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

In summary, GR is generally covariant, meaning that the Einstein's Field Equation can be solved in any coordinate system. However, for more complicated solutions, a clever choice of coordinates is crucial. The ADM formalism is often used for solving the EFE, along with a (1+3)-formalism for numerical simulations. The metric of spacetime is not irrelevant, as it is the solution to the ten independent differential equations. Gravity is considered to warp spacetime through the metric, rather than being a force. In solving the EFE, it is often helpful to use an arbitrary map to simplify the equations before mapping it back to the original frame.
  • #176
By "observed mass" I meant "relativistic mass", but since I've just learned that it's no longer a thing, I retract.
 
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  • #177
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
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.
 
  • #178
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.
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.
 
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  • #179
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.
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.
 
  • #180
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".
 
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  • #181
Orodruin said:
What would be measured by a radar measurement’s round trip time (multiplied by c/2) would be
$$
\int \sqrt{-\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$
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?
 
  • #182
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?
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.
 
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  • #183
Orodruin said:
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}##.
 
  • #184
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}##.
I always use units with ##c = 1##. It is just more ... natural.
 
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  • #185
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.
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.
 
  • #186
Yes
 
  • #187
Orodruin said:
I always use units with ##c = 1##. It is just more ... natural.
Of course, I was talking about MKS units.
 

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