Introducing MRI: Diffusion Imaging (49 of 56)

www.einstein.yu.edu - The forty-ninth chapter of Dr. Michael Lipton's MRI course covers Diffusion Imaging . Dr. Lipton is associate professor radiology at Albert Einstein College of Medicine and associate director of its Gruss Magnetic Resonance Research Center.

[@danielburritt5960]

I would pay so much money for this course! I can't believe this is free on youtube!

[@heshamtemraz603]

Actually you can read a lot of academic books, pdfs and articles about this topic BUT it is impossible to understand it as presented in this video in simple, clear and coherent manner. Thank you Dr Lipton for the incredible lectures that offered for radiologist and technologist all over the world by free.

[@incineration242]

WOW I wish all residency lectures were recorded like this.

[@sadafshahzad9983]

An amazing lecture. Thank you very much Sir for clarifying this with such simplicity and yet to such great details!

[@ridafatima1711]

you sir are amazing- thank you so so so much for making this widely available

[@riznafathima7991]

I can't thank you enough for creating this amazing content

[@luissilva6499]

thanks Dr. Michael ! Luís

[@h1h2h3h4jo]

46:15 I was discussing T2 Shine through with my attending today. He argued that a T2 shine through situation would always result in a high ADC. I am not sure thats true. My understanding is that ADC is merely a coefficient and contains no T2 information. Tissue which has high T2 signal but NO actual diffusion restriction or facilitation relative to normal brain should have just the same ADC value as normal (non high T2 signal) brain. Any help?

[@articuno1508]

Could this be due to the fact discussed later in the video, that it is a 2-point approximation of an exponential process? A tissue with higher native T2 signal would lose more signal, due to the exponential decay, than a tissue with lower native T2 signal. That leads me to believe that the 2-point approximation would overestimate the ADC based on the apparently higher slope (and proportionally to native T2 signal).

[@h1h2h3h4jo]

@@articuno1508 That makes a lot of sense to me, thank you so much for your reply!

[@user-jm4rt1kv8d]

Thank yoy Professor!

[@mjyoosefi5503]

ok