# Distance and the Hubble’s Constant

Gold Member
Response to #25

Marcus, informed comments are no intrusion and the points you raise do home in on any aspect on which I am slightly confused about. I realise that there is still some debate over whether k=0 or is just very close to zero, but I wanted to just focus on the case of k=0.

Watch out! the k=0 case refers to SPATIAL flatness it does not mean that you have 4D flatness or that you can fit the picture to a Minkowski frame----or even more radically to an Euclidean frame.
First, could I just clarify your terminology; are you defining an ‘Euclidean frame’ as 3-D plus absolute time, whilst a Minkowski frame is essentially 4-D spacetime of special relativity?

It does NOT mean that the FRW "reduces to the flat spacetime metric of SR"
Again, to clarify, the FRW metric is often presented in spherical coordinates as:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2} + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \, \mathrm{d}\phi^2 \right)$$

If I consider a radial path, I can reduce the complexity to:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2}\right)$$

If I set k=0, the equation appears to reduce to:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2$$

It was my understanding that today, a(t)=1, which seemed to imply that:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + \mathrm{d}r^2$$

Which appears equivalent to SR spacetime metric at a given point in time.

people are often confused about this and get the mistaken notion that they can fit all or part of a Friedmann model universe onto an SR frame, in cases when this is inappropriate. the confusion partly arises from there being two different meanings of the word flat, namely spatial flat and 4D flat
So my basic question is what is the difference between the formal definition of 'spatial flat' and '4D flat'?

Many thanks

George Jones
Staff Emeritus
Gold Member

Again, to clarify, the FRW metric is often presented in spherical coordinates as:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2} + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \, \mathrm{d}\phi^2 \right)$$

If I consider a radial path, I can reduce the complexity to:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2}\right)$$

If I set k=0, the equation appears to reduce to:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2$$

It was my understanding that today, a(t)=1, which seemed to imply that:

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + \mathrm{d}r^2$$

Which appears equivalent to SR spacetime metric at a given point in time.
If $a \left( t \right) = 1$ and $k = 0$, then the Scwharzschild metric is

$$g = - dt^2 + dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right),$$

which does indeed look like Minkowski spacetime in spherical coordinates. Spacetime curvature, however, is calculated using second derivatives of the metric. At a particular instant of time, $t_\mathrm{now}$, $a \left( t_\mathrm{now} \right) = 1$ is chosen as a scaling convention, but

$$\frac{d^2 a}{dt^2} \left( t_\mathrm{now} \right) \ne 0.$$

Consequently, even at the instant $t_\mathrm{now}$, spacetime curvature is non-zero, while the spacetime curvature of Minknoswki spacetime is zero at all spacetime events.
mysearch said:
So my basic question is what is the difference between the formal definition of 'spatial flat' and '4D flat'?
Spatially flat refers to the curvature of a 3-dimensional spatial section (hypersurace) of spacetime at any particular instant of time. If $t$ is fixed, then $dt=0$, and the spacetime metric induces the spatial metric (up to a scaling factor)

$$dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$$

Spatial curvature is zero, since spatial curvature is calculated using second derivatives of the spatial metric with respect to spatial coordinates. This can be easily seen after transforming to Cartesian coordinates.

Gold Member
Response to #27

George: thanks for the feedback, but could I just press for a few more details? While, the Schwarzschild metric can be presented in a number of forms, I am use to the following form containing the ratio of the Schwarzschild radius $$R_s = 2GM/c^2$$ over radius [r], which reflects the gravitational field strength with distance.

[1] $$c^2 {d \tau}^{2} = \left(1 - \frac{R_s}{r} \right) c^2 dt^2 - \left(1-\frac{R_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

However, if we assume r>>Rs and we only consider a radial expansion along an equatorial plane, we could reduce this to:

[2] $$c^2 {d \tau}^{2} = c^2 dt^2 - dr^2$$

Which I am assuming is essentially equivalent to the Minkowski or SR spacetime metric. So, in word, this is saying in the absence of any significant mass density, spacetime is flat. As such, this was what I was referring to when I said:

reduces to the flat spacetime metric of special relativity, where the complexity of GR curved spacetime is localised within much smaller regions of spacetime of higher gravitational potential, e.g. galaxies.
Additionally, on the assumption that k=0, I reduced the FRW metric to the form:

[3] $$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2$$

Now the similarities between [2] and [3] are obvious except for the term a(t).

At a particular instant of time, $t_\mathrm{now}$, $a \left( t_\mathrm{now} \right) = 1$ is chosen as a scaling convention, but
$$\frac{d^2 a}{dt^2} \left( t_\mathrm{now} \right) \ne 0.$$
As far as I can see this is simply saying that the rate of change of a(t) doesn’t have to be constant, i.e. it can be accelerating or decelerating. This seem reasonable enough, although I not sure whether there is any empirical verification of the rate of change of a(t) with time, which was essentially the question raised in post #24:

As a somewhat expansive question, can we plot the expansion factor (X) of the universe in conjunction with the falling value of (H) linked to gravity alone?
Finally, the point I seem to be missing is that I can understand how the expansion of the universe may be said to curved in sense that a(t) is accelerating, e.g. 1, 2, 4, but based on equation [3], does spacetime remain essentially flat at each instance in time under the assumption that k=0? Apologises if I have missed what might seem an obvious point in your response.

First, there are not many processes that can cause a shift in the position (central wavelength) of a spectral line ... at least, not in environments of interest to those who study the sky (astronomers, etc). Apart from relative motion and the cosmological redshift, there is only a gravitational redshift, which was predicted by Einstein's GR and observed in the lab by Pound and Rebka. Of course, if all you see is a point source, then in principle a range of other mechanisms for creating shifts could be in play; however, in astronomy all (?) extra-galactic point sources (as classes) have associated extended sources whose redshifts are essentially the same. For example, a GRB is a point source, but after it fades its 'parent galaxy' is often seen, and where that galaxy's redshift can be measured, it is similar to that of the GRB (so far anyway). Ditto a quasar or QSO.

Second, there are many processes that can result in a broadened spectral line (and some broadening may be asymmetric, so the central wavelength seems to shift a little), but this isn't what you're asking about, is it?

Third, the relevant theory of atoms (and more), which has been tested to the highest level of precision of any theory in physics*, allows the possibility of systematic spectral shifts, via changes in the fine structure constant, either by time or space (i.e. this constant is different in different parts of the universe, and/or at different times). It will come as no surprise to learn that firm discovery of such changes would be quite dramatic, suggesting perhaps that one or more of the conservation laws do not apply in at least some places (or times) in the observable universe. So quite a few attempts have been made to constrain any such variations. The net? Depends on how you read the various papers - some claim to have found a clear signal for a tiny, but non-zero, change over cosmological time; others no such change. To give you an idea of how this is done, read http://adsabs.harvard.edu/abs/2004ApJ...600..520B" by Bahcall et al.; it is a very elegant, simple, robust method (typical of Bahcall).

* it's actually the theory underlying atomic theory, QED
i don't quite understand how a constant can be different in different parts of the universe. surely it isn't a constant if it varies.

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i don't quite understand how a constant can be different in different parts of the universe. surely it isn't a constant if it varies.
It doesn't have to be a contant that varies, it may be its measurements.