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

• I
Mentor
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 ?

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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## ?

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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 ?

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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
$$\int \sqrt{-\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.$$

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Pyter, PeterDonis and cianfa72
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.

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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.

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## ?

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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.

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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.

PeroK
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.

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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.

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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.

Pyter
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).

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Pyter
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.

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Mentor
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##.

cianfa72, Orodruin and vanhees71
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.

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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.

cianfa72
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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.

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PeroK
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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.

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Pyter
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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.

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

vanhees71
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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.

vanhees71 and weirdoguy
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.

Mentor
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…

Mentor
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.