Talk:2027: Lightning Distance

Calculations I used:

t1=s/v1

t2=s/v2

Substract:

t1-t2=dt=s/v1-s/v2=(s*v2-s*v1)/(v1*v2)=s*(v2-v1)/(v1*v2)=s*dv/(v1*v2)

Therefore

s=(dt*v1*v2)/dv

I evaluated (v1*v2)/dv and it came to be 13.6 billion. Can someone verify it's correct? 172.68.51.112 13:08, 1 August 2018 (UTC)

The comic begins with the question "how many miles away", so converting to kilometers isn't the right calculation.172.69.71.24 17:06, 1 August 2018 (UTC)

I used refractive index for visible light of 1.000277 (air at STP as opposed to 0C 1atm) and arrived at around 7.9 billion instead. Refractive index of 1.000337 is then required for the radio waves for the comic to be correct. 172.68.11.221 13:46, 1 August 2018 (UTC)

Do you mean 7.9 billion to convert to miles or to kilometers? Because my 13.6 bilion is to kilometers.

I'm sure the actual comic is referring to miles and 5 billion was picked to match with the "divide by five" rule for miles. 172.69.70.131 13:59, 1 August 2018 (UTC)

I did mean kilometers. If we use miles, 1.000314 fits almost precisely! (5.04 billion) 172.68.11.17 14:42, 1 August 2018 (UTC)

If you can count several seconds, as is suggested in the comic, the flash is still billions of miles away, the widest possible distance between Earth and Neptun is about 5 billion km. Sebastian --172.68.110.40 14:51, 1 August 2018 (UTC)

Do you really need to know the spectrum of the flash? If we assume that a flash contains UV and X-ray radiation and that the visible light is generated at the same time as the UV or X-ray radiation then you only need to know the refractive index of light/UV/X-ray in air under the same temperature conditions and not the exact spectrum. Condor70 (talk)

I understand the joke Randall was going for, but have a problem with the wording. "Count the number of seconds" won't work for fractions of anything. "Measure" would work, but spoils the gag a bit. Counting numbers are integers; counting the seconds between the visible and radio frequency flashes will give you zero. 172.69.71.24 17:00, 1 August 2018 (UTC)

Assumptions on the medium properties sound?

Refractive index of *dry* air might be pretty close to 1 for both light and RF EM waves, but:

Let's assume that the air is humid, if not even full of water drops. After all, lightning.

Let's further assume that an air/water mixture or solution has electromagnetic properties between these two materials.

In water, refractive index for light is about $n_{\text{water, optical}}=1.33 n_{\text{air, optical}}$, (as easily demonstrated by the optical refractive effects); for RF, we typically use values of $\frac{n_{\text{water, RF}}^2}{\mu_r}=\epsilon\approx 80$. So, $n_{\text{water, RF}}\approx \sqrt{80}n_{\text{water, RF}}$.

Let's assume a 10⁻³ "EM-effective" water content in the comic air.

That would lead to $\frac{v_{\text{opt.}}}{v_{\text{RF}}} = \frac{\frac34}{\sqrt{80}^{-1}}= \frac34\sqrt{80}=6.7$.