## Is `T`_{2} necessarily shorter than `T`_{1}?

If you are familiar with MRI (or NMR in general), then probably also with the relaxation time constants `T`_{1} and `T`_{2}. These tissue-specific (or substance-specific) constants describe how fast the nuclear magnetization returns to its equilibrium value, `M`_{0}, after excitation by a pulsed radio-frequency (rf) field. Shortly summarized, `T`_{1} describes the exponential recovery of the longitudinal magnetization `M`_{L} (i. e., of the magnetization parallel to the external static magnetic field `B`_{0}), and `T`_{2} describes the exponential decay of the transverse magnetization `M`_{T} (that is precessing in the plane orthogonal to `B`_{0}).

Typically (as shown in the first figure), `T`_{2} values of tissue are considerably lower than `T`_{1} values, i. e., the transverse magnetization decays quicker than the longitudinal relaxation needs for recovery. For most tissues *in vivo*, `T`_{1} varies between about 300 ms and 3 s, while `T`_{2} varies between about 10 ms and 200 ms. Longer `T`_{2} relaxation times (up to about 3 s as well) are found for liquids.

So one may ask if there are good physical reasons for `T`_{2} values being shorter than or – at most – equal to `T`_{1}. As a physicist, I’d start with checking some extreme cases, e. g., assuming that `T`_{2} is *much* longer than `T`_{1}, i. e. `T`_{2} ≫ `T`_{1}. Then the longitudinal magnetization can fully recover while at the same time some transverse magnetization would be preserved. As a result, the magnitude of the total magnetization (`M`_{L}^{2} + `M`_{T}^{2})^{0.5} would become greater than the equilibrium value `M`_{0} – which is physically impossible.

To analyze these properties of `T`_{1} and `T`_{2} in more detail, longitudinal and transverse relaxation can also be plotted together in a diagram showing the transverse magnetization on the horizontal axis and the longitudinal magnetization on the vertical axis. These diagrams show the evolution of the magnetization (for experts: in a rotating frame of reference) after a 90° rf pulse (all trajectories start at the lower right corner of the diagram). The left-hand side of the following figure shows three cases `T`_{2} = `T`_{1}/2 (blue), `T`_{2} = `T`_{1} (green), and `T`_{2} = 2 `T`_{1} [sic!] (cyan); and all three curves show a “benign” behavior in that they lie in the shaded area below the black circle segment `M`_{L}^{2} + `M`_{T}^{2} = `M`_{0}^{2}. This means that the magnitude of the total magnetization vector is always smaller than `M`_{0}.

However, the right-hand side of this figure shows what’s happening if `T`_{2} becomes greater than 2 `T`_{1} – in this case, `T`_{2} = 3 `T`_{1} (red curve): Now the magnitude of the total magnetization vector increases above the physical limit of `M`_{0}, i. e., the red line crosses the black border of physically benign behavior!

In fact, it can be shown (see appendix if interested) that the maximum `T`_{2} value, for which the red curve stays always below the black line, is exactly `T`_{2} = 2 `T`_{1}. And as so often, almost everything that is physically possible is also realized in nature (although the case `T`_{1} < `T`_{2} < 2 `T`_{1} is really extremely rare), as described by Malcolm H. Levitt in his highly recommendable NMR text book “Spin dynamics” (2nd ed., section 11.9.2, note 13):

The case whereT_{2}>T_{1}is encountered when the spin relaxation is caused by fluctuating microscopic fields that are predominantly transverse rather than longitudinal. One mechanism which gives rise to fields of this form involves theantisymmetric component of the chemical shift tensor(not to be confused with the CSA). […] Molecular systems in which this mechanims is dominant are exceedingly rare (see F. A. L. Anet, D. J. O’Leary, C. G. Wade and R. D. Johnson,Chem. Phys. Lett.,171, 401 (1990)).

So, the answer to the title question is: No, `T`_{2} can in fact be greater than `T`_{1} in very special circumstances, but it can never be greater than 2 `T`_{1}.

### Appendix

The maximum `T`_{2} value, for which the red curve stays always below the black line, is exactly `T`_{2} = 2 `T`_{1}. This can be seen by analyzing the inequality

`M`

_{T}

^{2}+

`M`

_{L}

^{2}= (

`M`

_{0}exp(–

`t`/

`T`

_{2}))

^{2}+ (

`M`

_{0}(1 – exp(–

`t`/

`T`

_{1}))

^{2}≤

`M`

_{0}

^{2}.

First, we divide by `M`_{0}^{2} and set `T`_{2} = `α` `T`_{1} as well as `β` = exp(–`t`/`T`_{1}), yielding

`β`

^{1/α})

^{2}+ (1 –

`β`)

^{2}=

`β`

^{2/α}+ 1 – 2

`β`+

`β`

^{2}≤ 1

which is (after subtraction of 1 and division by `β`)

`β`

^{2/α – 1}– 2 +

`β`≤ 0 or

`β`

^{2/α – 1}≤ 2 –

`β`.

`β` is by definition (for positive `t`) between 0 and 1, so the right-hand side of the last inequality is a linear function descending from 2 to 1 (i. e. always ≤ 2). Its left-hand side has very different shapes depending on `α`: it is increasing from 0 to 1 for 0 < `α` ≤ 2 (since then the exponent 2/`α` – 1 ≥ 0); but it is going to infinity for `β` → 0 if `α` > 2 (since then the exponent 2/`α` – 1 < 0). So, the last inequality will not hold in the latter case for sufficiently small values of `β`, which means that non-physical behavior occurs if `α` > 2 or, using the definition from above, if `T`_{2} > 2 `T`_{1}.

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URL: <https://dtrx.de/od/mmm/mmm_20180117.html>