Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I wanted to know what the observable universe is so I was thinking and I thought, it must be age of the universe times 2.

Well I was wrong. I found on one website that it is 46B LY across in each direction. How does this make sense?

I get how the universe has expanded since then, but we should only be able to see light that is 13.7 billion LY old. Does this mean that the Universe is expanding faster than the speed of light? Or light from other objects is travelling to us faster than speed of light?

share|improve this question
    
    
I'm talking more about present day not the early beginings of the unvierse. –  Nick11111 Mar 20 '13 at 5:58
1  
The first link Qmechanic gives answers your question. We can only see 13.7Glyrs. We can calculate where those objects are now, but we can't see them there. The 46Glyr figure is entirely theoretical. –  John Rennie Mar 20 '13 at 6:55
    
@JohnRennie "We can only see 13.7 Glyrs" is a bit misleading. I cannot see why the 13.7Glyr limit would be less theoretical then the 46Glyr figure. In fact, I think we cab be pretty certain that the "distance" we can see is not 13.7 Glyr - the CMB was only some 40 Mlyrs away when it was emitted. It has travelled 13.7 Glyrs in its own frame, but that doesn't tell us much. How do you define how far "we can see"? In every day speech, we define it as the current distance to the object that emitted the light, don't we? –  Thriveth Jun 7 '13 at 11:23
add comment

marked as duplicate by John Rennie, Waffle's Crazy Peanut, Emilio Pisanty, dmckee Mar 21 '13 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

Explanatory Framework: Foliated Spacetime

In General Relativity, which used to describe the universe on cosmological length scales, spatial and temporal distances are no longer absolute quantities. Furthermore, in astronomy, there are several methods to determine a distance, which could disagree on cosmological length scales.

Therefore, it is convenient, to initially visualize the problem in a framework, which makes it easier to understand the involved quantities and the concrete question.

In the so called 3+1 formalism of General Relativity, spacetime is described as a foliation of spatial, i.e. 3-dimensional, hypersurfaces along the time-axis. In the figure, each slice $\Sigma_t$ represents 3-dimensional space at a given time $t$.

Foliation of Spacetime. Light is emitted from a distant object at time $t_e$ and reaches earth at time $t_0$.

Looking into the past

Since light, locally, is always traveling with the speed of light, $c$, on large length scales it becomes important that light that reaches the observer on earth at a time $t_0$ ("today") has been emitted at its origin at a time $t_e:<t_0$, i.e. earlier than $t_0$.

Expansion of the universe

Experimentally, by measuring the speed of distant objects relative to us and by measuring their distance, one finds that objects, which are farther away, move away from us with a greater velocity. This is called Hubbles Law.

$$ v=H_0 \cdot D \tag{Hubble's Law} $$

$v:\equiv\frac{dD}{dt}$ is the velocity of the object relative to us. $D$ is the so called proper distance to the object, which is the distance within one spatial slice at a given time (as seen in the figure). And $H_0$, for historical reasons, is called Hubble constant. But, actually, $H(t)$ is a function of time:

Parametrizing an expansion of the universe, where the proper distance $D(t)$ between two objects is some time-fixed distance, the so called comoving distance, $x$, multiplied with time-dependant scale factor $R(t)$, (i.e. all distances grow as the universe expands with growing scale factor $R(t)$),

$$ D(t) = x \cdot R(t) , $$

one finds, that the Hubble parameter $H(t)$ is actually the relative rate of expansion at time $t$:

$$ v(t) \equiv \frac{dD(t)}{dt} = \frac{d}{dt} \left( x \cdot R(t) \right) = x \cdot \frac{dR(t)}{dt} = \frac{D(t)}{R(t)} \cdot \frac{dR(t)}{dt} \equiv \frac{\dot R(t)}{R(t)} \cdot D(t) \equiv H(t) \cdot D(t) $$

Writing the very left and the very right expression together, $v(t) = H(t) \cdot D(t)$, one sees, that Hubble's Law, $v = H_0 \cdot D$, describes a special case, namely the situation "today": $v(t_0) = H(t_0) \cdot D(t_0)$, where $H(t_0)\equiv H_0 = \dot R(t_0) / R(t_0)$.

Superluminal velocity

Does this mean that the Universe is expanding faster than the speed of light?

In a way, that's absolutely right. The proper distance $D$ between a distant object and us can very well grow more rapidly than the speed of light. But, that is not because objects would move locally faster than the speed of light.

$$ \frac{dD}{dt} = \frac{dx\,R}{dt} + \frac{x\,dR}{dt} $$

In the formula above, the term $\frac{dx\,R}{dt}$ can be interpreted as the local velocity, or peculiar velocity, the term $\frac{x\,dR}{dt}$ as the part of the apparent velocity that is caused by the expansion of space.

In this formalism, the statement that nothing can move faster than the speed of light, would mean that nothing can move faster than the speed of light, locally:

$$ v_\text{local} := \frac{dx\,R}{dt} \leq c $$

But nothing prevents the universe from expanding more rapidly than the speed of light, i.e. prevent the scale factor $R(t)$ from growing.

Therefore, given that $v_\text{local}$ has to be smaller than the speed of light, the "velocity" $dD(t)/dt$, observed at earth, which is corresponding to the proper distance, $D(t):=x\cdot R(t)$, can still be larger than the speed of light.

Not entirely accurate, but for demonstration, one can visualize a balloon with coins glued to its surface. As you pump the balloon up, the proper distance between the coins grows, but, locally, the size of the coins stays the same.

Age of the universe

In cosmology, one can parametrize the Hubble parameter $H(t)$ by so called cosmological parameters, which can be measured, experimentally, by various methods. Therefore, one knows $H$ as a function of the scale factor $R(t)$ and those cosmological parameters.

$$ H=H(t)=H(R(t), \text{several cosmological parameters}) $$

Discussing Hubble's law, above, we've seen that $H(t)=\dot R(t)/R(t) = \frac{dR(t)}{dt}/R(t)$. Solved by $dt$, that reads $dt=dR \cdot \frac{1}{HR}$.

Integrating over each infinitesimal time interval $dt$ from the big bang ($t=0$) to today ($t=t_0$) gives the age of the universe $t_0$:

$$ t_0 = \int_0^{t_0} dt = \int_0^{R(t_0)} dR\,\frac{1}{R\,H(R, \text{cosm. param.})} \approx 13.7\,\text{Gyr} $$

Size of the observable universe

We can only observe the universe by looking at particles, e.g. photons, i.e. light, that reaches us. Since, locally, nothing can travel faster than the speed of light, the distance that light could have travelled within the age of the universe, $t_0$, determines the size of the observable universe.

The so called distance to the particle horizon, $r_p$, is the distance of an object that has emitted particles (light), which reach us today, and were emitted at $t=0$, i.e. one age of the universe ago.

Which kind of distance? Proper distance, $D:=R\,x$, or comoving distance, $x$?

A reasonable thing would be to ask for the size of the universe as it is today, i.e. to ask for the proper distance, as shown in the figure.

But, most commonly, the scale factor $R(t)$ is defined such that $R(t_0)=1$. Therefore, if we ask for today's particle horizon, $r_p$ ($t=t_0$), there is no difference.

$$ r_p := D_p(t_0) = x_p \, R(t_0), \ \ \ R(t_0)=1 $$

How to calculate the proper size of the universe

I found on one website that it is 46B LY across in each direction

This quantity refers to the distance to the particle horizon, $r_p$, i.e. to the radius of the observable universe. The diameter would be twice as large.

$$ \text{proper radius of the observable universe} = r_p \approx 46\,Gly $$

This can be calculated similarly to calculating the age of the universe as shown above.

$$ r_p = D_p(t_0) = R(t_0)\cdot x_p = R(t_0)\int_0^{x_p} dx $$

We have seen that $v_\text{local} := \frac{dx\,R}{dt} \leq c$. For light, we know, $v_\text{local}=c$. Therefore, for light, $dx = c\,dt / R$.

$$ r_p = R(t_0)\int_0^{x_p} dx = R(t_0)\int_0^{t_0} \frac{c\,dt}{R(t)} = R(t_0)\int_0^{R(t_0)} dR \frac{c}{R} \frac{1}{R\,H(R, \text{cosm. param.})} \approx 46\,Gly $$

Size of the rest of the universe

Since the particle horizon represents the limit from where any information can have reached us, we, in principle, cannot tell what is behind that horizon. Therefore, one cannot know the actual size of the rest of the universe. One cannot even know for sure whether there exists a rest of the universe---but it's a convenient assumption.

But models of the Inflationary Expansion of the universe of an early age suggest that the actual universe is considerably larger than the observable universe.

These models are strong in explaining structure formation of the universe, i.e. how and on which length scales clusters of galaxies etc. could have formed, despite the universe being homogeneous on larger scales. They also solve the so called cosmological fine-tuning problem, or flatness problem:

The universe appears flat as far as we can see (i.e. not geometrically curved, i.e. the sum of angles in a triangle is 180 degrees) on large observable scales, despite the fact that it would be more probable that the universe is curved.

The models of inflation solve this problem by suggesting that the actual universe could actually be curved if it would be large enough: Then, the observable part of the universe would be sufficiently small to appear flat. Like you, locally, can describe the earth's surface being flat even though the earth is a sphere.

But from this analogy one can see, that if these inflationary models are correct, the rest of the universe has to be significantly larger than the observable universe.

Further Reading

  1. Schneider, Introduction to Extragalactic Astronomy and Cosmology, Springer, ch. 4.
  2. d'Inverno, Introducing Einstein's Relativity, ch. 23.
  3. Hobson et al., General Relativity, An Introduction for Physicists, Cambridge University Press, ch. 14. In particular, chapter 14.11 for the different cosmological distance measures.
  4. Wikipedia article on distance measures in cosmology: Distance Measures (Cosmology)
  5. Wikipedia article on the size of the observable universe, including common misconceptions: Observable Universe, section Size.
share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.