1
$\begingroup$

This is a hypothetical question; considering both the earth and the sun as black bodies. If the temperature of the sun decreased N times, what would be the effect on the radiation intensity received on earth?

(i am inclined to say it would decrease N^4 times using the Law of Stefan-Boltzman)

$\endgroup$
1
$\begingroup$

Background

The specific intensity or brightness, $I_{\nu}$, is defined as: $$ I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1} $$ where $\nu$ is the frequency, $dE$ the differential energy, $dA$ the differential area, $dt$ the differential time, $d\Omega$ the differential solid angle, and $d\nu$ the differential frequency.

We can define a net flux at each frequency as: $$ F_{\nu} = \int d\Omega \ I_{\nu} \ \cos{\theta} \tag{2} $$ where we can get a total flux by integrating $F_{\nu}$ over all $\nu$.

For blackbody radiation, $I_{\nu}$ goes to the Planck function, $B_{\nu}$, defined as: $$ B_{\nu} \left( T \right) = \frac{ 2 \ h \ \nu^{3} }{ c^{2} } \left[ e^{h \ \nu/k_{B} T} - 1 \right]^{-1} \tag{3} $$ where $k_{B}$ is the Boltzmann constant, $c$ is the speed of light, $h$ is the Planck constant, and $T$ is the temperature.

According to the Stefan–Boltzmann law, the energy emitted per unit time per unit area, $\mathcal{F}$, is defined as: $$ \begin{align} \mathcal{F} & = \pi \ \int \ d\nu \ B_{\nu} \left( T \right) \tag{4a} \\ & = \sigma \ T^{4} \tag{4b} \end{align} $$ where $\sigma$ is the Stefan-Boltzmann constant.

If the temperature of the sun decreased N times, what would be the effect on the radiation intensity received on earth?

You are close in your assumption. The ratio of the emergent flux (i.e., Equations 4a,b) before to after would be $\propto N^{-4}$. However, the intensity at any given frequency and temperature is given by $B_{\nu} \left( T \right)$ (i.e., Equation 3).

For instance, if we define $T_{1}$ = 1 K, $T_{2}$ = 1/5 K, and $\nu_{o} = 10^{11}$ Hz, then the ratio $B_{\nu_{o}} \left( T_{1} \right)/B_{\nu_{o}} \left( T_{2} \right)$ would be $\sim 2.2 \times 10^{8}$, which is much larger than $5^{4}$ (= 625), where we have: $$ B_{\nu_{o}} \left( T_{1} \right) \sim 1.2 \times 10^{-19} \ J \ s^{-1} \ m^{-2} \ Hz^{-1} \ ster^{-1} \\ \sim 1.2 \times 10^{-16} \ erg \ s^{-1} \ cm^{-2} \ Hz^{-1} \ ster^{-1} \\ B_{\nu_{o}} \left( T_{2} \right) \sim 5.6 \times 10^{-28} \ J \ s^{-1} \ m^{-2} \ Hz^{-1} \ ster^{-1} \\ \sim 5.6 \times 10^{-25} \ erg \ s^{-1} \ cm^{-2} \ Hz^{-1} \ ster^{-1} $$

Below is a table of values illustrating the changes I mention.

Temperature [K] |     Ratio       |       Ratio
                |  [10^(11) Hz]   |    [10^(13) Hz]
------------------------------------------------------
  1.0000000     |   2.2 x 10^(8)  |    5.1 x 10^(833)
  10.000000     |     16.2        |    2.3 x 10^(83)
  100.00000     |      5.5        |    2.2 x 10^(8)
  1000.0000     |      5.0        |      16.2
  10000.000     |      5.0        |       5.5
  100000.00     |      5.0        |       5.0

If the temperature of the sun decreased N times, what would be the effect on the radiation intensity received on earth?

All of the above are intensities of light emitted by the blackbody source (i.e., the sun in this case). The power per unit area of radiation received at Earth's orbit scales as $\propto R_{s2a}^{-2}$, where $R_{s2a}$ is the distance from the source to the absorbing body. To calculate the total energy absorbed per unit time, one would need the following: $$ \begin{align} E_{ss} & = 4 \ \pi \ r_{s}^{2} \ \sigma \ T_{s}^{4} \tag{5a} \\ E_{sa} & = \frac{ E_{ss} }{ 4 \ \pi \ R_{s2a}^{2} } \tag{5b} \\ E_{aa} & = 4 \ \pi \ r_{a}^{2} \ \sigma \ T_{a}^{4} \tag{5c} \\ E_{as} & = \pi \ r_{a}^{2} \ E_{sa} \tag{5d} \end{align} $$ where $E_{ss}$ is the total energy emitted by source at the source surface, $E_{sa}$ is the total energy emitted by source at the absorbing bodies orbit, $E_{aa}$ is the total energy emitted by absorbing body at the absorbing body surface, $E_{as}$ is the total energy received over disk of absorbing body from the source, $r_{s(a)}$ is the radius of the source(absorbing body), and $T_{s(a)}$ is the blackbody temperature of the source(absorbing body).

To determine the change in the blackbody temperature of the the absorbing body, ignoring albedo effects, etc., one can just equate $E_{aa} = E_{as}$, or the energy absorbed equals the energy radiated (i.e., equilibrium).

References

  • G.B. Rybicki and A.P. Lightman, Radiative Processes in Astrophysics, First Edition, John Wiley & Sons, Inc., New York, NY, 1979.
$\endgroup$
  • $\begingroup$ Wow, this is really long :) But all that we need for the question (calculating the temperature if blackbody assumption is provided) is your formula (4b) $\endgroup$ – Ilja Apr 28 '16 at 21:16
  • $\begingroup$ @Ilja - No, that would only be the change in the emergent flux of the emitting body, not the intensity nor what the Earth would see. That is why I wrote all these details. $\endgroup$ – honeste_vivere Apr 28 '16 at 22:21
  • $\begingroup$ ehm, well, I don't agree. Or can you explain why the fraction of the emergent flux arriving on earth should change? $\endgroup$ – Ilja Apr 29 '16 at 5:01
  • $\begingroup$ The electromagnetic radiation that make it through the atmosphere is frequency-dependent. Changing the temperature changes the peak in the emission spectrum and thus changes what would be seen on the ground. Spacecraft could measure the total emergent flux if calibrated to cover the range of frequencies corresponding to the blackbody peak. The Earth's surface, however, would see very different things if the sun's temperature changed. $\endgroup$ – honeste_vivere Apr 29 '16 at 12:25
  • $\begingroup$ yes, I know this ... but you wrote yourself about "ignoring albedo etc"; and indeed the question, too, asked explicitly to treat earth as a black body. So I still don't agree :) $\endgroup$ – Ilja Apr 29 '16 at 12:34
0
$\begingroup$

Yes, the amount of radiation (i.e. power) would decrease $N^4$ times, since the whole geometry of the system does not change, only the power emitted by the sun.

But the temperature of the earth would decrease only $N$ times, since the ratio of the temperatures remains equal - again because of the unchanged geometry. The earth radiates a power proportional to its $T^4$ times $4\pi$ - the whole angle. The sun only hits the earth in a certain solid angle, depending on the distance and radius of the earth.
So the ratio of this angle and $4\pi$ has to be the ratio of the $T^4$, thus also the ratio of the $T$ does not change.

$\endgroup$
  • $\begingroup$ This is a much simpler answer, and this was what I was looking for, but honest_vivere's answer was much more complete (and complex). I selected his as the answer because I think it will be more helpful to other people :) $\endgroup$ – magamig Apr 30 '16 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.