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.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.
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 ?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.
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.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 ?
Do you mean a spaceship hovering w.r.t. the star at fixed ##r, \theta, \phi## ?Not in the form I mentioned, that assumed a stationary.
That is what stationary means, yes.Do you mean a spaceship hovering w.r.t. the star at fixed ##r, \theta, \phi## ?
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 ?That is what stationary means, yes.
No, as I mentioned in previous posts, ”distance” comes with a huge caveat in GR. What is typically defined as ”radial distance” is the integralok, 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 ?
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.What would be measured by a radar measurement’s round trip time (multiplied by c/2) would be
$$
\int \sqrt{\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$
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.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.
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.
Not so trivial to get ##\Delta t## from ##\Delta \tau## (your measured round-trip time) , since ##{\Delta \tau}/{\Delta t}## is not constant.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.
Yes it is.since Δτ/Δt is not constant.
Definitely not. That would involve going below the horizon of the Schwarzschild metric where r is no longer space-like.The r0 value (coordinate radius of the neutron star) is also unknown BTW, maybe we could set it to 0 as a first approximation.
Doesn't also ##g_{tt}## depend on r?Yes it is.
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.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.
Yes, but your r is fixed.Doesn't also ##g_{tt}## depend on r?
But that of the light/radio beam isn't.Yes, but your r is fixed.
That is irrelevant, it is not what you would be measuring.But that of the light/radio beam isn't.
So what? Equations like ##f(x) = g(x)## can many times be solved.it means that also the LHS (C) of the equation in #139 depends on the unknown r.
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##.The ##r_{0}## value (coordinate radius of the neutron star) is also unknown BTW
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.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##.
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.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.
Wrong. The ##r## coordinate is a coordinate. Coordinates are valid for any observer that uses that coordinate chart.But the measured ##r_0## matches the coordinate ##r_0##, needed for the integral, only for the observer at infinity.
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.Wrong. The ##r## coordinate is a coordinate. Coordinates are valid for any observer that uses that coordinate chart.
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.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.
1) You could measure the gravitational field strength (and thereby map out a circle using points of equal local acceleration).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.
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.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.
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.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 one way, but by no means the only viable way. See above.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.
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.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?
The ##M## in the metric is not relativistic mass. See above.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.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.
thus see a different relativistic mass.
Ok, you're measuring the effective coordinate through the measured circumference/area.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.
I was putting myself in the shoes of a spaceship pilot who needs to compute the metric tensor in practice, for astrogation purposes.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.
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?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.
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.@Pyter, you keep making intuitive guesses that are wrong. You should realize by now that your intuition is flawed and stop doing that.
It’s equal to the mass observed anywhere else too…Isn't the rest mass equal to the mass observed at infinity, where the space is flat?
It's not an "effective coordinate". It's the ##r## coordinate in the coordinate chart you've defined. There is no "effective" about it.you're measuring the effective coordinate through the measured circumference/area.
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.Isn't the rest mass equal to the mass observed at infinity, where the space is flat?