# Accelerating universe

1. Jun 6, 2007

### blumfeld0

We know through observations of distant SN that the deceleration parameter is negative and this implies that the universe is accelerating.
This is just me but when I think about acceleration, I think of a vector quantity with units of m/s^2 and has magnitude and direction. do we know the magntude and direction of the accleration of the universe?
i can see why there would be a magnitude maybe, but a direction? doesn't that violate the principle of isotropy and homogeneity?

I guess i don't really get what it means for the universe as a whole to be accelerating.

thanks!

Last edited: Jun 6, 2007
2. Jun 6, 2007

### marcus

no direction to the acceleration

another semantics problem!

what (nondirectional) word should cosmologists use instead?
================

I think what they are trying to communicate is that a"(t) > 0

It is simple if you say it in language of elementary differential calculus but if you try to translate it into English it immediately
gets vague and ambiguous.

a(t) is the scale factor in the FRW metric, a'(t) = da/dt is the time rate of change
a'(t)/a(t) is the hubble parameter
a''(t) is the second derivative. None of these things are vector quantitites or have direction. But they change.
How do you say "the rate of increase is increasing" without suggesting directionality?

I think we are stuck with saying something like "accelerating expansion".

Maybe: "accelerating growth of distances" (quickening growth makes explicit the nondirectionality)

Last edited: Jun 6, 2007
3. Jun 6, 2007

### blumfeld0

thanks. what about the magnitude of the acceleration?

4. Jun 6, 2007

### marcus

there is no obvious natural unit in which to express the scalefactor a(t)

hence no ABSOLUTE numbers can be attached to the growth rate or to its speeding up

the neat trick is to normalize
and write a'/a

then the choice of distance unit cancels out and one has only a unit of time (or its reciprocal) in the expression

also one can normalize the "acceleration" and write a''/a

then the distance unit again cancels, it does not matter if one originally measured in lightyears or inches
and one is left with the reciprocal of time squared (the square of a frequency, if you like)
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I don't know if I am answering your question. You asked about the magnitude. the magnitude depends on the choice of units, which is arbitrary.

for example. a'/a (present time) is the reciprocal of 13.7 billion years
that means that largescale distances are increasing by one percent every 137 million years.
that seems like an intuitive way to say it
but I don't happen to know an equally intuitive way to describe the magnitude of the acceleration.

Last edited: Jun 6, 2007
5. Jun 6, 2007

### MeJennifer

Let's assume for the moment that what you say is correct, then how could this factor in principle be falsified by experiment?

6. Jun 6, 2007

### Wallace

The 'absolute magnitude' of the expansion rate of the Universe is determined by measuring the Hubble parameter today. This is defined as

$$H = \frac{\dot{a}}{a} = \frac{v}{d}$$

where v is the recession speed of an object at distance d (just don't get into the discussion of how you measure those two quantities, it's a surprisingly long story...). Current measurements suggest that

$$H_0 \simeq 72 Kms^{-1} Mpc^{-1}$$

so for every Mpc (a unit of distance) that an object is from some other object, their relative velocities are an additional 72 Km/s away from each other.

Hence this does give a unitfull number to pin onto the expansion, which is otherwise described by the dimensionless parameter a(t).

By the way, and this is a bit of a semantic point, but the dimensionless quantity q that describes the acceleration rate is normally defined as

$$q = -\frac{a\dot{a}^2}{\ddot{a}}$$

since the more simple minded

$$\frac{\ddot{a}}{a}$$

is not dimensionless. Marcus had the right idea though, just clearing up the specifics.

Last edited: Jun 6, 2007
7. Jun 8, 2007

### damgo

To add to what others have said, you can think about the direction of the acceleration thusly:

When we say the universe is expanding at all, we mean it's just "stretching out" (like the black dots here: http://www.astro.ucla.edu/~wright/cphotons.gif ). All galaxies are getting further apart from one another. Any point in the universe looks like it's in the "center": all the galaxies seem to be moving away, and the further off they are the faster they're moving. The acceleration of the expansion just means this is happening at an increasing rate.

So the physical velocity of the expansion is different at different points. In our frame of reference, each galaxy is moving away from us, so their velocity always points directly outward from us. The further away from us they are, the higher this velocity. And the physical acceleration of the universe, similarly, always points directly away from us.

Its magnitude at a given point is gotten by

$$\ddot{R} = - \left( - \frac{R \ddot{R}}{\dot{R}^2} \right) \left( \frac{\dot{R}}{R} \right) ^2 R = - q_0 H_0^2 R$$

Best-fit for q0 currently is about -0.58, and Wallace gave H0 above. So just plug in the distance to the object R, and you get its rate of cosmological acceleration away from us,

$$\ddot{R} \approx 10^{-13} \frac{R}{1 megaparsec} m/s^2$$

8. Jun 9, 2007

### marcus

thanks for the reminder.

$$q = -\frac{a\ddot{a}}{\dot{a}^2}$$