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

[Kostik]

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"?

Yep, this is where the issue really lies. If we vary $$x^\mu \rightarrow \bar{x}^\mu = x^\mu + \delta x^\mu = x^\mu + b^\mu \,\, ,$$ and if $T(x)$ is a tensor field, then what EXACTLY is the resultant variation $\delta T(x)$?

It cannot be $T(x) - T(x-b)$, as I originally thought, because we cannot subtract vectors/tensors evaluated at different points. I need to decide exactly what $\delta T(x)$ means before computing it.

D'Inverno talks about dragging the tensor $T^{ab}$ from $P$ to $Q$ -- see his Figure 6.3. What does he mean exactly -- is this parallel transport?

Obviously, to get the right answer, I want $$\delta T(x) = T'(x') - T(x')$$ (or $\delta T(x) = T'(x) - T(x)$ ... does it matter?) where $T'$ is the transformed tensor under the coordinate change $x' = x + b$. (Thereby making $\delta T(x)$ equal to the opposite of the Lie derivative.) I just need to justify this to myself.

 

[Kostik]

I might have it figured out. Will post a coherent explanation if I do.

 

[JimWhoKnew]

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 "comma goes to semicolon" is optional, since Lie derivatives are covariant with it and without it (page 137 in Carroll's "Lecture Notes on GR"). There are applications where the manifestly covariant form (semicolons) is more convenient.

BTW:
I think there is a typo in Weinberg's equation (10.9.10). $T^\lambda{}_\mu$ should be $T_\mu{}^\lambda$ .

 

[Kostik]

@JimWhoKnew My derivation is shown below. It's a little verbose, but it helps me understand what's going on. I find it a little tricky keeping the $x$s and $x'$s straight, and also taking care with the $T$s and $T'$s and watching the error terms. Still, I think the approach is sound.

 

[Kostik]

@JimWhoKnew Something still bothers me here. Dirac went through all this trouble to establish (27.4), because he writes the matter action as $$I_m = -\int (p^\mu p_\mu)^{1/2} \, .$$ But then in his eq. (27.10), he writes the variation $\delta(I_g + I_m)$ as a sum of integrals, one with $\delta g_{\mu\nu}$ and one with $\delta p^\mu$. Then he uses (27.4) to write $\delta p^\mu$ in terms of $b^\mu$.

HOWEVER, you showed that $\delta g_{\mu\nu}=-\mathcal{L}_b \, g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ is also a function of $b^\mu$.

So, $\delta g_{\mu\nu}$ and $b^\mu$ are not independent. Hence, how can he justify setting both integrands equal to zero?

Can $b^\mu$ and $b_{\mu:\nu}+b_{\nu:\mu}$ be varied independently?

 

[JimWhoKnew]

@JimWhoKnew Something still bothers me here. Dirac went through all this trouble to establish (27.4), because he writes the matter action as $$I_m = -\int (p^\mu p_\mu)^{1/2} \, .$$ But then in his eq. (27.10), he writes the variation $\delta(I_g + I_m)$ as a sum of integrals, one with $\delta g_{\mu\nu}$ and one with $\delta p^\mu$. Then he uses (27.4) to write $\delta p^\mu$ in terms of $b^\mu$.

HOWEVER, you showed that $\delta g_{\mu\nu}=-\mathcal{L}_b \, g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ is also a function of $b^\mu$.

So, $\delta g_{\mu\nu}$ and $b^\mu$ are not independent. Hence, how can he justify setting both integrands equal to zero?

Can $b^\mu$ and $b_{\mu:\nu}+b_{\nu:\mu}$ be varied independently?

I noticed this problem a couple of days ago. The bottom line: I don't understand how Dirac gets (27.4).

I didn't show that $\delta g_{\mu\nu}=-\mathcal{L}_b \, g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ . All I showed is that by substituting it, we can recover (27.4) up to a term proportional to $p^\mu{}_{,\mu}$ . Big difference.

An idea that I have, but not sure of its validity:$$\delta_b g_{\mu\nu}\equiv -\mathcal{L}_b \, g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$$is the displacement-induced change that should be used in the derivation of $\delta p^\mu$ . To that, we are free to add an arbitrary $\delta_1 g_{\mu\nu}$ , so that$$\delta g_{\mu\nu}=\delta_b g_{\mu\nu}+\delta_1 g_{\mu\nu}$$ is completely arbitrary and not fixed by the displacement.

 

[Kostik]

Well, spurred on by your hint, I am comfortable that I have derived Dirac's (27.4) as shown in the notes pasted in #34 above. Now, as to the point that $\delta g_{\mu\nu} = -(b_{\mu:\nu}+b_{\nu:\mu})$, and the question whether $\delta g_{\mu\nu}$ and $b^\mu$ can be varied independently: I think I have an answer to that.

Write $b^\mu = ε B^\mu$. We can equally well choose $b^\mu = ε B^\mu + ε^2 A^\mu$ with a negligible effect on $b^\mu$. If we choose $A^\mu$ such that ${A^\mu}_{:\nu} \gg ε^{-2}$, then $$b_{\alpha:\beta}+b_{\beta:\alpha} = ε(B_{\alpha:\beta}+B_{\beta:\alpha}) + ε^2(A_{\alpha:\beta}+A_{\beta:\alpha}) \approx ε^2(A_{\alpha:\beta}+A_{\beta:\alpha})$$ which allows $b^\mu$ and $g_{\mu\nu}$ to be varied independently.

 

[dextercioby]

Why all this mumbo-jumbo? A vector function and its covariant derivative are functionally independent. Like x and xdot in elementary Lagrangian mechanics.

 

[Kostik]

Why all this mumbo-jumbo? A vector function and its covariant derivative are functionally independent. Like x and xdot in elementary Lagrangian mechanics.

What do you mean by "functionally independent"? It seems to me to require some justification, in order to carry out the final step after Dirac's (27.10) and set the coefficients of $\delta g_{\mu\nu}$ and $b^\mu$ equal to zero.

In the traditional principle of stationary action, we only vary $x^\mu$, not both $\dot{x}^\mu$. So, the situation here seems a little tricker.

 

[JimWhoKnew]

Why all this mumbo-jumbo? A vector function and its covariant derivative are functionally independent. Like x and xdot in elementary Lagrangian mechanics.

If we take Dirac's equation (27.4)$$
\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu}$$as provided, then it is a function of the variation variables $b^\mu$ and $b^\mu{}_{,\nu}$ . Taking $\delta g_{\mu\nu}$ as independent of these, we can happily apply it to the equation below (27.9) and get that the two terms in (27.10) have independent variations.
But if we attempt to justify (27.4) along the lines of post #20, we use$$\delta g_{\mu\nu}=-g_{\mu\nu,\rho}b^\rho-b_{\mu,\nu}-b_{\nu,\mu}$$which depends on the same variation variables as $\delta p^\mu$ . With this, the two RHS terms in (27.10) are no longer manifestly independent.
However, Dirac shows in equation (27.11) that the right term of (27.10) can be brought to a form that depends only on $b^\mu$ , so $\delta g_{\mu\nu}$ in the left term (of the RHS) can be regarded as independent (since it also contains $b^\mu{}_{,\nu}$ ). This is in accord with your (and @Kostik's) observation. Although that seems to settle the problem, I'm still not confident about the process I described in #20.

If we choose $A^\mu$ such that ${A^\mu}_{:\nu} \gg ε^{-2}$

With such a choice, $b^\mu{}_{;\nu}$ are no longer small.

 

[JimWhoKnew]

Why all this mumbo-jumbo? A vector function and its covariant derivative are functionally independent. Like x and xdot in elementary Lagrangian mechanics.

If in elementary Lagrangian mechanics we have a single particle Lagrangian$$\mathrm{L}\left(x,y,z,\dot x,\dot y ,\dot z \right)$$then for$$\delta\mathrm{I}=\int \delta \mathrm{L}dt$$we can vary $x$ arbitrarily while keeping $\delta y=\delta z=0$ (and similarly for the other cases). So $\delta x,\delta y,\delta z$ are mutually independent.
But that is not the case for $\delta x$ and $\delta \dot x$ . To derive the equations of motion we use$$\delta \dot x=\frac d{dt}(\delta x) \quad ,$$and if $\delta x$ vanishes during the entire action integration, so does $\delta \dot x$ .

Late edit:
Moreover, we end up with$$0=\int \left[\mathrm{~equation~~of~~motion~ }\right]_i \delta x^i dt \quad .$$$\delta \dot x ^i$ are eliminated. This is different from equation (27.10) for the case described in #40.

so $\delta g_{\mu\nu}$ in the left term (of the RHS) can be regarded as independent (since it also contains $b^\mu{}_{,\nu}$ ). This is in accord with your (and @Kostik's) observation. Although that seems to settle the problem,

Probably wrong.

 

[Kostik]

If we take Dirac's equation (27.4)$$
\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu}$$as provided, then it is a function of the variation variables $b^\mu$ and $b^\mu{}_{,\nu}$ . Taking $\delta g_{\mu\nu}$ as independent of these, we can happily apply it to the equation below (27.9) and get that the two terms in (27.10) have independent variations.
But if we attempt to justify (27.4) along the lines of post #20, we use$$\delta g_{\mu\nu}=-g_{\mu\nu,\rho}b^\rho-b_{\mu,\nu}-b_{\nu,\mu}$$which depends on the same variation variables as $\delta p^\mu$ . With this, the two RHS terms in (27.10) are no longer manifestly independent.
However, Dirac shows in equation (27.11) that the right term of (27.10) can be brought to a form that depends only on $b^\mu$ , so $\delta g_{\mu\nu}$ in the left term (of the RHS) can be regarded as independent (since it also contains $b^\mu{}_{,\nu}$ ). This is in accord with your (and @Kostik's) observation. Although that seems to settle the problem, I'm still not confident about the process I described in #20.


With such a choice, $b^\mu{}_{;\nu}$ are no longer small.

But $b^\mu$ is small, which is the assumption.

 

[JimWhoKnew]

But $b^\mu$ is small, which is the assumption.

Remember that in the derivation of $\mathrm{L}_\mathbf{b}g_{\mu\nu}$ , terms of second order or higher in $b^\mu$ and $b^\mu{}_{,\nu}$ are discarded. It makes no sense unless they are all small.

Edit:
Even worse, $\partial x'^\mu/\partial x^\nu$ may become singular.

 

[Kostik]

Remember that in the derivation of $\mathrm{L}_\mathbf{b}g_{\mu\nu}$ , terms of second order or higher in $b^\mu$ and $b^\mu{}_{,\nu}$ are discarded. It makes no sense unless they are all small.

Oh, that is quite right. So, my method fails -- and I am at a loss how to show that $\delta g_{\mu\nu}$ can be varied independently of $b^\mu$ in Dirac's (27.10).

From a physical point of view, the goal is to obtain $g_{\mu\nu}$ (Dirac's Eq (25.7)) from the stationary variation. (The geodesic equation is a bonus.) So it's natural that we should arbitrarily vary $g_{\mu\nu}$ in (27.10), just as we do in the vacuum case (26.8). It seems to be physically reasonable to vary $g_{\mu\nu}$ and $b^\mu$ independently. I just need to see, mathematically, how that is justifiable, given the relation $g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$.

EDIT: Hang on, maybe I'm getting hung up on $g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ and the apparent relation it imposes on $g_{\mu\nu}$ and $b^{\mu}$ in Dirac's (27.10). It seems that $g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ is simply the effect on $g_{\mu\nu}$ caused by $b^{\mu}$. But in (27.10), we are at liberty to vary $g_{\mu\nu}$ arbitrarily, since this is allowed to make an arbitrary variation $\delta(I_g + I_m)$. So I think equating the coefficients of $g_{\mu\nu}$ and $b^{\mu}$ in (27.10) and (27.11) is OK -- with no further explanation required!

 

[JimWhoKnew]

EDIT: Hang on, maybe I'm getting hung up on $g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ and the apparent relation it imposes on $g_{\mu\nu}$ and $b^{\mu}$ in Dirac's (27.10). It seems that $g_{\mu\nu}=-b_{\mu:\nu}-b_{\nu:\mu}$ is simply the effect on $g_{\mu\nu}$ caused by $b^{\mu}$. But in (27.10), we are at liberty to vary $g_{\mu\nu}$ arbitrarily, since this is allowed to make an arbitrary variation $\delta(I_g + I_m)$. So I think equating the coefficients of $g_{\mu\nu}$ and $b^{\mu}$ in (27.10) and (27.11) is OK -- with no further explanation required!

Now you are just repeating my speculations as if they were your own.

 

[Kostik]

Now you are just repeating my speculations as if they were your own.

I'm afraid not, what I wrote is 100% my own, original thought. You wrote "The bottom line: I don't understand how Dirac gets (27.4)"; however, I provided a detailed derivation of (27.4) in post #34.

I see now that you wrote "An idea that I have, but not sure of its validity", but I did not see that before; it was probably added by edit after I initially read the post. You might want to be more careful about making accusations of plagiarism. I have gone out of my way in previous posts to cite your contributions, in particular relating the $\delta P^\mu$ and $\delta g_{\mu\nu}$ variations to Lie derivatives, which was very helpful.

As an aside, why are you not sure of its validity? I think I explained that in #44.

 

[jbergman]

I read the passage around 27.4 and didn't understand Dirac's derivation either. I came up with the following.

First we have an active transformation where we move each particle in the dust from $$z^{\mu} \rightarrow z^{\mu} + b^{\mu}$$

Then given this active transformation $$p^{\prime\mu}(z^{\mu}) = p^{\mu}(z^{\mu} - b^{\mu})$$. And then $$ \delta p^0(z^{\mu}) = p^{\prime 0}(z^{\mu}) - p^0(z^{\mu}) =
p^0(z^{\mu}) - p^0(z^{\mu}),_{\nu} b^{\nu} - p^0(z^{\mu}) = -p^0,_{\nu} b^{\nu}$$

Using a Taylor expansion above. Now we use the conservation law that $p^{\mu},_{\mu} = 0$ or $p^0,_0=-p^r,_r$ we replace $p^0,_0$ in the previous equation and get $$\delta p^0 = p^r,_rb^0 - p^0,_r b^r$$.

I haven't convinced myself, but I believe a similar argument works for the other components of $\delta p$.

 

[JimWhoKnew]

I haven't convinced myself, but I believe a similar argument works for the other components of $\delta p$.

I believe that if the generalization of the equation below (27.3) to equation (27.4) could have been demonstrated within a short paragraph, Dirac would have shown it. It was in his nature. I think it was skipped because it is too long a detour for Dirac's packed book.

Using a Taylor expansion above. Now we use the conservation law that $p^{\mu},_{\mu} = 0$ or $p^0,_0=-p^r,_r$ we replace $p^0,_0$ in the previous equation and get $$\delta p^0 = p^r,_rb^0 - p^0,_r b^r$$.

Your result lacks the $-p^0 b^r{}_{,r}$ term of equation (27.3).
Note that $p^\mu$ is defined as a vector density. Since $b^r$ is not constant in general, the coordinate distance between two neighboring "particles" after the displacement is not necessarily the same as it was before. This should affect the density. To lowest order, the needed corrections should be linear in the $b^\mu{}_{,\nu}$ .

I read the passage around 27.4 and didn't understand Dirac's derivation either.

I can add some details that will make the derivation of (27.3) more clear (in my opinion). As for the generalization to (27.4), the best I have so far is post #20.

 

[jbergman]

I believe that if the generalization of the equation below (27.3) to equation (27.4) could have been demonstrated within a short paragraph, Dirac would have shown it. It was in his nature. I think it was skipped because it is too long a detour for Dirac's packed book.

Maybe, maybe not.

Your result lacks the $-p^0 b^r{}_{,r}$ term of equation (27.3).
Note that $p^\mu$ is defined as a vector density.

I was aware of this part.

Since $b^r$ is not constant in general, the coordinate distance between two neighboring "particles" after the displacement is not necessarily the same as it was before.

Ok, this part, I was wondering whether $b^r$ was constant or not. I haven't read the book other than this section. Thanks for the clarification. That definitely renders my derivation invalid. I will see if I can salvage it.

This should affect the density. To lowest order, the needed corrections should be linear in the $b^\mu{}_{,\nu}$ .


I can add some details that will make the derivation of (27.3) more clear (in my opinion). As for the generalization to (27.4), the best I have so far is post #20.

Look forward to it.

 

[jbergman]

I believe that if the generalization of the equation below (27.3) to equation (27.4) could have been demonstrated within a short paragraph, Dirac would have shown it. It was in his nature. I think it was skipped because it is too long a detour for Dirac's packed book.


Your result lacks the $-p^0 b^r{}_{,r}$ term of equation (27.3).
Note that $p^\mu$ is defined as a vector density. Since $b^r$ is not constant in general, the coordinate distance between two neighboring "particles" after the displacement is not necessarily the same as it was before. This should affect the density. To lowest order, the needed corrections should be linear in the $b^\mu{}_{,\nu}$ .


I can add some details that will make the derivation of (27.3) more clear (in my opinion). As for the generalization to (27.4), the best I have so far is post #20.

BTW, I found an old thread that mentions your Lie Derivative approach.

https://www.physicsforums.com/threads/unraveling-diracs-general-relativity-equation.734239/

That's an interesting idea and seems correct. Still trying to wrap my head around it since we are dealing with a density induced from dust particles and it wasn't obvious to me that moving particles was the same as flowing the resulting density along a vector field.

 

[Kostik]

I just noticed that Dirac actually derives the Lie derivative in chapter 30, on page 60. Here he calculates the variation $\delta g_{\mu\nu} = -\mathcal{L}_b \, g_{\mu\nu}$ in order to make the reverse argument that I provided in Post #34. There, to calculate the $\delta p^\mu$ in Dirac's (27.4), I took a point transformation and considered it as a coordinate transformation. In chapter 30, Dirac has a coordinate transformation, derives the Lie derivative $-\mathcal{L}_b \, g_{\mu\nu}$, and then considers it as a point transformation in order to find $\delta g_{\mu\nu}$.

The odd thing is, since he does this is chapter 30, he could have done it in chapter 27 to give a solid proof of (27.4).