Why is $\langle \partial_{\mu} f(x) \rangle=0$?

Question

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' $(x,y,z)$.

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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?

Answer 1

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.

Answer 2

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.

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.

Answer 3

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 $A_\mu=0$.