- #176

Pyter

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- #176

Pyter

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- #177

Pyter

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About that, I've found online a very simple and satisfying lecture directly from the horse's mouth: Gaussian coordinates explained by Einstein.Coordinates are arbitrary "maps" of spacetime points, which do not have a priori any physical meaning, because they change under general (local) coordinate transformations

In this chapter, he also explains why they're enough to describe any physical event.

- #178

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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.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.

- #179

Pyter

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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.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.

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.

- #180

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- #181

Pyter

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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?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.

$$

- #182

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

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

- #183

Pyter

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

- #184

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

- #185

cianfa72

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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.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.

- #186

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Yes

- #187

Pyter

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Of course, I was talking about MKS units.I always use units with ##c = 1##. It is just more ... natural.

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