## Friday, October 30, 2015

### 27 | Worksheet 8.1, Problem 4: Scale factor-comological redshift relation

There is a direct relationship between the scale factor of the universe during the epoch in which the light we receive from an object was originally emitted $$a$$ and the cosmological redshift $$z$$. One way to derive this is to consider a light ray with wavelength $$\lambda_e$$ that was emitted at time $$t_e$$ and traveling towards us, and track the change in its wavelength as it travels across space and time while the universe expands.

(a) Two observers along the path of this light ray are separated by a small distance $$\mathrm{d}r$$. If the light ray passes by them at time $$t$$, write down their relative velocity $$\mathrm{d}v$$ as a function of $$H(t)$$ and $$\mathrm{d}r$$. Then write their redshifts $$\mathrm{d}z = \mathrm{d}\lambda / \lambda$$ with respect to each other.

We can start by recalling the Hubble law:
$v = H(t) d$
Using this expression, the relative velocity over a small distance $$\mathrm{d}r$$ becomes:
$\mathrm{d}v = H(t) \mathrm{d}r$
Now let's deal with the redshifts. We have two expressions for the redshift. The first is given by:
$z = \frac{v}{c}$
Let's consider the redshift over a small change in velocity $$\mathrm{d}v$$. Our expression becomes:
$\mathrm{d}z = \frac{\mathrm{d}v}{c}$
The Hubble law gave us an expression for $$\mathrm{d}v$$ in terms of the Hubble constant and the distance. If we plug this in, we get:
$\mathrm{d}z = \frac{H(t) \mathrm{d}r}{c}$
The second expression for the redshift is given in the problem:
$\mathrm{d}z = \frac{\mathrm{d}\lambda}{\lambda}$
Now that we have two expressions for $$\mathrm{d}z$$, let's set them equal.
$\boxed{\frac{\mathrm{d}\lambda}{\lambda} = \frac{H(t) \mathrm{d}r}{c}}$

(b) Let's gradually work the time differential into the picture. Express the time it took light to travel between the two observers $$\mathrm{d}t$$ in terms of $$\mathrm{d}r$$ and $$c$$. Also, rearrange the definition of the Hubble constant to isolate for $$\mathrm{d}t$$.

The time it takes the light to travel between the observers can be found using the simple relationship that velocity is equal to distance over time. Using our differentials and the speed of light, this gives:
$\boxed{\mathrm{d}t = \frac{\mathrm{d}r}{c}}$
The Hubble constant is given by:
$H(t) = \frac{1}{a(t)}\frac{\mathrm{d}a}{\mathrm{d}t}$
Solving for the time differential and saying $$a = a(t)$$,
$\boxed{\mathrm{d}t = \frac{\mathrm{d}a}{H(t)a}}$

(c) Combine your answers in (b) with part (a) to show that $$\mathrm{d}\lambda / \lambda = \mathrm{d}a / a$$. This relation means that at time $$t$$ and over time interval $$\mathrm{d}t$$, the fractional change in the light ray's wavelength equals the fractional change in the scale factor.

In (a) we found:
$\frac{\mathrm{d}\lambda}{\lambda} = \frac{H(t) \mathrm{d}r}{c}$
Let's solve this for $$H(t)$$:
$H(t) = \frac{c \mathrm{d}\lambda}{\lambda \mathrm{d}r}$
In (b) we found the following two expressions for the time differential:
$\mathrm{d}t = \frac{\mathrm{d}r}{c}$
$\mathrm{d}t = \frac{\mathrm{d}a}{H(t)a}$
If we set these two equal and solve for $$H(t)$$, we get:
$H(t) = \frac{c \mathrm{d}a}{a\mathrm{d}r}$
Now we can combine our results from (a) and (b):
$\frac{c \mathrm{d}\lambda}{\lambda \mathrm{d}r} = \frac{c \mathrm{d}a}{a\mathrm{d}r}$
Sure enough, this simplifies to:
$\boxed{\frac{\mathrm{d}\lambda}{\lambda} = \frac{\mathrm{d}a}{a}}$

(d) Solve this differential equation for $$\lambda(a)$$ up to a constant $$C$$. This is the wavelength of the light ray as a function of the scale factor.

We found that:
$\frac{\mathrm{d}\lambda}{\lambda} = \frac{\mathrm{d}a}{a}$
To solve this differential equation, we can integrate both sides of our expression.
$\int \frac{\mathrm{d}\lambda}{\lambda} = \int \frac{\mathrm{d}a}{a}$
$\ln{\lambda} = \ln{a} + C'$
Now we can exponentiate both sides to get rid of the logarithms. This gives us:
$\lambda = \mathrm{e}^{\ln{a} + C'} = a\cdot \mathrm{e}^{C'}$
We can write this same expression as:
$\boxed{\lambda(a) = Ca}$

(e) Use appropriate boundary conditions on your solution above to show that $$C = \lambda_{\text{observed,today}}$$.

Let's say $$\lambda_0 = \lambda_{\text{observed,today}}$$. In the present, the value of the scale factor $$a$$ is, by definition, equal to 1. Since $$\lambda_0$$ is the observed wavelength at the present time,
$\lambda(a=1) = \lambda_0 = C(1) = C$
Therefore, $$C = \lambda{\text{observed,today}}$$.

(f) What is the original wavelength of the light $$\lambda_e$$ (i.e. what is the wavelength of the light ray at time of emission $$t_e$$ )?

Recall that the scale factor $$a$$ is a function of time $$t$$. As a result, we can use the expression we derived above to say:
$\lambda_e = Ca(t_e)$
However, we just showed that $$C = \lambda_0$$, where $$\lambda_0$$ is the observed wavelength in the present. Using this, our expression for the emitted wavelength becomes:
$\boxed{\lambda_e = \lambda_0 a(t_e)}$

(g) What is the measured redshift for the source of this light ray in terms of the scale factor of the universe at the time of emission?

The redshift is defined as:
$z = \frac{\lambda_o - \lambda_e}{\lambda_e}$
Where $$\lambda_o$$ is the observed wavelength and $$\lambda_e$$ is the emitted wavelength. In terms of our variables, $$\lambda_o = \lambda_0$$ and $$\lambda_e$$ is the same. So, our expression is:
$z = \frac{\lambda_0 - \lambda_e}{\lambda_e}$
In (f), we showed that $$\lambda_e = \lambda_0 a(t_e)$$. Using this, our redshift expression becomes:
$z = \frac{\lambda_0 - \lambda_0 a(t_e)}{\lambda_0 a(t_e)} = \frac{1 - a(t_e)}{a(t_e)}$
If we say represent the scale factor at time of emission as $$a_e = a(t_e)$$, we can express the measured redshift for the source of the light ray as:
$\boxed{z = \frac{1 - a_e}{a_e}}$

(h) You should have noticed that the redshift is a unique function of the scale factor. Alternatively, you can express the scale factor unambiguously in terms of the redshift. Interpret this finding. How large was the universe at the time the light from a galaxy at $$z=3$$ was emitted compared to the present-day?

To find this expression, we can simply rearrange the expression we just derived:
$z = \frac{1 - a_e}{a_e}$
Rearranging, we get:
$\boxed{a_e = \frac{1}{1+z}}$
This result means that a source with a larger observed redshift emitted it's light longer ago. Based on other relations we've shown, this also directly shows that a larger redshift represents light from an object that is farther away.

Let's use this expression to determine the scale factor of the universe at the time light was emitted from a source with an observed redshift of $$z = 3$$.
$a_e = \frac{1}{1+z} = \frac{1}{1+3} =\boxed{\frac{1}{4}}$
This means that at the time the light from this source was emitted, the observable universe had a Hubble length 1/4 the size it is today.