# The relationship between the diffusion coefficient of unit "sqrt(cm)" and "cm^2/s"

Dear experts,

The unit of transverse (or longitudinal) diffusion coefficient obtained from Garfield++ is “sqrt(cm)”, if the value is a, which means charges diffusion distance of a (um) per sqrt(d (cm)). In general, the unit of diffusion coefficient is cm^2/s, means charges diffusion area of cm^2 per second, caused by concentration gradient.
Could them be converted into each other？
Thanks very much for your help.

Best regard,
Jiechen

Hi,
thanks for your message. This is indeed a frequent source of confusion. See this explanation on the Q&A page.

@ hschindl
Thanks for your help, from my calculation, the equation is D=d^2/2*v, where the unit of D is cm^2/us, unit of d is sqrt(cm) from Garfield++, v is velocity (cm/us). However, I still don’t understand the derivation of coefficient value “2” (or “sqrt(2)”).
May I ask another question? If growing is not Gaussian distribution (Dt (transvers diffusion) is not equal to Dl (longitudinal diffusion)), how to express the diffusion coefficient of any position? I try to find some papers, the expression is a tensor ({D_xx,D_xy},{D_yx, D_yy})(suppose 2-D geometry), but I don’t know the concrete expression of the components.

Best regards,
Jiechen

You can find some background on the tensor description in this note by Rob Veenhof:
https://garfield.web.cern.ch/notes/diffusion.ps

This formalism has not (yet) been implemented in Garfield++ though.

Dear hschindl,

Thanks very much for your help.

Best regards,
Jiechen

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