# In turbulence theory, what happens if i take space average of fluctuating part?

According to Reynolds decomposition, velocity field is split into two time average and fluctuating parts:

$$u_{({\bf x},t)}=\overline u_{(\bf x)}+u'_{({\bf x},t)}$$

we know that time average of fluctuating part is zero $\overline u'=0$, but what if i take space average of fluctuating part?

• Do you want: 1) the space average of the fluctuating parts in the time-average decomposition or 2) the space average of the fluctuating part in a space average decomposition. I understand that you want 1) but looking at the other answer I'm not sure it is clear. – ucsky Aug 23 '13 at 15:54

You may define the space average over some domain in exactly the same manner as the time average. Call it $[U(t)]$. Then a Reynolds-like decomposition is always possible: $$U(x,t) = [U(t)] + U'(x,t),$$ where $U'(x,t)$ is the fluctuation around the spatial average of the field.
$$[U(x,t)] = [[U(t)]] + [U'(x,t)] \rightarrow [U(t)] = [U(t)] + [U'(x,t)] \rightarrow [U'(x,t)] = 0$$
Starting from the original statistical Reynolds decomposition where $\mathbf{u}$ is a random field, where $<...>_s$ denote the statistical average and where $\mathbf{u}^{(s)}$ is the random fluctuation field, it come the decomposition: \begin{equation} \mathbf{u}(\mathbf{x},t)=<u>_s(\mathbf{x},t)+\mathbf{u}^{(s)}(\mathbf{x},t) \end{equation} In your question it's look like you assume that the turbulence is stationary and that the ergodic hypothesis hold allowing to assimilate the statistical average to a time average and resulting of the decomposition: \begin{equation} \mathbf{u}(\mathbf{x},t)=<u>_t(\mathbf{x})+\mathbf{u}^{(t)}(\mathbf{x},t) \end{equation} where $<u>_s(\mathbf{x},t)=<u>_t(\mathbf{x})$ and the time average is defined by \begin{equation} <...>_t= \lim_{T \to +\infty} \frac{1}{T}\int_{t_0}^{t_0+T} ... dt \end{equation} Defining the spatial average over a bounded domain $\mathcal{D}$ of volume $V$: \begin{equation} <...>_{\mathbf{x}}=\frac{1}{V}\int_{\mathcal{D}} ... d\mathbf{x} \end{equation} For reply to your question, the space average of the previously defined fluctuating part is a random function of time $\mathbf{U}$ expressed \begin{equation} <\mathbf{u}^{(t)}(\mathbf{x},t)>_{\mathbf{x}}=\mathbf{U}(t)\in \mathbb{R}^3 \end{equation} Since $\mathcal{D}$ is bounded \begin{equation} lim_{T \to +\infty} \frac{1}{T}\int_{t_0}^{t_0+T} \frac{1}{V}\int_\mathcal{D}... d\mathbf{x} dt = \frac{1}{V} \int_\mathcal{D} ( lim_{T \to +\infty} \frac{1}{T}\int_{t_0}^{t_0+T} ... dt )d\mathbf{x} \end{equation} And it's result the property for $\mathbf{U}$: \begin{equation} <\mathbf{U}(t)>_t=\mathbf{0} \end{equation}