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I am looking for a clear definition for the power spectrum of a vector field. In the literature I've found this expression:

$$ \langle \tilde{\boldsymbol{v}}(\boldsymbol{k}) \cdot \tilde{\boldsymbol{v}}^\ast (\boldsymbol{k'}) \rangle = (2 \pi)^3 \delta^3 (\boldsymbol{k} - \boldsymbol{k'}) P_{\boldsymbol{v}}(k), $$

where $\boldsymbol{v}$ is a vector field defined in a 3D space and $\tilde{\boldsymbol{v}}$ the corresponding Fourier transform.

According to this, the power spectrum should be: $$ P_{\boldsymbol{v}}(k) = \tilde{v}_1^2+\tilde{v}_2^2+\tilde{v}_3^2, $$ where $\tilde{v}_1, \tilde{v}_2, \tilde{v}_3$ are the components of the vector $\tilde{\boldsymbol{v}}(\boldsymbol{k})$ and $k$ is the modulus of the wavenumber vector $\boldsymbol{k}$.

  1. Is this the correct definition?
  2. Is it different from the power spectrum of the scalar field given by the magnitude of the vector field in the real space?
  3. Are there any tools (in python or whatever) to perform a power spectrum of a vector field? I've been recently using the nbodykit library, but it is only designed for scalar fields.

Thank you for you help.

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1 Answer 1

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  1. Is this the correct definition?

It is one of the two power spectra for a statistically isotropic vector field. A more general expression is $$\langle v_i(\vec k) v_j^*(\vec k^\prime)\rangle\equiv (2\pi)^3\delta^3(\vec k-\vec k^\prime)P_{ij}(\vec k),$$ where $v_i$ is the $i$th component of $\vec v$, within some coordinate system. Note that your expression is just the trace of $P_{ij}(\vec k)$. In principle, $P_{ij}(\vec k)$ could also have 5 more degrees of freedom (a total of 6 since it is a symmetric matrix).

But if $\vec v$ is statistically isotropic, then $P_{ij}$ must be the sum of a term proportional to the Kronecker delta $\delta_{ij}$ and a term proportional to the outer product $k_i k_j$. This is because, in the absence of any preferred direction, the wavenumber $\vec k$ is the only vector that could possibly enter the expression for $\langle v_i(\vec k) v_j^*(\vec k^\prime)\rangle$.

One possibility is to write the decomposition as $$P_{ij}(\vec k)=\frac{\delta_{ij}}{3}P_1(k)+\left(\frac{k_i k_j}{k^2}-\frac{\delta_{ij}}{3}\right)P_2(k).$$ Then $P_1(k)$ is the power spectrum you gave, which describes whether vectors at different locations tend to be aligned (or anti-aligned) with each other. $P_2(k)$ accounts for residual tendency to align (or not) with the separation vector itself.

  1. Is it different from the power spectrum of the scalar field given by the magnitude of the vector field in the real space?

Yes, that would be $$\left\langle \sqrt{v_1(\vec k)^2+v_2(\vec k)^2+v_3(\vec k)^2} \sqrt{v_1^*(\vec k^\prime)^2+v_2^*(\vec k^\prime)^2+v_3^*(\vec k^\prime)^2}\right\rangle.$$

  1. Are there any tools (in python or whatever) to perform a power spectrum of a vector field? I've been recently using the nbodykit library, but it is only designed for scalar fields.

You could take the (cross-) power spectra of the scalar fields $v_i$ and $v_j$ to construct $P_{ij}$, and then decompose it into the sum of a term proportional to the identity matrix (the $P_1$ term) and a traceless matrix (the $P_2$ term). But note that you want to avoid spherically averaging $P_{ij}$.

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  • $\begingroup$ Thank you very much @Sten for your great answer. I just need you to make two points more clear: 1) what is the difference when you move from $P(k)$ and $P(\vec k)$? Is $k$ the modulus of $\vec k$? 2) What do you mean for spherically averaging $P_{ij}$? Thanks. $\endgroup$
    – Wil
    Jul 10, 2023 at 13:22
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    $\begingroup$ @Wil The power spectrum of a statistically isotropic field shouldn't depend on direction, so in principle $P(\vec k)=P(k)$, where $k=|\vec k|$. Thus you can suppress the statistical variance in the power spectrum by averaging over the sphere. This is the standard thing to do. But $P_{ij}$ isn't necessarily statistically isotropic (as is evident from how I decomposed it). Only $P_1$ and $P_2$ are, so you don't want to spherically average until after decomposing. $\endgroup$
    – Sten
    Jul 10, 2023 at 15:59
  • $\begingroup$ Thank you @Sten. Do you have references for these definitions? $\endgroup$
    – Wil
    Jul 10, 2023 at 16:10
  • $\begingroup$ @Wil Sorry, I've only encountered this situation in research and have never been able to find a reference... $\endgroup$
    – Sten
    Jul 10, 2023 at 16:13
  • $\begingroup$ I see. One last question: if $\vec v$ is isotropic, what does the decomposition you wrote reduce to? $\endgroup$
    – Wil
    Jul 10, 2023 at 16:19

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