I have a hw problem that I know involves calculus, but I'm stumped.
The number density of photons left over from the Big Bang has an energy dependence of the form n(E)=(E^2)/e^(E/T)−1, where E is the energy of the photon and T is the temperature of this relic radiation. (For your information, this temperature has been measured to be about -270 degrees Celsius, but don’t use this information for the questions below.)
The problem asks me for the derivative of n(E) with respect to E and to keep T fixed. I'm guessing I could break this up as (E^2)(e^(E/T)-1)^(-1) and I think I can use the product rule, but I'm not entirely sure. Please do explain how to go about the problem, I don't just want the answer. Thank you!
To find the derivative of n(E) with respect to E while keeping T fixed, you are correct that you can use the product rule.
The product rule states that if you have two functions, say f(x) and g(x), and you want to find the derivative of their product, then you can do so using the formula:
(fg)' = f'g + fg'
Now let's apply the product rule to the function n(E):
n(E) = (E^2)/(e^(E/T)−1)
To apply the product rule, we can consider f(E) = E^2 and g(E) = (e^(E/T)−1)^(-1).
Now let's find the derivatives of f(E) and g(E):
f'(E) = 2E (using the power rule for differentiation)
g'(E) = -[(e^(E/T)−1)^(-2)] * (e^(E/T)−1)' (using the chain rule)
To find (e^(E/T)−1)', we can use the derivative of the exponential function:
(e^(E/T)−1)' = (1/T) * e^(E/T) (using the chain rule)
Now let's substitute the derivatives back into the product rule formula:
n'(E) = f'(E) * g(E) + f(E) * g'(E)
n'(E) = 2E * (e^(E/T)−1)^(-1) + (E^2) * [-[(e^(E/T)−1)^(-2)] * (1/T) * e^(E/T)]
Simplifying this expression further, you can combine like terms and simplify the exponents:
n'(E) = 2E * (e^(E/T)−1)^(-1) - (E^2/T) * (e^(E/T)−1)^(-2) * e^(E/T)
This is the derivative of n(E) with respect to E, keeping T fixed.