Effect of 180°-Pulse

Why doesn't the 180° refocusing pulse also invert the longitudinal magnetization in addition to flipping over the spins in the transverse plane?  

  

The 180°-pulse flips over both the transverse and longitudinal components of magnetization.  However, the 180°-pulse is usually applied within a few milliseconds (specifically, at time TE/2) after the longitudinal magnetization has been entirely tipped into the transverse plane by the 90° pulse.  Thus at the time the 180°-pulse is applied, there is often little or no component of longitudinal magnetization left to be inverted.  As a first approximation, therefore, the 180°-pulse appears to be acting only upon the transverse components of magnetization.

The 180°-refocusing pulse in a SE sequence also inverts the small amount of longitudinal magnetization (M_z) that has developed in the interpulse interval. This detail is frequently omitted from most SE diagrams for simplicity.

This small effect cannot be entirely neglected, however. It will definitely affect signal strength and contrast in long TE sequences (where the interpulse time is long) and for tissues with short T1 values (where Mz regrows quickly).

Advanced Discussion (show/hide)»

In more complete analyses of the spin echo pulse sequence, one does need to account for the small amount of regrowth of longitudinal magnetization that occurs in the time interval between the 90° and 180° pulses. As we show in a later Q&A, the classic formula predicting the signal intensity (S) from a stationary tissue with spin density (ρ) and relaxation times (T1 and T2) subjected to a SE sequence with repetition time (TR) and echo time (TE) is given by

S = k ρ (1 − e^−TR/T1) e^−TE/T2

where k is a scaling factor. This equation is an approximate solution to the Bloch equations neglecting the inversion of the small amount of longitudinal magnetization that regrows in the interpulse interval. The complete equation, taking into account the longitudinal inverting property of the 180° pulse, is written

S = k ρ (1 − 2e^{−(TR−TE/2)/T1} + e^−TR/T1) e^−TE/T2

For TR>>TE/2, this equation reduces to the more simplified one above.

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References
     Elster AD, Burdette JH. Questions and Answers in MRI, 2nd ed. St. Louis: Mosby, 2001, pp 51-2.

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