Derivation of Karman-Howarth equation

TurbJet · March 13, 2022, 03:00

Greetings,

In the derivation of Karman-Howarth equation, a common assumption that being used is that homogeneity leads to the following equivalency:

$\frac{\partial}{\partial x_i} = \frac{\partial}{\partial r_i}\;,\;\frac{\partial}{\partial x_i'} = -\frac{\partial}{\partial r_i}$

where

being the spatial displacement.

(see e.g., the book by Davidson: Turbulence: An Introduction for Scientists and Engineers, on p.308)

Equivalently,

$\frac{\partial}{\partial x_i}\langle\cdot\rangle = \frac{\partial}{\partial r_i}\langle\cdot\rangle\;,\;\frac{\partial}{\partial x_i'}\langle\cdot\rangle = -\frac{\partial}{\partial r_i}\langle\cdot\rangle$

I really can't understand how this is arrived. Could anyone provide some fresh perspective?

Thanks.

FMDenaro · March 13, 2022, 05:31

Could you post the pages of Davidson?

I can remind the derivation proposed by Pope (he uses the reference of Hinze and Monin&Yaglom).

I remember the assumption of isotropy, thus independence of the statistics to any rotation.

TurbJet · March 13, 2022, 21:10

Quote:

Originally Posted by FMDenaro

Could you post the pages of Davidson?

I can remind the derivation proposed by Pope (he uses the reference of Hinze and Monin&Yaglom).

I remember the assumption of isotropy, thus independence of the statistics to any rotation.

Here it is; it is mentioned in point (ii).

davidson.png

I've seen a couple of authors made the same statement, including in Frisch's 1995 book.

March 13, 2022, 03:00	Derivation of Karman-Howarth equation	#1
TurbJet Senior Member Join Date: Oct 2017 Location: United States Posts: 233 Blog Entries: 1 Rep Power: 9	Greetings, In the derivation of Karman-Howarth equation, a common assumption that being used is that homogeneity leads to the following equivalency: $\frac{\partial}{\partial x_i} = \frac{\partial}{\partial r_i}\;,\;\frac{\partial}{\partial x_i'} = -\frac{\partial}{\partial r_i}$ where being the spatial displacement. (see e.g., the book by Davidson: Turbulence: An Introduction for Scientists and Engineers, on p.308) Equivalently, $\frac{\partial}{\partial x_i}\langle\cdot\rangle = \frac{\partial}{\partial r_i}\langle\cdot\rangle\;,\;\frac{\partial}{\partial x_i'}\langle\cdot\rangle = -\frac{\partial}{\partial r_i}\langle\cdot\rangle$ I really can't understand how this is arrived. Could anyone provide some fresh perspective? Thanks.

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March 13, 2022, 05:31		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	Could you post the pages of Davidson? I can remind the derivation proposed by Pope (he uses the reference of Hinze and Monin&Yaglom). I remember the assumption of isotropy, thus independence of the statistics to any rotation.