Answer
See answers.
Work Step by Step
The scale factor of the universe is inversely proportional to the temperature. This means that the scale factor multiplied by the temperature is a constant.
Define the scale factor today to be 1 to find the past size relative to today. The current temperature is about 3 K (figure 33-29), so the product of scale and temperature should be about 3.
$$(scale)(temperature)=3$$
We use figure 33–29 to roughly estimate the temperature at the times requested, to one significant figure. Your answers may vary due to different ways to read the figure.
a. At a time of t = $10^6y$, the temperature was about $10^3\;K$. Estimate the scale factor of the universe.
$$scale=\frac{3}{temperature} =\frac{3}{10^{3}}=3\times10^{-3}$$
Page 967 gives us a figure for the universe’s current size, about $14\times10^9\;ly$. Estimate the size back then.
$$Size\approx (3\times10^{-3})(14\times10^9\;ly)=4\times10^{7}ly$$
b. At a time of t = $1s$, the temperature was about $10^{10}\;K$. Estimate the scale factor of the universe.
$$scale=\frac{3}{temperature} =\frac{3}{10^{10}}=3\times10^{-10}$$
Page 967 gives us a figure for the universe’s current size, about $14\times10^9\;ly$. Estimate the size back then.
$$Size\approx (3\times10^{-10})(14\times10^9\;ly)=4\;ly$$
c. At a time of t = $10^{-6}s$, the temperature was about $10^{12}\;K$. Estimate the scale factor of the universe.
$$scale=\frac{3}{temperature} =\frac{3}{10^{12}}=3\times10^{-12}$$
Page 967 gives us a figure for the universe’s current size, about $14\times10^9\;ly$. Estimate the size back then.
$$Size\approx (3\times10^{-12})(14\times10^9\;ly)=4\times10^{-2}ly$$
d. At a time of t = $10^{-35}s$, the temperature was about $10^{26}\;K$. Estimate the scale factor of the universe.
$$scale=\frac{3}{temperature} =\frac{3}{10^{26}}=3\times10^{-26}$$
Page 967 gives us a figure for the universe’s current size, about $14\times10^9\;ly$. Estimate the size back then.
$$Size\approx (3\times10^{-26})(14\times10^9\;ly)=4\times10^{0-16}ly=4m$$
