at their Larmor frequency form the
basis of clinical magnetic resonance imag­
The gyromagnetic ratio (also called the ing (proton imaging).
magnetogyric ratio) is the ratio of the
The defining equation for the Larmor

NUCLEAR MAGNETIC RESONANCE 443

frequency shows that each type of nucleus across the object to be imaged. We will re­
will precess at a unique frequency in a turn to this imaging method later.
given magnetic field. Two different nuclei
will precess at two different frequencies in Figure 23-6B shows a nucleus (I = I;:;)
a given field. Therefore, the Larmor pre­
cession is a process that, for a given field, spinning in the spin-up orientation. Note
can distinguish between nuclear types. that in the spin-down orientation (not
We also need to note that a given type of shown), the nucleus, as far as the magnetic
nucleus will precess at different frequen­ field is concerned, has reversed its direc­
cies when in different magnetic fields. tion of spin. (Note the direction of the ar­

For example, a hydrogen nucleus (pro­ rows in Figure 23-5A and B.) The torque
ton) has a gyromagnetic ratio of 26.8 X 107
1/tesla-sec (Hz/tesla). In a magnetic field of on both spin-up and spin-down nuclei,
strength 0.2 tesla, the proton will precess however, is the same. Therefore, both spin­
at a frequency of about 8.5 MHz (8,500,000 up and spin-down nuclei precess in the
cycles/sec). Calculate this yourself using the same direction.
equation vL = -yH/21T, being careful to keep
track of units and powers of 10. Similar Energy States for Nuclear Spin Systems
calculations show that the Larmor fre­
quency for the proton in a 1-tesla field is We feel compelled to return to a discus­
about 42.58 MHz, as listed in Table 23-1.
We can find the resonance (Larmor) fre­ sion of the energy states associated with the
quency of any of the nuclei listed in Table
23-1 in any magnetic field strength simply spinning nucleus in a magnetic field. We
by multiplying the field strength (in tesla)
by the listed value of the Larmor frequency have already mentioned that there are two
in a 1-tesla field (last column of Table
23-1). Remember the conversion from energy states for an I = I;:; nucleus that
gauss to tesla: 1 tesla = 104 gauss. Tesla
and gauss are units of magnetic induction, finds itself in a magnetic field. Remember
but we are are going to use the term "mag­
netic field." that the spin-up orientation has the lower

Of course, the magnetic field need not energy.
vary much over a group of identical nuclei
for those nuclei to precess at detectably dif­ When this chapter was originally written,
ferent frequencies. This concept of iden­
tical nuclei in slightly different magnetic we included a calculation for the energy of
fields is the concept on which a great deal
of NMR work is based. The chemical a bar magnet when placed in a magnetic
shifts, measured so extensively in chemis­
try laboratories, are just instances of iden­ field. That calculation is not really difficult,
tical nuclei finding themselves in slightly
different fields because of local environ­ but we feel it would be sufficient for our
mental perturbations on the applied mag­
netic field. Of more interest in NMR imag­ purpose to give the results. When a bar
ing is the purposely varied field to help
establish different Larmor frequencies for magnet is aligned with a field, as in Figure
a given nuclear type (usually hydrogen)
23-4A, its energy (potential energy) is just

E - J.LH (J.L = MDM), while in the re­
=

versed orientation its energy is E = J.LH.

The difference in energy between these

two orientations is 2J.LH. This is also true

for the loop.

The energy of a nucleus may also be writ­

ten in exactly the same way. Figure 23-7A

shows the two energy states for the I = 1�

nucleus. Remember that spin angular mo­

mentum has direction as well as magni­

tude. We use the direction of spin to in­

dicate whether a nucleus is in the spin-up

or spin-down state. In this example, I = I;:;

indicates spin-up and I = -I;:; indicates

spin-down. Note that the spin-up (I = I;:;)

state is lower in energy than the spin-down

(I = - 14) state. The energy (E) of these
states (E1 = J.LH and E2 = - J.LH) may be

444 NUCLEAR MAGNETIC RESONANCE

�A. a nucleus flips from the lower to the higher
I=-h(Spin Down) n E----,- -_=___-- 1 =J-LH energy state, it must absorb this amount of
energy from somewhere. When it returns

6. E y H to the lower energy state, it must give up
this amount of energy. (We will see later

I=+'lf2(Spin Up)----'------ E2=-J-LH that the emission of a photon is not the
principal means of energy transfer when

the nucleus returns from the higher to the

B. lower energy state.) For x rays the emission
was electromagnetic radiation that we
1I= -1-------�------ E1 =1-LH called a photon. For nuclear transitions the
6.E=yllH energy is transformed to the lattice (the
material structure surrounding the nu­
1I= O-------+------ E2= 0 cleus). This type of transition is called a
6.E =y1\H radiationless transition, and usually heats
I= -t-1 --------.I.-._ --- E3=-I-L H the surrounding material (lattice).

Figure 23-7 Spin energy states In Chapter 1, we showed that the energy

of a photon was given by E = hv (Planck's

expressed in terms of the gyromagnetic ra­ constant multiplied by frequency). Because

tio ("'(). Because the equation defining the we have the energy difference between the

gyromagnetic ratio contains the terms I two nuclear spin states (E = "YiiH), we can

and 1-'- (MDM), such an expression can be find the frequency of the photon absorbed

obtained with a little simple algebra (i.e., to produce a spin state transition. That is,

E1 = j.LH = "YIIiH). With this substitution, the energy (hv) of the photon producing

the difference in the spin-up and spin­ the spin state transition must be exactly

down energy states, as illustrated in Figure equal to the difference in energy ("YiiH) be­

23-7A, may be expressed as .:lE = "'(hH. tween the two spin states. Let us write this

E1 - E2 = "YI11iH - "YI21iH. For E�> I1 = statement in the form of an equation, and

+ �; for E2, I2 = - �. Substituting these then solve the equation for the frequency

values for I, we obtain E1 - E2 = �"YiiH (v) of the photon producing the transition:

- (- �)"YiiH = "YiiH. E = hv = -yhH
Now let us use this expression to describe v = -yH/2'!1"

how Larmor frequency relates to the fre­ E = energy of photon absorbed
quency of the transition radiation between v = frequency of the photon
these two spin states. Transition radiation -y = gyromagnetic ratio
refers to the energy absorbed by the nu­ H = magnetic field
cleus when it flips from the lower energy h = Planck's constant (h = h/2'Tr)

state to the higher energy state. This is This equation for the frequency of the

analogous to the photoelectric effect when photon producing the transition should

electrons absorb energy and go from a look familiar. It is exactly the same as the

lower energy level to a higher energy level Larmor frequency of precession of nuclei
(e.g., from the K shell to the L shell).
about a magnetic field. The Larmor fre­
There is a reason for all this. If the nu­ quency of precession is exactly equal to
cleus flips from the lower energy state to the frequency of radiation absorbed in a
the higher, it must absorb an amount of transition from one spin state to another.

energy equal to the energy difference. We In the literature, statements about the ra­

have certainly seen this concept before. Re­ diation of transition are referred to as ra­

member that the production of character­ diation at the Larmor frequency. Now we

istic x rays looks like this. Consequently, if know why that is acceptable.

NUCLEAR MAGNETIC RESONANCE 445

As an example, the Larmor frequency of NMR PARAMETERS
a hydrogen nucleus (proton) in a 1-tesla
(104-gauss) magnetic field is about 42 MHz Magnetization Vector
(42,000,000 Hz). Let us compare this to
more familiar electromagnetic radiation: Recall that, in a normal-sized sample,
the frequency of 300-nm (3000 A) visible there is an extremely large number of nu­
light is about 109 MHz, and a 124-keV x-ray clei and, in a magnetic field, only a little
photon has about 1013 MHz. Inspection of more than half are in the spin-up state pre­
the electromagnetic radiation frequency cessing at the Larmor frequency. A little
spectrum reveals that 42 MHz falls within less than half are in the spin-down state,
the radio frequency range. All NMR stud­ also precessing at the Larmor frequency.
ies are performed in the radio frequency Remember that each nucleus has a MDM
range (about 1 KHz to 100 MHz). and an angular momentum that are par­
allel, and that each nucleus looks like a tiny
For interest, the energy states of an I = magnet. If we add up the MDM of all these
1 nucleus are shown in Figure 23-7B. But tiny magnets, the resulting sum is the mag­
be careful; there are three, one of which is netization, M, of the sample. Because the
zero. Only the two transitions shown are nuclei have a spin angular momentum par­
possible. The transition from the upper allel to the MDM, there will also be a net
state (I = - 1) to the lower state (I = 1) is angular momentum parallel to M.
impossible; it is one of those "forbidden"
transitions from quantum physics. Note In Figure 23-8A, we attempt to show
that the allowed transitions give an energy several of the large number of nuclei (these
are identical nuclei), some spin-up, some
difference just like the I = � nucleus. spin-down. Each MDM can be considered

Thus, we are back to Larmor frequency for A. H B.
the absorbed radiation.
C.
The basis of NMR is to induce transi­
tions between these energy states by the Figure 23-8 Effects of nuclear spins in an ex­
absorption and transfer of energy. What­ ternal field, H
ever description is used to picture NMR,
here or elsewhere, the whole concept ba­
sically deals with the transitions between
these spin states. NMR is nothing more
than the flooding of a sample with radia­
tion at the Larmor frequency, and then
measuring the Larmor frequency signal
coming from the sample.

Earlier we said that for a sample with
some 1023 nuclei, only about 1017 nuclei
would be involved in the NMR process
(these are the excess nuclei). Only 1017 nu­
clei! We cannot even image 107, let alone
1017, nuclei. It is impossible to visualize
what these nuclei are doing. Normally,
then, we replace all these individual nuclei
having individual MDMs by a single vector
termed the "magnetization," and give it
the symbol M. We will discuss magnetiza­
tion in the next section.

446 NUCLEAR MAGNETIC RESONANCE

to have one part along the field and one torque will be acting on M. This means that
part perpendicular to the field. (Because M cannot precess about H (a difficult con­
the MDM is a vector, it has components cept, because the individual nuclei that
parallel and perpendicular to the field.) make up M continue to precess). Without
precession we are unable to detect any spin
This is illustrated in Figure 23-SB, in angular momentum of the magnetization
vector, even though such spin angular mo­
which f.L1 is the component of the MDM mentum is present. If M were to be dis­
perpendicular to the magnetic field, H, and placed from H, M would precess about H
f.L2 is the component of the MDM parallel because there would now be a torque on
to the magnetic field. There are similar
components of the spin angular momen­ M. In the next sections we will see 1) how
tum. All the MDM components along the to displace M from H and 2) how to observe
field for the spin-up nuclei add together
and give a net result that is pointing up. and measure the resulting precessional
The sum of components along the field for motion of M.
the spin-down nuclei also add together, but
give a net result that is pointing down. RF Magnetic Field (Radio Frequency)
When these two results are added together,
there is a net result that points in the di­ We would like to displace M from its di­
rection of the field because there are more rection along H and watch M as it tries to
nuclei in the spin-up than in the spin-down go back to its alignment along H. In the
state. Effectively, then, the excess nuclei next few paragraphs we will discuss the
in the lower energy state give a net MDM method by which M can be displaced. It is
component along the field. not as easy as it sounds to make M move
away from H because, as soon as M is dis­
Figure 23-SC attempts to show the top placed, it starts precessing about whatever
view of Figure 23-SA. What we see are the magnetic field is present. If, by some
means, M were displaced a bit from the H
components (the f.L1 components of Fig. direction, M would precess about the H
field with the Larmor frequency (Fig.
23-SB) of the MDMs that are perpendic­
23-9A).
ular to the field. Even with this small sam­
ple of nuclei, the tendency for these com­ One way to displace M would be to apply
ponents to add up to zero (or to average a second magnetic field, H1, but this second
to zero) may be seen. For all nuclei, the field H1 would have to have a particular
perpendicular components of the MDM characteristic. Suppose H1 were a constant
for both spin-up and spin-down add up to (static) field. All we would see is the little
zero. dipole magnets realigning themselves and
beginning to precess about the net mag­
We can replace all the nuclei in the sam­ netic field, which would become the vector
ple by the single vector that is the sum of sum of H and H1• Of course, H1 could be
the components of the MDM for the excess turned on and off quickly, in which case
nuclei. This vector is called the magneti­ we would get a small displacement of M
zation, M, for the sample. The magneti­ and it would then precess about H and
zation behaves like a magnet that has a gradually realign itself along H (Fig.
spin angular momentum. That is, the mag­
netization can precess about a magnetic 23-9A).
field if the magnetization and the magnetic
field are not parallel. Or, we could have H1 rotating about H
at the Larmor frequency. In this case H1
Precession requires spin angular mo­ would follow M in the precessional motion
mentum and a torque perpendicular to the about H. At this point most authors intro­
angular momentum. Because the magnet­ duce a rotating coordinate system to dis­
cuss the motion of M and H1• We can il-
ization, M, as shown in Figure 23-SB, is

exactly parallel to the magnetic field, H, no

NUCLEAR MAGNETIC RESONANCE 447
z

Rotation of Magnetization Vector Record Player
A B

Figure 23-9 Rotating coordinate system compared to a record player

lustrate a rotating coordinate system by standing on the ground (a fixed coordinate
using mathematics, a merry-go-round, or system such as xyz in Figure 23-9B) would
a record player. We feel a record player see the little horses moving around and
would be best to use (because we can't draw around. When he stepped onto the merry­
horses). go-round (a rotating coordinate system
such as x1y1x1 in Fig. 23-9B), the little
Suppose we draw two perpendicular horses wouldn't seem to be moving (of
lines on a record and set the record on the course, the rest of the world would be whiz­
turntable. Now our coordinate system zing by).
would be the two perpendicular lines (x1
and yl) on the record, and the vertical line In NMR studies, H is normally much
through the spindle (z1) of the turntable larger than Hu so the precession of M
(Fig. 23-9B). When the record player is about H is much faster than M about H1•
turned on, x1 and y1 will rotate with the Remember that the Larmor frequency will
turntable at the angular velocity w (instead be greater in a larger magnetic field for
of the familiar rate of 33� rpm, we are identical nuclei: vL = )'H/2'1.1" Figure
talking about a few million revolutions per 23-lOA attempts to show the spiral path
second). Note that z1 is not rotating. Com­ that the tip of M will follow when both H
pare that to the fixed coordinate system (x, and the rotating H1 are present. This draw­
y, z) of Figure 23-9B. Of course, z and z1 ing represents the path seen from a fixed
do not change relative to each other, but coordinate system, which represents the
x1 and yl move relative to x and y. If we laboratory and the observer. The number
were to place z1 along the H field and rotate of spiral loops is determined by the sizes
the player at the Larmor frequency, we of H and H1•
would have a system that would rotate with
the same frequency as M, if M were slightly The reason for choosing this particular
displaced from H so that it could precess. rotating coordinate system (rotating at the
Therefore, M would not seem to be mov­ Larmor frequency) is that the effective
ing in the x1y1zi system while it precessed magnetic field appears to be zero (H ap­
about H. pears to disappear) prior to applying the
H1 field. When we apply the H1 field, it is
The situation with the merry-go-round the only magnetic field that the magneti­
analogy would be the same. A person zation vector sees. This is why the mag-

448 NUCLEAR MAGNETIC RESONANCE

AB

Figure 23-10 Beehive path of the magnetization vector, M, during a 90° pulse (A) and path of

M during a 90° pulse as seen from a rotating coordinate system (B)

netization vector, M, will leave its original coordinate system) the ball is no longer
moving. Because it was the force that the
direction (along H) and precess about the boy applied to the string that caused the
applied H1 field. In any other rotating co­ ball to move in a circle as you looked at it
ordinate system (with H1 rotating with it), from the ground, your new assumption
part of the original field H would still be must be that the boy is no longer applying
any force to the string. Of course, your
affecting M. At most, what we would see friends on the ground still see the ball
would be a perturbation of the magneti­ whirling around (and you now see them
whirling around while you stand still). By
zation precession about the H field. More whirling around at the same speed as the
ball, therefore, you no longer detect a force
than likely, M would continue to be along that makes the ball whirl. You really detect
no net force, but because the boy is still
H and we would see nothing new because applying tension by the string, some new
of the H1 field. force must have arisen to counteract that
tension. That new force is the fictitious
We had better reconsider the last para­ force, and appears only because you are on
graph. We said that if we rotate around a a rotating coordinate system. We may use
the term "fictitious," but you can certainly
magnetic field, H, at the Larmor frequency feel these forces. Jump on a merry-go­
round and right away you are trying to
of a precessing nucleus, the magnetic field slide away from the center; if you don't
will disappear as far as the precessing nu­ hold on you will indeed fall off.
cleus is concerned. That sounds like magic
We do the same sort of thing by standing
and, indeed, physicists use the term "ficti­ on the earth. The earth is moving rapidly,
tious forces" when discussing rotating co­ but we have the sensation of standing still.
ordinates (and you thought we weren't go­ This allows us to extend the analogy a bit
further. Standing on the moving and ro­
ing to get into science fiction). tating earth, we do not detect the force
Consider the analogy of a boy whirling causing the motion. If another force comes
along (such as a strong wind), however, we
a ball on a string. If you are standing near respond to the new force. Similarly, in our
the boy, you will see the ball going around coordinate system rotating at the Larmor
and around, faster if the boy applies more
force on the string (this force is normally
called tension) and slower with less tension.

The tension in the string is analogous to

the magnetic field, H. Now suppose you

can make yourself very small and that you

jump onto the ball (and have enough fric­
tional, not fictitious, force so that you will
stick to the ball). An amazing thing hap­
pens; from your new vantage point (your

NUCLEAR MAGNETIC RESONANCE 449

frequency, the nuclear magnets no longer mountain. Rather, the jeep must circle
detect the H field, so they are able to re­ down and around and around the moun­
spond to the new H1 field as if it were the
only one present. Note that if the speed of tain in a gradual descent. Similarly, M
rotation is not at the exact Larmor fre­ leaves H and moves along a spiral path
quency of the nucleus in question, some of until it reaches the plane perpendicular to
the H field will still be detected by its in­ H and parallel to H1 (for 90° motion, at
teraction with the magnetization vector, M. least). A little later we will allow M to return
parallel to H; it will move "up the moun­
You might think of this detectable part tain" in a spiral path, but the path will be
of H in terms of chasing the whirling ball
a different spiral from that used on the
in a tiny airplane. If your speed is exactly downhill trip (see Fig. 23-llA).
the same as that of the ball (i.e., if your
When M finally reaches the xy plane, we
airplane is moving in a circle with the ball), have an interesting condition. There is no
the ball will appear stationary relative to
magnetization component along the z axis,
you. If your circular speed is a little slower
which means that the spin-up-spin-down
or faster than the ball, however, the ball
states are equally populated. The entire M
will appear to move ahead or behind you.
vector is a result of the little magnets (nu­
This motion of the ball would be your de­
clear magnetic dipole moments) grouping
tection of a portion of the original tension together as they precess about H. In Figure
23-SC, this would be seen as an excess
(force) applied to the string. Now back to number of nuclei on one side of the circle.

the magnetization vector. We might call this grouping together as be­
Because M cannot detect H in a coor­ ing "in phase." Note that the transverse

dinate system rotating at the Larmor fre­ component of M (that component perpen­
quency, M is free to interact with H1 as if dicular to H) is produced by the in-phase
H1 were the only magnetic field present. If
H1 is also rotating at the Larmor frequency, grouping of the dipoles. The parallel com­
H1 will appear to be a constant (unchang­ ponent of M (that component parallel to
ing) magnetic field. If H1 is applied so that H) is produced by the population ratio of
it is perpendicular to H, then Figure the two spin states. (We will return to a
23-lOB represents the precession motion more complete discussion of in phase later.)
of M about H1• This precession motion is
much easier to see than is the spiral motion We need to determine how to produce
H1 so that it will rotate at the Larmor fre­
of M as seen in the fixed laboratory system. quency about H and be perpendicular to
H. Fortunately, a sinusoidally varying mag­
We must remember, though, that the co­ netic field that is perpendicular to H will
look exactly like a constant field in the ro­
ordinate system x1y1zl and the entire Fig­
ure 23-lOB is rotating about z1 at the Lar­ tating coordinate system, and this is what
mor frequency vL = -yH/271". All that said we want. We can then produce the H1 field
and done, it is still within the fixed coor­ by a conductor with electrons flowing

dinate system that we actually observe the through it if we design the conductor con­
figuration in such a way that H1 is perpen­
motion of M. Indeed, we will see M move dicular to H. (We will present more on this
structure later; see "Radio Frequency
in the "beehive" motion shown in Figure Coils.") Remember that H1 must rotate at
23-lOA. the Larmor frequency, which is in the radio

The spiral path concept of M moving frequency range. Therefore, the electron
from being parallel to H to being perpen­
dicular to H is hard to visualize, so we flow in the RF coil must vary at the Larmor
frequency (thus the term RF [radio fre­
would like to give an analogy before con­
quency] magnetic field).
tinuing. Consider a jeep perched on top of
a steep mountain. Obviously, you cannot

drive the jeep straight down the side of the

450 NUCLEAR MAGNETIC RESONANCE

If this discussion of the precession of M her that M can be considered to be a mag­
about H1 is sufficient, we can define two
more terms before describing NMR pa­ net. When the end of the M magnet sweeps
rameters. If H1 is on long enough to rotate past the windings of the RF coil, a current
M by goo, we have what is called a 90° pulse.
If H1 stays on twice as long, so that the (electron flow) is induced in the coil. For
precession of M carries M all the way to all jeep drivers, you may visualize Figure
the - z axis, we have rotated M by 180°. 23-llA as though you were in a helicopter
The RF pulse necessary to produce the hovering exactly over the top of a moun­
180° rotation is called a 180° pulse. tain while watching the jeep make its spiral
climb back up the mountain. In Figure
Free Induction Decay 23-11B, M at 1 is not inducing a current;
at 2 the induced current has a maximum
We are now ready to discuss NMR pa­ value; it is zero at 3; and at 4 the induced
rameters, which are those quantities that current is again a maximum but in the op­
are measurable with the NMR technique. posite direction from that at 2. It is the
It is important here to state that the NMR opposite direction because M is sweeping
signal cannot be described in terms of pho­ the loops in opposite directions at 2 and 4.
ton emission alone, as we do with x-rays. Remember that M is getting closer to H
Rather, we must move M away from the H between 1 and 4 (the jeep is getting closer
direction and observe M moving back to the top of the mountain, so it will appear
along H. We are going to describe the pa­ to be closer to the center top of the moun­
rameters first and then discuss some pos­ tain), so the current at 4 has a smaller max­
sible reasons for observing what we do. imum than at 2. For each trip of M around
H we have the same shape of induced cur­
Recall that we can rotate M away from rent, with the maximum values getting
H by applying an RF field at the Larmor smaller. This decreasing induction con­
frequency. If M precesses about the varying tinues until M is along the H direction.
RF field by goo, we have applied a goo pulse
(see Fig. 23-lOB). Suppose we apply a goo The total signal induced in the coil is
pulse; the RF field is on until M is entirely shown in Figure 23-l2A. What we see is
in the plane perpendicular to H, and then an alternating voltage (produced by the in­
H1 is turned off. Thus, we see that a goo duced current) that has the Larmor fre­
pulse is simply the length of time that H1
is turned on (the time will be different for quency. It also decays to zero (at least, to
different values of H1). When H1 is turned background noise) exponentially. (X-ray
off, M will continue precessing about H. At beams decay exponentially. At least, a mon­
the same time, the dipole magnets (spin­
ning nuclei) will start returning to the equi­ ochromatic beam would be attenuated ex­
librium distribution (i.e., the distribution ponentially by some attenuating material.)
that existed before H1 was turned on). In This signal produced by the free return of
other literature you may encounter the M to the H direction is called the free in­
term thermal equilibrium; this is exactly duction decay (FID).
the same thing that we have called equilib­
rium distribution. When we discuss MRI in the next chap­
ter, we will use Mxr to indicate the mag­
So, we have M precessing about H and netization vector in the xy imaging plane.
moving back toward the direction of H. In the current discussion it is more correct
Figure 23-llA shows the path along which to use the term Mr to indicate the goo ro­
M returns to H. The coil shown in Figure tated M vector in a rotating coordinate sys­
23-llB is an RF coil (more than likely the tem.
one used to produce the field HJ). Remem-
The return of M to the H direction is
accomplished by two different mecha­
nisms. Note that as M returns to the H
direction, the component of M along yJ

NUCLEAR MAGNETIC RESONANCE 451

AB

Figure 23-11 A. Beehive path of M during free induction decay. B. Generation of the FID signal

(called MY in Figure 23-12B) decreases to after the goo pulse.) Going in phase may
zero. The Mz component (the component be compared to closing one of those old­
of M along H), however, increases from fashioned folding fans that high-society
zero during the FID. We will need to dis­ matrons used to carry to the opera, and
cuss these two components as separate going out of phase is analogous to opening
topics. They produce two other parame­ the fan. In fact, some authors use the term
ters of NMR, namely T1 and T2, the re­ "fanning out" instead of going out of
laxation times. phase.

Spin-Spin Relaxation Time (T2). Having When the goo pulse is ended, the little
mentioned the FID, it would be appropri­ dipole magnets start precessing about H,
ate here to introduce the concept of spin­ each with a Larmor frequency. The Lar­
spin relaxation time, T2, because T2 is mor frequency for each little magnet is not
closely related to the FID. Before explain­ necessarily the same because of local mag­
ing what T2 represents, we need to discuss netic fields that are produced by the en­
what happens after a goo pulse to get M vironment (the material structure sur­
pointing back along H. rounding the dipole) and by poor H-field
uniformity. Remember, the Larmor fre­
Refer back to Figure 23-SC, which quency will vary if the magnetic field varies.
showed that the little magnets (the nuclear The in-phase dipole magnets start to sep­
MDM) were precessing about H so that arate, because some precess faster than
their components perpendicular to the H others (we will later call these fast and slow
field added up to zero. Obviously , this is dipoles). When the M vector is back along
not the case after a goo pulse. We can vi­ H, we again have a random distribution of
sualize the little magnets as being placed in the dipoles about H. The return to random
phase by the goo pulse. distribution from the in-phase distribution
is an exponential function of time. T2 is the
Figure 23-12B and C attempt to show time constant for this exponential function
when only the interaction between the di­
how the nuclei are in phase at the end of poles is considered. The exponential func­
a goo pulse when My is greatest. As My de­ tion is presented in Figure 23-12A as the
creases with time, the nuclei are resuming dashed line connecting to the tops of the
peaks. We would like to draw that function
the random distribution that existed before again in Figure 23-13.
the goo pulse (H1) was applied. The draw­
ing represents an extreme case in which all Note that T� is introduced in Figure

the nuclei group together in phase, which
is hardly to be expected in actuality. (The
next paragraph will explain why the nuclei

(dipoles) return to a random distribution

452 NUCLEAR MAGNETIC RESONANCE

Time-

z' B.
My Time-
(1)

.:!:!.l:
a_
0

0 My
Q)

t=O

Figure 23-12 A. FID signal. B. Components of magnetization along y' and z' axes following a
90° pulse. C. MY decreases with increasing dephasing

23-13 as a constant in the exponential por­

tion of the equation

.05 My .018 My My;= Mye-tfT2
5% 2% My = component of M along the y axis
My = component of M along y at the end of the
3T{
goo pulse
Figure 23-13 Graphic representation of the t = time following the goo pulse
Ti = decay constant for randomization of the
decay constant T!
in-phase dipoles

We do exactly the same thing in x-ray
attenuation where we introduced the at­
tenuation coefficient, and we do the same

NUCLEAR MAGNETIC RESONANCE 453

thing in radioactive decay by introducing response to inhomogeneity in the magnetic
the disintegration constant. With these two field, and T2 is a characteristic of the ma­
constants we then proceeded to find the terial under investigation. In general, T2 of
half-value layer and the half-life. We would abnormal tissue is longer than T2 of normal
like to do something slightly different here. tissue.
Consider the exponential expression from
Figure 23-13, e-'/T� If time (t) is made During the dephasing of the dipoles, en­
equal to T�, then 'IT� = 1 and the expo­ ergy may be transferred from one dipole
nential becomes e-1 (remember that e-1 = to another, but no energy is transferred to
1/e). Therefore, at t = T�, the value of the the surrounding material. The transfer of
magnetization in the y1 direction has de­ energy to the surrounding environment
creased to lie of its maximum value at t = brings about the return of the nuclear di­
0. (In this paragraph, e is the base of the pole to the lower energy spin state. There
natural log; it is not the electronic charge. is another time constant for the return of
lie has the value of 0.37.) In Figure 23-13, the dipoles to the lower energy spin state.
T� is the time required to reduce the value
of magnetization along the y1 axis (the Spin-Lattice Relaxation Time (T1). Re­
transverse magnetization) to lie of its call that at equilibrium in a magnetic field,
the spin-up (lower energy) population of
value following the 90° pulse. nuclei contain about 1 in 1,000,000 more
nuclei than are contained in the spin-down
Therefore, T� is the decay constant for (higher energy) population. The excess nu­
randomization of the in-phase dipoles. T� clei in the spin-up state are responsible for
relates to randomization analogous to the producing the component of the magnet­
way in which an isotope's half-life relates
to the decay of the radioactive sample. It ization vector, M, that is parallel to the
takes several T� periods for the in-phase
dipoles to approach randomization. The magnetic field, H.
dispersion of the dipoles is determined by
the local Larmor frequency, which is de­ At the end of a goo pulse, the two spin
termined by the local magnetic field. The
local magnetic field depends on, first of all, states of the nuclei are equally populated,
the applied field, H. The variations in the and there is no component of the magnet­
local field, however, depend on the fluc­
tuating fields produced by neighboring ization vector (Mz) in the H direction. This
charged particles and on permanent in­
homogeneity in the applied field, H. The means that some (half ) of the excess nuclei
permanent inhomogeneity in the H field in the lower energy state have absorbed
simply means that it is impossible to make energy and have flipped to the higher en­
a perfect magnet. If the applied field, H,
were perfectly uniform (no inhomogene­ ergy state. At the end of the goo pulse, those
ity), the T� would be exactly equal to the
spin-spin relaxation time, T2. Because this nuclei (or their relatives) transfer energy
is never the case, we will see later (see to the structure (called the lattice) that sur­
"Spin-Echo Technique") that T2 is meas­ rounds them, and return to the lower en­
ured experimentally by procedures de­ ergy state. As they return to the lower state,
signed to eliminate the effects of the field
inhomogeneity. T� is very much shorter the component of M along H is reestab­
than T2. The T2 relaxation time of most
tissues ranges from about 50 to 350 ms, lished as the equilibrium distribution of the
while T� is only a few ms. Thus, T� is a nuclei is reestablished. Here again, we find
that the number of excess nuclei remaining
in the higher energy state as time passes is
an exponential function of time. (Remem­
ber that the number of x-ray photons re­
maining in a beam is an exponential func­
tion of absorber thickness.) Therefore, the
number of nuclei that have returned to the
lower state is obtained by subtracting those
remaining in the higher energy state from
the total number of excess nuclei that were

454 NUCLEAR MAGNETIC RESONANCE

in the higher state at the end of the goo terms of the component of M along H, Mz.

pulse. (This is exactly analogous to the Note that we are close to the equilibrium
number of x-ray photons removed from
the x-ray beam.) population after a time of 4 to 5 T1 has

As always, we have a constant for this passed.
exponential function (like the attenuation
coefficient for x rays) that we call TH the Things have been going too well in the
spin-lattice relaxation time. T1 is the time
discussion of T1 and T2• So it's time for a
for 63% of the nuclei to return to the lower
energy state following a goo pulse. The small problem. The return of the nuclei to

equation for the return to equilibrium fol­ equilibrium distribution does not give an

lowing a goo pulse is NMR signal (Remember that T� could be

Mz = Mz (1 - e-VTl) measured by the FID signal produced by
Mz = component of M along H at time t
M, = component of M along H at equilibrium the transverse My.) This means that T1 can­

t = time following a goo pulse not be measured directly by NMR tech­
T, = spin-lattice relaxation time
niques. But, more importantly, it means
When t is equal to T1, we have e-ur, = e-1
that at the end of the FID signal, there may
= 0.37.
still be some excess nuclei in the higher
One more interpretation of T1 might be
in order. T1 is a rate constant. The rate energy state. In general, the nuclei will de­
(the number of nuclei per unit of time) at
which the nuclei are flipping from the phase (remember the T2 discussion) before
higher to lower energy state depends on
the number of nuclei available for the tran­ all the nuclei return to the equilibrium dis­
sition and on some constant, Tu that is a
function of the environment in which the tribution following a goo pulse, or time T1
nuclei find themselves. Thus, the number
of nuclei per unit of time that undergo a will be longer than time T2• Sometimes T1
transition approaches zero because fewer
nuclei are available to make the transition. and T2 are nearly the same, as in liquids,
(This again is completely analogous to
x-ray attenuation or radioactive decay.) and sometimes T1 is much longer than T2•

Figure 23-14 shows the return to equi­ For example, in a 1-tesla field, cerebral spi­
librium population following a goo pulse in
nal fluid has a T1 of 3000 ms and T2 of
Mz 2000 ms, whereas muscle has a T1 of 750
ms and a T2 of 55 ms.
I Mz at}
There are several ways to measure T1•
\}- Mz (1-e11T1) Equilibrium
Probably the simplest is to use an RF pulse
: I
II sequence of a goo pulse followed by a short
I I latent period, followed by another goo
I pulse (Fig. 23-15A). Pulse sequences are

II used extensively in NMR, with each de­

t=O r, signed to provide some particular bit of

Figure 23-14 Graphic representation of the information. The sequence given above
time constant T,
may be written as goo-,.- goo. (A sequence

is normally notated by some shorthand no­

tation such as this.) In the goo-,.- goo se­
quence, the first goo pulse rotates the M

vector. During the time, 'T, relaxation oc­

curs as some of the nuclei flip from the

higher to the lower energy state. Another

goo pulse is used to produce an FID that is

made by fewer nuclei (because we are not

at thermal equilibrium at the start of this

pulse). The ratio of the FID amplitudes

immediately after the two goo pulses is a

measure of how many nuclei flipped from

the higher to lower state during the time,
'T, between the pulses. This procedure is

repeated for various values of 'T, and T1

NUCLEAR MAGNETIC RESONANCE 455
A.

B.

Figure 23-15 T1 can be measured by a 90°-T-90° sequence

can then be calculated from the data (Fig. tribution before applying the goo pulse to
23-15B). Many people may prefer using a evoke the FID. Usually, for NMR meas­
180° pulse sequence: 180°- T- goo. urements, we use more than one pulse to
help improve the signal-to-noise ratio
Spin Density. The density of the spin­ (S:N). That means that a repetitive pulse
ning nuclei is proportional to the ampli­ sequence is normally used to form the final
tude of the FID. Density here means the NMR signal.
number of identical nuclei per unit volume
that are found in an identical environment MECHANISMS FOR RELAXATION
or, if you prefer, the number of nuclei per
unit volume that have the same Larmor We have said that photon emission alone
frequency. We would expect the magneti­ cannot explain the spin-lattice relaxation
zation to increase with an increased num­ time, T1• (This is different from the pro­
ber of nuclei simply because there would duction of characteristic x rays, which we
be more dipoles to help form the magnet­ described entirely as photon emission.)
ization. As the magnetization vector sweeps Consequently, we must discuss what meth­
past the RF coil to produce the FID, we ods can be used by the nuclei to get rid of
have already seen that the induced signal energy so that they can flip from the higher
size depends on the size of the transverse to the lower energy state.
component of the magnetization. There­
fore, we would expect larger values of mag­ For a proton in a 1-tesla magnetic field,
netization to produce larger FID ampli­ the energy separation between the two spin
tudes. states is only 1.75g X 10-7 eV. We generally
think in terms of kiloelectron volts when
Because the number of nuclei at the Lar­ dealing with x rays which means that the
mor frequency determines the size of the energy state separation for proton spins is
FID signal, we need only measure the FID some 10-10 smaller than the energy sepa­
amplitude to obtain the spin density. We ration of the inner atomic electrons . (You
must ensure that no other effect is chang­ can try this calculation if you use 1 tesla =
ing the amplitude of the FID. For example, 1 kg/C · sec, 1 MHz/tesla = 1 C/kg, 1 eV
the nuclei must be in the equilibrium dis- = 1.6 X 10-t9 J, and h = 6.6 X 10-34 J ·

456 NUCLEAR MAGNETIC RESONANCE

sec). Look up the gyromagnetic ratio of a the H-field inhomogeneity is built into the
magnet and is undesirable but unfortu­
proton in Table 23-1, and remember the nately quite permanent. Furthermore, the
formula llE = -yhH. field inhomogeneity varies from point to
point within the H field. The spin-echo
When there are two quantum energy technique was developed to remove the
effect of the H-field inhomogeneity.
states (either spin states or electron en­
The spin-echo process starts with a goo
ergy shells), there is competition between pulse along the x1 axis (Fig. 23- 16A) so that

photon emission and radiationless trans­ M will precess about x1 to the yi axis. Re­
member that x1 and y1 are rotating at the
fer of energy to the environment (some­ Larmor frequency. Of course, at the end
of the pulse, dephasing starts among the
times called the lattice). For the nuclear spin dipoles. Some dipoles will be in a field
larger than the H field because of local
spin flips, most transitions occur by radia­ fields adding to the H field. They will pre­
cess faster than the M vector, which is pre­
tionless transfer of energy to the lattice. cessing at the Larmor frequency. Some will
be slower than M because they are in a
There are several ways, depending on smaller field than H. Remember, the fre­
quency of precession is related to the
the lattice structure in which the nuclei find strength of H:

themselves, by which the nuclei may trans­ 'YH

fer their energy to the lattice. The main v =-

requirement is that the nuclei be in the 27T

presence of locally fluctuating magnetic We can refer to these dipoles as fast and

fields produced by the environment. slow dipoles. In Figure 23-16B, the x1y1

Therefore, any means by which varying coordinate is rotating at the Larmor fre­
quency and, therefore, M appears not to
magnetic fields can be produced can be a be rotating. Those slow dipoles will then
appear to be rotating in the - y1 direction
means to transfer energy. The larger the (i.e., they appear to be falling behind M)
while the fast dipoles appear to be rotating
field, the more quickly the energy can be in the + y1 direction (they appear to be
pulling away from M). That is, the fast and
transferred. slow dipoles are rotating in opposite direc­
tions relative to M.
In no way do we intend to leave the im­
So, we have the goo pulse. After a time,
pression that the T1 interactions are simple T, a 180° pulse is applied along the x1 axis.
and easy to catalog. Much of the work in
(Remember again that x1y1 are rotating at
NMR is in the area of trying to understand the Larmor frequency. Our electronics
must keep track of how far the xi axis has
the interactions that produce the T1 time.
rotated during the time, T, so the 180° pulse
We should note that the time T1 is an in­ can be applied along the x1 axis.) The 180°

dication of the environment in which the pulse simply inverts the magnetization
components. For the spin-echo technique,
dipole is located. This is useful in NMR the component along H (MJ is of no con­
sequence. The spin grouping, however, is
iamboaugtin2g0. 0Inmtsisfsouresf,atT, 1ocraans be as short as inverted about the x1 axis for the transverse
long
as 3,000

ms for cerebrospinal fluid and free water.

In the brain, gray matter and white matter

differ in water content by only about 10%,

but their T1 times differ by a factor of al­

most 1.5 (in a 1-T field, the T1 of white
about 3g0 ms and gray matter
matter is

. about 520 ms). In general, the T1 of ab-

normal tissue is longer than the T1 of nor­

mal tissues.

Spin-Echo Technique

We have seen that the dephasing (T�) of

the nuclear dipoles following a goo pulse is
determined by two quantities, 1) the spin­
spin relaxation time and 2) the H-field in­

homogeneity. We should keep in mind that

NUCLEAR MAGNETIC RESONANCE 457

z• z' z' z'
A. B. C. D.

t=O I=T t = T + 180° Pulse t = 2T + 180° Pulse
Figure 23-16 Spin-echo sequence

components by precession about the x1 23-17). This back-to-hack FID is the spin­
echo. Because we have used a 180° pulse,
aXlS.
we effectively remove the H-field inho­
It is noteworthy that, after the 180° mogeneity from the dephasing.
pulse, the fast dipoles are behind M and
the slow ones are ahead of M (Fig. The dephasing caused by spin-spin re­
23-16C). After a time, -r (the same as be­ laxation, however, would continue. The
fore), the fast dipoles will catch up with M amplitude of the echo is smaller than the
at the same time that M catches up with initial FID. The decrease in amplitude is
an exponential function of T2, the spin­
the slow dipoles. Everything will again be spin relaxation time. T2 can be measured

in phase, and the magnetization, M, along by repeating the 180° pulse every 2-r and
the.yJ axis will be maximal (Fig. 23-16D).
by measuring the time constant of the de­
The effect is similar to that of a group crease of the amplitude of the echoes. This
type of pulse sequence has the name Carr­
of runners starting a mile run. Because Purcell echo sequence, and it (or some
they run at different speeds, the runners modification) is usually used to measure T2•

spread out as they get farther from the Fourier Transforms

starting line. If, at an arbitrary time, -r, they We have previously mentioned Fourier
transforms (FT) in connection with the
are suddenly told that they are going the modulation transfer function and com­
puted tomography reconstruction meth­
wrong way, they will turn around and head ods. The application of the FT to magnetic
back for the starting line. The fast runners resonance is very straightforward, perhaps
will be farther from the starting line than somewhat easier to understand than our
previous examples. The signal received
the slow runners. Because the runners who
from NMR (the FID, for instance) is in the
are the farthest away are the fastest, all will
arrive back at the starting line together at Figure 23-17 Spin-echo signal

a time equal to 2T (assuming that everyone

runs back at a constant speed). This anal­

ogy would be more accurate if we kept the

runners going in the same direction all the

time. By magic, at time t we would inter­

change the fast and slow runners. Then, at

time 2T, the fast runners would exactly

catch up with their slower companions. Of

course, the dipoles (as would the runners)
immediately start dephasing again as the

fast dipoles pass the slow ones. As the di­

poles come together (we get a reverse FID)

and separate (another FID), we get a pulse

that looks like 2 FIDs back to back (Fig.

458 NUCLEAR MAGNETIC RESONANCE

form of amplitude versus time. The FT of frequency decaying sine curve. We might
a signal of this type is in the form of signal expect the local environmental magnetic
strength versus frequency. fields to cause the dipoles to precess at
slightly different frequencies. Further­
For instance, consider a simple sine more, we might expect that those slightly
wave, which has a time for one complete different frequencies would show up in the
cycle period of T (Fig. 23-18A). This sine FT of the FID.
wave has only one frequency (f), so the FT
of the sine wave is only a single line in the Figure 23-19 gives an FID and its FT.
plot with a height that is proportional to There are some points to note about the
the total strength of the sine wave (Fig. FT curve. First, it is a Lorentzian shape.
23-18B). (Of course, in any practical situ­ (We make this observation only because
ation, we cannot obtain a true single-fre­ you may have read this elsewhere.) The
quency sine wave because additional fre­ term "Lorentzian" is used to provide in­
quencies result from starting and stopping formation (to mathematicians) about the
the wave form.) It is interesting to note that symmetry of the line about the Larmor fre­
both curves of Figure 23-18 contain the quency. The second point is that the line
same information and that, given one, we width at half its maximum value (called the
can get the other by means of the FT. Note FWHM, the full width at half-maximum)
that two sine waves with different frequen­ is 2/T�. This result is verifiable by mathe­
cies would add together to give a compli­ matics but is a terrible task involving inte­
cated-looking signal in the "time domain" gral calculus. Remember that 2/T� has the
(before the FT of the signal), but would unit of 1/time, which is the same unit that
simply be two lines in the "frequency do­ the frequency has.
main" (after the FT of the signal) whose
heights would represent the relative Suppose we have a sample with two
strengths of the sine waves. groups of nonidentical nuclei. If their gy­
romagnetic ratios were different, there
The sine curve looks something like an would be two Larmor frequencies. (Re­
FID, except that the FID decays to zero.
Thus, we might expect the FT of the FID fmreeqmubeenrc,yv,L')'==')'gHyIr2o1mT, awghneerteicvrLat=ioL, aanrmd oHr
to be something like the FT for the sine
curve. If the FID is composed of dephasing = applied magnetic field.) In such a case,
nuclei in similar environments, the Larmor the FID would be the sum of the two ex­
frequency of the FID is essentially a single- ponentially decaying sine waves, one at
each of the Larmor frequencies. If, by
A. chance, we had two groups of identical
nuclei in different magnetic fields, we
I would also see the composite FID of the
I two Larmor frequencies. We will depict the
II two Larmor frequencies (Fig. 23-20A), the

�:

.··

B.

---0+.l: -Frequency
(w)

Figure 23-18 Fourier transform of a single­ Figure 23-19 Fourier transform of a single

frequency sine wave FID