Dirac on localization of energy in the gravitational field

[Kostik]

TL;DR

Dirac claims that in a plane wave, one can determine the local energy content of the gravitational field, because the energy-momentum pseudo-tensor behaves like a tensor. Either he is not explaining his point with sufficient clarity, or he is simply wrong?

Question concerns a remark by Dirac in his "General Theory of Relativity", chap. 33, p. 65-66. Dirac is working with plane waves, so the metric $$g_{\mu\nu}(x) = g_{\mu\nu}(l_\sigma x^\sigma)$$ where $l_\sigma$ is the wave vector. Writing $\xi = l_\sigma x^\sigma$, Dirac defines $$u_{\mu\nu}(\xi)=\frac{dg_{\mu\nu}(\xi)}{d\xi}$$ Thus, $g_{\mu\nu,\sigma} = u_{\mu\nu} l_\sigma .$ It should be noted that $u_{\mu\nu}$ is not a tensor. If it were, then $g_{\mu\nu,\sigma}$ would be a tensor (since $l_\sigma$ is a vector), which it is not. Indeed, $g'_{\mu\nu}$ will not be a function of the variable $\xi'=\xi$ under an arbitrary coordinate transformation; hence, $u'_{\mu\nu}$ is not even defined.

Dirac shows that the (Einstein-Hilbert) energy-momentum pseudo-tensor has the form $$16\pi t^{\mu\nu} = \frac{1}{2}\left( u_{\alpha\beta}u^{\alpha\beta} - \frac{1}{2}u^2 \right) l^\mu l^\nu $$ where $u = g^{\alpha\beta}u_{\alpha\beta} .$

Dirac says, "We have a result for $t^{\mu\nu}$ that looks like a tensor. This means that $t^{\mu\nu}$ transforms like a tensor under those transformations that preserve the character of the field of consisting only of waves moving in the direction $l_\sigma$, so that the $g_{\mu\nu}$ remain functions of the single variable $l_\sigma x^\sigma$. Such transformations must consist only in the introduction of coordinate waves moving in the direction $l_\sigma$, of the form $$x^{\mu'} = x^\mu + b^\mu$$ where $b^\mu$ is a function only of $l_\sigma x^\sigma$. With the restriction that we have waves moving only in one direction, gravitational energy can be localized."

First, once again, $u_{\mu\nu}$ is not a tensor, and $u$ is not a scalar, so it's not clear what Dirac means by "We have a result for $t^{\mu\nu}$ that looks like a tensor." Second, it is well known and understood that the pseudo-tensor cannot be a tensor, since then it would have to vanish by changing to locally inertial coordinates, where there is no gravitational field. Thus, the energy-momentum pseudo-tensor is coordinate dependent; hence, we cannot have a concept of local energy-momentum in the gravitational field.

What exactly is Dirac saying?

 

[anuttarasammyak]

TL;DR Summary: Dirac claims that in a plane wave, one can determine the local energy content of the gravitational field, because the energy-momentum pseudo-tensor behaves like a tensor. Either he is not explaining his point with sufficient clarity, or he is simply wrong?

With the restriction that we have waves moving only in one direction, gravitational energy can be localized."

By the context I read his text with supplement that

With the restriction that we have waves moving only in one direction 【and limit coordinates transformation to be used as mentioned in the above equation】, gravitational energy can be localized.

I think the allowed coordinates are abundant enough for the observation of gravitational waves on the Earth, e.g., LIGO project.

 

[Kostik]

By way of additional detail. Under the coordinate transformation $x^{\mu'} = x^\mu + b^\mu$ where $|b^\mu_{\,,\nu}| \ll 1$, the metric transforms (neglecting second order terms): $$g'_{\mu\nu} = g_{\mu\nu} - b_{\mu ,\nu} - b_{\nu , \mu} = g_{\mu\nu} - \dot{b}_\mu l_{\nu} - \dot{b}_\nu l_{\mu} \,\,. \quad\quad(*)$$ where $\dot{b}^\mu = d b^\mu/d\xi$. Additionally, this coordinate transformation gives $$\frac{\partial x^\alpha}{\partial x'^\mu} = \delta^\alpha_\mu - b^\alpha_{,\mu} = \delta^\alpha_\mu - \dot{b}^\alpha l_{\mu} \,\,.\quad\quad(**)$$ Using $(*)$ and $(**)$, you cannot show that $u = g^{\mu\nu}u_{\mu\nu}$ or $u^{\mu\nu}u_{\mu\nu}$ are scalars ... try it!

Therefore, Dirac has no basis for saying "We have a result for $t^{\mu\nu}$ that looks like a tensor" even under the specific coordinate transformation he indicates!

 

[renormalize]

$$g'_{\mu\nu} = g_{\mu\nu} - b^\mu_{\,,\nu} - b^\nu_{\,,\mu} = g_{\mu\nu} - \dot{b}^\mu l_{\nu} - \dot{b}^\nu l_{\mu} \,\,. \quad\quad(*)$$ where $\dot{b}^\mu = d b^\mu/d\xi$.

Your indices are unbalanced in eq.($*$). Can you rewrite it?

 

[Kostik]

Your indices are unbalanced in eq.($*$). Can you rewrite it?

Oops, fixed it.

 

[Ibix]

Did we not discuss this before?

 

[Kostik]

Did we not discuss this before?

I reviewed that thread. My question was not well posed; I have clarified the question, emphasizing the fact that $u$ and $u^{\mu\nu}u_{\mu\nu}$ are not scalars.

 

[renormalize]

you cannot show that $u = g^{\mu\nu}u_{\mu\nu}$ or $u^{\mu\nu}u_{\mu\nu}$ are scalars ... try it!

But that's not what Dirac is implying. When he says that the pseudo-tensor $16\pi\, t_{\mu\nu}=\frac{1}{2}\left(u_{\alpha\beta}\,u^{\alpha\beta}-u^{2}\right)l_{\mu}l_{\nu}$ "looks like a tensor" he is referring strictly to the symmetric rank-2 tensor piece $l_{\mu}l_{\nu}$ made from the finite null-vector $l_{\mu}\,$, which transforms according to $l_{\mu}\rightarrow\acute{l}_{\mu}=l_{\mu}-\dot{b}^{\alpha}l_{\alpha}l_{\mu}$ under the infinitesimal coordinate-transformation $\frac{\partial x^{\alpha}}{\partial\acute{x}^{\mu}}=\delta_{\mu}^{\alpha}-b_{,\mu}^{\alpha}=\delta_{\mu}^{\alpha}-\dot{b}^{\alpha}l_{\mu}\,$, where $\left\Vert \dot{b}^{\alpha}\right\Vert \ll1$. In contrast, the pseudo-tensor factor $S\equiv u_{\alpha\beta}\,u^{\alpha\beta}-u^{2}$ is already quadratic in the infinitesimal quantity $\left\Vert u_{\alpha\beta}\right\Vert \ll1$ and hence to lowest order $S$ is invariant under the infinitesimal coordinate change: $\acute{u}_{\mu\nu}\equiv u_{\mu\nu}+\mathcal{O}\left(u\times\dot{b}\right)\simeq u_{\mu\nu}$. So the pseudo-tensor can be written in the manifest tensor-form $16\pi\,t_{\mu\nu}=\frac{1}{2}S\,l_{\mu}l_{\nu}\,$, where $S$ is a scalar under the given class of transforms. In other words, Dirac is correct.

 

[Kostik]

@renormalize How do you get $$u'_{\mu\nu}=u_{\mu\nu}+O(u\times \dot{b}) \,\,?$$ I get
$$u'_{\mu\nu}=\frac{d}{d\xi}(g_{\mu\nu} - \dot{b}_\mu l_\nu - \dot{b}_\nu l_\mu) = u_{\mu\nu} - \ddot{b}_\mu l_\nu - \ddot{b}_\nu l_\mu = u_{\mu\nu} + O(\ddot{b}) \,\,.$$

 

[renormalize]

How do you get $$u'_{\mu\nu}=u_{\mu\nu}+O(u\times \dot{b}) \,\,?$$

I'm simply treating $u_{\mu\nu}$ as an infinitesimal standard covariant tensor under coordinate transforms.

I get
$$u'_{\mu\nu}=\frac{d}{d\xi}(g_{\mu\nu} - \dot{b}_\mu l_\nu - \dot{b}_\nu l_\mu) = u_{\mu\nu} - \ddot{b}_\mu l_\nu - \ddot{b}_\nu l_\mu = u_{\mu\nu} + O(\ddot{b}) \,\,.$$

But this ignores the fact that $\xi\equiv l_{\mu}x^{\mu}$ is not a scalar because the coordinate $x^{\mu}$ doesn't transform like a contravariant vector:
$$\acute{l}_{\mu}\acute{x}^{\mu}=l_{\alpha}\left(\delta_{\mu}^{\alpha}-\dot{b}^{\alpha}l_{\mu}\right)\left(x^{\mu}+b^{\mu}\right)=l_{\mu}x^{\mu}\left(1-\dot{b}^{\alpha}l_{\alpha}\right)+b^{\mu}l_{\mu}+\mathcal{O}\left(b\dot{b}\right)$$Or are you assuming that $b^{\mu}$ must be chosen so that $b^{\mu}l_{\mu}$ vanishes?

 

[anuttarasammyak]

Let me state my understanding. In a simple case of
\xi=x^0-x^3
Non-zero components of {t} are :
t_0^{\ 0}=t_3^{\ 0}=-t_0^{\ 3}=-t_3^{\ 3}=Q(x^0-x^3)
as a function of $x^0-x^3$ where
Q(x^0-x^3)=\frac{1}{32 \pi }(u_{\alpha\beta}u^{\alpha\beta}-\frac{1}{2}u^2)
As far as we stay in the world of transformations that every time-space point keeps its $x^0-x^3$ value, which appears as phase of waves usually, we identify local energy, momentum of x^3 direction and stress there. Am I wrong in this primitive view ?

 

[Kostik]

I think I have a solution, although I need to impose an extra condition that Dirac does not.

$\quad\quad$ We make a coordinate change $${x'}^\mu = x^\mu + b^\mu \, , \quad\quad |b_{\mu,\rho}|,|b_{\mu,\rho}| \ll 1 \quad\quad\quad (*)$$ so that (neglecting second-order terms in small quantities) $$\frac{\partial x^\alpha}{\partial x'^\mu } = \delta^\alpha_\mu - b^\alpha_{,\mu}$$ and which results in a transformation of the metric $$g'_{\mu\nu} = g'_{\mu\nu} - b_{\mu,\nu} - b_{\nu,\mu}$$ We also assume $b^\mu = b^\mu (\xi)$, writing $\xi = l_\sigma x^\sigma$, so that $b_{\mu,\nu} = \dot{b}_\mu k_\nu$. Thus, $g'_{\mu\nu}$ is a function of $\xi$.

$\quad\quad$ Moreover, since $\xi' = \xi$ (a scalar): $$\frac{dg'_{\mu\nu}}{d\xi} = \frac{d}{d\xi}\left( g_{\alpha\beta}\frac{\partial x^\alpha}{\partial {x'}^\mu }\frac{\partial x^\beta}{\partial {x'}^\nu} \right) = \frac{dg_{\alpha\beta}}{d\xi}\left( \frac{\partial x^\alpha}{\partial {x'}^\mu }\frac{\partial x^\beta}{\partial {x'}^\nu} \right)-2\ddot{b}_\beta k_\mu (\delta^\beta_\nu - \dot{b^\beta k_\nu})$$ hence $$u'_{\mu\nu} = u_{\alpha\beta} \frac{\partial x^\alpha}{\partial {x'}^\mu }\frac{\partial x^\beta}{\partial {x'}^\nu} + O(\ddot{b}_\sigma)$$ Note that $b_{\mu,\rho}=\dot{b}_\mu k_\rho$ and $b_{\mu,\rho\sigma}=\ddot{b}_\mu l_\rho l_\sigma$. Hence, the assumptions $|b_{\mu,\rho}|, |b_{\mu,\rho\sigma}| \ll 1$ are equivalent to $|\dot{b}_\mu|, |\ddot{b}_\mu| \ll 1$. Also, of course, $u_{\mu\nu} \ll 1$ since $g_{\mu\nu,\sigma} = u_{\mu\nu} l_\sigma$, and we assume $g_{\mu\nu,\sigma} \ll 1$ in the weak-field approximation.

Thus, we may consider that $u_{\mu\nu}$ transforms as a tensor, providing the conditions of $(*)$ hold and also the condition $$|\ddot{b}_\sigma| \ll |u_{\mu\nu}| \,. \quad\quad\quad(**)$$ Therefore, $u^2$ and $u^{\mu\nu}u_{\mu\nu}$ can be considered scalars. And therefore $t^{\mu\nu}$ can be considered a tensor.

This is what Dirac is saying, except he doesn't impose the condition $(**)$.

 

[Kostik]

I'm simply treating $u_{\mu\nu}$ as an infinitesimal standard covariant tensor under coordinate transforms.

But this ignores the fact that $\xi\equiv l_{\mu}x^{\mu}$ is not a scalar because the coordinate $x^{\mu}$ doesn't transform like a contravariant vector:
$$\acute{l}_{\mu}\acute{x}^{\mu}=l_{\alpha}\left(\delta_{\mu}^{\alpha}-\dot{b}^{\alpha}l_{\mu}\right)\left(x^{\mu}+b^{\mu}\right)=l_{\mu}x^{\mu}\left(1-\dot{b}^{\alpha}l_{\alpha}\right)+b^{\mu}l_{\mu}+\mathcal{O}\left(b\dot{b}\right)$$Or are you assuming that $b^{\mu}$ must be chosen so that $b^{\mu}l_{\mu}$ vanishes?

@renormalize: How can you treat $u_{\mu\nu}$ as a tensor? Actually, I believe this is (approximately) correct, but this requires a calculation; please see my reply to the original post.

I agree $\xi\equiv l_{\mu}x^{\mu}$ is not a scalar in curved spacetime, because the wave vector $l_\sigma = \xi_{,\sigma}$ is only defined in flat spacetime, where the phase $\xi$ of a plane-wave is a well-defined scalar. However, Dirac is working in the weak-field approximation, so I am assuming that $l_\sigma$ can be considered (approximately) to be a vector.

(Note that you are treating $l_\sigma$ as a vector, to show that $\xi$ is not a scalar, but the very definition of $l_\sigma$ comes from $l_\sigma \equiv \xi_{,\sigma}$, so I don't follow your logic.)

Please examine the posting I made, showing that $u_{\mu\nu}$ transforms (approximately) as a tensor, which means that $u^2$ and $u^{\mu\nu}u_{\mu\nu}$ can be considered as scalars. Does this look reasonable?

 

[renormalize]

Note that $b_{\mu,\rho}=\dot{b}_\mu k_\rho$ and $b_{\mu,\rho\sigma}=\ddot{b}_\mu l_\rho l_\sigma$. Hence, the assumptions $|b_{\mu,\rho}|, |b_{\mu,\rho\sigma}| \ll 1$ are equivalent to $|\dot{b}_\mu|, |\ddot{b}_\mu| \ll 1$. Also, of course, $u_{\mu\nu} \ll 1$ since $g_{\mu\nu,\sigma} = u_{\mu\nu} l_\sigma$, and we assume $g_{\mu\nu,\sigma} \ll 1$ in the weak-field approximation.

Thus, we may consider that $u_{\mu\nu}$ transforms as a tensor, providing the conditions of $(*)$ hold and also the condition $$|\ddot{b}_\sigma| \ll |u_{\mu\nu}| \,. \quad\quad\quad(**)$$

Your two conditions $u_{\mu\nu} \ll 1$ and $|\ddot{b}_\sigma| \ll |u_{\mu\nu}|$ combine to give $|\ddot{b}_\sigma|\ll\ll 1$. What does this mean? How do you define one quantity as being infinitesimal relative to a second infinitesimal quantity?
In my view, your proof demonstrates that $\ddot{b}_\sigma=0$ must hold. That is, the most general infinitesimal coordinate-transform satisfying Dirac's requirements must be linear in $\xi\,$:$$\acute{x}^{\mu}=x^{\mu}+\beta_0^{\mu}+\beta_1^{\mu}\,l_{\alpha}x^{\alpha}$$

 

[Kostik]

Your two conditions $u_{\mu\nu} \ll 1$ and $|\ddot{b}_\sigma| \ll |u_{\mu\nu}|$ combine to give $|\ddot{b}_\sigma|\ll\ll 1$. What does this mean? How do you define one quantity as being infinitesimal relative to a second infinitesimal quantity?
In my view, your proof demonstrates that $\ddot{b}_\sigma=0$ must hold. That is, the most general infinitesimal coordinate-transform satisfying Dirac's requirements must be linear in $\xi\,$:$$\acute{x}^{\mu}=x^{\mu}+\beta_0^{\mu}+\beta_1^{\mu}\,l_{\alpha}x^{\alpha}$$

@renormalize Thanks, your comments have been very helpful. I would suggest that $|\ddot{b}_\sigma| \ll |u_{\mu\nu}|$ means that $|\ddot{b}_\sigma| / |u_{\mu\nu}| \ll 1$ for each triplet of components $(\sigma,\mu,\nu)$. However, the stricter condition that $b_{\mu,\rho\sigma}=\ddot{b_\mu}=0$ clearly makes $u_{\mu\nu}$ transform like a tensor under ${x'}^\mu = x^\mu + b^\mu(\xi)$.

The question I'm wondering is this: it's well known that the pseudo-tensor is not a tensor, since in locally inertial coordinates of a free particle (on a geodesic) there is no gravitational field, hence the pseudo-tensor would vanish in those coordinates and hence in all coordinate systems. What is the point of introducing a restricted class of coordinate transformations, such as ${x'}^\mu = x^\mu + b^\mu(\xi)$ (with or without the restriction that $\ddot{b}_\mu=0$) under which $t_\mu^\nu$ transforms as a tensor? It isn't a tensor, and therefore, there does not exist a concept of local energy-momentum of the gravitational field. What has Dirac achieved by showing that the pseudo-tensor transforms "like a tensor" under a very limited category of coordinate transformations?

 

[anuttarasammyak]

Paul Dirac
Gravitational Waves
Category: Lectures
Date: 1 July 1959
Duration: 38 min
Quality: HD MD SD
Subtitles: EN DE
https://mediatheque.lindau-nobel.org/recordings/31422/gravitational-waves-1959
In this audio record of Dirac's lecture, he referred to the wave of x^3 direction at the end but the record ends suddenly !

 

[anuttarasammyak]

In Landau-Lifshitz Classical Theory of Fields

Wave coordnates are introduced to eliminate non-essential components of h.

In slides #14 and #17 of Introduction to General Relativity and
Gravitational Waves,Patrick J. Sutton https://static.sif.it/SIF/resources/public/files/va2017/Sutton1.pdf
Introduciton of initesimal transformation, though I am not sure it is wave coorinate or not, is expected to the equation satisfy Lorenz gauge.
tells the details.

So I think wave coordinates is introduced to reveal some essential feature of plane garavitational wave which is shared in base flat spacetime.But I must confess I do not clearly distinguish harmonic coordinates and wave coordinates how they work for what fruit.