# Einstein tensor of a gravitational source

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1. Jan 12, 2016

### CharlesJQuarra

In section 4.4 of gravitational radiation chapter in Wald's general relativity, eq.4.4.49 shows the far-field generated by a variable mass quadrupole:

$\gamma_{\mu \nu}(t,r)=\frac{2}{3R} \frac{d^2 q_{\mu \nu}}{dt^2} \bigg|_{t'=t-R/c}$

I have the following field from a rotating binary

$\gamma_{\mu \nu} = \frac{2 M}{3 R} \begin{bmatrix} 2 & - r_o \omega(t \omega \cos{\omega t} + 2 \sin{\omega t} )& - r_0 ( \omega t \sin{\omega t} - 2 \cos{\omega t}) & 0 \\ - r_o \omega(t \omega \cos{\omega t} + 2 \sin{\omega t} ) & -2 r_o^2 \omega^2 \cos{2 \omega t} & - 2 r_o^2 \omega^2 \sin{2 \omega t} & 0 \\ - r_0 ( \omega t \sin{\omega t} - 2 \cos{\omega t}) & - 2 r_o^2 \omega^2 \sin{2 \omega t} & r_o^2 \omega^2 \cos{2 \omega t} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

With $r_o$ being the orbital radius. I've computed (and double-checked) the Einstein tensor of this radiation field, taking

$$g_{\mu \nu} = \eta_{\mu \nu} + \gamma_{\mu \nu}(t',r) \Bigg|_{t'=t-R/c}$$

And the resulting Einstein tensor has the following components:

$$G^{\mu \nu} = \frac{M}{c^2} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & \frac{2 x^2 - y^2 - z^2}{ R^5} & \frac{3 x y}{R^5} & \frac{3 x z}{R^5} \\ 0 & \frac{3 x y}{R^5} & \frac{2 y^2 - x^2 - z^2}{ R^5} & \frac{3 y z}{R^5} \\ 0 & \frac{3 x z}{R^5} & \frac{3 y z}{R^5} & \frac{2 z^2 - x^2 - y^2}{ R^5} \end{bmatrix}$$

Now, I'm trying to interpret the fact that the Einstein tensor is not zero outside the binary, as one would expect from the Einstein's equation

$$G_{\mu \nu} = \frac{G}{c^4} T_{\mu \nu}$$

Question: why is the Einstein tensor non-zero for this metric?

EDIT:
if you have Mathematica available, here is a notebook with the detailed calculation, with explanations:

https://github.com/CharlesJQuarra/GravitationCalcs/blob/master/GravWaveAnalysis.nb

If someone wants to improve on it or propose changes, please, send me a pull request

Last edited: Jan 12, 2016
2. Jan 12, 2016

### Staff: Mentor

I'm confused about how you did it, since your Einstein tensor is expressed in different coordinates than Wald's radiation field. (His has $\gamma$ as a function of $r$, but yours has $x$, $y$, and $z$.)

Also, these computations are notoriously complicated, and it's easy to think you've double (and triple and quadruple) checked everything and still end up finding a mistake. One way to make it easier is to use a computer program like Maxima. But even then, you need to be clear on exactly what coordinates you're using. If you just assumed that $x^2 + y^2 + z^2 = r^2$, then it's no wonder you got an answer that doesn't make sense; coordinate transformations in curved spacetime are not that simple.

I would suggest either trying the computation again using Maxima or a similar tool, or posting more details about how you did the computation. (Or both. )

3. Jan 12, 2016

### CharlesJQuarra

if you have Mathematica available, here is a notebook with the detailed calculation, with explanatory comments:

https://github.com/CharlesJQuarra/GravitationCalcs/blob/master/GravWaveAnalysis.nb

If someone wants to improve on it or propose changes, please, send me a pull request

4. Jan 12, 2016

### Staff: Mentor

I'm not seeing any comments, and the structure of the code isn't really telling me anything. Are you using any packages in Mathematica that are specifically designed to do GR calculations? In Maxima there is a package called ctensor that does that; you input the metric as a matrix and it automatically computes all of the key GR quantities: the Christoffel symbols, the Riemann tensor, the Ricci tensor, the Einstein tensor, and others.

5. Jan 12, 2016

### CharlesJQuarra

You should open it as a Mathematica notebook in order to see the structured code and the comments. Let me see if I can export it to a non-interactive text version

6. Jan 12, 2016

### CharlesJQuarra

Here is the notebook content as a pdf

#### Attached Files:

• ###### GravWaveAnalysis.pdf
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7. Jan 13, 2016

### Staff: Mentor

Ah, ok. On pg. 2 I see that you are assuming $R = \sqrt{x^2 + y^2 + z^2}$. That is not a valid assumption in a curved spacetime.

Cartesian coordinates generally aren't a good choice for these cases anyway; spherical coordinates work much better. IIRC Wald is using spherical coordinates, which means you not only are incorrectly assuming that $R = \sqrt{x^2 + y^2 + z^2}$, you are also incorrectly writing down the components of the tensors for the coordinates you are using. Wald's tensor components are not $tt, tx, ty, tz$, and so on; they are $tt, tr, t\theta, t\varphi$, and so on. So to even use Cartesian coordinates, you would first have to work out the correct coordinate transformation (which will not be the same as the ordinary transformation from spherical to Cartesian coordinates in Euclidean space), and then transform Wald's expressions using it. That is a lot of unnecessary work compared to just using spherical coordinates as Wald does.

8. Jan 13, 2016

### Staff: Mentor

One other comment: you can't assume that the coordinate speed of light is $c$ in a curved spacetime. So the correct retarded time at which to evaluate might not be $t - R / c$.

9. Jan 13, 2016

### pervect

Staff Emeritus
I don't have any idea if the calculation is right or not. But if you've got a system that's radiating gravitational radiation, why would you expect the Einstein tensor to be zero?

10. Jan 13, 2016

### CharlesJQuarra

The Einstein tensor (the geometric "marble") is proportional only to stress-energy tensor (the material "wood"). Einstein equations does not treat gravitational radiation on equal footing with matter or electromagnetic radiation

11. Jan 13, 2016

### CharlesJQuarra

I would hope that to first order should be correct, in any case I'm not sure what you think it should be replaced with.

After eq. 4.4.45, it makes a simplification where $e^{I \omega | x' - x |}/|x'-x|$ becomes simplified to $e^{I \omega R}/R$. I'm not certain, but it seems to me that any error on the assumption $R = \sqrt{x^2 + y^2 + z^2}$ should be smaller than the error in the paraxial approximation above

I can try changing the coordinates and rerunning the notebook, but if there is a particular reason why the Einstein tensor is not becoming zero, it ought be non-zero also on spherical coordinates

12. Jan 13, 2016

### Staff: Mentor

What do you mean by "to first order"? First order in what?

It's not a question of "replacing" it with anything. Once again, you are implicitly doing a coordinate transformation from the spherical coordinates Wald is using, to the Cartesian coordinates you want to use. I don't understand why you want to do that in the first place, since it just makes the math a lot more cluttered and doesn't change the physics at all. But if you're going to do it, you have to do it with the correct geometry of the underlying manifold (spacetime). Since the underlying manifold is not flat, you can't use a coordinate transformation that only works in a flat manifold.

Error in what? Remember, $x$, $y$, and $z$ are not physical measurements; they're coordinates. $R$ is a physical measurement; it's the circumference of the object emitting the radiation, divided by $2 \pi$. (This is assuming that Wald is using the Schwarzschild definition of the radial coordinate $r$, which IIRC is what he uses in this example.) So the assumption $R = \sqrt{x^2 + y^2 + z^2}$ is not an assumption about how different physical measurements are related; it's an assumption about how your coordinate labels $x$, $y$, and $z$ relate to a physical measurement. (It also only applies to the $x$, $y$, and $z$ coordinates that are used to label the surface of the emitting object; it doesn't apply anywhere else, but $x$, $y$, and $z$, as coordinates, need to cover all spatial locations, not just the surface of the object.) Any "error" in this assumption is an error in how you've written down all of your equations; it's not something you can assess by comparing measurements.

13. Jan 13, 2016

### CharlesJQuarra

that is $r_o$ in the notebook, $R$ is the distance to the source, as it is explained in the snippet I quoted from Wald:

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14. Jan 13, 2016

### Staff: Mentor

Ah, ok, that makes more sense, and it means that the retarded time $t - R / c$ is correct. But it still does not make it correct to assume $R = \sqrt{x^2 + y^2 + z^2}$, because the geometry of space is not Euclidean.

15. Jan 13, 2016

### CharlesJQuarra

Did you saw the quote of the approximation $e^{I \omega | x' - x |}/| x' - x| \approx e^{I \omega R}/R$? This approximation assumes roughly that $r_0 \ll R$, so we are already simplifying the dependence on the distance and ignore the size of the source. It would seem to me to be a much coarser simplification that assuming $R \approx \sqrt{x^2 + y^2 + z^2}$

But let's suppose your hypothesis is correct, and the source of the non-zero Einstein tensor quantity is caused by an error in the coordinate dependence of $R$. If that is true, if I replace $R -> \sqrt{x^2 + y^2 + z^2} + \xi(t,x,y,z)$, where $\xi(t,x,y,z)$ is essentially the difference between the "real" $R$ and $\sqrt{x^2+y^2+z^2}$, our approximating function, then the Einstein tensor non-zero components should have all to be functions of $\xi(t,x,y,z)$ and its derivatives. Otherwise, even with an exact $R$ expression, the Einstein tensor would evaluate to non-zero components

Do you agree?

Last edited: Jan 13, 2016
16. Jan 13, 2016

### Staff: Mentor

You're missing my point. The approximation involving $e^{I \omega}$ etc. is a physical approximation. (Note that $x$ in this approximation is, I believe, a vector--the "position vector" of the location where the function is being evaluated. It's not the coordinate $x$ that you are trying to use.) The error in this approximation is, as you note, of order $r_0 / R$; we can estimate that because both $r_0$ and $R$ are physical measurements, whose ratio is a physical quantity, an invariant, independent of our choice of coordinates.

The assumption that $R \approx \sqrt{x^2 + y^2 + z^2}$ is not a physical approximation; it's an error in your definition of the coordinates $x$, $y$, and $z$. You can't even tell how wrong that makes your calculation unless you know the correct coordinate transformation from spherical coordinates to Cartesian coordinates for this particular curved spacetime. Do you?

17. Jan 13, 2016

### Staff: Mentor

In fact, I can state this in another way that doesn't even involve Cartesian coordinates. Suppose we're using spherical coordinates, and the point at which we're evaluating the metric tensor has radial coordinate $r$. What value of $r$ corresponds to the physical distance $R$ from the center? The answer is not unique; there are different ways of constructing spherical coordinates that give different answers, and those ways correspond to different expressions for $\gamma_{\mu \nu}$ in terms of the coordinates.

I'm pretty sure, as I mentioned before, that the definition of spherical coordinates that Wald is using is the Schwarzschild one, in which the radial coordinate $r$ is the circumference of a circle at that radius (i.e., a circle composed of spatial points all lying in the same plane and all at the same distance from the center), divided by $2 \pi$. With that convention, it is not true that $R = r$, i.e., $r$ is not the physical distance from the center. The physical distance from the center is given by an integral:

$$R = \int_0^r \sqrt{g_{rr}} dr'$$

where $g_{rr}$ is the radial metric coefficient (possibly with a sign change to make the quantity under the square root positive, depending on what metric sign convention you are using). For the metric Wald is using, where $g_{\mu \nu} = \eta_{\mu \nu} + \gamma_{\mu \nu}$, this integral would look like

$$R = \int_0^r \sqrt{1 + 2 r_o^2(t'(r'), r') \omega^2 \cos 2 \omega t'(r')} dr'$$

where I have put the functional dependencies in to make clear that, at each value of the radial variable $r'$, we have to evaluate the factors under the square root at that value of $r'$ and at the "retarded" value of $t$ corresponding to that $r'$, which will be different for each $r'$. Also, the value of $t'$ for a given $r'$ will not just be $t' = t - r / c$ (where $t$ with no prime symbol is the time at which we are evaluating the integral), because, again, $r$ is not the radial distance to the center. (Also, for small values of $r$, we can no longer ignore $r_o$, since it won't be much smaller than $r'$.) What's more, we can't even assume that the difference between $r$ and the actual physical distance from the center is small.