Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved!...

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1 Lecture #11 Relaxation in the Rotating Frame • Topics The spin-lock experiment Physics of T 1ρ Mechanisms contributing to T 1ρ – Applications • References Kowalewski, Chpt 4.3, pp 89-94.

Transcript of Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved!...

Page 1: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

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Lecture #11 ���Relaxation in the Rotating Frame

•  Topics–  The spin-lock experiment–  Physics of T1ρ–  Mechanisms contributing to T1ρ–  Applications

•  References–  Kowalewski, Chpt 4.3, pp 89-94.

Page 2: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

NMR Relaxation Times

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•  T2 probes spectral density function at J(0) and J(ω0)•  T1 probes spectral density function at J(ω0)

3T

liquid( )

in between( )

solid ( )

•  T1ρ , “spin-lattice relaxation in the rotating frame”, is said to be more sensitive to slow fluctuations, typically 100 Hz to a few KHz, characteristic of multiple in vivo processes such as chemical exchange with macromolecules.

; most efficient at ω0=1/τc

fast

slow

fast

slow€

min T1 @ ω0τ c =1

Page 3: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Relaxation in the rotating frame?

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•  Haven’t we being doing almost all of our NMR calculations in the rotating frame?

•  We explicitly showed:

•  Question: so how does “spin-lattice relaxation in the rotating frame” differ from T1?

•  Answer: it doesn’t. -  Rather, “spin-lattice relaxation in the rotation frame” is a not a great name

for T1ρ.-  T1ρ is more accurately described as relaxation in the presence of a

radiofrequency field.

ˆ !Γ = ˆΓ

Relaxation superoperator in the rotating frame

Relaxation superoperator in the laboratory frame

Is there also a T2ρ?

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z

x yRotation axis

Magnetization

T1ρ relaxation by random fields

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Bx2 = By

2 = Bz2 = B2 .whereΔ

!B t( ) = Bx t( )

!x +By t( )!y +Bz t( )

!z

B0!z

•  Consider a system of spin-½ particles subjects to a large main magnetic field plus small independent randomly fluctuating magnetic fields, , given by:Δ

!B t( ) << B0

•  We wish to perform a spin-lock experiment and measure relaxation during the application of a constant Rf pulse.

π2!

"#

$

%&−y

Rfacquire

t

Spin lock

On resonance with amplitude γB1 = ω1, Rf along x-axis.

t

T2* decay

t

vs

T1ρ decay

TSL

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The laboratory frame

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•  The general Hamiltonian in the laboratory frame is…

Hlab t( ) = −γB0 Iz −γB1 I x cosω0t − I y sinω0t( )−γBz t( ) Iz −γBx t( ) I x −γBy t( ) I y

Hlab = H0 + HRf t( )+ H1 t( )

= −ω0 Iz −ω1 I x cosω0t − I y sinω0t( )−γBz t( ) Iz −γBx t( ) I x −γBy t( ) I y

•  In the presence of an on-resonance spin-lock Rf field, the Hamiltonian becomes:

Page 6: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

The rotating frame

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•  Let’s switch to the interaction frame, i.e. H = ei ˆH0t Hlab

•  It’s easier at this point if we switch from the product operators to the eigenoperators of ˆH0 = −ω0

ˆIzIz,  I+ = I x + iI y, and  I− = I x − iI ywhich are

H t( ) = −γB1I x −γBz t( ) Iz −γe−iω0tˆIz Bx t( ) I x +By t( ) I y( )

•  After a little algebra and fun with commutators, we get:

ˆH0 I+ = −ω0 I+, and  ˆH0 I− =ω0 I−ˆH0 Iz = 0 ⋅ Iz,

γ BxIx +ByIy( ) = γ2 Bx + iBy( ) I+ + γ

2 Bx − iBy( ) I−,Noting that

H t( ) = −ω1I x −γBz t( ) Iz − 12 γ Bx t( )+ iBy t( )( )e−iω0t I+ − 1

2 γ Bx t( )− iBy t( )( )eiω0t I−

(good exercise for the reader)

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The doubly rotating frame

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ˆ !σ t( ) = e−iω1t ˆIx e−iω0tˆIzσ lab t( ) = e−iω0t

ˆIxσ t( )

ˆ !H t( ) = −γBz t( )e−iω1tˆIx Iz − 1

2 γ Bx t( )+ iBy t( )( )e−iω0te−iω1t ˆIx I+ − 12 γ Bx t( )− iBy t( )( )eiω0te−iω1t ˆIx I−

•  First note that this Hamiltonian looks a lot like part (a), namely where B0 has been replace by B1. ˆ !H t( ) = −ω1I x + H pert t( )

•  Proceed by switching to a doubly rotating frame, i.e. rotating about the axis (as determined by the spin-lock Rf pulse).I x

B0

x

y

zLaboratory frame

x

y

zRotating frame

B1H0 = −γB0 Iz ˆ !H0 = −γB1I x

which yields…

Page 8: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

A change of coordinates..

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x

y

zRotating frame

B1ˆ !H0 = −γB1I x

ˆ !H t( ) = −γBz t( )e−iω1tˆIx Iz − 1

2 γ Bx t( )+ iBy t( )( )e−iω0te−iω1t ˆIx I+ − 12 γ Bx t( )− iBy t( )( )eiω0te−iω1t ˆIx I−

•  But these commutators are not so straightforward…•  Solution: make a simple change of coordinates.

equivalent to

Rotating frame

B1

!z

!x

!y This looks just like a relaxation problem we’ve already solved!H t( ) = −ω1I "z + pert.

•  In this new doubly rotated frame, the new “longitudinal” axis is and the new “transverse” axes are

!z ,!x  and  !y .

!z = x,  !y = z, and  !x = y.where

•  Now, calling T1ρ the “spin-lattice relaxation time in the rotating frame” makes more sense, as we are looking for 1

T1ρ=ddt

I !z .

Page 9: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Computing the relaxation superoperator

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•  Like before, we want to express the Hamiltonian in terms of the eigenoperators of ˆI !z .

ˆ !H t( ) = −γBz t( )e−iω1tˆI !z i2 I !+ − I !−( )

            − 12 γ Bx t( )+ iBy t( )( )e−iω0te−iω1t ˆI !z I !z + i

2 I !+ + I !−( )( )            − 1

2 γ Bx t( )− iBy t( )( )eiω0te−iω1t ˆI !z I !z − i2 I !+ + I !−( )( )

•  Substituting…

ˆ !H t( ) = −γBz t( )e−iω1tˆIx Iz − 1

2 γ Bx t( )+ iBy t( )( )e−iω0te−iω1t ˆIx I+ − 12 γ Bx t( )− iBy t( )( )eiω0te−iω1t ˆIx I−

Iz = i2 I !+ − I !−( ) I+ = I !z + i

2 I !+ + I !−( ) I− = I "z − i2 I "+ + I "−( )

I !z ,  I !+ = I !x + iI !y , and  I !− = I !x − iI !y .

Page 10: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Computing the relaxation superoperator

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•  Taking the commutators and grouping terms, we have…ˆ !H t( ) = − 1

2 γ Bx t( )+ iBy t( )( )e−iω0t + Bx t( )− iBy t( )( )eiω0t( ) I "z

+ iγ2 Bz t( )e−iω1t − 1

2 Bx t( )+ iBy t( )( )e−i ω0+ω1( )t + 12 Bx t( )− iBy t( )( )e−i −ω0−ω1( )t( ) I "+

− iγ2 Bz t( )eiω1t + 1

2 Bx t( )+ iBy t( )( )e−i ω0+ω1( )t − 12 Bx t( )− iBy t( )( )e−i −ω0+ω1( )t( ) I "−

ˆ !H t( ) = !F0 t( ) ˆ!A0 + !F1 t( ) ˆ!A1 + !F−1 t( ) ˆ!A−1

•  Which is just the form we are looking for, namely

•  The relaxation superoperator in this doubly rotating frame is:

ˆ !Γ = !Fq t( ) !F−q !t( )!t =0

∫t=0

∫q=−1

1

∑ ˆ!Aq ˆ!A−q

with ˆ!A0 = I !z  and  ˆ!A±1 = I !±

Page 11: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

The correlation functions

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•  To proceed, we need to find the expressions for the correlation functions and corresponding spectral densities.

Δ!B t( ) = Bx t( )

!x +By t( )!y +Bz t( )

!z

•  Exponential correlation function: G(τ ) = B2 e−τ τ c

J(ω)B2

=1B2

G(τ )e−iωτ dτ0

∫ =τ c

1+ω 2τ c2

Spectral density function

Bp t( )Bq t −τ( ) =0 if  p ≠ q

B2 e−τ τ c  if  p = q

!"#

$# for  p,q ∈ x, y, z{ }

•  Given:

Δ!B t( ) << B0,where and

Page 12: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Rotating frame relaxation superoperator

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•  Now, making the usual secular approximations and computing the correlation functions, we arrive at (after much algebra):

ˆ !Γ ≈ γ 2

2 B2 J ω0( ) ˆI !zˆI !z +

γ 2

2 B2 J ω1( )+ J ω0( )( ) ˆI !xˆI !x + ˆI !y

ˆI !y( )•  Assumingω0 >>ω1

•  1/T1ρ is simply “longitudinal” relaxation rate in this doubly rotating frame, and is given by the deceptively simple equation:

1T1ρ

= I !zˆ !Γ I !z = γ 2 B2 J ω1( )+ J ω0( )( )

ˆ !Γ = γ 2

2 B2 J ω0( ) ˆI !zˆI !z + J ω1( )+ 1

2 J ω0 +ω1( )+ 12 J ω0 −ω1( )( ) ˆI !x

ˆI !x + ˆI !yˆI !y( )$

%&'()

ˆ !Γ = γ 2

4 Bx2 + By

2( )J ω0( ) ˆI !zˆI !z  +

γ 2

4 Bz2 J ω1( )+ 1

4 Bx2 + By

2( )J ω0 +ω1( )+ 14 Bx

2 + By2( )J ω0 −ω1( )( ) ˆI !+

ˆI !− + ˆI !−ˆI !+( )

Page 13: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

T1ρ versus T2

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•  T1ρ differs from T2 in that:-  probes density function at J(γB1) rather than J(0)-  User selects γB1 value of interest, typically 0.1-3 KHz.

limω1→0

1T1ρ

= γ 2 B2 J 0( )+ J ω0( )( ) = 1T2•  Study of the change in relaxation rates versus magnetic field (called dispersion) gives

insight into different relaxation mechanisms.

•  Whereas the J(0) term dominates T2, T1ρ is ideal for studying physical processes with relatively slow to intermediate correlation or exchange times.

•  However, studying slower processes by lowering B0 results in an unacceptably low SNR.

•  Rapid processes can be studied with magnets at different fields strengths.

•  In the limit of ω1 = 0, 1/T1ρ just equal 1/T2 (which is expected since this case simply become relaxation along the axis.I x

Key idea: T1ρ is maximally sensitive to physical processes with time variations at frequency γB1.

Can T1ρ ever approach T1?

Page 14: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

The T1ρ experiment

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•  Variety of approaches for generating T1ρ contrast

•  Two basic measurement approaches

π2!

"#

$

%&−y

Rfacquire

t

Spin lock

γB1

TSL

map T1ρ

measure T1ρ dispersion

π2!

"#

$

%&−y

π2!

"#

$

%&y

RfFavorite fast imaging sequence

t

Spin lock

π2!

"#

$

%&−y

π2!

"#

$

%&y

π y

RfFavorite fast imaging sequence

t

SLx SL-x etc.

-  Fix B1 and increment TSL -  Fix TSL and increment B1

Page 15: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

SAR considerations

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•  A B1 pulse amplitude in the range of 500 Hz is generally chosen to “coincide with the dominant macromolecular frequencies of interaction in many tissues”.

•  However other B1 amplitudes may be more appropriate depending on the application.

•  SAR can be a limiting factor for both max B1 and max TSL.

•  Some T1ρ approaches for reduced SAR:-  Just apply spin-lock during acquisition of central k-space lines (i.e. like

fast spin echo)-  Off-resonance spin-lock. Remember we are looking at relaxation along

the effective field:

Heff = − ω0 −ω( ) Iz −ω1I x Beff = ω0 −ω( )2 +ω12

x

z

1γ ω0 −ω( )

1γ ω1

!Z

Page 16: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Magic angle effects

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•  The spin-lock techniques can reduce dipolar interaction at any angle with respect to B0.

•  If γB1 >> RDI, then magic angle effects disappear.

•  However, SAR restrictions typically limit maximum spin-lock frequencies for in vivo studies.

•  In which case, there is a reduction in magic angle effects, but they don’t disappear entirely.

•  Residual dipole interactions (RDI) appear as an angularly dependent off-resonance term.

Page 17: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

T1ρ dispersion

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•  T2 is typically dominated by the J(0) term, and hence largely independent of field.

•  T1ρ, however, has an important dispersive component.J(ω1)∝

τ c1+ω1

2τ c2

1

1

1

1

1T1ρ

1 T1ρ

1 T2

γB1 = 500Hz

10-6 10-5 10-4 10-3 10-2 10-1

Rel

axat

ion

rate

!τ c =1 2π ⋅500Hz( )

Page 18: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Dipolar relaxation

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•  The main interaction responsible for T1ρ relaxation are:-  dipolar interactions

-  scalar coupling interactions-  chemical exchange processes

•  Equations describing T1ρ relaxation differ among these sources.

•  Example: T1ρ due to dipolar coupling (for everything on-resonance):

1T1ρ,DD

=µ04π!

"#

$

%&2 3γ 4!2

10π 2r632 J 2ω1( )+ 5

2 J ω0( )+ J 2ω0( )( )Probes spectral density at 2ω1!

-  diffusion effects

=µ04π!

"#

$

%&2 3γ 4!2

20π 2r63τ c

1+ 4ω12τ c

2 +5τ c

1+ω02τ c

2 +2τ c

1+ 4ω02τ c

2

!

"#

$

%&

Page 19: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

Chemical exchange

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•  Consider a fast exchange between sites A and B, with molar fractions pA and pB, chemical shift difference δ, and exchange rate τex.

•  For the case of free water interacting with a bound pool, the observed T1ρ will be due to a combination of chemical exchange and dipolar coupling at sites A and B:

•  For the “on-resonance” case, i.e. ω1 >> δ,

1T1ρ,obs

=1

T1ρ,ex+

pAT1ρ,A,DD

+pB

T1ρ,B,DD

w/ correlation time τA

w/ correlation time τB

1T1ρ,ex

=pApBδ

2τ ex1+ω1

2τ ex2

Page 20: Lecture #11 Relaxation in the Rotating Frame · relaxation problem we’ve already solved! Hˆ(t)=−ω 1 Iˆ z" +pert. • In this new doubly rotated frame, the new “longitudinal”

T1ρ applications

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•  The most ubiquitous use of T1ρ contrast has been for the study of cartilage and osteoarthritis. Example: Borthakur, et al., NMR Biomed. 2006; 19: 781–821.

•  We’ll devote a whole lecture to the relaxation properties of cartilage, so for the rest of this lecture, I’ll just present some less common, but no less interesting, applications of T1ρ imaging.

•  In general, tissue T1ρ is a combination of multiple factors:

1T1ρ

!

"##

$

%&&tissue

=1

T1ρ,DD+1

T1ρ,ex+

1T1ρ,SC

+1

T1ρ,Diff

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Black circles, high; dark gray circles, moderate; and light gray circles, low treatment response groups; white circles, control animals.

“Observed changes included increased correlation time of water associated with macromolecules and a decreased fractional population of this pool.”

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•  T1ρ relaxation is sensitive to macromolecule-water interactions in protein solutions and in biological tissues.

•  T1ρ relaxation may serve as a biomarker for biological processes associated with alterations in the macromolecular content properties of tissues.

•  Liver fibrosis involves the accumulation of collagen. proteoglycans, and other macromolecules in the extracellular matrix, and thus T1ρ relaxation could be a candidate biomarker for liver fibrosis

T1ρ 80 (ms)

0

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“T1ρ is related to slow motional interaction between bulk water and extracellular matrix molecules.”

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•  T1ρ is potentially sensitive to HIFU-induced protein denaturation.

•  Tissue T1ρ values are predominantly influenced by protein-water interactions.

•  Significant changes in tumor T1ρ were observed after HIFU treatment. These T1ρ changes were distinctly more pronounced than HIFU-induced changes in T2.

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Porcine model Clinical patients

“In ex vivo myocardial infarction (MI) tissue, it has been shown that a significantly higher T1ρ is found in the MI region, and studies in animal models of chronic MI showed the first in vivo evidence for the ability to detect myocardial fibrosis with native T1ρ-mapping. In this study we aimed to translate and validate T1ρ-mapping for endogenous detection of chronic MI in patients.”

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“Relaxation Along a Fictitious Field (RAFF) in the second rotating frame of reference (RAFF2) that can measure relaxation dispersion based on fictitious fields using decreased SAR [vs T1ρ].”

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“…The T1ρ abnormalities may very well reflect abnormal metabolism rather than inflammation or cellular loss. However, at present, we cannot rule out local changes in tissue structure or water content. These possibilities could be investigated in future studies by using complementary mapping of T2, adiabatic T1ρ and T2ρ, and T1ρ dispersion…

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Movement Disorders, Vol. 00, No. 00, 2015

“…These data suggest that quantitative T1ρ mapping may provide a useful marker for assessing disease progression in HD. The absence of T2 changes suggests that the T1ρ abnormalities are unlikely owing to altered water content or tissue structure. The established sensitivity of T1ρ to pH and glucose suggests that these factors are altered in HD perhaps owing to abnormal mitochondrial function…”

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•  By suppression of T1ρ dispersion, iron oxide nanoparticles cause enhanced T1ρ contrast compared to T2.

•  The underlying mechanism appears to be loss of lock. •  Spin-lock MR is therefore a promising technique for sensitive

detection of iron oxide contrast agents.”

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Next Lecture: ���Guest Speaker: Dr. Michael Moseley!���

“MRI Contrast Agents”