# Finding turbulent velocity Fourier mode amplitudes from kinetic energy spectrum

A random vector field, such as a turbulent flow, can be decomposed into Fourier modes. Taking a snapshot in time (say an initial condition) we have that the randomly fluctuating component of the flow can be described by the sum of fourier modes as follows:

\begin{equation} \mathbf{u}(\mathbf{x}) = \sum_\mathbf{k} \hat{\mathbf{u}}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x} } \end{equation}

where $\mathbf{k}$ is the wavenumber vector and $\mathbf{u} = (u_1,u_2,u_3)$ and $\hat{\mathbf{u}}$ is the amplitude of one Fourier mode.

I now want to use a simple model of a turbulent kinetic energy spectrum that can be found in reality (for example the turbulence that develops in the boundary layer of a duct), and from this turbulent spectrum deduce the amplitudes of the 3D Fourier modes.

As an example: One model spectrum I have encountered is that in Pope- Turbulent Flows pg. 232:

\begin{equation} E(\kappa) = C\epsilon^{2/3}\kappa^{-5/3}f_L(\kappa L)f_{\eta}(\kappa \eta) \end{equation}

where

\begin{equation} f_L(\kappa L) = \left ( \frac{\kappa L}{((\kappa L)^2+c_L)^{1/2}} \right )^{5/3+p_0} \end{equation}

\begin{equation} f_{\eta}(\kappa \eta) = \exp(-\beta((\kappa \eta)^4+c_\eta^4)^{1/4}-c_{\eta}) \end{equation}

Where $\kappa = \mathbf{|k|}$ and there are a bunch of constants I have not defined, but are available in the literature. Now I know that $\frac{1}{2}<u_iu_i> = \int_0^{\infty} E(\kappa) d\kappa$ i.e. the area under the graph is equal to the turbulent kinetic energy and the area of a bin centered at a $\kappa$ is the energy at that wavenumber, but I am not sure how I go about determining the amplitudes of the Fourier modes from this.

• I am confused. You are starting with the energy spectrum and looking for the Fourier coefficients (Note: $\hat{\mathbf{u}}_{k}$ is not necessarily a mode of the system for real data)? In any case, you should look at the reference in my next comment... – honeste_vivere Jan 29 '15 at 23:46
• J.C. Samson and J.V. Olson "Some comments on the descriptions of the polarization states of waves," Geophys. J. R. astr. Soc. 61, pp. 115-129, 1980. – honeste_vivere Jan 29 '15 at 23:48
• @honeste_vivere I want to decompose a velocity field with a given energy spectrum as above into a sum of fourier modes. Would this not be possible? – Dipole Feb 2 '15 at 19:43
• What do you already know, $< u_{i} \ u_{j} >$ or the velocity field components? If you know $E(\kappa)$, then you already have the power as a function of wavenumber. Remember, you do not need to care about the $e^{ikx}$ as you use the magnitude of the transform in a spectrum. So unless I am completely misunderstanding your goal, I think $E(\kappa_{i})$ will be your Fourier amplitude at wavenumber $\kappa_{i}$. – honeste_vivere Feb 3 '15 at 0:08
• @honeste_vivere Thanks for your reply. I am assuming I have a model $E(\kappa)$ available as above, and from this spectrum I want to deduce the amplitudes of the fourier series which constitute the unsteady, turbulent part of my flow. Could you perhaps show me how you arrive at the fact that the $E(\kappa_i)$ is the amplitude at wavenumber $\kappa_i$? I would accept that answer. – Dipole Feb 3 '15 at 17:01

$$\mathbf{u}(\mathbf{x}) = \sum_{i} \hat{\mathbf{u}}(\mathbf{k}_{i}) e^{i \ \mathbf{k}_{i} \cdot \mathbf{x}}$$ which has the inverse given by: $$\hat{\mathbf{u}}(\mathbf{k}) = \sum_{j} \mathbf{u}(\mathbf{x}_{j}) e^{- i \ \mathbf{k} \cdot \mathbf{x}_{j}}$$ Therefore, you can see, in one dimension, that: $$E(\kappa) = \sum_{j} E(x_{j}) e^{- i \ \kappa \ x_{j}}$$ or, in the form you are looking for: $$E(x) = \sum_{i} E(\kappa_{i}) e^{i \ \kappa_{i} \ x}$$ Thus, the values $E(\kappa_{i})$ represent the Fourier amplitudes at wavenumbers $\kappa_{i}$ (within a factor of $2 \ \pi$, depending on choice). This is explicitly discussed in the introduction section of the Wikipedia page for Fourier transforms.
• I'm not sure I'm following this- you are saying $E(x)$ is the dual of $E(k)$ and that therefore implies that since $u(x)$ is the dual of $u(k)$ then $u(k_i) = E(k_i)$? – Dipole Feb 4 '15 at 16:50
• I am not sure what you mean by "dual" here, but I am only showing the relationships between Fourier transforms and illustrating that the amplitudes are independent of the exponent term. Thus, you already have the amplitudes in the form of $E(\kappa)$. – honeste_vivere Feb 4 '15 at 21:24
• Sorry I mean dual as in the fourier transform pair $E(k)$ and $E(x)$. I get that $E(k_i)$ are the Fourier amplitudes of the function $E(x)$, but what I want are the Fourier amplitudes $\hat{u}(k)$ of the function $u(x)$, given knowledge of $E(k)$. You seem to be suggesting that $\hat{u}(k) = E(k)$? – Dipole Feb 5 '15 at 13:57
• I am not entirely sure what you are asking but I think the answer is yes. I just used subscripts for $x$ and $k$ as a matter of choice, but could have just as easily summed over $k$. – honeste_vivere Feb 5 '15 at 17:30