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I'm analysing a bunch of simulated galaxies, and one of the properties I'm looking at is their velocity dispersion (which is the same thing as the standard deviation of their speeds as far as I know).

Right now I'm comparing the values for velocity dispersion in the rest frame of the simulation with that found in the rest frame of the galaxy. Both our inertial frames and I would expect the values to be the same, but to my surprise I'm finding the velocity dispersion is HIGHER in the galaxies CoM frame.

So my question is how would you expect velocity dispersion to vary with a change of inertial reference frame?

n.b. all frames should be non-relativistic, but see below for a very good answer for the case when frames are relativistic

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2 Answers 2

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This is to be expected from the relativistic velocity addition formula, which has a way of "compressing" differences in speed.

Consider the simple case of two stars with velocities $u_1$ and $u_2$ in the galaxy's rest frame. (For definiteness let's say both $u_1$ and $u_2$ are positive.) Their speeds $s_1$ and $s_2$ relative to an observer moving with velocity $v$ relative to the galaxy are $$ s_1 = {v + u_1 \over 1 + vu_1/c^2 }, \hbox{ } s_2 = {v+u_2\over 1 + vu_2/c^2}.$$

The question can then be asked: Is $s_2 - s_1 < u_2 - u_1$?

If you evaluate $s_2-s_1$ you'll find it's equal to $$ (u_2 - u_1){ 1 -v^2/c^2 \over \left( 1 + vu_2/c^2 \right)\left( 1+vu_1/c^2 \right)}$$.

The numerator in that fraction is less than 1, whereas the denomintor is greater than 1. Thus indeed $s_2 - s_1 < u_2 - u_1$.

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  • $\begingroup$ I agree this would explain the phenomenon, thanks! but as wolphram jonny surmised I'm in a non-relativistic regime $\endgroup$
    – zephyr
    Commented Nov 27, 2014 at 15:37
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    $\begingroup$ Oh ok! When I hear "velocity dispersion" and "galaxies" I think "doppler shift" and thus relativity. Note that even if $u_1$ and $u_2$ are small ("nonrelativistic") it's pretty common for $v$ to be large, e.g. redshifts of 1.5 and higher, and this velocity-addition rule still applies. You might consider including this in your code for brownie points. ;-) Also, you could compute that fraction using the data in your simulations and see how close it is to 1, for a measure of how good your Newtonian approximation is working! Good luck. One other handy check: Is angular momentum being conserved? $\endgroup$
    – sh37211
    Commented Nov 27, 2014 at 16:54
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No, the velocity dispersion cannot change because you are always using a diference of velocities to compute the dispersion, and the difference does not change when you change to another intertial frame because all velocities are added the same speed and cancels. If you want you can post the code to see where is the error. Assuming you use $\sigma^2=\Sigma|v_i- v_{average}|^2$ with v the vectors, I dont know to to put the upper arrow

Note: I assume you are using nonrelativistic approach, otherwise see the other answer

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  • $\begingroup$ so I would definitely agree with this in 1 dimension, when velocity and speed are effectively interchangeable, but seeing as we find the relative velocities in any one of the three cartesian dimensions by subtracting off the CoM velocity from the original velocity then it seems like sum(v^2 - v(com)^2) will differ from sum|v(rel)| where the sums are over all stars in the galaxy (n.b. sorry for the awful notation but can't do LaTeX in comments can I?) $\endgroup$
    – zephyr
    Commented Nov 27, 2014 at 15:32
  • $\begingroup$ there are different ways to define it, but if you use your definition onviously it will differ. $\endgroup$
    – user65081
    Commented Nov 27, 2014 at 15:49
  • $\begingroup$ What other ways are there to define it? As far as I can see the two ways I've presented are just the most fundamental way of calculating velocity dispersion in the "lab" and CoM frame respectively, but maybe I've missed something? $\endgroup$
    – zephyr
    Commented Nov 28, 2014 at 12:26

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