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Offline macman104

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T1 and T2 Times
« on: July 30, 2008, 12:23:19 PM »
Hi everyone!

So, I would really appreciate if someone could help explain some aspects of T1 and T2.

T1:

In one experiment, you do a 180° pulse, and flip the spins of the nuclei.  Then as time goes on they start to decay back to the aligned spin of the magnet.  The time it takes for 69% of the spin to decay back to the beginning is your T1 time.

T2:

This is where I'm a little lost.  A 90° pulse is sent that knocks the spins into the XY plane.  Then there is some sort of dephasing of the spins?  Then a 180° echo?  This is the part I'm kind of lost on.  Wiki says:

T2 characterizes the rate at which the Mxy component of the magnetization vector decays in the transverse magnetic plane. It is the time it takes for the transverse signal to reach 37% (1/e) of its initial value after flipping into the magnetic transverse plane.

So, the spins dephase in the XY plane?  I'm just kind of confused.

If anyone could assist in further understanding how the T2 works and verify that I have a correct understanding of T1 I would appreciate it.

Offline Yggdrasil

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Re: T1 and T2 Times
« Reply #1 on: July 30, 2008, 08:24:09 PM »
Hi everyone!

So, I would really appreciate if someone could help explain some aspects of T1 and T2.

T1:

In one experiment, you do a 180° pulse, and flip the spins of the nuclei.  Then as time goes on they start to decay back to the aligned spin of the magnet.  The time it takes for 69% of the spin to decay back to the beginning is your T1 time.

That's a decent explanation of how to find T1.

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T2:

This is where I'm a little lost.  A 90° pulse is sent that knocks the spins into the XY plane.  Then there is some sort of dephasing of the spins?  Then a 180° echo?  This is the part I'm kind of lost on.  Wiki says:

T2 characterizes the rate at which the Mxy component of the magnetization vector decays in the transverse magnetic plane. It is the time it takes for the transverse signal to reach 37% (1/e) of its initial value after flipping into the magnetic transverse plane.

So, the spins dephase in the XY plane?  I'm just kind of confused.

As you correctly mentioned, T2 relaxation comes from dephasing of the spins.  Essentially, identical nuclei can experience slightly different magnetic fields, and these differences will cause each spin to precess at a slightly different frequency.  Over time, these differences in frequencies will cause the spins to precess out of phase from each other, so that the signals from each nuclei will constructively interfere and you will see no overall signal.  When you examine the signal (called an FID, free induction decay) from nuclei precessing in the xy plane, the time constant of the decay of the FID is called T2*.

It turns out that there are two components to T2*: T2 and T2(ΔB).  These two components reflect two different sources of transverse relaxation.  T2 relaxation (aka spin-spin relaxation) causes dephasing because identical nuclei will experience changes in their local magnetic fields due to coupling with nearby spins (hence the name, spin-spin relaxation).  This coupling combined with the random tumbling of the molecules in solution will cause random fluctuations in the larmor frequency of the spins and lead to dephasing.  T2(ΔB) relaxation comes from inhomogeneities in the applied magnetic field.  Despite all the shimming you do on the NMR machine to create a uniform magnetic field, the inevitable small differences in magnetic field at different points in the sample will cause nuclei to precess at different frequencies and dephase. 

So, how do we go about measuring T2?  To do this we need some way of separating T2 decay from T2(ΔB) decay.  Luckily, unlike spin-spin relaxation, this T2(ΔB) relaxation is not random and one can use a trick called a "spin-echo" to refocus signals that have dephased through magnetic field inhomogeneities (this is where the 180o echo comes in).  By using the spin echo to remove the effects of T2(ΔB) decay, one can directly measure the T2 relaxation constant.  Measuring T2 instead of T2* is advantageous because T2 relates directly to molecular properties while T2* does not relate directly to these properties.


In summary, NMR is really hard to explain (and understand!).  I hope this at least made some sense to you...

Offline macman104

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Re: T1 and T2 Times
« Reply #2 on: July 31, 2008, 09:36:36 AM »
As you correctly mentioned, T2 relaxation comes from dephasing of the spins.  Essentially, identical nuclei can experience slightly different magnetic fields, and these differences will cause each spin to precess at a slightly different frequency.  Over time, these differences in frequencies will cause the spins to precess out of phase from each other, so that the signals from each nuclei will constructively interfere and you will see no overall signal.
And this precessing/dephasing is just in the XY plane?  Since the magnetic field is along the Z-axis, do they start to dephase from all being aligned in the XY plane to dephase towards the Z axis as well?  But we're just interested in the level of dephasing in the XY plane.
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When you examine the signal (called an FID, free induction decay) from nuclei precessing in the xy plane, the time constant of the decay of the FID is called T2*.
Can you elaborate what you mean by the phrase "time constant of the decay"?
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It turns out that there are two components to T2*: T2 and T2(ΔB).  These two components reflect two different sources of transverse relaxation.  T2 relaxation (aka spin-spin relaxation) causes dephasing because identical nuclei will experience changes in their local magnetic fields due to coupling with nearby spins (hence the name, spin-spin relaxation).  This coupling combined with the random tumbling of the molecules in solution will cause random fluctuations in the larmor frequency of the spins and lead to dephasing.
Make sense
Quote
T2(ΔB) relaxation comes from inhomogeneities in the applied magnetic field.  Despite all the shimming you do on the NMR machine to create a uniform magnetic field, the inevitable small differences in magnetic field at different points in the sample will cause nuclei to precess at different frequencies and dephase.

So, how do we go about measuring T2?  To do this we need some way of separating T2 decay from T2(ΔB) decay.  Luckily, unlike spin-spin relaxation, this T2(ΔB) relaxation is not random and one can use a trick called a "spin-echo" to refocus signals that have dephased through magnetic field inhomogeneities (this is where the 180o echo comes in).  By using the spin echo to remove the effects of T2(ΔB) decay, one can directly measure the T2 relaxation constant.  Measuring T2 instead of T2* is advantageous because T2 relates directly to molecular properties while T2* does not relate directly to these properties.
Awesome this last part made perfect sense!  Thank you so much!
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In summary, NMR is really hard to explain (and understand!).  I hope this at least made some sense to you...
It did.  A grad student in our lab said she has a book that has a whole chapter on T1 and T2 that she might bring in.  We do a lot of work with molecular magnets and MRI agents and measuring their relaxivities, so my professor wants us to have a thorough understanding behind the theory/physics surrounding T1 and T2.

I think I pretty much understand how the T2 "works", I'm having a little trouble visualizing it, but besides that.

Thanks again Yggdrasil!

Offline Yggdrasil

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Re: T1 and T2 Times
« Reply #3 on: July 31, 2008, 08:06:08 PM »
And this precessing/dephasing is just in the XY plane?  Since the magnetic field is along the Z-axis, do they start to dephase from all being aligned in the XY plane to dephase towards the Z axis as well?  But we're just interested in the level of dephasing in the XY plane.

An external magnetic field in the z-direction will exert a torque on magnetic dipoles with an xy-component, causing them to precess.  The magnitude of the torque is related to the cross-product of the external magnetic field and the magnetic moment, so when the external magnetic field and magnetic moment are both along the same axis, there will be no torque and your spin won't precess.

At the same time the dephasing, you will also see T1 decay (the gradual return of a signal in the xy-plane to the z-axis).

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Can you elaborate what you mean by the phrase "time constant of the decay"?

Any quantity that decays exponentially can be fit to the following equation:

X(t) = X(0)e-t/τ

As you can see from the equation, τ has units of time, so we call this the time constant for the decay (it can also be called a characteristic time for the decay.  Note that this value is also the reciprocal of the rate constant for the decay).  To see what this value represents physically, let's consider the value of X at time t = τ.

X(τ) = X(o)e-τ/τ = X(0)/e ~ 0.37X(0)

In other words, this is why the the time constant for any exponential decay is the time taken for 63% of the signal to decay.


If your friend's book doesn't have any good pictures, I'll see if I can try to scan some useful diagrams for explanation.  Having visuals definitely helps.

Offline macman104

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Re: T1 and T2 Times
« Reply #4 on: July 31, 2008, 08:15:48 PM »
At the same time the dephasing, you will also see T1 decay (the gradual return of a signal in the xy-plane to the z-axis).
Ah!  Ok, I suppose that makes sense.  I shouldn't have expected the two to exist in a vacuum from eachother.  That explains the diagram I was seeing (it was decaying in both the xy and the z direction
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Quote
Can you elaborate what you mean by the phrase "time constant of the decay"?
Any quantity that decays exponentially can be fit to the following equation:

X(t) = X(0)e-t/τ

As you can see from the equation, τ has units of time, so we call this the time constant for the decay (it can also be called a characteristic time for the decay.  Note that this value is also the reciprocal of the rate constant for the decay).  To see what this value represents physically, let's consider the value of X at time t = τ.

X(τ) = X(o)e-τ/τ = X(0)/e ~ 0.37X(0)

In other words, this is why the the time constant for any exponential decay is the time taken for 63% of the signal to decay.
Understood.
Quote
If your friend's book doesn't have any good pictures, I'll see if I can try to scan some useful diagrams for explanation.  Having visuals definitely helps.
Thanks for the offer.  I'll have to see what she brings in.  You've explained things very well so far.

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