I Introduction to Relaxation Theory I James Keeler Irelaxation; estimation of distance Introduction...

EUROMAR

Zurich, 2014

Introduction toRelaxation Theory

James Keeler

University of CambridgeDepartment of Chemistry

1

What is relaxation? Why might it be interesting?

I relaxation is the process which drives the spins to equilibrium(equilibrium z-magnetization, no xy-magnetization)

I a natural phenomenon, driven by molecular motion

I unusually slow in NMR

I useful probe of molecular motion

I the Nuclear Overhauser Effect (NOE) arises because ofrelaxation; estimation of distance

Introduction and outline 2

Further reading

I for more detail: James Keeler, Understanding NMRSpectroscopy, 2nd edit., Wiley 2010 (Chapter 9)

I a PDF of this presentation is available to download atwww-keeler.ch.cam.ac.uk


Outline

1. How relaxation arises

2. Describing random motion – the correlation time

3. Motional regimes

4. Relaxation in terms of populations

5. Relaxation of a spin pair

6. Solomon equations and relaxation rate constants

7. Nuclear overhauser effect (NOE)

8. Transverse relaxation


Behaviour of individual magnetic moments

z

xy

I bulk magnetization result of sum ofmagnetic moments of each spin

I each moment behaves as the overallmagnetization i.e. precesses about z,rotated away from z by transverse fieldsoscillating at Larmor frequency

I pulse affects all the same way, butrelaxation caused by local fields whichare different for each spin

How relaxation arises 5

Local fields

z

r

A

B

Bloc

I e.g. spin B generates magneticfield at A

I a local field only significant over ashort distance

I local field varies in size anddirection according to length andorientation of A–B vector

I local field is random


How random fields drive the system to equilibrium

random fields fluctuating at close to the Larmor frequency shoulddrive spins to equilibrium

(a) (b) (c) (d)

+1

0

-1

z-c

om

po

ne

nt

time

Mz = −20 Mz = −16 Mz = −12 Mz = 0How relaxation arises 7

Coming to equilibrium with the lattice

I random fields would appear to drive the z-magnetization tozero

I incorrect: equilibrium z-magnetization is finite

I since ‘surroundings’ are large and at equilibrium, greaterchance of loosing energy to surroundings than gaining energy

I result is finite z-magnetization at equilibrium

I process is called longitudinal or spin–lattice relaxation


Relaxation mechanisms

I dipolar: local field goes as γ1γ2/r3

z

r

A

B

Bloc

I chemical shift anisotropy (CSA): local field goes as B0 andtypically depends on shift range

Bloc

B0

I paramagnetic species (e.g. dissolved oxygen)


Describing random motion – the correlation time

I in solution, molecular collisions change orientation on aboutright timescale for relaxation

I each collision only alters orientation by a small amount –rotational diffusion

I correlation time, τc, is average time it takes a molecule tomove through 1 radian

I correlation time describes the timescale of the random motion

Describing random motion – the correlation time 10

Rotational diffusion

representative molecules undergoing rotational diffusion

molecule 1 molecule 2 molecule 3 molecule 4(a)

(b)

0

1

-1

0

1

-1

θ

θ

time

(a) has longer correlation time than (b)


The correlation function

G(t, τ) =1N

[Bloc,1(t)Bloc,1(t + τ) + Bloc,2(t)Bloc,2(t + τ) + . . .

]=

1N

N∑i=1

Bloc,i(t)Bloc,i(t + τ)

= Bloc(t)Bloc(t + τ)

I average of product of local field at time t with that at time(t + τ)

I usually only depends on τ: stationary random function


The correlation function

(a) (b) (c)

Blo

c(0

)B

loc(τ

)B

loc(0

) ×

Blo

c(τ

)

τ < τc τ >> τcτ = 0

+1

-1

0

+1

-1

0

+1

-1

0


Typical correlation function

I maximum at τ = 0

G(0) = Bloc(t)Bloc(t)

= B2loc

I simplest form is an exponential

G(τ) = B2loc exp (−|τ|/τc)

I typical behaviour

τ

00

G(τ

)

B2loc

τc = τmin

τc = 2τmin

τc = 4τmin


Reduced correlation function

I time dependent part (max value 1)

g(τ) = exp (−|τ|/τc)

I hence

G(τ) = B2locg(τ)

I recall that we need motion at the Larmor frequency to causerelaxation; g(τ) describes time dependence


The spectral density

I need to know amount of motion at Larmor frequency

I Fourier transform of function of time, G(τ), gives function offrequency, J(ω)

G(τ)Fourier transform−−−−−−−−−−−−−→ J(ω)

I J(ω) gives amount of motion at frequency ω

I

B2loc exp (−|τ|/τc)

FT−−→ B2

loc2τc

1 + ω2τ2c

J(ω) = B2loc

2τc

1 + ω2τ2c


Spectral density: interpretation

ω

J(ω

)

00

τc = τmin

τc = 2τmin

τc = 4τmin

I area under curve is independent of τc

I the shorter τc, the higher the frequency present in motion

I always has maximum value at zero frequency


Spectral density at Larmor frequency

0

J(ω

0)

τcτc = 1/ω0

J(ω0) plotted against τc

I maximum when τc = 1/ω0; fastest relaxation with this value

I reduced spectral density, j(ω)

g(τ)FT−−→ j(ω)

exp (−|τ|/τc)FT−−→

2τc

1 + ω2τ2c

I hence J(ω) = B2loc j(ω)


Motional regimes

j(ω0) =2τc

1 + ω20τ

2c

I fast motion ω0τc << 1

fast motion: j(ω0) = 2τc

j(ω0) independent of frequency

I slow motion ω0τc >> 1

slow motion: j(ω0) =2

ω20τc

or j(ω0) =j(0)ω2

0τ2c

I small molecules: τc ≈ 10 ps – fast motion

small protein: τc ≈ 10 ns – slow motion

Motional regimes 19

Summary

I rotational diffusion gives motion suitable for NMR relaxation

I rotational diffusion characterised by the correlation time τc

I spectral density gives the frequency distribution of the motione.g.

J(ω) = B2loc

2τc

1 + ω2τ2c

J(ω) depends on τc

I rate of longitudinal relaxation depends on spectral density atω0; max. when ω0τc = 1

Motional regimes 20

Populations

I useful to think in terms of ‘populations’ of the spin states(energy levels) α (‘spin up’) and β (‘spin down’)

I z-magnetization due to a population difference between thesetwo states

Mz = 12~γ(nα − nβ)

nα population of α state; nβ population of β state

I Boltzmann distribution gives equilibrium magnetization

M0z =

γ2~2NB0

4kBT

I omit constants

Mz = nα − nβ and M0z = n0

α − n0β

0 indicates equilibrium valuesRelaxation in terms of populations 21

Rate equations

Wα→β

Wβ→α

nα

nβ

. . . just like chemical kinetics

W are rate constants

I rate from α to β = Wα→β nα

I rate from β to α = Wβ→α nβ

I rate of change of nα = +Wβ→α nβ︸︷︷︸increase in nα

−Wα→β nα︸︷︷︸decrease in nα

I rate of change of nβ = +Wα→β nα︸︷︷︸increase in nβ

−Wβ→α nβ︸︷︷︸decrease in nβ

Relaxation in terms of populations 22

Problem with the rate equations

Wα→β

Wβ→α

nα

nβ

I at equilibrium, no change in population with time

0 = Wβ→α n0β −Wα→β n0

α 0 = +Wα→β n0α −Wβ→α n0

β

I hence

n0α

n0β

=Wβ→α

Wα→β

I simple theory predicts Wβ→α = Wα→β, hence n0α = n0

β, which iswrong

I need more advanced theory, or . . .


Modifying the rate equations

I instead of

rate of change of nα = +Wβ→α nβ−Wα→β nαrate of change of nβ = +Wα→β nα−Wβ→α nβ

I set Wβ→α and Wα→β equal to Wαβ and write

rate of change of nα =Wαβ (nβ − n0β) −Wαβ (nα − n0

α)

rate of change of nβ = −Wαβ (nβ − n0β) + Wαβ (nα − n0

α)

I rate depends on deviation from equilibrium population


Relaxation in terms of populations

I recall Mz = nα − nβ, so

rate of change of Mz = rate of change of nα−rate of change of nβ

I using

rate of change of nα =Wαβ (nβ − n0β) −Wαβ (nα − n0

α)

rate of change of nβ = −Wαβ (nβ − n0β) + Wαβ (nα − n0

α)

I gives

rate of change of Mz = 2Wαβ(nβ − n0β) − 2Wαβ(nα − n0

α)

= −2Wαβ

[(nα − nβ) − (n0

α − n0β)]

= −2Wαβ(Mz −M0z )


Writing the magnetization in terms of the populations

I we have

rate of change of Mz = −2Wαβ(Mz −M0z )

I usually written

dMz(t)dt

= −Rz[Mz(t) −M0

z

]with Rz = 2Wαβ, longitudinal relaxation rate constant

I or

dMz(t)dt

= −1T1

[Mz(t) −M0

z

]with T1 = 1/Rz, time constant for longitudinal relaxation


Longitudinal relaxation

I we have

dMz(t)dt

= −Rz[Mz(t) −M0

z

]

I implies that the rate of change of Mz is proportional to thedeviation of Mz from the equilibrium value M0

z

I implies that Mz tends to M0z

I can integrate using Mz = Mz(0) at time t = 0 to give

Mz(t) =[Mz(0) −M0

z

]exp (−Rzt) + M0

z


Longitudinal relaxation

different initial conditions different rate constants

-1.0

-0.5

0.0

0.5

1.0

timeMz(t

)/M

z0

time

(a) (b)

note Mz always tends to equilibrium value


Two spins: energy levels and transition rates

αα

αβ

ββ

βα

1

2 3

4

W1(2,α)

W1(2,β)

W2

W0

W1(1,α)

W1(1,β)

I two spins, four energy levels

I dipolar interaction causes relaxation-induced transitionsbetween any two levels

I rate constants W∆M, ∆M gives change in M

I note W (1,α)1 and W(1,β)

1Relaxation of a spin pair 29

Rate equations for the populations and z-magn.

αα

αβ

ββ

βα

1

2 3

4

W1(2,α)

W1(2,β)

W2

W0

W1(1,α)

W1(1,β)

I as before, look at gain and loss processes for level 1

dn1

dt= −W (2,α)

1 (n1 − n01) −W(1,α)

1 (n1 − n01) −W2(n1 − n0

1)︸︷︷︸loss from level 1

+ W(2,α)1 (n2 − n0

2)︸︷︷︸gain from level 2

+ W (1,α)1 (n3 − n0

3)︸︷︷︸gain from level 3

+ W2(n4 − n04)︸︷︷︸

gain from level 4

Relaxation of a spin pair 30


I define z-magn. for spin 1 as population difference across thespin-1 transitions 1–3 and 2–4

I1z = (n1 − n3) + (n2 − n4)

I similarly for spin 2

I2z = (n1 − n2) + (n3 − n4)

I also need the difference in the population difference acrossthe spin 1 levels

2I1zI2z = (n1 − n3)− (n2 − n4) or 2I1zI2z = (n1 − n2)− (n3 − n4)

I and equilibrium values (2I1zI2z = 0 at equil.)

I01z = n0

1 − n03 + n0

2 − n04 I0

2z = n01 − n0

2 + n03 − n0

4



I after much algebra

dI1z

dt= − R(1)

z (I1z − I01z) − σ12(I2z − I0

2z) − ∆(1) 2I1zI2z

dI2z

dt= − σ12(I1z − I0

1z) − R(2)z (I2z − I0

2z) − ∆(2) 2I1zI2z

d 2I1zI2z

dt= − ∆(1)(I1z − I0

1z) − ∆(2)(I2z − I02z) − R(1,2)

z 2I1zI2z

I rate constants in terms of the W

R(1)z = W(1,α)

1 + W(1,β)1 + W2 + W0

R(2)z = W(2,α)

1 + W(2,β)1 + W2 + W0

σ12 = W2 −W0

∆(1) = W(1,α)1 −W (1,β)

1

∆(2) = W(2,α)1 −W (2,β)

1

R(1,2)z = W(1,α)

1 + W (1,β)1 + W(2,α)

1 + W (2,β)1



I1z

Rz(1)

Rz(1,2)

σ12

∆(1)

∆(2)

Rz(2)

2I1zI2z

I2z

dI1z

dt= − R(1)

z (I1z − I01z) − σ12(I2z − I0

2z) − ∆(1) 2I1zI2z

dI2z

dt= − σ12(I1z − I0

1z) − R(2)z (I2z − I0

2z) − ∆(2) 2I1zI2z

d 2I1zI2z

dt= − ∆(1)(I1z − I0

1z) − ∆(2)(I2z − I02z) − R(1,2)

z 2I1zI2z



I1z

Rz(1)

Rz(1,2)

σ12

∆(1)

∆(2)

Rz(2)

2I1zI2z

I2z

I pure dipolar relaxation: W(1,α)1 = W (1,β)

1 , W (2,α)1 = W(2,β)

1

I hence ∆(1) = 0 and ∆(2) = 0

I 2I1zI2z not connected to I1z or I2z


Solomon equations

dI1z

dt= −R(1)

z (I1z − I01z) − σ12(I2z − I0

2z)

dI2z

dt= −σ12(I1z − I0

1z) − R(2)z (I2z − I0

2z)

d 2I1zI2z

dt= −R(1,2)

z 2I1zI2z

with

R(1)z = 2W(1)

1 + W2 + W0

R(2)z = 2W(2)

1 + W2 + W0

σ12 = W2 −W0

R(1,2)z = 2W(1)

1 + 2W(2)1

Solomon equations and relaxation rate constants 35

Solomon equations

dI1z

dt= −R(1)

z (I1z − I01z) − σ12(I2z − I0

2z)

dI2z

dt= −σ12(I1z − I0

1z) − R(2)z (I2z − I0

2z)

d 2I1zI2z

dt= −R(1,2)

z 2I1zI2z

I R(1)z is rate constant for self relaxation of spin 1

I likewise R(2)z for spin 2

I σ12 is cross-relaxation rate constant between spins 1 and 2

I cross relaxation connects the z-magnetizations of the twospins


Relaxation rate constants

I detailed theory shows rate constants Wij always given byexpression of the form

Wij = Aij × Y2 × j(ωij)

I Aij is a quantum mechanical factor

I Y2 relates to magnitude of local field (always squared)depends on e.g. distance between spins, γ

I j(ωij) reduced spectral density at ωij, the transition frequencybetween the two levels


Relaxation rate constants: dipolar

I dipolar only

W (1)1 = 3

40 b2j(ω0,1) W (2)1 = 3

40 b2j(ω0,2)

W2 = 310 b2j(ω0,1 + ω0,2) W0 = 1

20 b2j(ω0,1 − ω0,2)

magnitude factor

b =µ0γ1γ2~

4πr3

I hence

R(1)z = b2

[3

20 j(ω0,1) + 310 j(ω0,1 + ω0,2) + 1

20 j(ω0,1 − ω0,2)]

R(2)z = b2

[3

20 j(ω0,2) + 310 j(ω0,1 + ω0,2) + 1

20 j(ω0,1 − ω0,2)]

σ12 = b2[

310 j(ω0,1 + ω0,2) − 1

20 j(ω0,1 − ω0,2)]

R(1,2)z = b2

[3

20 j(ω0,1) + 320 j(ω0,2)

]Solomon equations and relaxation rate constants 38

Cross relaxation in the two motional regimes

homonuclear: ω0,1 = ω0,2 → ω0

I σ12 = b2 310 j(2ω0)︸︷︷︸

W2

− b2 120 j(0)︸︷︷︸W0

I fast motion j(ω) = 2τc

fast motion: σ12 = b2 310 j(2ω0)︸︷︷︸

W2

− b2 120 j(0)︸︷︷︸W0

= b2 310 2τc − b2 1

20 2τc

= 12 b2τc

I σ12 positive in this limit


Cross relaxation in the two motional regimes

I σ12 = b2 310 j(2ω0)︸︷︷︸

W2

− b2 120 j(0)︸︷︷︸W0

I slow motion j(0) = 2τc, j(2ω0) negligible in comparison

slow motion: σ12 = b2 310 j(2ω0)︸︷︷︸

W2

− b2 120 j(0)︸︷︷︸W0

= 0 − b2 120 2τc

= − 110 b2τc

I σ12 negative in this limit


Cross relaxation as a function of τc

τc / ps

σ1

2

100 200 300 500 6000

400

I computed for proton at 500 MHz

I cross over at

ω0τc =

√54

I 360 ps in this case


The nuclear overhauser effect (NOE)

I Solomon equation

dI1z

dt= −R(1)

z (I1z − I01z) − σ12(I2z − I0

2z)

I implies that if spin 2 not at equilibrium, spin 1 will be affected

I but only if cross-relaxation rate constant σ12 , 0

I cross relaxation is a feature of dipolar relaxation, hencedetection of cross relaxation implies dipolar relaxationi.e. nearby spins

I origin of Nuclear Overhauser Effect

Nuclear overhauser effect (NOE) 42

The transient NOE experiment

τ

spin 2

180°(a)

(b)

I (a) perturb spin 2 with a selective inversion pulse

I wait time τ for cross relaxation to occur

I 90 pulse to give observable signal

I (b) repeat without inversion pulse (reference spectrum)

I compute difference spectrum (a) − (b) to reveal changes

I can analyze the experiment using the Solomon equationsNuclear overhauser effect (NOE) 43

Difference spectroscopy reveals the NOE

Ω1 Ω2

(a)

(b)

(c) = (a) - (b)

difference reveals NOENuclear overhauser effect (NOE) 44

Transverse relaxation: non-secular contribution

I transverse relaxation is decay of xy-components ofmagnetization; determines rate of decay of FID and hencelinewidth

I transverse local fields, oscillating near to the Larmorfrequency, cause longitudinal relaxation

I such fields can also affect the x- and y-components ofindividual magnetic moments, and therefore also causetransverse relaxation

I called the non-secular contribution to transverse relaxation

Transverse relaxation 45

Transverse relaxation: secular contribution

I the z-component of the local field will cause a change in the(local) Larmor frequency

I individual magnetic moments will precess at slightly differentLarmor frequencies and so get out of step with one another

I result is a decay in the transverse magnetization

I called the secular contribution to transverse relaxation


Transverse relaxation

I non-secular contribution to transverse relaxation: descriptionsimilar to longitudinal relaxation; rate depends on j(ω0)

I secular contribution to transverse relaxation: rate depends onspectral density at zero frequency, j(0)

I chemical exchange is a useful analogy for the secularcontribution


Chemical exchange

Fax

Feq

I Fax and Feq have different shifts: frequency differencebetween two resonances

I if rate constant for exchange is much less than the frequencydifference→ two lines (slow exchange)

I if rate constant for exchange is much greater than thefrequency difference→ one line (fast exchange)


Two-site chemical exchange

× 1

× 1

× 5

× 5

× 5

A B

kex = 0 s–1

kex = 10 s–1

kex = 100 s–1

kex = 250 s–1

kex = 1 000 s–1

kex = 1 000 s–1

kex = 5 000 s–1

kex = 20 000 s–1

kex = 100 00 s–1

kex = 500 000 s–1

× 1/2

× 1/2

× 1/2

× 1/2

× 1/2

I shift difference 160 Hz

I initial broadening and then coalescence when exchange rateconstant is 350 s−1

I further increase in exchange rate constant results in anarrower line: exchange narrowing


Exchange processes from the point of view of singlespins

I consider behaviour of individual spins

I either in environment A or B

I the spin jumps between them randomly

I the larger the exchange rate constant, the more frequent thejumps


Simulation of two-site exchange

(b) (c)(a)

time

frequency

spin 1

spin 2

spin 3

spin 4

FID

spin 1

spin 2

spin 3

spin 4

FID

A B

slow intermediate fast


Slow and fast exchange: interpretation

0.0 0.1

0.0 1.0

time / s

time / s

(a)

(b)

I two cosine waves at 10.0 and 10.5 Hz; ∆ = 0.5 Hz

I (a): observing for 0.1 s, can hardly see that the waves are atdifferent frequencies

I (b): observing for 1 s, difference is clear

I phase difference over period τ is 2π∆ × τ; must be significantif frequencies are to be distinguished


Slow and fast exchange: interpretation

0.0 0.1

0.0 1.0

time / s

time / s

(a)

(b)

I phase difference over period τex is 2π∆ × τex

I for significant phase difference τex >> (1/∆)

I can only distinguish frequencies if τex >> (1/∆);since τex = 1/kex the condition for slow exchange is kex << ∆

I frequencies indistinguishable if τex << (1/∆);the condition for fast exchange is kex >> ∆


The secular contribution to transverse relaxation

frequency

(a) (b) (e)(d)(c)

I continuous range of Larmor frequencies due to spread of localfields

I square profile in absence of motion: (a)

I molecular motion is fast compared to range of Larmorfrequencies, so line is exchange narrowed: (e)


The secular contribution to transverse relaxation

I for A B, in fast exchange limit

linewidth ≈π∆2

AB

2kex

I molecular motion: kex → 1/τc; ∆AB → W, width of distributionof Larmor frequencies

width of narrowed line ≈ W2 × τc

I W = 100 kHz, τc = 100 ps gives width of 1 Hz

I j(0) = 2τc hence

width of narrowed line ≈ 12 W2 × j(0)


Secular and non-secular contribution to relaxation

Summary

I non-secular: due to transverse local fields oscillating near tothe Larmor frequency

I the same fluctuations cause longitudinal relaxation

I secular: due to a distribution of local fields along z giving aspread in Larmor frequencies

I molecular motion results in a linewidth very much less thanthe spread of Larmor frequencies

I non-secular depends on j(ω0); secular depends on j(0)


Relaxation by random fields

I assume a randomly varying field with mean square value B2loc

in all three directions

I Rz = γ2B2loc j(ω0)

I transverse relaxation

Rxy = 12γ

2B2loc j(0)︸︷︷︸

secular

+ 12γ

2B2loc j(ω0)︸︷︷︸

non-secular

I comparing

Rxy = 12γ

2B2loc j(0)︸︷︷︸

secular

+ 12 Rz︸︷︷︸

non-secular

i.e. non-secular part is precisely half of the overall longitudinalrate constant


Relaxation by random fields

τc

R Rxy

Rz

ω0τc = 1

I Fast motion: Rz = Rxy

I Slow motion: Rxy continues to increase due to secular term:J(0) ∝ τc, but Rz decreases as spectral density at Larmorfrequency decreases


EUROMAR

Zurich, 2014

Introduction toRelaxation Theory

The End

Want more on the basic theory of NMR?Search for “ANZMAG” on YouTube

The end 59

I Introduction to Relaxation Theory I James Keeler Irelaxation; estimation of distance Introduction...

Documents

Transcript of I Introduction to Relaxation Theory I James Keeler Irelaxation; estimation of distance Introduction...