Chapter 22: Problem 34
Why is particle physics important for understanding the early universe?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 22: Problem 34
Why is particle physics important for understanding the early universe?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeThe principal difference between normal matter and dark matter is that a. normal matter interacts with light, while dark matter does not. b. normal matter has gravity, while dark matter does not. c. things made of normal matter are larger when they are more massive; things made of dark matter are smaller. d. there is no difference; dark matter was just discovered later.
One GUT theory predicts that a proton will decay in about \(10^{31}\) years, which means if you have \(10^{\text {si }}\) protons, you should see one decay per year. The Super-Kamiokande observatory in Japan holds about 20 million kg of water in its main detector, and it did not see any decays in 5 years of continuous operation. What limit does this observation place on proton decay and on the GUT theory described here?
Go to the website for the Dark Energy Survey, an international project beginning in 2012 (https: \(/ /\) www.darkenergysurvey org/index.shtml). What observations will be made for this project? What will it tell scientists about dark energy? Click on "News." What is the status of this project? Are there any results yet?
Astronomers will never directly observe the first few minutes of the universe, because a. the universe was opaque at that time. b. the universe is too large now. c. there were no particles or other matter to see. d. there were no photons.
The universe today has an average density \(\rho_{0}=3 \times 10^{-28} \mathrm{kg} / \mathrm{m}^{3}\) Assuming that the average density depends on the scale factor, as \(\rho=\rho_{0} / R_{\mathrm{U}}^{3},\) what was the scale factor of the universe when its average density was about the same as Earth's atmosphere at sea level \(\left(\rho=1.23 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)
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