00:01
For this problem on the topic of quarks, we want to answer various questions on cosmological redshift and the expansion of the universe.
00:09
Firstly, in part a, we're told that during the time interval delta t, the light emitted from galaxy a in the problem has traveled a distance c -data -t.
00:19
Meanwhile, the distance between earth and the galaxy has expanded from r to r -prime, which is equal to r plus r -alpha delta -t.
00:31
We let c delta t equal to r prime, and this is equal to r plus r alpha delta t, which gives us the time interval delta t to be r divided by c minus r alpha as required.
00:54
Now in part b of the problem, the detected a wavelength lambda prime is longer than lambda by lambda alpha delta t due to the expansion of the universe.
01:02
And so lambda prime is equal to lambda plus lambda alpha delta t which means delta lambda over lambda is equal to lambda prime minus lambda over lambda and from the equation above this is alpha delta t which we can write as alpha r divided by c minus alpha r.
01:40
For part c, we can make use of the binomial expansion formula, and that is one plus or minus x all to the power n is equal to one plus or minus nx over one factorial plus n into n minus one times x squared over two factorial and so on for x squared less than one and we obtain for delta lambda over lambda which is alpha r over c minus alpha r which you can write as alpha r over c into one minus alpha r over c all to the minus one minus alpha r over c all to the into 1 plus minus 1 over 1 factorial into minus alpha r over c plus minus 1 times minus 2 over 2 factorial into minus alpha r over c all squared and higher order terms.
03:10
And so we can approximate this to alpha r over c plus alpha r over c squared plus alpha r over c cubed for the first three terms.
03:37
Now for part d, when only the first term in the expansion for delta lambda over lambda is considered, we have this expression delta lambda over lambda to simply be approximately alpha r over c.
03:58
For part e, we'll set delta lambda over lambda to be v over c, which is equal to hr over c.
04:10
And if we compare this to the result of part d, we get alpha equal to h.
04:18
For part f, we'll use delta lambda over lambda equal to alpha r over c minus alpha r to solve for r.
04:38
And we get r to be c times delta lambda over lambda over alpha into one plus delta lambda over lambda.
04:53
Putting in values, we get this to be 2 .998 times 10 to the 8 meters per second times the ratio delta lambda over lambda 0 .05 divided by alpha 0 .0218 meter per second light year multiplied by 1 plus 0 .05.
05:25
This gives a distance r of 6 .548 times 10 to the 8, or 6 .5 times 10 to the power 8 light years...