<p>The brackets refer to the average, so <span class="math-container">$\left&lt; x^2 \right&gt;^{1/2}$</span> is the root-mean-square (RMS) of <span class="math-container">$x$</span>.  That is the square root of the mean (or average) of the square of multiple <span class="math-container">$x$</span>s.</p>
<p>The RMS average is useful when a quantity can be either negative or positive.  For instance, a sine or cosine wave has an average of zero over one cycle, but its RMS average is proportional to its amplitude:</p>
<p><span class="math-container">$$\left&lt; A\sin x \right&gt; = 0, \quad\quad \left&lt; (A\sin x)^2 \right&gt;^{1/2} = \frac{A}{\sqrt{2}}.$$</span></p>

CC BY-SA 4.0
2021-09-07T21:28:48.257Z
what further advantage does it have over the average of the absolute values?
njzk2
CC BY-SA 4.0
2021-09-07T23:08:47.653Z
@njzk2 : You might benefit from reading about the [RMS](https://en.wikipedia.org/wiki/Root_mean_square), learning, for instance, that it is related to the standard deviation of a (0-mean, or centered) sample.
Eric Towers
CC BY-SA 4.0
2021-12-14T21:28:53.493Z
Another example of where taking the square root of squared values, because of possible negative values is when calculating variance, in statistics.
Fred
CC BY-SA 4.0
2021-09-07T13:53:18.233Z
2021-09-08T07:03:12.707Z
2021-09-08T07:03:12.707Z
Paul T.
<p>From Sabine Hossenfelder's recent video, <a href="https://www.youtube.com/watch?v=JETGS64kTys" rel="nofollow noreferrer">New Evidence AGAINST Standard Cosmology</a>:</p>
<p><a href="https://i.sstatic.net/CLTjW.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CLTjW.jpg" alt="enter image description here" /></a></p>
<p>And her <a href="https://academic.oup.com/mnras/article/401/1/547/1008219" rel="nofollow noreferrer">source</a>....</p>
<p><a href="https://i.sstatic.net/5UneZ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/5UneZ.png" alt="enter image description here" /></a></p>
<blockquote>
<p>Figure 2. The variation with increasing void radius of the variance of the Hubble parameter, the density contrast, the density parameter and the peculiar velocity for the CDM power-law and CHDM bump models, given the <em>WMAP5</em> and SDSS data (with 1σ limits).</p>
</blockquote>
<p>I don't get why a graph would show a quantity that is squared, then immediately 'square-rooted' such as <span class="math-container">$\langle \delta^2_H \rangle^{1/2}, \langle \delta^2 \rangle^{1/2}, \langle \delta^2_{\Omega} \rangle^{1/2} \ \text{and} \ \langle v^2 \rangle^{1/2}$</span>.</p>
<p>Also, the lowercase delta does stand for the Hubble constant difference or contrast, correct?</p>

CC BY-SA 4.0
2021-09-07T18:45:39.870Z
The plot she is showing averages over $\delta H$, not $H$, i.e. a fluctuation field.
AtmosphericPrisonEscape
<observational-astronomy><cosmology><observable-universe><dark-energy><hubble-constant>
Why would a quantity like the 'Hubble contrast' be squared, then have its square root taken?
Kurt Hikes
# Why would a quantity like the 'Hubble contrast' be squared, then have its square root taken?

(Asked by: username_0 on 2021-09-07)

From Sabine Hossenfelder's recent video, [New Evidence AGAINST Standard Cosmology](https://www.youtube.com/watch?v=JETGS64kTys):

[![enter image description here](https://i.sstatic.net/CLTjW.jpg)](https://i.sstatic.net/CLTjW.jpg)

And her [source](https://academic.oup.com/mnras/article/401/1/547/1008219)....

[![enter image description here](https://i.sstatic.net/5UneZ.png)](https://i.sstatic.net/5UneZ.png)

> Figure 2. The variation with increasing void radius of the variance of the Hubble parameter, the density contrast, the density parameter and the peculiar velocity for the CDM power-law and CHDM bump models, given the *WMAP5* and SDSS data (with 1σ limits).

I don't get why a graph would show a quantity that is squared, then immediately 'square-rooted' such as $\langle \delta^2\_H \rangle^{1/2}, \langle \delta^2 \rangle^{1/2}, \langle \delta^2\_{\Omega} \rangle^{1/2} \ \text{and} \ \langle v^2 \rangle^{1/2}$.

Also, the lowercase delta does stand for the Hubble constant difference or contrast, correct?

**Tags:** observational-astronomy, cosmology, observable-universe, dark-energy, hubble-constant
> **Comments:**
> **username_1**: The plot she is showing averages over $\delta H$, not $H$, i.e. a fluctuation field.  

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## 1 answers

### Answer by username_2 on 2021-09-07. Score: 8 | Accepted answer

The brackets refer to the average, so $\left< x^2 \right>^{1/2}$ is the root-mean-square (RMS) of $x$. That is the square root of the mean (or average) of the square of multiple $x$s.

The RMS average is useful when a quantity can be either negative or positive. For instance, a sine or cosine wave has an average of zero over one cycle, but its RMS average is proportional to its amplitude:

$$\left< A\sin x \right> = 0, \quad\quad \left< (A\sin x)^2 \right>^{1/2} = \frac{A}{\sqrt{2}}.$$

> **Comments:**
> **username_3**: what further advantage does it have over the average of the absolute values?  
> **username_4**: @njzk2 : You might benefit from reading about the [RMS](https://en.wikipedia.org/wiki/Root_mean_square), learning, for instance, that it is related to the standard deviation of a (0-mean, or centered) sample.  
> **username_5**: Another example of where taking the square root of squared values, because of possible negative values is when calculating variance, in statistics.  

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