Let's break down the solution step-by-step as a helpful tutor: --- ## **Step 1: Write the Scale Factor** The scale factor is given as: \[ a(t) = a_ t^n \] where \(a_\) and \(n\) are constants, and \(n < 1\). --- ## **Step 2: Calculate the Hubble Parameter** The Hubble parameter is defined as: \[ H(t) = \frac{\dot{a}}{a} \] Calculate \(\dot{a}\): \[ \dot{a}(t) = \frac{d}{dt}(a_ t^n) = a_ n t^{n-1} \] So, \[ H(t) = \frac{a_ n t^{n-1}}{a_ t^n} = \frac{n}{t} \] --- ## **Step 3: Write the Particle Horizon Formula** The comoving particle horizon is: \[ R_h(t) = a(t) \int_{}^{t} \frac{d\tilde{t}}{a(\tilde{t})} \] --- ## **Step 4: Substitute the Scale Factor** \[ R_h(t) = a_ t^n \int_{}^{t} \frac{d\tilde{t}}{a_ \tilde{t}^n} = t^n \int_{}^{t} \tilde{t}^{-n} d\tilde{t} \] --- ## **Step 5: Integrate** \[ \int_{}^{t} \tilde{t}^{-n} d\tilde{t} = \left[ \frac{\tilde{t}^{1-n}}{1-n} \right]_{}^{t} = \frac{t^{1-n}}{1-n} - \] (Because \(n < 1\), the lower limit at does not diverge.) --- ## **Step 6: Combine Results** \[ R_h(t) = t^n \cdot \frac{t^{1-n}}{1-n} = \frac{t}{1-n} \] --- ## **Step 7: Final Answer** \[ \boxed{R_h(t) = \frac{t}{1-n}} \] --- ## **Explanation** - The particle horizon \(R_h(t)\) tells us the maximum distance from which particles could have traveled to the observer in the age of the universe at time \(t\). - For power-law expansion (\(a(t) \propto t^n\)), the horizon grows as \(t/(1-n)\). - The result is inversely proportional to the Hubble parameter, \(H(t)\), since \(H(t) = n/t\). --- **Summary:** The value of the particle horizon for a scale factor \(a(t) = a_ t^n\) is \(\boxed{R_h(t) = \frac{t}{1-n}}\), assuming \(n < 1\).