According to Big Bang theory and The Red shift theory "space" is "expanding". Keeping this notion of "space" as physically something in and of itself(space -time did not exist before The Big Bang); what are the supposed theoretical answers for the omnidirectional expansion of this "fabric"–noting that no matter where you are in the universe "space" is expanding away from you??

# [Physics] How does “space” expand?

cosmologyspace-expansionspacetimeuniverse

#### Related Solutions

There are quite a few common misconceptions about the expansion of the universe, even among professional physicists. I will try to clarify a few of these issues; for more information, I highly recommend the article "Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe" from Tamara M. Davis and Charles H. Lineweaver.

I will assume a standard ΛCDM-model, with $$ \begin{align} H_0 &= 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} &= 9.24\times 10^{-5},\\ \Omega_{M,0} &= 0.315,\\ \Omega_{\Lambda,0} &= 0.685,\\ \Omega_{K,0} &= 1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0} = 0. \end{align} $$

The expansion of the universe can be described by a *scale factor* $a(t)$, which can be thought of as the length of an imaginary ruler that expands along with the universe, relative to the present day, i.e. $a(t_0)=1$ where $t_0$ is the present age of the universe.

From the standard equations, one can derive the *Hubble parameter*
$$
H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}},
$$
such that $H(1)=H_0$ is the *Hubble constant*. In a previous post, I showed that the age of the universe, as a function of $a$, is
$$
t(a) = \frac{1}{H_0}\int_0^a\frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}},
$$
which can be numerically inverted to yield $a(t)$, and consequently $H(t)$. It also follows that the present age of the universe is $t_0=t(1)=13.8$ billion years.

Now, another consequence of the Big Bang models is *Hubble's Law*,
$$
v_\text{rec}(t_\text{ob}) = H(t_\text{ob})\,D(t_\text{ob}),
$$
describing the relation between the *recession velocity* $v_\text{rec}(t_\text{ob})$ of a light source and its *proper distance* $D(t_\text{ob})$, at a time $t_\text{ob}$. In fact, this follows immediately from the definition of $H(t_\text{ob})$, since $v_\text{rec}(t_\text{ob})$ is proportional to $\dot{a}$ and $D(t_\text{ob})$ is proportional to $a$.

However, it should be noted that this is a theoretical relation: neither $v_\text{rec}(t_\text{ob})$ nor $D(t_\text{ob})$ can be observed directly. The recession velocity is not a "true" velocity, in the sense that it is not an actual motion in a local inertial frame; clusters of galaxies are locally at rest. The distance between them increases as the universe expands, which can be expressed as $v_\text{rec}(t_\text{ob})$. Some cosmologists therefore prefer to think of $v_\text{rec}(t_\text{ob})$ as an *apparent* velocity, a theoretical quantity with little physical meaning.

A related quantity that *is* observable is the *redshift* of a light source, which is the cumulative increase in wavelength of the photons as they travel through the expanding space between source and observer. There is a simple relation between the scale factor and the redshift of a source, observed at a time $t_\text{ob}$:
$$
1 + z(t_\text{ob}) = \frac{a(t_\text{ob})}{a(t_\text{em})},
$$
such that the observed redshift of a photon immediately gives the time $t_\text{em}$ at which the photon was emitted.

The proper distance $D(t_\text{ob})$ of a source is also a theoretical quantity. It's an "instantaneous" distance, which can be thought of as the distance you would obtain with a (very long!) measuring tape if you were able to "stop" the expansion of the universe. It can however be derived from observable quantities,
such as the luminosity distance or the angular diameter distance. The proper distance to a source, observed at time $t_\text{ob}$ with a redshift $z_\text{ob}$ is
$$
D(z_\text{ob},t_\text{ob}) = a_\text{ob}\frac{c}{H_0}\int_{a_\text{ob}/(1+z_\text{ob})}^{a_\text{ob}}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}},
$$
with $a_\text{ob} = a(t_\text{ob})$. The furthest objects that we theoretically can observe have infinite redshift; they mark the edge of the *observable universe*, also known as the *particle horizon*. Ignoring inflation, we get:
$$
D_\text{ph}(t_\text{ob}) = a_\text{ob}\frac{c}{H_0}\int_0^{a_\text{ob}}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}.
$$
In practice though, the furthest we can see is the CMB, which has a current redshift $z_\text{CMB}(t_0)\approx 1090$.

A source that has a recession velocity $v_\text{rec}(t_\text{ob})=c$ has a corresponding distance
$$
D_\text{H}(t_\text{ob})=\frac{c}{H(t_\text{ob})}.
$$
This is called the *Hubble distance*.

Almost there, just a few more quantities need to be defined. The photons that we observe at a time $t_\text{ob}$ have travelled on a null geodesic called the *past light cone*. It can be defined as the proper distance that a light source had at a time $t_\text{em}$ when it emitted the photons that we observe at $t_\text{ob}$:
$$
D_\text{lc}(t_\text{em},t_\text{ob})= a_\text{em}\frac{c}{H_0}\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}.
$$
There are two special cases: for $t_\text{ob}=t_0$ we have our present-day past light cone (i.e. the photons that we are observing right now), and for $t_\text{ob}=\infty$ we get the so-called *cosmic event horizon*:
$$
D_\text{eh}(t_\text{em})= a_\text{em}\frac{c}{H_0}\int_{a_\text{em}}^\infty\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}.
$$
For light emitted today, $t_\text{em}=t_0$, this has a special significance: if a source closer to us than $D_\text{eh}(t_0)$ emits photons today, then we will be able to observe those at some point in the future. In contrast, we will never observe photons emitted today by sources further than $D_\text{eh}(t_0)$.

One final definition: instead of proper distances, we can use co-moving distances. These are distances defined in a co-ordinate system that expands with the universe. In other words, the co-moving distance of a source that moves away from us along with the Hubble flow, remains constant. The relation between co-moving and proper distance is simply $$ D_c(t) = \frac{D(t)}{a(t)}, $$ so that both are the same at the present day $a(t_0)=1$. Thus $$ \begin{align} D_\text{c,ph}(t_\text{ob}) &= \frac{D_\text{ph}(t_\text{ob})}{a_\text{ob}},\\ D_\text{c,lc}(t_\text{em},t_\text{ob}) &= \frac{D_\text{lc}(t_\text{em},t_\text{ob})}{a_\text{em}},\\ D_\text{c,H}(t_\text{ob}) &= \frac{D_\text{H}(t_\text{ob})}{a_\text{ob}}. \end{align} $$ In fact, it would have been more convenient to start with co-moving distances instead of proper distances; in case you've been wondering where all the above integrals come from, those can be derived from the null geodesic of the FLRW metric: $$ 0 = c^2\text{d}t^2 - a^2(t)\text{d}\ell^2, $$ such that $$ \text{d}\ell = \frac{c\,\text{d}t}{a(t)} = \frac{c\,\text{d}a}{a\,\dot{a}} = \frac{c\,\text{d}a}{a^2\,H(a)}, $$ and $\text{d}\ell$ is the infinitesimal co-moving distance.

So, what can we do with all these tedious calculations? Well, we can draw a graph of the evolution of the expanding universe (after inflation). Inspired by a similar plot in the article from Davis & Lineweaver, I made the following diagram:

This graph contains a lot of information. On the horizontal axis, we have the co-moving distance of light sources, in Gigalightyears (bottom) and the corresponding Gigaparsecs (top). The vertical axis shows the age of the universe (left) and the corresponding scale factor $a$ (right). The horizontal thick black line marks the current age of the universe (13.8 billion years). Co-moving sources have a constant co-moving distance, so that their world lines are vertical lines (the black dotted lines correspond with sources at 10, 20, 30, etc Gly). Of course, our own world line is the thick black vertical line, and we are currently situated at the intersection of the horizontal and vertical black line.

The yellow lines are null geodesics, i.e. the paths of photons. The scale of the time axis is such that these photon paths are straight lines at 45° angles. The orange line is our current past light cone. This is the cross-section of the universe that we currently observe: all the photons that we receive now have travelled on this path. The path extends to the orange dashed line, which is our future light cone. The particle horizon, i.e. the edge of our observable universe, is given by the blue line; note that this is also a null geodesic. The red line is our event horizon: photons emitted outside the event horizon will never reach us.

The purple dashed curves are distances corresponding with particular redshift values $z(t_\text{ob})$, in particular $z(t_\text{ob}) = 1, 3, 10, 50, 1000$. Finally, the green curves are lines of constant recession velocity, in particular $v_\text{rec}(t_\text{ob}) = c, 2c, 3c, 4c$. Of course, the curve $v_\text{rec}(t_\text{ob}) = c$ is nothing else than the Hubble distance.

What can we learn from all this? Quite a lot:

**The current (co-moving) distance of the edge of the observable universe is 46.2 billion ly**. Of course, the*total*universe can be much bigger, and is possibly infinite. The observable universe will keep expanding to a**finite**maximum co-moving distance at cosmic time $t = \infty$, which is 62.9 billion ly. We will never observe any source located beyond that distance.- Curves of constant recession velocity expand to a maximum co-moving distance, at $t_\text{acc} = 7.7$ billion years, and then converge again. This time $t_\text{acc}$, indicated by the horizontal black dashed line, is in fact the moment at which the expansion of the universe began to accelerate.
- Curves of constant redshift also expand first, and converge when $t$ becomes very large. This means that a given source, which moves along a vertical line, will be observed with an infinite redshift when it enters the particle horizon, after which its redshift will decrease to a mimimum value, and finally increase again to infinity at $t = \infty$. In other words, every galaxy outside our local cluster will eventually be redshifted to infinity when the universe becomes very old. This is due to the dominance of dark energy at late cosmic times. Photons that we currently observe of sources at co-moving distances of 10, 20, 30 and 40 Gly have redshifts of 0.87, 2.63, 8.20 and 53.22 respectively.
**The edge of the observable universe is receding from us with a recession velocity of more than 3 times the speed of light.**$3.18c$, to be exact.**In other words, we can observe sources that are moving away from us faster than the speed of light.**Sources at co-moving distances of 10, 20, 30 and 40 Gly are receding from us at 0.69, 1.38, 2.06 and 2.75 times the speed of light, respectively.**Sources outside our particle horizon are moving away even faster.**There is no a priori limit to the maximum recession velocity: it is proportional to the size of the total universe, which could be infinite.**The Hubble distance lies completely inside the event horizon**. It will asymptotically approach the event horizon (as well as the curve of constant redshift 1) as $t$ goes to infinity. The current Hubble distance is 14.5 Gly (corresponding with $z=1.48$) , while the current distance to the event horizon is 16.7 Gly ($z=1.87$).**Photons emitted today by sources that are located between these two distances will still reach us at some time in the future.**- Although the difference between the Hubble distance and the event horizon today is rather small, this difference was much larger in the past. Consider for example the photons that we observe today, emitted by a source at a co-moving distance of 30 Gly. It emitted those photons at $t=0.62$ Gy, when the source was moving away from us at $3.5c$. The source continued its path along the vertical dotted line, while the photons moved on our past light cone. At $t=0.83, 1.64, 4.06$ Gy those photons passed regions that were moving away from us at $3c, 2c, c$ respectively. Along the way, those photons accumulated a total redshift of 53.22.

From all the above, it should be clear that **the Hubble distance is not a horizon.** I should stress again that all these calculations are only valid for the standard ΛCDM-model.

Apologies for the very lengthy post, but I hope it has clarified a few things.

The simple answer is that no, time is not expanding or contracting.

The complicated answer is that when we're describing the universe we start with the assumption that time isn't expanding or contracting. That is, we choose our coordinate system to make the time dimension non-changing.

You don't say whether you're at school or college or whatever, but I'm guessing you've heard of Pythagoras' theorem for calculating the distance, $s$, between two points $(0, 0, 0)$ and $(x, y, z)$:

$$ s^2 = x^2 + y^2 + z^2 $$

Well in special relativity we have to include time in the equation to get a spacetime distance:

$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$

and in general relativity the equation becomes even more complicated because we have to multiply the $dt^2$, $dx^2$, etc by factors determined by a quantity called the metric, and usually denoted by $g$:

$$ ds^2 = g_{00}dt^2 + g_{11}dx^2 + g_{22}dy^2 + ... etc $$

where the $... etc$ can include cross terms like $g_{01}dtdx$, so it can all get very hairy. To be able to do the calculations we normally look for ways to simplify the expression, and in the particular case of the expanding universe we assume that the equation has the form:

$$ ds^2 = -dt^2 + a(t)^2 d\Sigma^2 $$

where the $d\Sigma$ includes all the spatial terms. The function $a(t)$ is a scale factor i.e. it scales up or down the contribution from the $dx$, $dy$ and $dz$, and it's a function of time so the scale factor changes with time. And this is where we get the expanding universe. It's because when you solve the Einstein equations for a homogenous isotropic universe you can calculate $a(t)$ and you find it increases with time, and that's what we mean by the expansion.

However the $dt$ term is not scaled, so time is not expanding (or contracting).

## Best Answer

Space itself was once concentrated in an infinitesimally small point. During the Bang of the Big Bang all distances between points got bigger. If you try to measure the expansion of the universe from any point you will draw the conclusion that the expansion started from that point. It seems that the expansion happened everywhere, and nowhere at the same time.

Think of it like this, if space itself was a tiny dot, then the bang happened in all space, its just that that tiny space is now a lot bigger.