- #176
Pyter
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By "observed mass" I meant "relativistic mass", but since I've just learned that it's no longer a thing, I retract.
About that, I've found online a very simple and satisfying lecture directly from the horse's mouth: Gaussian coordinates explained by Einstein.vanhees71 said:Coordinates are arbitrary "maps" of spacetime points, which do not have a priori any physical meaning, because they change under general (local) coordinate transformations
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 said: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.
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.Orodruin said: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.
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 said:What would be measured by a radar measurement’s round trip time (multiplied by c/2) would be
$$
\int \sqrt{-\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$
Yes. What you want is to integrate dt to obtain the global time difference.Pyter said: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?
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 said:Note: It is ##ds^2 = g_{tt} dt^2 + g_{rr} dr^2##. The signs are included in the metric components.
I always use units with ##c = 1##. It is just more ... natural.Pyter said: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}##.
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 said: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.
Of course, I was talking about MKS units.Orodruin said:I always use units with ##c = 1##. It is just more ... natural.