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

[cianfa72]

Btw is the curve $(r, \theta_0, \phi_0,t_0)$ with $r$ in the interval $r_0-r_1$ a geodesic of the spacelike hypersurface $t=t_0$ ?

In other words is the integral (i.e. the definition of "distance")
$$
\int \sqrt{g_{rr}} dr
$$ the spacetime length of a geodesic of the spacelike hypersurface $t=0$ ?

 

[Orodruin]

Yes. It is relatively easy to conclude this from symmetry.

 

[Pyter]

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

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.

Not so trivial to get $\Delta t$ from $\Delta \tau$ (your measured round-trip time) , since ${\Delta \tau}/{\Delta t}$ is not constant.
And once you somehow get your "measured" $c\Delta t/2 = C(onstant)$, should you solve numerically the equation:
$$ C = \int^{r_1}_{r_0} g_{rr} dr $$
for $r_1$, then plug it into the metric to evaluate it?
The $r_{0}$ value (coordinate radius of the neutron star) is also unknown BTW, maybe we could set it to 0 as a first approximation.

 

[Orodruin]

since Δτ/Δt is not constant.

Yes it is.

The r0 value (coordinate radius of the neutron star) is also unknown BTW, maybe we could set it to 0 as a first approximation.

Definitely not. That would involve going below the horizon of the Schwarzschild metric where r is no longer space-like.

If you know the mass of the star, the easiest way to get the r coordinate would be to measure the proper acceleration required to remain stationary. (If you are stationary.)

Alternatively measure the proper time for a full orbit if you are in circular orbit.

 

[Pyter]

Yes it is.

Doesn't also $g_{tt}$ depend on r?

If you know the mass of the star, the easiest way to get the r coordinate would be to measure the proper acceleration required to remain stationary. (If you are stationary.)

Alternatively measure the proper time for a full orbit if you are in circular orbit.

That would be equivalent to "measure" g**, then work backwards to obtain r. But if I already have determined g**, I don't need r.

 

[Orodruin]

Doesn't also $g_{tt}$ depend on r?

Yes, but your r is fixed.

 

[Pyter]

Yes, but your r is fixed.

But that of the light/radio beam isn't.
Anyway, even if I should only consider my fixed r for computing $\Delta t$, it means that also the LHS (C) of the equation in #139 depends on the unknown r.

 

[Orodruin]

But that of the light/radio beam isn't.

That is irrelevant, it is not what you would be measuring.

it means that also the LHS (C) of the equation in #139 depends on the unknown r.

So what? Equations like $f(x) = g(x)$ can many times be solved.

 

[cianfa72]

As far as I can understand the round-trip time measurement in term of coordinate time $\Delta t$ from the hovering spaceship at coordinate radius $r_1$ from the star using a bouncing light beam, is actually the integral (divided by c/2)

$$
\int \sqrt{-\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$
evaluated between $r_0$ and $r_1$ ($r_0$ is the coordinate radius of the star itself).

 

[Pyter]

@cianfa72 see the post #153.

 

[Pyter]

So you need at least an estimate of M to compute $r_0$, provided the density of neutronium is an invariant. But is it? I doubt it, because M certainly is, but the measured radius/volume of the star isn't.

EDIT: or better, the rest mass of the star is surely an invariant. The $\gamma$ factor changes with the observer, so the relativistic mass isn't.

 

[PeterDonis]

The $r_{0}$ value (coordinate radius of the neutron star) is also unknown BTW

You can measure $r_0$ by measuring the star's surface area or circumference and using the formulas $C = 2 \pi r$ or $A = \pi r^2$.

 

[Pyter]

You can measure $r_0$ by measuring the star's surface area or circumference and using the formulas $C = 2 \pi r$ or $A = \pi r^2$.

But the measured $r_0$ matches the coordinate $r_0$, needed for the integral, only for the observer at infinity. You'd need at least another equation to link them.

 

[Orodruin]

But the measured $r_0$ matches the coordinate $r_0$, needed for the integral, only for the observer at infinity. You'd need at least another equation to link them.

No. The $r_0$ measured in the way described (area of sphere or circumference of circle) is by definition the $r$ coordinate of the surface.

 

[PeterDonis]

But the measured $r_0$ matches the coordinate $r_0$, needed for the integral, only for the observer at infinity.

Wrong. The $r$ coordinate is a coordinate. Coordinates are valid for any observer that uses that coordinate chart.

 

[Pyter]

Wrong. The $r$ coordinate is a coordinate. Coordinates are valid for any observer that uses that coordinate chart.

Yes and that's just the point of these whole operations. To match the (absolute) coordinates with the (relative) physical distances measured by an arbitrary observer.

No. The $r_0$ measured in the way described (area of sphere or circumference of circle) is by definition the $r$ coordinate of the surface.

Now I guess that to send a probe with a rod to the surface of the star is out if discussion, unless we can prevent the tidal effects from tearing it to smithereens.
The only viable way to measure the apparent diameter could be from the visual angle covered by the star disc from a known measured distance, and from that compute $r_0$ using the metric.

EDIT: I'm also doubting that the observed $M$ is an absolute. Isn't that in the metric the mass of the star observed at infinity? Any other observer would have a $\gamma <> 1$ and thus see a different relativistic mass.

 

[PeroK]

The only viable way to measure the apparent diameter could be from the visual angle covered by the star disc from a known measured distance, and from that compute $r_0$ using the metric.

1) You could measure the gravitational field strength (and thereby map out a circle using points of equal local acceleration).

2) You could use local light rays to map out an n-gon with a large number of sides. That would be a very good approximation of a circle.

That would also give you an estimate for the mass of the star.

PS 3) Something clever using gyroscopes.

 

[PeterDonis]

Yes and that's just the point of these whole operations. To match the (absolute) coordinates with the (relative) physical distances measured by an arbitrary observer.

That doesn't change the fact that your claim in post #163 was wrong, as I responded. The $r$ coordinate in Schwarzschild coordinates does directly represent tangential distances (i.e., distances around a circle at that radial coordinate); that's obvious from its definition. It just doesn't directly represent radial distances.

Now I guess that to send a probe with a rod to the surface of the star is out if discussion, unless we can prevent the tidal effects from tearing it to smithereens.

This is a thought experiment, so you can do anything that's consistent with the laws of physics. The laws of physics don't forbid building a probe that can withstand being on a neutron star's surface.

The only viable way to measure the apparent diameter could be from the visual angle covered by the star disc from a known measured distance, and from that compute $r_0$ using the metric.

This is one way, but by no means the only viable way. See above.

EDIT: I'm also doubting that the observed $M$ is an absolute. Isn't that in the metric the mass of the star observed at infinity?

No. It's an invariant parameter in the metric, which is equal to the star's rest mass. Any observer can measure it in a number of different ways; for example, by putting a test object in orbit around the mass (the "test object" could be the observer's own spaceship) and measuring its orbital parameters.

Any other observer would have a $\gamma <> 1$ and thus see a different relativistic mass.

The $M$ in the metric is not relativistic mass. See above.

 

[PeterDonis]

@Pyter, you keep making intuitive guesses that are wrong. You should realize by now that your intuition is flawed and stop doing that.

 

[PeroK]

EDIT: I'm also doubting that the observed $M$ is an absolute. Isn't that in the metric the mass of the star observed at infinity? Any other observer would have a $\gamma <> 1$ and thus see a different relativistic mass.

If you are still using relativistic mass, then I'm not sure how much sense GR is going to make.

 

[weirdoguy]

thus see a different relativistic mass.

This article is worth reading:
https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[Pyter]

That doesn't change the fact that your claim in post #163 was wrong, as I responded. The $r$ coordinate in Schwarzschild coordinates does directly represent tangential distances (i.e., distances around a circle at that radial coordinate); that's obvious from its definition. It just doesn't directly represent radial distances.

Ok, you're measuring the effective coordinate through the measured circumference/area.

This is a thought experiment, so you can do anything that's consistent with the laws of physics. The laws of physics don't forbid building a probe that can withstand being on a neutron star's surface.

I was putting myself in the shoes of a spaceship pilot who needs to compute the metric tensor in practice, for astrogation purposes.

No. It's an invariant parameter in the metric, which is equal to the star's rest mass. Any observer can measure it in a number of different ways; for example, by putting a test object in orbit around the mass (the "test object" could be the observer's own spaceship) and measuring its orbital parameters.

The $M$ in the metric is not relativistic mass. See above.

Who said the contrary? I was talking about the observed mass from some arbitrary point. Isn't the rest mass equal to the mass observed at infinity, where the space is flat?

@Pyter, you keep making intuitive guesses that are wrong. You should realize by now that your intuition is flawed and stop doing that.

I partially disagree, some of them proved right. I don't pretend to be right 100% of the times anyway, I'm here to learn. Thanks for your corrections.

 

[Nugatory]

Isn't the rest mass equal to the mass observed at infinity, where the space is flat?

It’s equal to the mass observed anywhere else too…

 

[PeterDonis]

you're measuring the effective coordinate through the measured circumference/area.

It's not an "effective coordinate". It's the $r$ coordinate in the coordinate chart you've defined. There is no "effective" about it.

Isn't the rest mass equal to the mass observed at infinity, where the space is flat?

No, it's equal to the rest mass, period. There is no such thing as "rest mass observed at a particular point" that changes from point to point or observer to observer. Rest mass is an invariant.

 

[vanhees71]

It is very simple to remember, what's measured by a given local observer in GR. Coordinates are arbitrary "maps" of spacetime points, which do not have a priori any physical meaning, because they change under general (local) coordinate transformations (local diffeomorphism invariance), and thus this is a gauge symmetry, and that's why the Einstein field equations for the pseudometric are only solvable modulo these general coordinate transformations.

To know what a local observer measures, you have to use local scalar quantities referring to this observer. Formally the local observer moves along a time-like world line. Then you can define a tetrad along this world line for this observer by choosing at an arbitrary point at this world line the time-like normalized tangent, pointing to the future, to the world line and then three space-like Minkowski-orthonormal vectors. Then you Fermi-Walker transport this tetrad further along the worldline to have a locally non-rotating local inertial frame for this observer. Then all local observables can be defined wrt. this reference frame as in special relativity.

 

[Pyter]

By "observed mass" I meant "relativistic mass", but since I've just learned that it's no longer a thing, I retract.

 

[Pyter]

Coordinates are arbitrary "maps" of spacetime points, which do not have a priori any physical meaning, because they change under general (local) coordinate transformations

About that, I've found online a very simple and satisfying lecture directly from the horse's mouth: Gaussian coordinates explained by Einstein.
In this chapter, he also explains why they're enough to describe any physical event.

 

[Orodruin]

About that, I've found online a very simple and satisfying lecture directly from the horse's mouth: Gaussian coordinates explained by Einstein.
In this chapter, he also explains why they're enough to describe any physical event.

General pro tip: Modern textbooks will usually do a better job in terms of pedagogy, didactics, and up to date treatment than anything written by Einstein.

 

[Pyter]

General pro tip: Modern textbooks will usually do a better job in terms of pedagogy, didactics, and up to date treatment than anything written by Einstein.

I've noticed, the chapter before that where he makes the analogy with the marble table and the metal rods is quite confusing, I couldn't tell if the local temperature changes only affected the rods or also the table.
But the chapters I've cited helped me "visualize" the physical meaning of the GR, because that particular book is written in layman terms, without too much math and topology.

 

[vanhees71]

When it comes to relativistic thermodynamics and statistical physics, anything written before 1968 is confusing, because the foundations haven't been understood before van Kampen et al. Nowadays, every intrinsic property (among them the intensive quantities of thermodynamics like, temperature, chemical potential, densities of various other thermodynamical potential, and entropy density) related to some medium are defined as scalar (field) quantities as measured in the (local) rest frame of a "fluid cell".

 

[Pyter]

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

Did you get that formula from the metric $ds^2 = g_{tt}\, (c\,dt)^2 - g_{rr}\,dr^2$, by setting $ds^2 = 0$ for the light-like path?

 

[Orodruin]

Did you get that formula from the metric $ds^2 = g_{tt}\, (c\,dt)^2 - g_{rr}\,dr^2$, by setting $ds^2 = 0$ for the light-like path?

Yes. What you want is to integrate dt to obtain the global time difference.

Note: It is $ds^2 = g_{tt} dt^2 + g_{rr} dr^2$. The signs are included in the metric components.

 

[Pyter]

Note: It is $ds^2 = g_{tt} dt^2 + g_{rr} dr^2$. The signs are included in the metric components.

Right. But not $c^2$, I guess? Otherwise either the RHS of the line element is not dimensionally correct, or $g_{tt}$ has a dimension different from $g_{rr}$.

 

[Orodruin]

Right. But not $c^2$, I guess? Otherwise either the RHS of the line element is not dimensionally correct, or $g_{tt}$ has a dimension different from $g_{rr}$.

I always use units with $c = 1$. It is just more ... natural.

 

[cianfa72]

Yes. What you want is to integrate dt to obtain the global time difference.

Note: It is $ds^2 = g_{tt} dt^2 + g_{rr} dr^2$. The signs are included in the metric components.

So for the round-trip coordinate time $\Delta t$ of the complete null path $ds=0$ from the spaceship hovering at fixed $(\theta, \phi, r_1)$ to the star surface at $(\theta, \phi, r_0)$ and back we get the value of that integral multiplied by 2.

 

[Orodruin]

Yes

 

[Pyter]

I always use units with $c = 1$. It is just more ... natural.

Of course, I was talking about MKS units.