What does g00 element of the Schwarzschild metric tensor tell us?

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SUMMARY

The discussion centers on the interpretation of the g00 element of the Schwarzschild metric tensor, emphasizing that Schwarzschild coordinate time aligns with proper time only at infinite distances from the center of mass. It is established that within the Schwarzschild radius, g00 serves merely as a coordinate label without physical correspondence, while outside the horizon, the square root of |g00| indicates the time dilation ratio between local hovering clocks and those at infinity. The conversation also contrasts Schwarzschild spacetime with FLRW spacetime, highlighting significant differences in gravitational potential and time dilation implications.

PREREQUISITES
  • Understanding of the Schwarzschild metric and its implications in general relativity.
  • Familiarity with the concept of time dilation in gravitational fields.
  • Knowledge of the Friedmann-Lemaître-Robertson-Walker (FLRW) metric and its applications in cosmology.
  • Basic grasp of the equivalence principle in general relativity.
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  • Study the implications of gravitational redshift in both Schwarzschild and FLRW spacetimes.
  • Explore the mathematical derivation of the Schwarzschild metric and its applications in astrophysics.
  • Investigate Birkhoff's theorem and its relevance to vacuum spacetimes.
  • Learn about the Einstein tensor and its relationship to the stress-energy tensor in cosmological models.
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Physicists, cosmologists, and students of general relativity seeking to deepen their understanding of spacetime metrics and their implications on time dilation and gravitational effects.

ongoer
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It's a coordinate, just a number used to label events. It doesn't correspond to anything anywhere except at infinite distance, where it is equal to the proper time along the worldline of a stationary observer - we defined it to be that way.
 
Nugatory said:
It's a coordinate, just a number used to label events. It doesn't correspond to anything anywhere except at infinite distance, where it is equal to the proper time along the worldline of a stationary observer - we defined it to be that way.
SchwarzschildForNugatory.webp
 
If you are using Schwarzschild coordinates (which you seem to be assuming) and are outside the horizon then ##\sqrt{|g_{00}|}## gives the ratio between tick rates of local hovering clocks and Schwarzschild coordinate time. Since Schwarzschild coordinate time ticks synchronously with clocks at infinity (at least for one reasonable definition of synchronous) then this leads you to the time dilation ratio between a local hovering clock and one at infinity.

Note the lengthy list of caveats and special case restrictions.
 
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A gravitational redshift can also equivalently be interpreted as gravitational time dilation at the source of the radiation:[8][2] if two oscillators (attached to transmitters producing electromagnetic radiation) are operating at different gravitational potentials, the oscillator at the higher gravitational potential (farther from the attracting body) will tick faster; that is, when observed from the same location, it will have a higher measured frequency than the oscillator at the lower gravitational potential (closer to the attracting body).
Screenshot_20250511_183947_Chrome.webp

https://en.wikipedia.org/wiki/Gravitational_redshift#Spherically_symmetric_gravitational_field
 
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The Wiki page fails to note that Birkhoff's theorem applies to a vacuum spacetime (although the Birkhoff's theorem page it links to does so). Otherwise that seems fine.

Is there any reason you are quoting large chunks of text without any questions?
 
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Ibix said:
The Wiki page fails to note that Birkhoff's theorem applies to a vacuum spacetime (although the Birkhoff's theorem page it links to does so). Otherwise that seems fine.

Is there any reason you are quoting large chunks of text without any questions?
That's the reason:
She's not talking crap.
 
So this is still related to your questions about cosmological spacetimes, which are not static, not vacuum, and where you are trying to compare clock rates of timelike separated clocks? In that case I don't see why you're quoting results that apply to spacelike-separated clocks in static vacuum spacetime.
 
Ibix said:
So this is still related to your questions about cosmological spacetimes, which are not static, not vacuum, and where you are trying to compare clock rates of timelike separated clocks? In that case I don't see why you're quoting results that apply to spacelike-separated clocks in static vacuum spacetime.
Because I'm convinced there is no difference between them in terms of the redshift and its corresponding time dilation.
 
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  • #10
ongoer said:
I'm convinced there is no difference between them in terms of the redshift and its corresponding time dilation
One big difference is that the Schwarzschild spacetime has a gravitational potential and the FLRW spacetime does not.

All of the other differences already mentioned are also present.
 
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  • #11
Dale said:
One big difference is that the Schwarzschild spacetime has a gravitational potential and the FLRW spacetime does not.

All of the other differences already mentioned are also important.
Equivalence principle - the effects of gravity and acceleration are indistinguishable. Expansion has never been linear.

I also wrote "in terms of the redshift and its corresponding time dilation."
 
  • #12
ongoer said:
Because I'm convinced there is no difference between them in terms of the redshift and its corresponding time dilation.
Ok. Let's put aside whether that makes any sense. Say gravitational redshift in FLRW spacetime is the ratio of ##\sqrt{|g_{00}|}## terms as it is in Schwarzschild spacetime. What is ##g_{00}## in the case of the FLRW metric?
ongoer said:
Also, is the vacuum with the gravitational field above the atmosphere no longer vacuum?
On the large scale, space has a more or less uniform density of matter. That's part of why you only start to see expansion and redshift on very large scales and don't see it on galactic and sub-galactic scales where the deviation from the average density is important.
ongoer said:
Equivalence principle - the effects of gravity and acceleration are indistinguishable.
No, the equivalence principle says that free-falling in a gravitational field is, on a small scale, indistinguishable from free fall outside a gravitational field. More formally, it says that the first derivatives of the metric can be made to vanish by an appropriate choice of coordinates. It does not say that acceleration and gravity are indistinguishable.
 
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  • #13
ongoer said:
I also wrote "in terms of the redshift and its corresponding time dilation."
In a static spacetime, like Schwarzschild, the redshift is related to a potential. And that potential is related to time dilation.

But in a non-static spacetime, like FLRW, there is no potential. So the connection is broken. Two comoving observers, neither of which are time dilated in FLRW coordinates, will still be redshifted wrt each other.

ongoer said:
Equivalence principle - the effects of gravity and acceleration are indistinguishable.
Note that for FLRW neither of two comoving observers are accelerating. But for Schwarzschild hovering observers only the one at infinity is not accelerating.
 
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  • #14
ongoer said:
Equivalence principle - the effects of gravity and acceleration are indistinguishable.
That's not what the equivalence principle says. It equates a uniform gravitational field and an appropriate accelerating reference frame.

Moreover, the equivalence principle deals with local experiments, over a small enough region of space and a short enough period of time.
 
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  • #15
Ibix said:
Ok. Let's put aside whether that makes any sense. Say gravitational redshift in FLRW spacetime is the ratio of ##\sqrt{|g_{00}|}## terms as it is in Schwarzschild spacetime. What is ##g_{00}## in the case of the FLRW metric?
##g_{00}=1## and it's wrong, because it doesn't account for cosmological time dilation. FLRW metric is wrong. In case of the expanding spacetime it should be the same as for the spatial diagonal elements, that is ##a(t)^2## with the opposite sign.
 
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  • #16
Ibix said:
No, the equivalence principle says that free-falling in a gravitational field is, on a small scale, indistinguishable from free fall outside a gravitational field.
PeroK said:
That's not what the equivalence principle says. It equates a uniform gravitational field and an appropriate accelerating reference frame.
Just to note, these are two different ways of looking at the same thing.
 
  • #17
ongoer said:
In case of the expanding spacetime it should be the same as for the spatial diagonal elements, that is ##a(t)^2## with the opposite sing.
Compute the curvature tensor of that metric. You'll find it's all zeroes - you are describing flat spacetime in silly coordinates.
 
  • #19
ongoer said:
FLRW metric is wrong.
The FLRW metric is well established both theoretically and experimentally. We are not going to entertain personal speculation. Please review the forum rules before posting again
 
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  • #20
Ibix said:
Compute the curvature tensor of that metric. You'll find it's all zeroes - you are describing flat spacetime in silly coordinates.
I did, this expanding spacetime IS flat both spatially and temporally, and it's still expanding. Why are you calling these coordinates silly?
 
  • #21
ongoer said:
I did, this expanding spacetime IS flat both spatially and temporally, and it's still expanding.
The FLRW spacetime is curved.

However, in the comoving coordinates the spatial foliation can be flat or curved. The current evidence is consistent with both cases.
 
  • #22
Dale said:
The FLRW spacetime is curved.

However, in the comoving coordinates the spatial foliation can be flat or curved. The current evidence is consistent with both cases.
As I've said that I reject FLRW. Please, tell me how OUR spacetime can be curved spatially assuming large scale homogeneity, or how can it be curved temporally, if all the observers resting in their CMB reference frames experience the same flow of time, if we neglect the local effects of gravity?
 
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  • #23
ongoer said:
I did, this expanding spacetime IS flat both spatially and temporally, and it's still expanding.
It's just Milne's coordinates on flat Minkowski spacetime. Your coordinates are curved; nothing more. And that space is everywhere vacuum. There is no matter, no CMB, nothing.

This is not consistent with our own spacetime where the density is close to critical.
 
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  • #24
Ibix said:
It's just Milne's coordinates on flat Minkowski spacetime. Your coordinates are curved; nothing more. And that space is everywhere vacuum. There is no matter, no CMB, nothing.
Do you think, that FLRW without the explicit function of the scale factor from the Friedmann equations tell you anything about matter and radiation?
 
  • #25
ongoer said:
Do you think, that FLRW without the explicit function of the scale factor from the Friedmann equations tell you anything about matter and radiation?
If the curvature tensor is zero then the spacetime is vacuum. If not, then it may not be vacuum (depends on the Einstein tensor). That much is absolutely certain. And FLRW has non-zero curvature and Einstein tensors if the density is non-zero.

I can't really make sense of what you're trying to ask, so if that doesn't answer your question, please rephrase.
 
  • #26
Ibix said:
If the curvature tensor is zero then the spacetime is vacuum. If not, then it may not be vacuum (depends on the Einstein tensor). That much is absolutely certain. And FLRW has non-zero curvature and Einstein tensors if the density is non-zero.

I can't really make sense of what you're trying to ask, so if that doesn't answer your question, please rephrase.
You can have zero Einstein tensor, SEM tensor with the CMB radiation density and pressue, and the metric tensor with the diagonal ##g_{\mu\mu}=a(t)^2(1, -1, -1, -1)## proportional to the SEM tensor diagonal ##T_{\mu\mu}=(\rho, -p, -p, -p)## where ##\rho=-p##.
 
  • #27
ongoer said:
You can have zero Einstein tensor, SEM tensor with the CMB radiation density and pressue, and the metric tensor with the diagonal ##g_{\mu\mu}=a(t)^2(1, -1, -1, -1)## proportional to the SEM tensor diagonal ##T_{\mu\mu}=(\rho, -p, -p, -p)## where ##\rho=-p##.
The Einstein tensor is equal to the stress-energy tensor times ##8\pi G/c^4##. So no, you cannot.

You seem to be confusing the metric, ##g_{ab}## and the Einstein tensor, ##G_{ab}##. They are not the same.
 
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  • #28
ongoer said:
That's the reason:
She's not talking crap.
You were already told in that previous thread not to rely on the source you are referring to here, and why.
 
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  • #29
ongoer said:
I reject FLRW.
And you have just received another warning for continuing to post this misinformation after being explicitly told not to.

As far as I can tell, everything you think you know about the FLRW metric is wrong.

This thread is closed.
 
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