Shape of Vacuum Space: Constant Spatial Volume with Time

[TGlad]

If we have a small 4-volume of empty spacetime of boxlike dimensions t, x, y, z, then according to the vacuum field equations the change of the shape of this box with respect to time is (I think): $$\frac{d (xyz)}{dt}=0$$ or equally: $$yz\frac{dx}{dt}+xz\frac{dy}{dt}+xy\frac{dz}{dt}=0$$
in other words, the spatial volume is constant with time.

What is the equation with respect to a spatial coordinate? (say x), is it:
$$\frac{d (yz/t)}{dx}=0$$, or is it perhaps $$zt\frac{dy}{dx}+yt\frac{dz}{dx}-yz\frac{dt}{dx}=0$$?

Many thanks,
Tom.

 

[haushofer]

The volume V of a piece of Minkowski-spacetime with Cartesian sides $\Delta x,\Delta y, \Delta z, \Delta t$ is $V=\Delta x \Delta y \Delta z \Delta t$. These sides are constants, so partial differentiating $V$ with respect to any coordinate gives zero. You have to be careful how to coordinatize your spacetime volume.

Of course, if you take other solutions to the Einstein Field equations, you have to involve the determinant $g$ of the metric, which can give you coordinate dependancy on the volume (e.g. it can become time-dependent):
$$V = \int \sqrt{|g|} dx dy dz dt$$

In your case we get

$$V = \int_{0}^{\Delta x} \int_{0}^{\Delta y} \int_{0}^{\Delta z} \int_{0}^{\Delta t} \sqrt{|\eta|} dx dy dz dt = \Delta x \Delta y \Delta z \Delta t$$

The (absolute value of the) determinant of the Minkowski metric $\eta_{\mu\nu}$ in Cartesian coordinates is one.

Hope this helps.

 

[PeterDonis]

If we have a small 4-volume of empty spacetime of boxlike dimensions t, x, y, z, then according to the vacuum field equations the change of the shape of this box with respect to time

...doesn't make sense. A 4-volume is a 4-volume; it's a small piece of spacetime, that doesn't "move" or "change", it just is, in the same way that 4-dimensional spacetime itself doesn't "move" or "change", it just is.

What the field equations tell you, roughly speaking, is how the "shape" of some 4-volume is determined by the stress-energy tensor present within that 4-volume. (Strictly speaking, the 4-volume in question should be infinitesimal in size, since the field equations are differential equations.) "Vacuum" means the stress-energy tensor is zero, and flat spacetime is indeed one solution of the vacuum field equations. But it's not the only one--a vacuum region of spacetime can still be curved (for example, Schwarzschild spacetime, which describes the vacuum region surrounding a spherically symmetric object, is curved). In any case, though, the 4-dimensional spacetime does not "change" and 4-volumes do not "move" within it.

 

[PeterDonis]

in other words, the spatial volume is constant with time.

The spatial volume is not a 4-volume; it's a 3-volume. And any such statement about how spatial volume changes with time will be coordinate-dependent, since splitting up 4-d spacetime into 3-d "space" and time requires a choice of coordinates. It just so happens that, in flat Minkowski spacetime, you can pick any set of inertial coordinates and still get the answer you got, that the spatial volume is constant with time. There are also curved spacetimes in which you can find particular choices of coordinates that have the same property (these spacetimes are called "stationary"). But even in the case of flat spacetime, you can pick non-inertial coordinates in which the spatial volume will not necessarily be constant with time.

 

[TGlad]

Sorry, badly (and wrongly) written question, I will try to be clearer.

Let us say we have a Ricci-flat 3+1D spacetime, but which is not Minkowski space. In other words it has some curvature but is a vacuum spacetime.

Now at a particular event $e=(t, x, y, z)$ we look at a small 4-volume $\delta t,\delta x,\delta y,\delta z$, and we are in the 'rest frame' such that the metric is $\eta_{\mu\nu}$ exactly at event $e$.

For a curved spacetime the dimensions of the 4-volume change with respect to each coordinate. I'm not talking about the volume physically moving, just how the dimensions change in each direction. So, with respect to time, the requirement that the space is Ricci flat means that:
$$\left . \frac{d^2 (\delta x \delta y \delta z)}{dt^2}\right|_e=0$$
i.e. its spatial 3-volume is constant. As seen in equation 2 here when no matter present.

I am wondering how the dimensions change with space. Is it simply:
$$\left .\frac{d^2 (\delta t \delta y \delta z)}{dx^2}\right|_e=0$$
Or does the Lorentzian signature cause a different sort of constraint? perhaps like this:
$$\left .\frac{d^2 (\delta y \delta z / \delta t)}{dx^2}\right|_e=0$$

 

[PeterDonis]

For a curved spacetime the dimensions of the 4-volume change with respect to each coordinate. I'm not talking about the volume physically moving, just how the dimensions change in each direction.

Sorry but this still doesn't make sense. What do you mean by "the dimensions"? What sort of actual physical measurements are you talking about?

with respect to time, the requirement that the space is Ricci flat means that:

$$
\left . \frac{d^2 (\delta x \delta y \delta z)}{dt^2}\right|_e=0
$$​

i.e. its spatial 3-volume is constant. As seen in equation 2 here when no matter present.

The equation you wrote is not the same as equation (2) in the article you linked to. Your equation is in terms of coordinates. Equation (2) in the article is in terms of the physical volume of a small ball of test particles, and its rate of change (actually second derivative) with respect to proper time along the worldline of the particle at the center of the ball. That is a very important difference.

I am wondering how the dimensions change with space.

Again, you need to first explain what you mean by "dimensions". But I think it would be even better if you forget the term "dimensions" altogether, and rethink your questions in terms of an actual physical observable, such as is described above in reference to the article you linked to. I would recommend reading and re-reading that article, carefully, so you understand exactly what physical situation it is describing. Note carefully, for example, that there are no coordinates anywhere in the article; $V$, as noted above, is the physical volume of a small ball of test particles, and $t$ is the proper time of the particle at the center of the ball.

 

[TGlad]

Yes, Baez talks about the volume of a ball of test particles. But if there is no matter present there can be no particles. Never-the-less, if it is clearer then let's talk about some imaginary test particles that have zero mass (or as mass tends to zero if that is preferred). Instead of a ball of them, we have a 4D block of particles with dimensions (defined as the distance from one side to the other) of $\delta t, \delta x, \delta y, \delta z$. Then the second differential of the spatial 3-volume with respect to proper time (or coordinate time) is as given in my formula above.

I am wondering how this 4D block changes (the second differential) with respect to a spatial direction (such as $x$) at the specific point in spacetime $e$. We could call x the proper distance or the coordinate distance along the axis that has block width $\delta x$, it shouldn't matter at $e$ because that is the rest frame of the particles, so they coincide.

If this still is unclear then no worries, I will do more reading up to try and better understand the Ricci-flat (zero covariant divergence) constraint. I just haven't found the answer yet.

 

[PeterDonis]

if there is no matter present there can be no particles.

Test particles don't count as "matter"; they are particles that are so small that their stress-energy is negligible, so they don't contribute to spacetime curvature. This term occurs all the time in GR textbooks and papers, so it's important to understand what it means and how it is used.

if it is clearer then let's talk about some imaginary test particles that have zero mass (or as mass tends to zero if that is preferred).

That what I was talking about. See above.

Instead of a ball of them, we have a 4D block of particles with dimensions (defined as the distance from one side to the other) of δt,δx,δy,δz\delta t, \delta x, \delta y, \delta z.

No, you don't. You are still not grasping a key point: coordinate intervals are not distances or times. That is a very important aspect of GR that many people find hard to grasp.

Once again, go back and read the Baez article you linked to, carefully. It doesn't mention coordinates once. Even $t$, which might otherwise be mistaken for a coordinate, is carefully defined to be the proper time of the particle at the center of the ball. Proper time is actual, measured clock time, not coordinate time.

Also, the point of Baez' article is to show you how to translate the Ricci tensor, or more specifically the component $R_{00}$ of the Ricci tensor, into something physical which is easily visualized. That requires considering, not just one infinitesimal 4-d volume of spacetime, but a whole series of them, describing the history of the ball of test particles--what happens to it over time (proper time of the particle at the center of the ball) because of the effects of $R_{00}$. In other words, what Baez is describing, in 4-d spacetime terms, is a "world tube"--a 4-d volume that is like a cylinder or tube, with a small spatial "width" and running along the "time" direction (but here, again, "time" means proper time, not coordinate time).

We could call x the proper distance or the coordinate distance along the axis that has block width $\delta x$, it shouldn't matter at $e$ because that is the rest frame of the particles, so they coincide.

No, they don't.

Also, $e$ is an event, not a frame. They are not the same thing.

It looks to me like you lack some important background in GR. What textbooks, papers, or other resources have you studied.

I will do more reading up to try and better understand the Ricci-flat (zero covariant divergence) constraint.

These are not the same thing. The condition of zero covariant divergence applies to the Einstein tensor, not the Ricci tensor, and it applies in all spacetimes, regardless of their geometry; it is a geometric identity called the Bianchi identity.

Ricci flat means the Ricci tensor is zero. Only certain spacetimes satisfy that property.

 

[PeterDonis]

I am wondering how this 4D block changes (the second differential) with respect to a spatial direction (such as $x$) at the specific point in spacetime $e$.

There is another confusion embodied here. The quantity $\ddot{V} / V$ that appears in Baez' article (and which, if you look at the further page on mathematical details, is equal to $- R_{00}$) is not the "change in volume of spacetime". It is the change in volume of a small ball of test particles. The concept of "change in volume of spacetime" doesn't make any sense, as I said in post #3.

 

[TGlad]

Thanks, I agree with you on these points. I think I'll read up some more before any attempt to rephrase the question. And perhaps in my studying I might find the answer. It would be better to get a cleaner and more correct question as a new post anyway.