I What's the underlying frame of the Einstein's Field Equation?

[Pyter]

Since this distance has no physical meaning I would not call that space “Cartesian”.

Maybe "Euclidean" would be more appropriate. Since we have n coordinates, each $\in E_1$, we have an $E_n$ space. Every point in this space is identified by a n-ple of values.

No, they are in in the sense that they have values in the real numbers. It has nothing to do with rectlinear, which doesn't make sense in genereal.

"Rectilinear" in the sense that you could consider each one a coordinate axis in an $E_n$ space.

That is not what a metric is!

How else would you map the points of the manifold to the "local coordinates" (so-called even if they may cover the whole manifold) space $E_n$, if you don't have the charts' explicit functions?

No, r is not a parameter of the solution. It is one of the coordinates. For the Sun-Earth system, it is to very good approximation given by the distance from the Sun.

Right, r and t are the "local coordinates" (according to the definition above). As it was stated in a previous post, they exactly match the physical quantities for the observer at infinity only. If you plug in the observed distance from the Sun for r you do, as you said, an approximation.

 

[Orodruin]

If you plug in the observed distance from the Sun for r you do, as you said, an approximation.

In the Sun-Earth system that approximation is much more precise than the variations in the distance due to the orbital characteristics. Accurate enough for the difference not to matter.

Furthermore, when the difference becomes relevant, the entire concept of distance needs a big caveat.

 

[Pyter]

Furthermore, when the difference becomes relevant, the entire concept of distance needs a big caveat.

Exactly, for instance in situations where the curvature gradient is very steep, like in proximity of a neutron star.

EDIT: but in such a situation I could still "observe" a distance from the star, maybe with telemetry, imagining that the star's surface reflects a laser beam back.

 

[Dale]

Maybe "Euclidean" would be more appropriate. Since we have n coordinates, each ∈E1, we have an En space. Every point in this space is identified by a n-ple of values.

Yes, every event in an open region of spacetime is identified by a 4-tuple of values. So I would simply say that: every event in the spacetime manifold is identified by a point in $\mathbb{R}^4$. I wouldn't characterize $\mathbb{R}^4$ further. Anything else, Cartesian or Euclidean, seems to add a flat geometry that is not physically meaningful in the coordinate space.

Does it need to have a name for some reason? Is $\mathbb{R}^4$ not sufficient for some reason?

 

[PeterDonis]

That's what I was going to ask next: you have to plug in the observable quantities into the metric to make it "usable". Not only M but also r, I believe, since it's a parameter too

Strictly speaking, $r$ is a coordinate, not a parameter.

If you wanted to compute the Schwa. metric induced by the Sun at the Earth's orbit, you can't simply replace r with Earth's observed distance from the Sun.

Correct, you have to compute $r$ from the observed distance. (Although in the case of the solar system the correction is very small.)

Do you find this value by solving other differential equations, this time involving the observations?

No. You find it by knowing the relationship between radial distance and the $r$ coordinate in Schwarzschild spacetime, which is given by the metric that you've already solved for.

 

[PeterDonis]

Maybe "Euclidean" would be more appropriate. Since we have n coordinates, each $\in E_1$, we have an $E_n$ space. Every point in this space is identified by a n-ple of values.

No, "Euclidean" is not appropriate, at least not if you want to use standard terminology. The standard terminology for "the space of 4-tuples of real numbers" with no other structure is what @Dale said, $\mathbb{R}^4$. "Euclidean n-space" means you have a particular metric. (The coordinates are not individually in "Euclidean 1-space" either; that would imply a particular metric on the real numbers, which is not valid in the general case.)

 

[Pyter]

Does it need to have a name for some reason? Is not sufficient for some reason?

I used $E_4$ just because the cited definition of manifold involved n coordinates $\in E_1$, implying now that I think about it that only the metric of the single coordinate is Euclidean. But we agree that the only metric that matters is on the manifold.

You find it by knowing the relationship between radial distance and the coordinate in Schwarzschild spacetime, which is given by the metric that you've already solved for.

By "radial distance" you mean the "observed" distance, measured perhaps through telemetry? Is it connected to r through the geodesic equation?

 

[ergospherical]

By "radial distance" you mean the "observed" distance, measured perhaps through telemetry? Is it connected to r through the geodesic equation?

The Schwarzschild coordinate $r$ of a point $p$ is the area radius function $r = \sqrt{A(p)/4\pi}$, with $A(p)$ the area of the spherical symmetry orbit through $p$. Meanwhile the radial distance comes through $dR =\sqrt{ g_{rr}} dr$.

 

[Pyter]

The Schwarzschild coordinate $r$ of a point $p$ is the area radius function $r = \sqrt{A(p)/4\pi}$, with $A(p)$ the area of the spherical symmetry orbit through $p$. Meanwhile the radial distance comes through $dR =\sqrt{ g_{rr}} dr$.

So the measured distance $R = \int g_{rr}\,dr$? What are the extremes of integration and how do you express r in function of the measured distance?

 

[ergospherical]

So the measured distance $R = \int g_{rr}\,dr$? What are the extremes of integration and how do you express r in function of the measured distance?

what'd you do with my precious square root

 

[PeterDonis]

By "radial distance" you mean the "observed" distance, measured perhaps through telemetry?

Yes.

Is it connected to r through the geodesic equation?

No, it's connected to $r$ through the metric. A radial line of coordinate increment $dr$ covers an actual physical distance of $dr / \sqrt{ 1 - 2M / r }$.

So the measured distance $R = \int g_{rr}\,dr$?

With a square root over $g_{rr}$, as @ergospherical says, yes.

What are the extremes of integration

The extremes of integration are the starting and ending $r$ values. Since $r$ is an "areal radius", a circular orbit at radius $r$ has circumference $2 \pi r$, and that is an actual physical distance, so you could, for example, obtain the radial distance between two circular orbits by measuring both their circumferences $C_1$ and $C_2$ and inverting $C = 2 \pi r$ to obtain $r_1$ and $r_2$.

how do you express r in function of the measured distance?

If you're doing the integral that has been described, you don't, at least not in terms of measured radial distance.

 

[PeterDonis]

I used $E_4$ just because the cited definition of manifold involved n coordinates $\in E_1$,

No, it doesn't. The coordinates are members of $\mathbb{R}$, but that is not the same as $E_1$ because $E_1$ implies a Euclidean metric whereas $\mathbb{R}$ does not.

implying now that I think about it that only the metric of the single coordinate is Euclidean.

Not even that is true. See above. There is no "metric of the single coordinate".

 

[Orodruin]

Yes

No. This will not be the distance measured by the method suggested by the OP (see eg #123) - which was bouncing a light signal and measuring the round trip time - due to Shapiro delay.

Hence my call for caveats in #122.

Ironically, if my early morning brain is not playing jokes on me, that distance would be given by the integral without the square root (but only because $g_{tt} = 1/g_{rr}$) ...

 

[vanhees71]

Yes, every event in an open region of spacetime is identified by a 4-tuple of values. So I would simply say that: every event in the spacetime manifold is identified by a point in $\mathbb{R}^4$. I wouldn't characterize $\mathbb{R}^4$ further. Anything else, Cartesian or Euclidean, seems to add a flat geometry that is not physically meaningful in the coordinate space.

Does it need to have a name for some reason? Is $\mathbb{R}^4$ not sufficient for some reason?

Well, mathematically a differentiable manifold is defined by an equivalence class of complete atlasses of compatible local maps of a Hausdorff topological space to $\mathbb{R}^4$ (as a topological space with the usual topology). That's why you can define derivatives, tangent and cotangent spaces etc. by inheriting these notions from standard tensor analysis in $\mathbb{R}^4$ via the compatible maps of the atlasses to the Hausdorff topological space, making it in this way to a differentiable manifold.

Then you can go further and define more structures on this space by inheriting the corresponding constructions in $\mathbb{R}^4$. For GR you need a connection and a non-degenerate but indefinite fundamental form ("pseudo-metric") with signature (1,3) (or equivalently (3,1)) and a metric compatible connection. In standard GR in addition you assume that the connection is torsion free, and thus being uniquely determined by the usual symmetric Christoffel symbols. This makes the GR spacetime model a pseudo-Riemannian manifold.

 

[Pyter]

what'd you do with my precious square root

Sorry, I hope you've got the gist of my question anyway.

No, it doesn't. The coordinates are members of $\mathbb{R}$, but that is not the same as $E_1$ because $E_1$ implies a Euclidean metric whereas $\mathbb{R}$ does not.
[...]
Not even that is true. See above. There is no "metric of the single coordinate".

In the cited definition in post #117, the $x^n$ charts all map to $E_1$. Doesn't that mean a 1-D Euclidean space?

No, it's connected to through the metric. A radial line of coordinate increment $dr$ covers an actual physical distance of $dr / \sqrt{ 1 - 2M / r }$.

How do you get that formula? Is it the $ds$ with $dt = d\theta = d\phi = 0$?

Since $r$ is an "areal radius", a circular orbit at radius $r$ has circumference $2 \pi r$, and that is an actual physical distance, so you could, for example, obtain the radial distance between two circular orbits by measuring both their circumferences $C_1$ and $C_2$ and inverting $C = 2 \pi r$ to obtain $r_1$ and $r_2$.

So if the Earth was orbiting around a neutron star and I wanted an r value to plug into the metric, I could measure the length of Earth's orbit (how so?) and divide by $2\pi$?
But what if I'm on a spaceship and I only have a radar blip of the estimated distance from the star?

 

[Orodruin]

Doesn't that mean a 1-D Euclidean space?

”Euclidean space” includes the presumption of the existence of a metric. $\mathbb R$ does not.

How do you get that formula? Is it the ds with dt=dθ=dϕ=0?

Yes, it is the radial distance along the hypersurface of constant t-coordinate. This is often taken as the ”distance” by definition. Note that it is not what you would measure as a ”distance” if you try to measure it by bouncing a light signal off the object and multiplying the round trip time by c/2 (see my previous post) due to Shapiro delay.

But what if I'm on a spaceship and I only have a radar blip of the estimated distance from the star?

Then you will have to do the math to get r correctly.

 

[Dale]

a differentiable manifold is defined by an equivalence class of complete atlasses of compatible local maps of a Hausdorff topological space to R4 (as a topological space with the usual topology).

Right. To my understanding, calling a space “Cartesian” or “Euclidean” introduces a notion of distance that is explicitly absent in a topological space.

 

[vanhees71]

The standard topology of $\mathbb{R}^d$ is, however, induced by all $\mathrm{L}_p$-norms.

 

[Pyter]

Then you will have to do the math to get r correctly.

How would you compute r from the measured radar distance?

 

[Orodruin]

How would you compute r from the measured radar distance?

With the integral without the square root.

 

[PeterDonis]

his will not be the distance measured by the method suggested by the OP (see eg #123) - which was bouncing a light signal and measuring the round trip time - due to Shapiro delay.

Ah, yes, I didn't take not of the "by telemetry" in the post I was responding to. In principle there are ways of obtaining the radial distance by measurements, but it won't be anything as simple as just computing it directly from round-trip light travel times.

 

[cianfa72]

No. This will not be the distance measured by the method suggested by the OP (see eg #123) - which was bouncing a light signal and measuring the round trip time - due to Shapiro delay.

Does Shapiro delay basically take in account the displacement of the spaceship w.r.t. the star (same $r, \phi$ but different $\theta$) during the process of sending and getting back the light signal off the orbiting spaceship to the star ?

 

[Orodruin]

Does Shapiro delay basically take in account the displacement of the spaceship w.r.t. the star (same $r$ but different $\theta, \phi$) during the process of sending and getting back the light signal off the orbiting spaceship to the star ?

Not in the form I mentioned, that assumed a stationary. Shapiro delay itself of course will exist regardless, but the general case will be more convoluted.

 

[cianfa72]

Not in the form I mentioned, that assumed a stationary.

Do you mean a spaceship hovering w.r.t. the star at fixed $r, \theta, \phi$ ?

 

[Orodruin]

Do you mean a spaceship hovering w.r.t. the star at fixed $r, \theta, \phi$ ?

That is what stationary means, yes.

 

[cianfa72]

That is what stationary means, yes.

ok, so in case of stationary spaceship (i.e. hovering at fixed $r, \theta, \phi$) can we assume as "physical distance" from the star c times the average round-trip time for a bouncing light signal as measured from the spaceship ?

 

[Orodruin]

ok, so in case of stationary spaceship (i.e. hovering at fixed $r, \theta, \phi$) can we assume as "physical distance" from the star c times the average round-trip time for a bouncing light signal as measured from the spaceship ?

No, as I mentioned in previous posts, ”distance” comes with a huge caveat in GR. What is typically defined as ”radial distance” is the integral
$$
\int \sqrt{g_{rr}} dr
$$
with appropriate boundary conditions. This corresponds to the distance along a hypersurface of constant $t$-coordinate.

What would be measured by a radar measurement’s round trip time (multiplied by c/2) would be
$$
\int \sqrt{-\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$

 

[cianfa72]

What would be measured by a radar measurement’s round trip time (multiplied by c/2) would be
$$
\int \sqrt{\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$

ah, I believe we get this formula considering a lightlike geodesic starting from the event $(r,\theta,\phi,t_0)$, bouncing at the center of the star and back to the event $(r,\theta,\phi,t_1)$. Then from the difference $\Delta t$ we derive the interval of proper time $\Delta \tau$ elapsed at point $r,\theta,\phi$ and multiplying for c/2 we get the "distance" radar measurement.

 

[Orodruin]

There is another adjustment to consider of course: The integrals give $\Delta t$, but the locally measured time is of course gravitationally time dilated relative to this.

 

[cianfa72]

There is another adjustment to consider of course: The integrals give $\Delta t$, but the locally measured time is of course gravitationally time dilated relative to this.

I take it as for the lightlike path described in #148 the $\Delta t$ is given by the integral as in your #147. Then to get the proper time (since gravitational time dilation) that difference in coordinate time $\Delta t$ has to be rescaled accordingly.