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I'm reading page 488 of Hobson, Efstathiou & Lasenby, and I don't understand something they write... so I came here.

The concept they describe is in linearised general relativity. In particular, they are describing the averaging process over small spacetime regions that makes it possible to define the energy-momentum tensor of gravitatonal waves.

They say that since we are averaging of all directions at each point, first derivatives average to zero. That is, for any function of position $f(x)$, we have $\langle \partial_{\mu} f(x) \rangle=0$. Here $x = (x^1,x^2,x^3)$, which is an index-notation version of the 'standard' $(x,y,z)$.


Edit

By 'average', I assume they mean an integral. So, if we take the simplest possible case for starters, we would have a 1D spacetime, say the $x$ lined. So, all functions of position, are just $f(x)$.

So, I would interpret \begin{equation} \langle f(x) \rangle = \int f(x) dx. \end{equation} Then, \begin{equation} \begin{split} \langle \partial_{x} f(x) \rangle &= \partial_{x} \int f(x) dx \\ &=f(x). \end{split} \end{equation} So, how can this be zero?

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Please do not migrate this to Maths.SE –  user12345 Feb 17 '13 at 21:01
    
Isn't this just another way of saying that we're in a flat background (considering the fact that it's linearized gravity you mentioned)? But I'm not sure to be honest :) Cheers, a friendly helper –  user21091 Feb 17 '13 at 21:52
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I don't own the books so I can't really determine the context. Could it have to do with the fact that the average amplitude of a sinusoidal wave is zero? –  elfmotat Feb 17 '13 at 22:25
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In the book they consider small spacetime regions, which we can consider flat. And $\partial_{\mu}a(x)$ is just the gradient of that function. Having a flat space and averaging over all direction gives you zero. –  nijankowski Feb 17 '13 at 22:43
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@MichaelBrown I don't understand. Taking the case of a 1D spacetime. If $f(x) = x^{2}$ and we choose to position ourselves and $x=0$, we see that $f$ increases if we look in the $+x$ direction but it also increases if we look in the $-x$ direction. –  user12345 Feb 18 '13 at 20:34

3 Answers 3

The derivative of a function $f$ in the direction $\vec{n}$ is defined as

$$ \nabla_{\vec{n}}f \equiv \vec{n}\cdot\nabla f $$

where $\vec{n}$ is a unit vector. Integrating this over $\vec{n}$ gives

$$ \left< \nabla f \right> \equiv \int \mathrm{d}\vec{n}\ \nabla_{\vec{n}}f = \int \mathrm{d}\vec{n}\ \vec{n}\cdot\nabla f = \nabla f \cdot \int \mathrm{d}\vec{n}\ \vec{n} = 0 $$

If you need to see a particular example pick some ridiculous function like this one I just made up (in two dimensions, just cause):

$$ f(x,y) = \frac{\cos(x y)}{2 y^2} - x \sin(x^2 + y^2) $$

and take the average of the gradient at the point $(x,y)=(3,2)$, pictured as the red dot below.

a ridiculous function

Mathematica tells me that the gradient of this function evaluated at the point of interest is, in (x,y) components,

$$(-(\sin(6)/4)-\sin(13)-18 \cos(13),\ -((3 \sin(6))/8)-\cos(6)/8-12 \cos(13))$$

So in general the derivative of $f$ in the direction of the unit vector $\vec{n} = (\cos(\theta), \sin(\theta))$ is

$$ \vec{n}\cdot(-(\sin(6)/4)-\sin(13)-18 \cos(13),\ -((3 \sin(6))/8)-\cos(6)/8-12 \cos(13))$$

which works out to some stupid sinusoidal function of $\theta$.

Now it should be easy to convince yourself that the average over directions is

$$ \int \mathrm{d}\theta\ (\cdots) = 0 $$

It should be obvious that the choice of function is completely arbitrary, so long as it is differentiable at the point of interest.

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Thanks, this seems to make sense. I think my problem was that, to me, the result seems counter-intuitive. Nice graph btw :). Also, if you didn't see, I have edited the link in my question. –  user12345 Feb 19 '13 at 16:14
    
@user16307 The link goes to an error page for me... I think I might have the book anyway, I just need to dig it out. Hope this helps. :) –  Michael Brown Feb 19 '13 at 16:19
    
Weird. If I use Chrome, the link (physics.usyd.edu.au/~fvon4093/introgr.pdf) works, but with IE it doesn't. Maybe my browser has cached it and it's now been removed. –  user12345 Feb 19 '13 at 16:29
    
@MichaelBrown Nice Polynomials! –  Hal Swyers Feb 20 '13 at 0:01

Well, I cannot find this book, but it is most probably a standard symmetry argument -- a nonzero average would mean a preferred space-time direction.

More formally. Let us introduce two variables: $$x_\rho \quad \mbox{and} \quad x'_\sigma=\Lambda^\rho_\sigma x_\rho$$ With arbitrary Loretnz transformation $\Lambda$.

I'm pretty sure that the average you are talking about shouldn't care about what variable are we using to average over: $$\langle f(x_\rho)\rangle=\langle f(x'_\sigma)\rangle$$ Let us differentiate both sides with $\partial_\mu = \frac{\partial}{\partial x_\mu}$ $$\langle \partial_\mu f(x_\rho)\rangle=\langle \partial_\mu f(\Lambda^\rho_\sigma x_\rho)\rangle$$ $$\langle \partial_\mu f(x_\rho)\rangle=\langle \Lambda^\sigma_\mu \partial_\sigma f( x_\rho)\rangle$$

Using linearity and denoting $\langle \partial_\mu f(x)\rangle = A_\mu$:

$$A_\mu=\Lambda^\sigma_\mu A_\sigma$$ For arbitrary $\Lambda$ that could be satisfied only if $A_\mu=0$.

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I follow this all the way down to the last line, where I don't understand why you say that $A_{\mu}$ must be zero. P.s. I have edited the link so you will definitely be able to access the book. –  user12345 Feb 19 '13 at 16:12
    
@user16307 umm.. it is quite strange that this step is difficult for you... Ok, can you provide a nonzero 3D vector which stays the same for arbitrary 3D rotations? –  Kostya Feb 19 '13 at 20:40
    
Of course not. But to me, it looks like you start of with some equation in $x$ and then re-write this equation in $A$, and nothing more. I can't see why you need $A_{\mu}=0$ from this, anymore than you would need $x^{\prime}_{\sigma}=0$ in your first equation. –  user12345 Feb 19 '13 at 21:21
    
@user16307 Have you seen, say, how the Euler-Lagrange equation are derived? –  Kostya Feb 19 '13 at 21:37
    
I have seen a derivation, yes. –  user12345 Feb 19 '13 at 21:43

We agree with OP that Ref. 1 does not explain the averaging procedure $\langle \cdots \rangle$ adequately. For reference, the relevant section of Ref. 1 reads:

One way of circumventing this problem is to take seriously the fact that the energy–momentum of a gravitational field at a point in spacetime has no real meaning in general relativity, since at any particular event one can always transform to a free-falling frame in which gravitational effects disappear. This suggests that, at each point in spacetime, one should average $G^{(2)}_{\mu\nu}$ over a small region in order to probe the physical curvature of the spacetime, which gives a gauge-invariant measure of the gravitational field strength. Denoting this averaging process by $\langle \cdots \rangle$, one should thus replace (17.55) by

$$\tag{17.57} t_{\mu\nu}~\equiv~\frac{c^4}{8\pi G} \langle G^{(2)}_{\mu\nu}\rangle.$$

Having made this identification, our task is now an algebraic one of determining the form of $\langle G^{(2)}_{\mu\nu}\rangle$ as a function of $h_{\mu\nu}$. This is rather a cumbersome calculation, but the job is made somewhat easier by averaging over small spacetime regions. Since we are averaging over all directions at each point, first derivatives average to zero. Thus, for any function of position $a(x)$, we have $\langle \partial_{\mu}a(x)\rangle$.

Here the superscript $(2)$ refers to terms that are second-order in $h_{\mu\nu}$.

The averaging procedure is explained in greater detail in Ref. 2 as part of the shortwave approximation/formalism in the limit of small typical wave amplitude $A\ll 1$ and typical wavelength $\lambda \ll R$ much smaller than the typical radius $R$ of curvature.

This is essentially a Wilsonian effective theory, where one integrates out UV modes to be left with IR modes. However, one should keep in mind that Ref. 2 are considering the UV modes as classical configurations (as opposed to quantum fluctuations), so rather than integrating out in path integral sense, Ref. 2 is averaging out.

Technically, the averaging procedure $\langle \ldots \rangle$ in Ref. 2. is called the Brill-Hartle average.

To convey the main idea in an oversimplified manner, the average procedure $\langle \ldots \rangle$ is taken over several wavelengths of a plane gravitational wave of the form

$$\tag{1} a(x)~=~f\left(c(x)+A\sin(k\cdot x)\right),$$

where the IR part $c(x)$ depends so slowly on position $x$, that it can be treated as a constant in the average procedure, and the derivative $\partial_{\mu}c(x)$ is negligible. So we are essentially just averaging the first derivative

$$\tag{2} \partial_{\mu}a(x)~=~f^{\prime}\left(c(x)+A\sin(k\cdot x)\right)~\partial_{\mu}\left(c(x)+A\sin(k\cdot x)\right)~\approx~f^{\prime}\left(c(x)+A\sin(k\cdot x)\right)~k_{\mu}A\cos(k\cdot x)$$

over several wavelengths and getting zero. The detailed form $(1)$ of $a(x)$ does not matter as long as it is approximately periodic.

References:

  1. M.P. Hobson, G.P. Efstathiou, and A.N. Lasenby, General Relativity: An Introduction for Physicists, 2005, p. 488.

  2. C.W. Misner, K.S. Thorne, and J.A. Wheeler, Gravitation, 1973, Section 35.13-35.15, p. 964-973.

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Okay, now I'm confused by the language "averaging over all directions at each point," which doesn't seem to be what they are actually doing. This is averaging an approximately periodic function over positions, not arbitrary functions over directions. *head scratch* I'm happy to have my answer removed or un-accepted (is that possible?) if it's irrelevant. –  Michael Brown Feb 20 '13 at 3:17
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Very cool, I particularly like the slowly changing constant! –  Hal Swyers Feb 20 '13 at 10:44
    
@Qmechanic So, are you saying that the average is zero because $a$ is periodic? Also, what physical entity is represented by the $c$ that you use? I've not seen a wave with this form before. On p.969 of MTW, they say that the error incured from the averaging gradients to zero is $O(\lambda/R)$, where R is the length-scale over which the background metric tensor changes. I was wondering where this error would come from, specifically if it was from $a$. Finally, thanks. –  user12345 Feb 20 '13 at 11:09

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