# I T2 relaxation and Spin Echo of protons

1. Apr 4, 2017

### JohnnyGui

I’ve been reading on the physics of how a Magnetic Resonance Imaging (MRI) works from this link regarding the T1 and T2 relaxation time of protons.
So far I understand the concept except what they’re trying to explain in the last 2 paragraphs about spin echo.

This is what I got from these last 2 paragraphs, please correct me if I misunderstood something:
-The T2 is the time in which the magnetization vector in the XY plane has dropped by 63.2%.

- In reality however, there are several external factors (called fixed factors in the link) that make this T2 time differ even if it concerns the same amount of protons in a tissue. Such factors can be that the magnetic field differ in homogeinity between MRI’s, patients having some type of metal in them, etc. Thus, because of those external factors, waiting a T2 time doesn’t necessarily make the magnetization vector in the XY plane drop to 63.2%

- To be able to correct for these external factors, there’s a certain technique called the spin echo. Right after an RF pulse is turned off, one waits half a T2 time for the protons to relax/dephase. At ½ T2 time a 180 degrees RF pulse is given so that the magnetization vector of the protons again tips to the direction of the XY plane and rephase again, but now at the other side.
The time given for the protons to rephase at the other side of the XY plane is also ½ T2. In that time the protons rephase together again by that 180 degrees RF pulse, creating a magnetization vector at the other side, which is the spin echo.

Here are my questions regarding this:

1. Why does a MRI really need to know the exact real T2 time of the protons in a tissue to measure the magnetization vector and form an image? Isn’t it enough just to measure the rate per any other time unit in which a certain amount of protons relax? Since different tissues have different relaxation rates overall, T2 isn’t the only relaxation time that differs among them.

2. How does the spin echo actually determine the exact real T2 time by making the protons rephase together again at the other side of the XY plane? The ½ T2 time itself is also influenced by the external factors in the MRI. What makes waiting a ½ T2 time give the exact magnetization vector decrease for that ½ T2 time, without influence of the fixed external factors, while waiting a full T2 time won’t give the exact 63.2% decrease because of the external fixed factors?

Hope someone could enlighten me on this.

2. Apr 4, 2017

### Staff: Mentor

You don't need the exact T2 for most clinical purposes. Usually you just need a somewhat T2 weighted image.

Yes, most clinical exams also include T1 weighted images and often other contrasts as well

You misunderstood slightly. You wait 1/2 the TE time, not 1/2 the T2 time. Since there are always multiple tissues with different T2 it is not possible to do that for an entire image.

3. Apr 5, 2017

### JohnnyGui

What I meant is why does an MRI have to measure the time when the magnetizing vector has decreased (in the XY plane, T2) or increased (in the Z plane, T1) by 63.2% specifically. Doesn't any other fixed percentage have different time durations among tissues as well? Or is the 63.2% time duration difference among diffrent tissues more distinguishable than time durations at other percentages?
But even if it's another fixed TE time, doesn't that time get influenced by the same external factors as the T2 time? So that TE also differ among patients or certain MRIs even if it concerns the same type of tissue? I think I don't get what the advantage is of spin echo over determining the T2 time of tissues.

4. Apr 5, 2017

### f95toli

The 63.5% simply comes from the assumption that the decay is exponential; i.e. that T2 is a time constant (63.5%=1-1/e).
It is no different than say the decay time of a radioactive isotope.

Hence, the 63.5% is just a convention(albeit a very sensible one); there is nothing "special" about it, it just math.

Also, I think you've misunderstood the point of a spin-echo. it is a technique that it basically measures something "intrinsic" about the system; the re-focusing essentially compensates for the fact that the phase of the spins is randomized by external noise (in the simplest case).

5. Apr 5, 2017

### JohnnyGui

So one could say, for example, determine the time for any other fixed percentage and use that time as well to determine different tissues? The 63.2% is chosen just because the difference in time duration at that percentage is strong among different tissues?

That's exactly what I don't get. Why and how does re-focussing compensate for external noise?

6. Apr 5, 2017

### f95toli

No, it is not "chosen" as such. T2 is a time constant for an exponential decay(the function is proportional to exp^(-t/T2) ); and as such you will have that t=T2 when it has decayed to 63% of it full value (0.63=1-exp(-t/T2) when t=T2). This has absolutely nothing to do with MRI as such; the same convention is used or ALL physical processes that decay exponentially in physics and engineering.

I'm not sure there is an easy way to explain this in words. You basically have to think about it in terms of the Bloch sphere. Have you looked at the wiki?

7. Apr 5, 2017

### JohnnyGui

What I mean is that, an MRI distinguishes different tissues based on the difference in T2 among these tissues. So in theory, you could also let the MRI choose to determine the time at which a magnetizing vector in the XY plane decrease by 50% instead of 63.2% for example, since the time for 50% decay also differ among different tissues.

I'll look up the wiki and let you know if I understand it. Thanks!

8. Apr 5, 2017

### f95toli

Sure, but why would you? The T2 time constant is a measure of how quickly the system dephases and can in the simplest case be. determined by curve- fitting an exponential function to the time-domain data. Hence, you can use whatever percentage you want but it won't give you any more information (you are just multiplying T2 by some fixed constant), it all comes down the SHAPE of the curve, not a single value.
The only time it would make sense to change the "percentage" would be if you were using a system where T2 was actually measured by triggering at a single point (i.e. not by curve fitting but by extracting T2 by measuring a single point) but I seriously doubt any modern system is that crude; you would be throwing away a lot of data.

(for the record: I have no practical experience of MRI. However, I do work with ESR as well as artificial quantum systems and the physics is identical)

9. Apr 5, 2017

### Staff: Mentor

That isn't how it works. The T2 is a property of the tissue, and it varies from tissue to tissue within an image. On the MRI system what you set is the TE. At that TE some tissues will have decayed 50%, some will have decayed 80%, and some will have decayed 99%. All of those will be visible in the same image, and it is precisely that difference that is called "T2 weighting".

10. Apr 8, 2017

### JohnnyGui

Ah, so TE is a fixed time duration that MRI's use in general? I thought MRI's measure the time duration until the XY plane magnitizing vector has decreased by 63.2%.

If it's the TE that MRIs use, and TE duration is different from T2, why are corresponding images called "T2 weighed" images?

11. Apr 8, 2017

### Staff: Mentor

The TE is a user adjustable parameter, but once the user has set it then it will apply for all tissues in the image. You cannot set one TE for one tissue and another TE for another tissue.

That is not usually what you do, but it is possible. What you do is to acquire several images with different TE values. Then you look at the decay in the signal over time, and fit it to an exponential decay curve. The time constant of the exponential decay is the T2.

T2 weighted means that tissues with different T2 values have visibly different signal intensities.

12. Apr 9, 2017

### cosmik debris

I think in simple terms the spin echo technique cancels out the external factors by phasing the RF pulse 180 degrees. Say you're in a plane aiming for some island in the South Pacific and you have errors in your navigation system that make you miss by 10% to the right, if you fly the reverse way around the Earth you miss by 10% to the left, the average puts you on target canceling the error in the nav equipment. That may not be a completely accurate analogy but...

Cheers

13. Apr 13, 2017

### JohnnyGui

That analogy makes sense, but I can't really implement it in the spin echo technique.
Here's why;
So the spin echo first let's protons dephase for half a TE time, then after that 1/2 TE time a 180 RF pulse is emitted and another 1/2 TE is given to let the protons rephase again on the other side. The first 1/2 TE time in which the protons dephase is influenced by external noise. The second 1/2 TE in which the protons rephase is also influenced by external noise.

If you now choose a different MRI with a different amount of external noise for the very same tissue as before, then the protons will get influenced differently during their de- and rephasing. The result; waiting the same 1/2TE's time will lead to a different amount of dephase and rephasing.

14. Apr 13, 2017

### Staff: Mentor

But you don't care if the amount of dephasing and rephasing is different on a different system. You only care that the amount of rephasing is equal to the amount of dephasing, regardless of how much it is.

15. Apr 13, 2017

### JohnnyGui

I see. Sorry for my stubborness but I still can't grasp how the external noise during rephasing cancels out the external noise during dephasing.

Here's what I understand:
The first 1/2 TE includes the influence of external noise along with the pure dephasing time, so that: $\frac{1}{2}TE = \frac{1}{2} (T_{dephase} + T_{noise})$.
The second 1/2 TE includes the pure rephasing time along with the influence of external noise, so that: $\frac{1}{2}TE = \frac{1}{2} (T_{rephase} + T_{noise})$
If I add both 1/2 TE's up, I'll end up with $TE = T_{rephase} + T_{dephase} + T_{noise}$. How can the $T_{noise}$'s cancel each other out?

16. Apr 25, 2017

### JohnnyGui

Anyone who could answer this question for me?

17. Apr 25, 2017

### Staff: Mentor

I don't know where you got this expression from, but I wouldn't describe it that way at all. I cannot answer your question directly because the basis for the question just seems wrong.

Here is how I describe it when I am teaching the subject:

We assume that the field is completely uniform, at least over a single voxel. By the Larmor equation that means that all of the spins precess at the same frequency. If they all precess at the same frequency then they are phase coherent, meaning that their relative phase stays the same. That is the ideal assumption.

The reality is different. Due to each proton's individual micro environment they experience a slightly different field. By the Larmor equation that slight difference in field means a slight difference in frequency which, over time, leads to a slightly different phase and therefore a loss of phase coherence.

You can roughly split the field deviations into two categories: those that are non random and those that are random. An example of the non random ones would be small constant local variations in the field due to susceptibility from nearby objects. An example of the random ones would be small variations in field due to thermal motion towards and away from other spins.

Does this make sense so far?

More later...

18. Apr 25, 2017

### Staff: Mentor

So the total effect on the signal is a loss of phase coherence that is driven by both random and non random processes. There exist RF pulses which reverse the phase.

So if you have a constant frequency offset of say 1 Hz then you will lose 90 deg of phase coherence in 250 ms. If you flip that phase to -90 deg, and then let it continue to precess with the same frequency offset then in another 250 ms it will be back to 0 phase.

Similarly, if you have a constant frequency offset of 0.5 Hz then over the 250 ms it will dephase by 45 deg. The RF pulse flips that to -45 deg and again it will be back to 0 phase after another 250 ms.

So it doesn't matter how much the frequency offset is, all that matters is that it is constant in time. By allowing it to dephase over TE/2, inverting the phase, and allowing it to rephase over another TE/2, any constant rate of dephasing is corrected.

This is the spin echo phenomenon. Note that it does not correct any of the random sources of dephasing. That is T2 decay, and it is not recoverable by any known technique. It also does not recover sources of dephasing which are not random, but also not constant in time (there are other techniques to do that).

19. Apr 25, 2017

### JohnnyGui

Great expanation! I think I've got it now. So by flipping you're basically giving for example a "slower" precessing proton a head start to let the faster precessing proton approach it and eventually rephase again. I know that this description is very simplistic but it kind of helps me understand it.

I think my confusion was right there. So by flipping the protons with an RF pulse you are left with a dephasing rate that is determined by random processes (and non-random processes that are not constant in time)?

However, this leads me to the following question; why not just determine the dephasing rate without flipping it? If you're using the same system, external noise won't really matter would it? Or is it because the amount of external noise differ in certain areas of the same system?

20. Apr 25, 2017

### Staff: Mentor

You can do that, it is called T2*. It includes both the systematic and also the random dephasing effects.