<p>This is a very clever question. To answer it, consider a patch of space-time before and after a train of gravitational waves comes to it.</p>
<p>In vacuum, and in the absence of gravity waves, the Einstein equations read:
<span class="math-container">$$
G_{\alpha\beta}(\gamma) = 0\;\,,\qquad\qquad\qquad\qquad (1)
$$</span>
<span class="math-container">$\gamma$</span> being a vacuum metric.</p>
<p>In the presence of waves, the metric becomes
<span class="math-container">$$
g = \gamma + h\;\,,\qquad\;\qquad\;\qquad\;\qquad (2)
$$</span>
and the Einstein equations acquire the form of
<span class="math-container">$$
G_{\alpha\beta}(g) = 0\;\,,\qquad\qquad\qquad\qquad (3)
$$</span>
The right-hand side of both equations (1) and (3) is zero, because both are written for vacuum.</p>
<p>To endow the gravity waves with an energy-momentum tensor, we have to split equation (3), leaving the smooth, average part on the left and moving the rest to the right, with a &quot;minus&quot; sign.</p>
<p>A wrong way of doing this would be to plug (2) into (3) and arrive at
<span class="math-container">$$
0 = G_{\alpha\beta}(g) = G_{\alpha\beta}(\gamma) + ... h + ... h^2 + O(h^3)\;\,,\qquad\qquad\qquad (4)
$$</span>
in hope to reshape it as
<span class="math-container">$$
G_{\alpha\beta}(\gamma)  = - ... h - ... h^2 + O(h^3)\;\,,\qquad\qquad\qquad\qquad\qquad\;\; (5)
$$</span>
and to average out the terms linear in <span class="math-container">$h$</span>, to end up with
<span class="math-container">$$
G_{\alpha\beta}(\gamma)  = - ... h^2 + O(h^4)\qquad\qquad\qquad\qquad\qquad\qquad\qquad (6)
$$</span>
and to approximate it with
<span class="math-container">$$
G_{\alpha\beta}(\gamma)  = - ... h^2 \qquad\quad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\qquad (7)
$$</span>
[For brevity, I am using <span class="math-container">$h$</span> for all terms linear in <span class="math-container">$h$</span> and its derivatives; the same for <span class="math-container">$h^2$</span>.]</p>
<p>That the transition from (6) to (7) is plainly wrong follows from (1). If <span class="math-container">$\gamma$</span> is a vacuum metric, then the right-hand side of (6) must be exactly zero.</p>
<p>A right way to endow gravity waves with a stress-energy tensor is to realise that a gravity-wave train is acting as a physical field and is thus shifting the background. So, in the presence of waves, the smooth average background becomes not <span class="math-container">$\gamma$</span> but <span class="math-container">$\gamma + b$</span>, where <span class="math-container">$b$</span> is a smooth shift.</p>
<p>Accordingly, while <span class="math-container">$h$</span> is the ripple w.r.t <span class="math-container">$\gamma$</span>, the actual ripple on the average metric <span class="math-container">$\gamma + b$</span> is <span class="math-container">$h' = h - b$</span>. Consequently, the expansion of the Einstein equation
<span class="math-container">$$
G_{\alpha\beta}(g) = 0\quad\Longleftrightarrow\quad G_{\alpha\beta}(\,(\gamma+b) + (h-b)\,) =0\qquad\qquad\;\; (8)
$$</span>
about the mean metric <span class="math-container">$\gamma + b$</span> looks, after averaging, as
<span class="math-container">$$
G_{\alpha\beta}(\,(\gamma+b)\,) = - ... {h^{\,\prime}}^{\,2}\;\;, \qquad\qquad\qquad\qquad\qquad\qquad\qquad (9)
$$</span>
where, again, <span class="math-container">${h^{\,\prime}}^{\,2}$</span> comprises also terms with derivatives.</p>
<p>In practical calculations, we may substitute on the r.h.s <span class="math-container">$h'$</span> with <span class="math-container">$h$</span>, in the situations where <span class="math-container">$b = O(h^2)$</span>. It is, however, important to keep in mind that on the l.h.s. we have not the (identically vanishing) <span class="math-container">$G_{\alpha\beta}(\gamma)$</span> but <span class="math-container">$G_{\alpha\beta}(\gamma+b) $</span>, with the shift <span class="math-container">$b$</span> present.</p>
<p>Whether the shift is always <span class="math-container">$O({h^{\,\prime}}^{\,2})$</span> is a nontrivial question. I can provide references, if someone is interested.</p>

CC BY-SA 4.0
2023-04-11T18:33:24.107Z
2023-04-11T21:25:17.680Z
2023-04-11T21:25:17.680Z
Michael_1812
<p>Massless light beams produce a surrounding gravity field. A massive particle will move towards it and at the same time be dragged along in the direction of the beam's velocity.<br />
But what will happen around a gravity &quot;beam&quot;? Will spacetime get distorted in the same way as around a light beam with comparable energy? In the light of general relativity, the wave contains no particles (it's just a distortion of spacetime traveling along), so how can it contain energy? Or is the energy contained in the curvature of spacetime itself?<br />
So maybe my question should be, does empty (classical) curved spacetime contain energy?</p>

<general-relativity><gravitational-waves>
Do gravitational waves produce a surrounding gravity field?
Deschele Schilder
# Do gravitational waves produce a surrounding gravity field?

(Asked by: username_0 on 2021-06-10)

Massless light beams produce a surrounding gravity field. A massive particle will move towards it and at the same time be dragged along in the direction of the beam's velocity.  
But what will happen around a gravity "beam"? Will spacetime get distorted in the same way as around a light beam with comparable energy? In the light of general relativity, the wave contains no particles (it's just a distortion of spacetime traveling along), so how can it contain energy? Or is the energy contained in the curvature of spacetime itself?  
So maybe my question should be, does empty (classical) curved spacetime contain energy?

**Tags:** general-relativity, gravitational-waves


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## 1 answers

### Answer by username_1 on 2023-04-11. Score: 2

This is a very clever question. To answer it, consider a patch of space-time before and after a train of gravitational waves comes to it.

In vacuum, and in the absence of gravity waves, the Einstein equations read:
$$
G\_{\alpha\beta}(\gamma) = 0\;\,,\qquad\qquad\qquad\qquad (1)
$$
$\gamma$ being a vacuum metric.

In the presence of waves, the metric becomes
$$
g = \gamma + h\;\,,\qquad\;\qquad\;\qquad\;\qquad (2)
$$
and the Einstein equations acquire the form of
$$
G\_{\alpha\beta}(g) = 0\;\,,\qquad\qquad\qquad\qquad (3)
$$
The right-hand side of both equations (1) and (3) is zero, because both are written for vacuum.

To endow the gravity waves with an energy-momentum tensor, we have to split equation (3), leaving the smooth, average part on the left and moving the rest to the right, with a "minus" sign.

A wrong way of doing this would be to plug (2) into (3) and arrive at
$$
0 = G\_{\alpha\beta}(g) = G\_{\alpha\beta}(\gamma) + ... h + ... h^2 + O(h^3)\;\,,\qquad\qquad\qquad (4)
$$
in hope to reshape it as
$$
G\_{\alpha\beta}(\gamma) = - ... h - ... h^2 + O(h^3)\;\,,\qquad\qquad\qquad\qquad\qquad\;\; (5)
$$
and to average out the terms linear in $h$, to end up with
$$
G\_{\alpha\beta}(\gamma) = - ... h^2 + O(h^4)\qquad\qquad\qquad\qquad\qquad\qquad\qquad (6)
$$
and to approximate it with
$$
G\_{\alpha\beta}(\gamma) = - ... h^2 \qquad\quad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\qquad (7)
$$
[For brevity, I am using $h$ for all terms linear in $h$ and its derivatives; the same for $h^2$.]

That the transition from (6) to (7) is plainly wrong follows from (1). If $\gamma$ is a vacuum metric, then the right-hand side of (6) must be exactly zero.

A right way to endow gravity waves with a stress-energy tensor is to realise that a gravity-wave train is acting as a physical field and is thus shifting the background. So, in the presence of waves, the smooth average background becomes not $\gamma$ but $\gamma + b$, where $b$ is a smooth shift.

Accordingly, while $h$ is the ripple w.r.t $\gamma$, the actual ripple on the average metric $\gamma + b$ is $h' = h - b$. Consequently, the expansion of the Einstein equation
$$
G\_{\alpha\beta}(g) = 0\quad\Longleftrightarrow\quad G\_{\alpha\beta}(\,(\gamma+b) + (h-b)\,) =0\qquad\qquad\;\; (8)
$$
about the mean metric $\gamma + b$ looks, after averaging, as
$$
G\_{\alpha\beta}(\,(\gamma+b)\,) = - ... {h^{\,\prime}}^{\,2}\;\;, \qquad\qquad\qquad\qquad\qquad\qquad\qquad (9)
$$
where, again, ${h^{\,\prime}}^{\,2}$ comprises also terms with derivatives.

In practical calculations, we may substitute on the r.h.s $h'$ with $h$, in the situations where $b = O(h^2)$. It is, however, important to keep in mind that on the l.h.s. we have not the (identically vanishing) $G\_{\alpha\beta}(\gamma)$ but $G\_{\alpha\beta}(\gamma+b) $, with the shift $b$ present.

Whether the shift is always $O({h^{\,\prime}}^{\,2})$ is a nontrivial question. I can provide references, if someone is interested.


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