A Dirac's "GTR" Eq (27.4): how momentum ##p^\mu## varies

[Kostik]

TL;DR

Dirac derives Einstein's equation for a dust $G^{\mu\nu}=-8\pi \rho v^\mu v^\nu$ from the principle of stationary action using the appropriate Lagrangian for the gravitational field and a dust. First, he "infers" an important relation between an arbitrary variation of an element of the matter field and the resultant change in the momentum vector density. How to derive this rigorously?

In Dirac's "General Theory of Relativity", chapters 26-30, he builds up various action principles from which Einstein's equation $G^{\mu\nu}=-8\pi T^{\mu\nu}$ can be obtained. In chapter 27, he extends the result of chapter 26 (Einstein's vacuum equation) to the case of a dust, where $$T^{\mu\nu}=\rho v^\mu v^\nu \,\, .$$ Here, $\rho$ is the mass density in the MCRF ("rest frame") of a fluid element of the dust.

The momentum density is $$p^\mu = \rho v^\mu \sqrt{-g} \, .$$ Dirac makes arbitrary changes $b^\mu$ in the position of the dust, and uses his Eq. (27.4) $$\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu )_{,\nu} \qquad\qquad(*)$$ to find the corresponding change in $p^\mu$.

Using $(*)$, Dirac does indeed derive Einstein's field equation and the geodesic equation for the motion of the matter. I'm not asking about that, and won't mention the details; Dirac's derivation is straightforward and easy.

My question is about $(*)$, his Eq. (27.4). He infers this, but does not derive it. I am attaching below my notes, in which I had hoped to derive $(*)$. (A clearer MSWord doc is also attached.) I start by considering the dust as a collection of individual matter particles. The first result is to show that the velocity field $v^\mu$ is divergenceless, i.e., $v^\mu_{\,\, ,\mu}=0$ (my Eq. (27.5)). Using this fact and $p^\mu_{\,\, ,\mu}=0$, I obtain the desired result.

Unfortunately, if both $v^\mu_{\,\, ,\mu}=0$ and $p^\mu_{\,\, ,\mu}=0$, then the proportionality factor $\rho \sqrt{-g}$ must be constant. But $\rho \sqrt{-g}$ definitely is not a constant.

In fact, $v^\mu_{\,\, ,\mu}=0$ cannot be right -- this would only be true if $\rho$ were constant, which again is not the case. The only conservation law is $p^\mu_{\,\, ,\mu}=0$.

Is my entire derivation pure rubbish? I have a feeling that it's not. I must be missing something; something connected with the fact that this is a dust, and I am working with a sum over individual particles (which can be supposed to have mass $m$). Somewhere, I am not using the full power of the delta functions in the derivation.

I think the fact that spacetime may be curved, hence the correct conservation law is $p^\mu_{\,\, ,\mu}= (\rho v^\mu \sqrt{-g})_{\,\, ,\mu} =0$, has nothing whatever to do with proving $(*)$. Dirac does not even bring in the metric until after his Eq. (27.4). (In fact, he merely states that the momentum density $p^\mu$ "lies in the same direction" as the velocity vector.)

Any help, please?

 

[dextercioby]

In his derivation there's no mention of being in a curved spacetime. There's no √|g|, nor any semicolon. So what's wrong with his argument leading to (27.4)?

 

[Kostik]

In his derivation there's no mention of being in a curved spacetime. There's no √|g|, nor any semicolon. So what's wrong with his argument leading to (27.4)?

Dirac infers (27.4) but doesn't derive it. I was looking to derive (27.4).

 

[JimWhoKnew]

Dirac infers (27.4) but doesn't derive it. I was looking to derive (27.4).

Are you satisfied with Dirac's arguments that led him to the equation below (27.3) ?
If yes, note that it is equal to $\delta p^0$ in (27.4). For the other components $\delta p^i$ , I think you can re-use Dirac's arguments: Starting with the equation above (27.3), this time with$$p^1 dx^0 dx^2 dx^3$$(for example). The main difference from before is that $p^1$ may vanish now, leaving us short of a constraint. To remedy that, we may try adding an infinitesimal $\epsilon$ component to $p^1$ , demand vanishing up to $O(\epsilon)$ when $b^1=0$ or when $b^\mu$ is parallel to $p^\mu$ (as before), and expect continuity when $\epsilon \rightarrow 0$ (I haven't worked it out all the way).

Late edit: Note that $\delta p^\mu$ is a vector function of $p^\mu$ and $b^\mu$ which is expected to be linear in both. We can also argue that it should not include any second order (or higher) derivatives. These restrictions considerably narrow down the possibilities.

 

[Kostik]

Are you satisfied with Dirac's arguments that led him to the equation below (27.3) ?

Heck, no. Dirac says that if $b^\mu \propto p^\mu \propto u^\mu$, then at $x=z(\lambda)$: $$z^\mu (\lambda) + b^\mu (x) = z^\mu (\lambda) + \epsilon \frac{dz^\mu}{d\lambda} = z^\mu (\lambda + \epsilon)$$ where $\epsilon$ is small since we assume $b^\mu$ is small. Thus, $b^\mu$ displaces the element of matter along its worldline, and therefore $\delta p^\mu = 0$ . (This is where he uses the assumption that the matter distribution is a dust: all matter particles in the fluid element have the same velocity vector $u^\mu$, hence the same $p^\mu$.)

But he does not derive the equation $$\delta p^0 = (p^r b^0 - p^0 b^r)_{,r} \,\, ,$$ he merely observes that it satisfies (27.3) when $b^0 = 0$ and satisfies $\delta p^0 = 0$ when $b^\mu \propto p^\mu$.

 

[Kostik]

If yes, note that it is equal to $\delta p^0$ in (27.4). For the other components $\delta p^i$ , I think you can re-use Dirac's arguments: Starting with the equation above (27.3), this time with$$p^1 dx^0 dx^2 dx^3$$(for example). The main difference from before is that $p^1$ may vanish now, leaving us short of a constraint. To remedy that, we may try adding an infinitesimal $\epsilon$ component to $p^1$ , demand vanishing up to $O(\epsilon)$ when $b^1=0$ or when $b^\mu$ is parallel to $p^\mu$ (as before), and expect continuity when $\epsilon \rightarrow 0$ (I haven't worked it out all the way).

It's not clear that the argument above (27.3) for $p^0$ applies for $p^1$. The integral $$\int_V p^0 dx^1 dx^2 dx^3$$ is the mass within $V$ at a fixed time. Why is $$\delta \left( \int_V p^1 dx^0 dx^2 dx^3 \right) = -\int p^1 b^r \, dS_r \,\, ?$$ Not obvious at all.

$\qquad$ I'm inclined to think that Dirac should have made use of ${p^\mu}_{,\mu}=0$ to conclude directly that $\delta p^\mu$ must be $$(\text{something antisymmetric in } \mu \text{ and } \nu)_{,\nu} \,\, .$$ But even that would still be inferring (27.4), not deriving it.

 

[JimWhoKnew]

(This is where he uses the assumption that the matter distribution is a dust: all matter particles in the fluid element have the same velocity vector $v^\mu$, hence the same $p^\mu$.)

For dust, all matter particles in the infinitesimal fluid element have the same velocity vector $v^\mu$, hence the same $p^\mu$. The vector field defined by $v^\mu$ is not constant nor uniform in general.

It's not clear that the argument above (27.3) for $p^0$ applies for $p^1$. The integral $$\int_V p^0 dx^1 dx^2 dx^3$$ is the mass within $V$ at a fixed time. Why is $$\delta \left( \int_V p^1 dx^0 dx^2 dx^3 \right) = -\int p^1 b^r \, dS_r \,\, ?$$ Not obvious at all.

$p^0$ is positive and bound below. In all other aspects, the 4D treatment by Dirac is symmetric with respect to the roles of the components of $p^\mu$ . The conservation assumption (27.2)$$p^\mu{}_{,\mu}=0$$says that the geodesics of the congruence in 4D are analogous to the force lines (in vacuum) in electrostatics: nothing gets accumulated. Dirac mentions that $$p^1 dx^0 dx^2 dx^3$$is the amount (of matter) flowing through a volume element $dx^2 dx^3$ in the time interval $dx^0$ .

To visualize $$\delta \left( \int_V p^1 dx^0 dx^2 dx^3 \right) = -\int p^1 b^r \, dS_r $$we may try to suppress the $x^3$ component. In the $x^0-x^2$ plane, draw a closed curve, representing the volume and its boundary. The $x^1$ direction points into the page. We may choose $b^r$ , with $r=(0,2)$ , to be small enough, so that changes in $p^\mu$ over the "length" $b^r$ vanish to lowest order. Now, Dirac is considering an active displacement of the dust by $b^r$ , but we can equally well consider the passive displacement by $(-b^r)$, where the dust "stays put" and the volume is displaced. Try to see what you get for $$\delta \int p^1 dx^0 dx^2$$when $b^r=\epsilon \delta^r_2$ (no boundary deformation) and when $b^r=\epsilon(x) \delta^r_2$ (boundary is deformed).

But he does not derive the equation $$\delta p^0 = (p^r b^0 - p^0 b^r)_{,r} \,\, ,$$ he merely observes that it satisfies (27.3) when $b^0 = 0$ and satisfies $\delta p^0 = 0$ when $b^\mu \propto p^\mu$.

So we see that$$ (p^r b^0 - p^0 b^r)_{,r}$$satisfies all the requirements. I would not argue with your standards for derivation, but if it can be argued that it is the unique acceptable possibility that does so, I'll consider it a proof.

 

[JimWhoKnew]

$p^0$ is positive and bound below. In all other aspects, the 4D treatment by Dirac is symmetric with respect to the roles of the components of $p^\mu$ . The conservation assumption (27.2)$$p^\mu{}_{,\mu}=0$$says that the geodesics of the congruence in 4D are analogous to the force lines (in vacuum) in electrostatics: nothing gets accumulated. Dirac mentions that $$p^1 dx^0 dx^2 dx^3$$is the amount (of matter) flowing through a volume element $dx^2 dx^3$ in the time interval $dx^0$ .

To visualize $$\delta \left( \int_V p^1 dx^0 dx^2 dx^3 \right) = -\int p^1 b^r \, dS_r $$we may try to suppress the $x^3$ component. In the $x^0-x^2$ plane, draw a closed curve, representing the volume and its boundary. The $x^1$ direction points into the page. We may choose $b^r$ , with $r=(0,2)$ , to be small enough, so that changes in $p^\mu$ over the "length" $b^r$ vanish to lowest order. Now, Dirac is considering an active displacement of the dust by $b^r$ , but we can equally well consider the passive displacement by $(-b^r)$, where the dust "stays put" and the volume is displaced. Try to see what you get for $$\delta \int p^1 dx^0 dx^2$$when $b^r=\epsilon \delta^r_2$ (no boundary deformation) and when $b^r=\epsilon(x) \delta^r_2$ (boundary is deformed).

I thought some more about it and found that my argument is insufficient.
An arbitrary $b^\mu (x)$ displacement introduces density changes. In the $\delta p^0$ case, where the volume is spacelike and $b^0=0$ , each particle is counted once, along with its displacement-invariant contribution to $p^0$. So the difference between the integrals comes entirely from the boundary, as in Dirac's book. In the case of$$
\delta \left( \int_V p^1 dx^0 dx^2 dx^3 \right)$$where $b^1=0$ , each worldline passes through infinitely many infinitesimal volume elements, so (currently) I can't argue that all the difference must come solely from the boundary.

 

[Kostik]

I thought some more about it and found that my argument is insufficient.
An arbitrary $b^\mu (x)$ displacement introduces density changes. In the $\delta p^0$ case, where the volume is spacelike and $b^0=0$ , each particle is counted once, along with its displacement-invariant contribution to $p^0$. So the difference between the integrals comes entirely from the boundary, as in Dirac's book. In the case of$$
\delta \left( \int_V p^1 dx^0 dx^2 dx^3 \right)$$where $b^1=0$ , each worldline passes through infinitely many infinitesimal volume elements, so (currently) I can't argue that all the difference must come solely from the boundary.

I found a reasonable proof that Dirac's (27.3) holds for $p^k$ (still in the $b^0=0$ case). Dirac should actually have done this before moving on to the $b^0 \neq 0$ case:

 

[Kostik]

Dirac simply guesses that, for $b^0 \neq 0$, the formula is $$\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu} $$ and verifies that it gives Eq. (27.3) when $b^0 = 0$, and also that $\delta p^0 = 0$ when $b^\mu$ is proportional to $p^\mu$.

He could also have rationalized this guess by showing that it satisfies ${p^\mu}_{,\mu}=0$. But either way, he is simply rationalizing a guess.

Which brings me back to my original post ... I think that the derivation is correct, I may just need to convince myself of this.

 

[JimWhoKnew]

since $u^0=dt/d\lambda$ has no spatial dependence

Read again the first sentences of chapters 25 and 27 in Dirac's book.

 

[Kostik]

Read again the first sentences of chapters 25 and 27 in Dirac's book.

I think you have misunderstood. In Ch. 27, Dirac defers introduction of the metric, so the trajectory of a matter element (fluid element) is $z(\lambda)$, where $\lambda$ is any parameter. In the "laboratory frame", $t=x^0$ is time. He could easily have taken $\lambda=t$, in which case $dz/d\lambda = dz/dt$. It makes no difference; but at this stage, since there is no metric, there is no spacetime interval $ds$.

$dt/d\lambda$ is just the ratio of the parameters: it does not have any spatial dependence. You can choose $\lambda=t$ if you like, and then $dt/d\lambda=1$.

 

[JimWhoKnew]

The displacement field $b^\mu$ can not be fully arbitrary, since we want a diffeomorphism (we don't want 2 initially distinct points to land on the same target one, nor to cross each other's "path"). A way to guarantee safety is to demand $b^\mu{}_{,\mu}=0$ ("force lines don't cross").
So with $b^\mu{}_{,\mu}=0$ and $p^\mu{}_{,\mu}=0$ , equation (27.4) is just the negative of the Lie derivative$$\mathrm{L}_\mathbf{b}p^\mu=\left[\mathbf{b},\mathbf{p}\right]^\mu$$

 

[JimWhoKnew]

I think you have misunderstood. In Ch. 27, Dirac defers introduction of the metric, so the trajectory of a matter element (fluid element) is $z(\lambda)$, where $\lambda$ is any parameter. In the "laboratory frame", $t=x^0$ is time. He could easily have taken $\lambda=t$, in which case $dz/d\lambda = dz/dt$. It makes no difference; but at this stage, since there is no metric, there is no spacetime interval $ds$.

$dt/d\lambda$ is just the ratio of the parameters: it does not have any spatial dependence. You can choose $\lambda=t$ if you like, and then $dt/d\lambda=1$.

Dirac's discussion doesn't forbid the choice that $u^0$ is spatially independent. But then, if you add the assumption $u^k{}_{,n}=0$, you run into troubles with the possibility of $v^\mu$ to vary from one point to a neighboring one, when the metric is re-introduced.
(For example: if at a single point $u^\mu \propto \delta^\mu_0$ , then $v^\mu=\delta^\mu_0$ everywhere on the timeslice)

Note that $u^\mu$ drops out from the final results, as it should.

 

[Kostik]

Dirac's discussion doesn't forbid the choice that $u^0$ is spatially independent. But then, if you add the assumption $u^k{}_{,n}=0$, you run into troubles with the possibility of $v^\mu$ to vary from one point to a neighboring one, when the metric is re-introduced.
(For example: if at a single point $u^\mu \propto \delta^\mu_0$ , then $v^\mu=\delta^\mu_0$ everywhere on the timeslice)

Note that $u^\mu$ drops out from the final results, as it should.

Just take $\lambda=t$ and the issue goes away. Of course, $u^\mu$ varies throughout the matter field, but the whole point of the matter field being a dust is that, within a sufficiently small "matter (volume) element", the tangent vector $u^\mu$ (or 4-velocity $v^\mu$) is constant. (Equivalently, all the matter particles are at rest in the Momentarily Co-moving Reference Frame of the matter particle at a particular point $x$.) Therefore, choosing $b^n$ smaller than the size of the fluid element, we have $\delta u^k = 0$, and also ${u^k}_{,n} = 0$.

 

[JimWhoKnew]

Just take $\lambda=t$ and the issue goes away.

That is choosing, as I said above. It can be done, but Dirac does not impose it.

Of course, $u^\mu$ varies throughout the matter field, but the whole point of the matter field being a dust is that, within a sufficiently small "matter (volume) element", the tangent vector $u^\mu$ (or 4-velocity $v^\mu$) is constant. (Equivalently, all the matter particles are at rest in the Momentarily Co-moving Reference Frame of the matter particle at a particular point $x$.) Therefore, choosing $b^n$ smaller than the size of the fluid element, we have $\delta u^k = 0$, and also ${u^k}_{,n} = 0$.

Well... It seems that Dirac implicitly does something in that spirit below equation (25.8), so maybe his $u^\mu$ is indeed piecewise spatially constant. In chapter 27 he claims that his approach "makes the discussion clearer". It is certainly not clear enough for me.

 

[Kostik]

If Dirac is not going to actually derive (27.4) (which I think I have correctly done in the original post), but rather to simple infer its reasonableness, he might as well simply have shown -- as he does -- that using (27.4), the principle of stationary action yields Einstein's equations for a dust (25.7) starting from the appropriate action $I$ given in (27.9). Since we expect the action principle with the correct action for any particular matter field to yield Einstein’s equation with the energy-momentum tensor corresponding to the field, we must infer the correctness of (27.4).

 

[Kostik]

Well... It seems that Dirac implicitly does something in that spirit below equation (25.8), so maybe his $u^\mu$ is indeed piecewise spatially constant. In chapter 27 he claims that his approach "makes the discussion clearer". It is certainly not clear enough for me.

$u^\mu$ must be constant within a sufficiently small volume element of the fluid. That's what it means for all matter particles to be at rest within the momentarily co-moving reference frame of the fluid at a point. If not, then the particles would have to be interacting with one another, creating pressure and/or shear forces (or the particles would be under the influence of other external forces).

 

[Kostik]

Just take $\lambda=t$ and the issue goes away. Of course, $u^\mu$ varies throughout the matter field, but the whole point of the matter field being a dust is that, within a sufficiently small "matter (volume) element", the tangent vector $u^\mu$ (or 4-velocity $v^\mu$) is constant. (Equivalently, all the matter particles are at rest in the Momentarily Co-moving Reference Frame of the matter particle at a particular point $x$.) Therefore, choosing $b^n$ smaller than the size of the fluid element, we have $\delta u^k = 0$, and also ${u^k}_{,n} = 0$.

I'm thinking about this. I am comfortable with ${u^k}_{,n} = 0$, but I'm having doubts about $\delta u^k = 0$. The displacement $b^\mu$ applies to the matter element; and if you move an entire matter element from one place to another nearby place, I don't see why $u^\mu$ stays the same. Alternatively, $$\delta u^k = \delta \left( \frac{dz^k}{d\lambda} \right) = \frac{d(\delta z^k)}{d\lambda} = \frac{db^k}{d\lambda} \,\, .$$ I don't see why this vanishes. The displacement field $b^k$ is arbitrary (but small).

Hang on: maybe $db^k/d\lambda$ vanishes simply because $b^k$ doesn't depend on ##lambda#??

 

[JimWhoKnew]

The displacement field $b^\mu$ can not be fully arbitrary, since we want a diffeomorphism (we don't want 2 initially distinct points to land on the same target one, nor to cross each other's "path"). A way to guarantee safety is to demand $b^\mu{}_{,\mu}=0$ ("force lines don't cross").
So with $b^\mu{}_{,\mu}=0$ and $p^\mu{}_{,\mu}=0$ , equation (27.4) is just the negative of the Lie derivative$$\mathrm{L}_\mathbf{b}p^\mu=\left[\mathbf{b},\mathbf{p}\right]^\mu$$

Since I joined this thread, I was puzzled by the question why doesn't $\delta p^\mu$ in equation (27.4) just equal the Lie derivative (or its negative, depending on chosen convention). Today I noticed something that I've missed due to lack of sufficient attention. In chapter 27 Dirac defines $p^\mu$ to be a vector density, rather than a vector field. Now, if we set$$p^\mu\equiv \sqrt{g} P^\mu$$where $P^\mu$ is a vector field, we get from equation (21.1) that the assumption $p^\mu{}_{,\mu}=0$ implies $P^\mu{}_{;\mu}=0$ .
The active displacement of matter can be viewed as a passive displacement of the coordinates in the opposite directions:$$x'^\mu=x^\mu-b^\mu \quad .$$Using the conventions and results in sections 4.7 and 10.9 in Weinberg's book:$$\delta p^\mu = \delta \sqrt{g} P^\mu+\sqrt{g} \delta P^\mu=\frac{\sqrt{g}}2 g^{\rho\sigma}\delta g_{\rho\sigma} P^\mu+\sqrt{g} \delta P^\mu \quad .$$Setting$$\delta P^\mu=-\mathrm{L}_\mathbf{b}P^\mu=-b^\nu P^\mu{}_{,\nu}+P^\nu b^\mu{}_{,\nu}$$and $$\delta g_{\mu\nu}=-\mathrm{L}_\mathbf{b}g_{\mu\nu}=-b_{\mu ; \nu}-b_{\nu ; \mu} \quad ,$$along with Dirac's constraint $p^\mu{}_{,\mu}=0$ and $$(p^\mu b^\nu)_{,\nu}=[P^\mu(\sqrt{g}b^\nu)]_{,\nu}=P^\mu{}_{,\nu}\sqrt{g}b^\nu+p^\mu b^\nu{}_{;\nu}$$ to eliminate the connections with equation (21.1), I get equation (27.4) $$\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu} \quad .$$

(no further assumptions added this time)

 

[Kostik]

post under construction. please dont reply

Since I joined this thread, I was puzzled by the question why doesn't $\delta p^\mu$ in equation (27.4) just equal the Lie derivative (or its negative, depending on chosen convention). Today I noticed something that I've missed due to lack of sufficient attention. In chapter 27 Dirac defines $p^\mu$ to be a vector density, rather than a vector field. Now, if we set$$p^\mu\equiv \sqrt{g} P^\mu$$where $P^\mu$ is a vector field, we get from equation (21.1) that the assumption $p^\mu{}_{,\mu}=0$ implies $P^\mu{}_{;\mu}=0$ .
The active displacement of matter can be viewed as a passive displacement of the coordinates in the opposite directions:$$x'^\mu=x^\mu-b^\mu \quad .$$Using the conventions and results in sections 4.7 and 10.9 in Weinberg's book:$$\delta p^\mu = \delta \sqrt{g} P^\mu+\sqrt{g} \delta P^\mu=\frac{\sqrt{g}}2 g^{\rho\sigma}\delta g_{\rho\sigma} P^\mu+\sqrt{g} \delta P^\mu \quad .$$Setting$$\delta P^\mu=-\mathrm{L}_\mathbf{b}P^\mu=-b^\nu P^\mu{}_{,\nu}+P^\nu b^\mu{}_{,\nu}$$and $$\delta g_{\mu\nu}=-\mathrm{L}_\mathbf{b}g_{\mu\nu}=-b_{\mu ; \nu}-b_{\nu ; \mu} \quad ,$$along with Dirac's constraint $p^\mu{}_{,\mu}=0$ and $(p^\mu b^\nu)_{,\nu}=[P^\mu(\sqrt{g}b^\nu)]_{,\nu}$ to eliminate the connections with equation (21.1), I get equation (27.4) $$\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu} \quad .$$

(no further assumptions added this time)

We know $p^\mu = \rho v^\mu \sqrt{-g}$. Your $P^\mu$ is $\rho v^\mu$.

Your equation below "Using the conventions and results in sections 4.7 and 10.9 in Weinberg's book" follows immediately from Dirac's equation (26.10).

I don't see how Weinberg's (10.9.9) is the same as the equation immediately above (10.9.7). Not even remotely. Can you help with this?

Then I think I see what you are doing, using Weinberg's (10.9.9) and the equation on p. 292 for $\Delta_\epsilon U^\mu$.

 

[JimWhoKnew]

We know $p^\mu = \rho v^\mu \sqrt{-g}$. Your $P^\mu$ is $\rho v^\mu$.

Your equation below "Using the conventions and results in sections 4.7 and 10.9 in Weinberg's book" follows immediately from Dirac's equation (26.10).

I don't see how Weinberg's (10.9.9) is the same as the equation immediately above (10.9.7). Not even remotely. Can you help with this?

Then I think I see what you are doing, using Weinberg's (10.9.9) and the equation on p. 292 for $\Delta_\epsilon U^\mu$.

I explicitly asked "please dont reply yet" (check the quotation in your reply #21). The LaTeX preview issue causes some of us to post unfinished versions, especially when many equations are involved. Common courtesy will be appreciated.

Your equation below "Using the conventions and results in sections 4.7 and 10.9 in Weinberg's book" follows immediately from Dirac's equation (26.10).

So??

Edit:
Equation (10.9.9) in Weinberg's follows immediately from the general case (10.9.10). A nice treatment can be found in d'Inverno's book (new edition, and likely in the old one too), although he declares an active displacement but appears to be using a passive one without changing the sign.

Additional edit:
Ok, now I think I see what's bothering you. In the line above (10.9.7) Weinberg writes "In covariant terms" and applies the "comma goes to semicolon" rule to the equation above (10.9.7). Replacing $g_{\mu\nu}$ by $T_{\mu\nu}$ in the equation above (10.9.7), and then "covarianting", gives you (10.9.10).

 

[Kostik]

@JimWhoKnew No, I am perfectly happy with (10.9.9) being a special case of (10.9.10). I get to Weinberg's equation above (10.9.7) alright, and then (10.9.7) says that $\Delta_\epsilon g_{\mu\nu}(x)$ is equal to the last three terms in the equation above (10.9.7). But how the heck are these three terms equal to (10.9.9)?

(PS. Apologies for the premature post above.)

 

[JimWhoKnew]

@JimWhoKnew No, I am perfectly happy with (10.9.9) being a special case of (10.9.10). I get to Weinberg's equation above (10.9.7) alright, and then (10.9.7) says that $\Delta_\epsilon g_{\mu\nu}(x)$ is equal to the last three terms in the equation above (10.9.7). But how the heck are these three terms equal to (10.9.9)?

(PS. Apologies for the premature post above.)

Given the "additional edit" in post #22, I don't understand what is still bothering you. So I'll repeat the same argument, but with some more details.
The 3 difference terms above (10.9.7) are$$g_{\rho\nu}\epsilon^\rho{}_{,\mu}+g_{\rho\mu}\epsilon^\rho{}_{,\nu}+g_{\mu\nu,\rho}\epsilon^\rho \quad .$$Now, on the line between the equations, Weinberg says "in covariant terms, we conclude that". I interpret it as an implicit application of the "comma goes to semicolon" practice that guarantees covariance. The 3 terms will become$$\Delta_\epsilon g_{\mu\nu}\equiv g_{\rho\nu}\epsilon^\rho{}_{;\mu}+g_{\rho\mu}\epsilon^\rho{}_{;\nu}+g_{\mu\nu;\rho}\epsilon^\rho \quad ,$$which is the same as (10.9.9).

Does that answer your question? If not, what is missing?

 

[Kostik]

Does that answer your question? If not, what is missing?

Sorry, brain freeze. Yes, it's obvious. Thanks, this has been really useful, and a much better way to show Dirac's relation without mucking about with all those delta functions. (Interestingly, for a guy who invented the delta function, he never uses it once in his "GTR" book, opting to work only with matter fields treated as continua.)

 

[Kostik]

@JimWhoKnew How do you get $$\delta P^\mu = -\mathcal{L}_b \, P^\mu$$ and $$\delta g_{\mu\nu} = -\mathcal{L}_b \, g_{\mu\nu} \quad ?$$ I get $+$ signs. Yes, here $\delta P^\mu = P'^\mu - P^\mu$ (and likewise for $\delta g_{\mu\nu}$), which is opposite to the usual Lie derivative definition, but also the coordinate transformation is $x'^\mu = x^\mu - b^\mu$. Accordingly, I get the opposite of Dirac's eq. (27.4).

To be clear, if Dirac applies a displacement field $$x \rightarrow x+\delta x = x+b$$ then The components of a tensor $T$ at $x$ change to $T(x-b)$. Thus, the coordinate change here is $x' = x - b$.

Here is my logic; could you please identify the mistake?

 

[JimWhoKnew]

@JimWhoKnew How do you get $$\delta P^\mu = -\mathcal{L}_b \, P^\mu$$ and $$\delta g_{\mu\nu} = -\mathcal{L}_b \, g_{\mu\nu} \quad ?$$ I get $+$ signs. Yes, here $\delta P^\mu = P'^\mu - P^\mu$ (and likewise for $\delta g_{\mu\nu}$), which is opposite to the usual Lie derivative definition, but also the coordinate transformation is $x'^\mu = x^\mu - b^\mu$. Accordingly, I get the opposite of Dirac's eq. (27.4).

To be clear, if Dirac applies a displacement field $$x \rightarrow x+\delta x = x+b$$ then The components of a tensor $T$ at $x$ change to $T(x-b)$. Thus, the coordinate change here is $x' = x - b$.

Here is my logic; could you please identify the mistake?

View attachment 363216

With my physics background, I operate on the belief that signs will always take care of themselves and that one should never confront his calculations with reality unless absolutely necessary.

The present case doesn't qualify...

I mentioned "conventions" because these cases are tricky. Compare the equation above (10.9.7) in Weinberg's with equation (6.7) in d'Inverno's. In case you don't have access, I'll mention that d'Inverno uses $T(x')-T'(x')$ . They both get the same expressions for the Lie derivatives because d'Inverno uses $x'=x+b$ , unlike Weinberg's (10.9.6). This teaches us that we should consider all the relevant assumptions, rather than just the final expression. The second equation in your attachment contains an implicit assumption about the sign of $\delta$ .

So now we have to reconstruct Dirac's "meaning". Maybe you've read it in one of the earlier chapters, I didn't. But anyway, we can use the paragraph leading to (27.3) for this purpose. Since we are only interested in the sign, we can choose $b^\mu=\epsilon \delta^\mu_1$ and assume that $p^0{}_{,1}$ and $P^0{}_{;1}$ are both positive. With that,$$\Delta_b g_{\mu\nu}=0 \quad ,$$$$\delta \sqrt{g}=0 \quad , $$and$$\Delta_b P^\mu = \epsilon P^\mu{}_{;1} \quad .$$Since all $~\partial x'^\mu / \partial x^\nu-\delta^\mu_\nu~$ vanish with our choice, we don't have to worry about their contributions.
Following the example in the equation above (10.9.7) in Weinberg's, we'll get for this specific case $$P'^0(x)-P^0(x)=P^0(x+b)-P^0(x)+O(\epsilon^2) \quad (>0) \quad . $$But Dirac seems to use $$p^0(x-b)-p^0(x) \quad (<0) $$instead (actively moving the vector density "to the right" and subtracting the values of the vector density that was at that point before), hence the extra minus sign in his definition of $\delta p^0$ . To align my results with Dirac's, I chose the negative of Weinberg's Lie derivatives in #20.

BTW:
Note that the variation that uses Lie derivatives, as in #20, is general and can be applied to any tensor. Applied to $p^\mu$, it will be equal to equation (27.4) only because of the assumption $p^\mu{}_{,\mu}=0$ .

 

[Kostik]

@JimWhoKnew I think Dirac is using $\delta$ is the most basic way it is used when varying a path or dynamical coordinate, to implement the principle of stationary action. Using the action $I_g + I_m$, Dirac must vary $g_{\mu\nu}$ and -- in view of Eq. (27.9) -- $p^\mu$. Thus, the varied path or variation is $$\bar{g}_{\mu\nu} = g_{\mu\nu} + \delta g_{\mu\nu}$$ $$\bar{p}^{\mu} = p^{\mu} + \delta p^{\mu} \, .$$ My error isn't in the definition of the Lie derivative. I am actually using the computations from Ohanian & Ruffini's "Gravitation and Spacetime", and I am carrying them through carefully, line by line. And as I said, I get $$\delta p^\mu = \mathcal{L}_b \, p^\mu$$ $$\delta g_{\mu\nu} = \mathcal{L}_b \, g_{\mu\nu}$$ which gives me the opposite of Dirac's Eq. (27.4).

Dirac's simple derivation of $\delta p^0$ in the base $b^0=0$ (his Eq. (27.3)) is very clear and gives us the correct signs.

My mistake must be something conceptual in the way I have set up my picture of the active vs. passive transformation, the definition of $\delta p^\mu$ in terms of $p^\mu$ and ${p'}^\mu$, etc.

I have D'Inverno & Vickers (2nd Ed) and will read their treatment carefully. I just need to get myself off on the right foot in terms of clearly describing how the displacement of the matter field $$z^\mu \longrightarrow z^\mu + \delta z^\mu = z^\mu + b^\mu$$ implies $$p^\mu \longrightarrow p^\mu + \delta p^\mu$$ is equivalent to a change of coordinates $x \rightarrow x'$, with the appropriate formula for $\delta p^\mu$ in terms of $p^\mu$ and ${p'}^\mu$. (Note that D'Inverno essentially does the same thing, when he writes $x^a \rightarrow x^a + \delta u X^a(x)$.)

My work is shown below:

 

[JimWhoKnew]

I think Dirac is using $\delta$ is the most basic way it is used when varying a path or dynamical coordinate, to implement the principle of stationary action.

In general we have$$\delta= ~^"\mathrm{new}^"~ - ~^"\mathrm{old}^" \quad .$$Apparently this should be straightforward: we know who the old is, we can figure out the new, go ahead. So why "apparently"?

Let's consider a simplified example:
Consider a scalar field $\phi(x)$ , since we currently have no need in the headache of transforming components. Let's assume that $g_{\mu\nu}=\eta_{\mu\nu}+O(\epsilon^2)$ , so we can sloppily ignore the difference between $\phi$ and its associated scalar density. Assume further $b^\mu=\epsilon \delta^\mu_1$ .
Case 1:
We actively displace $\phi$ by $b$ . The new value at the point $x$ is the old $\phi(x-b)$ so$$\delta \phi=\phi(x-b)-\phi(x)=-\epsilon \phi_{,1} \quad . $$The same result is obtained if we passively move the spatial volume element in the $(-b)$ direction.
Case 2:
We apply the corresponding passive displacement $x'=x-b$ . Naturally the new value is $\phi'(x')$ . But to get $\delta\phi$ we have to compare the values at the same nominal point $x^\mu$ , so$$\delta\phi= \phi'(x)-\phi(x)=\phi'(x'+b)-\phi(x)=\phi'(x')+\epsilon \phi'_{,1}-\phi(x)=\epsilon \phi_{,1} \quad .$$
So which one is the correct $\delta \phi$ , and why?

I think Dirac is using $\delta$ is the most basic way it is used when varying a path or dynamical coordinate, to implement the principle of stationary action. Using the action $I_g + I_m$, Dirac must vary $g_{\mu\nu}$ and -- in view of Eq. (27.9) -- $p^\mu$. Thus, the varied path or variation is $$\bar{g}_{\mu\nu} = g_{\mu\nu} + \delta g_{\mu\nu}$$ $$\bar{p}^{\mu} = p^{\mu} + \delta p^{\mu} \, .$$ My error isn't in the definition of the Lie derivative. I am actually using the computations from Ohanian & Ruffini's "Gravitation and Spacetime", and I am carrying them through carefully, line by line. And as I said, I get $$\delta p^\mu = \mathcal{L}_b \, p^\mu$$ $$\delta g_{\mu\nu} = \mathcal{L}_b \, g_{\mu\nu}$$ which gives me the opposite of Dirac's Eq. (27.4).

The derivation of the geodesic equation from equation (27.11) will not be affected by a change of sign $\delta p^\mu \rightarrow -\delta p^\mu$ in (27.4).

I am actually using the computations from Ohanian & Ruffini's "Gravitation and Spacetime", and I am carrying them through carefully, line by line.

What sections?

My work is shown below:

Your attached files are just what I did in post #20, except for my insertion "by hand" of the minus signs to fit Dirac's (27.4).

 

[Kostik]

What sections?

"Gravitation and Spacetime", 3rd Ed., Ohanian & Ruffini, pp. 262-264. I like how he computes the Lie derivative, keeping careful track of the error terms in exchanging $x$ and $x'$, and $T_{\mu\nu}$ and $T'_{\mu\nu}$.