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What's the physical difference between the quantities $\left\langle v_{i}v_{j}\right\rangle $ and $\left\langle v_{i}\right\rangle \left\langle v_{j}\right\rangle $?

Where $\left\langle v_{i}\right\rangle$ is the mean value of the $v_{i}$ velocity component with $i=1,2,3$.

(Same story with $v_{j}$.)

This is in the context of stars' velocities in galaxies.

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2 Answers

up vote 6 down vote accepted

think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result.

When the functions are arbitrary, the average of the product quantifies statistical correlation between the two functions/variables. This correlation gives a rough measure of how causally connected are both of the variables.

This correlation information is completely lost when one takes the averages of the functions individually.

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I would say that $\left\langle v_i \right\rangle\left\langle v_j \right\rangle$ is the product of averages, whereas $\left\langle v_i v_j \right\rangle$ is the average of a product. Hope you get the difference.

Take an example of $v_i = v_j = v$, then $\left\langle v_i \right\rangle\left\langle v_j \right\rangle = \left\langle v \right\rangle^2$, while $\left\langle v_i v_j \right\rangle = \left\langle v^2 \right\rangle$. It should be trivial now, that the quantities are not the same. lurscher's example also confirms this. You can find another example in statistics: consider a dataset of \begin{array}{cccccc} v & 0 & 1 & 1.5 & 0.75 & 0.5 \end{array} then $\left\langle v \right\rangle = (0+1+1.5+0.75+0.5)/5 = 0.95$, hence $\left\langle v \right\rangle^2 = 0.9025$, but $\left\langle v^2 \right\rangle = (0+1+2.25+0.5625+0.25)/5 = 0.8125$, which is by no mean equal to $0.9025$.

This was pure mathematical explanation, and there is no difference, which functions or statistical quantities you consider. Hope that now you will be able to explain yourself the physical difference of particular quantities you are working with.

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