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So I have been given this question:

A typical Type Ia Supernova emits $10^{44}\ \mathrm{J}$. All nuclear weapons tested by humans have released a total of about $2\times10^{6}\ \mathrm{TJ}$ of energy. Assume the Supernova energy is released at the distance of the Sun. Determine how close to your retina the nuclear weapons energy would have to be released to make both appear to have the same apparent bolometric magnitude.

The issue I'm have is normally apparent magnitudes I would have to have joules per second where this is in just total joules.

So is it that with bolometric magnitude there is an automatic assumption that when total power is mention you assume joules a second?

Could someone please advise?

I would just like to point out I am not looking for a direct answer to this question, I am just confused with why total energy has been used. I have read posts on wiki which seems to confuse me even more. I have just used the above question as an example from a book.

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    $\begingroup$ You have been given a terrible question. Without knowing a characteristic timescale over which the energy is released in each case, then it is not answerable. Note that the timescale for energy release in the case of supernovae and bombs are different by many orders of magnitude. $\endgroup$
    – ProfRob
    Commented Jan 26, 2018 at 7:24
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    $\begingroup$ @RobJeffries in that vein, the nuclear weapons energy release has occurred over 78 years now :-) . -- oops, I see you mentioned that in your answer. Twisted minds think alike $\endgroup$ Commented Jan 26, 2018 at 15:21

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You have been given a terrible question and your confusion is totally justified.

The magnitude system is a measure of the brightness of objects which is a measure of the energy received per unit time interval.

Without knowing a characteristic timescale over which the energy is released in each case, then it is not answerable. Note that the timescale for energy release in the case of supernovae and bombs are different by many orders of magnitude - a type Ia supernova has a bolometric light curve (e.g. see Fig.6 of Scalzo et al. 2014) that lasts 2-3 weeks, peaking at around $10^{36}$ Watts.

Given this, we can compare with the Sun's bolometric luminosity of $3.8\times 10^{26}$ Watts and apparent bolometric magnitude of -26.8, to estimate that the peak apparent bolometric magnitude of a type Ia supernova at the distance of the Sun would be -50.3.

You cannot do the same thing with nuclear explosions. They last a fraction of a second (in terms of their light output). Or are you meant to assume that you take the average flux over the 70 years that mankind has had nuclear weapons?? The question is a nonsense unless you are given some sort of timescale over which this explosive energy is released. About the only sane things you can do are (a) use ~70 years (actually, most of the energy was released during the 1950s and early 1960s) (b) use the duration of a single nuclear explosion or (c) (and I imagine what was intended) use a similar duration as for a typical type Ia supernovae so that the question should have begun - "Imagine that X and Y amounts of total energy are emitted over a similar timescale..."

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  • $\begingroup$ 41.2k377124 Thank you for detailed response, I spoke to my lecture who said for the use the light curve for the supernova, but I am slightly confused by the because most curve I have looked at approx 350 days, I mean that how I read it, I am assuming that total energy is the time take for the star to loose all it energy, where you have said 2-3 weeks? I have in the meantime been looking for evidence of the duration of a nuclear bomb which I have come up with 1ns to 10ns weather this is true I do not know. $\endgroup$ Commented Jan 27, 2018 at 12:45
  • $\begingroup$ Most of the energy is released in 2-3 weeks. What matters is the peak power output and I have given you a reference for that. And for representative bolometric light curves @JasonTaylor $\endgroup$
    – ProfRob
    Commented Jan 27, 2018 at 14:28
  • $\begingroup$ Many thanks ofr the link, which is kinda making better sense, but may I ask why peak output is important? I only ask this because from me trolling round the internet today, a supernova total energy is 10^44 but I am confused as to weather that means total light intensity or total energy released until it dies. $\endgroup$ Commented Jan 27, 2018 at 15:41
  • $\begingroup$ @JasonTaylor What is important is power, not total energy. That determines how bright something is. That is the basic message of both answers here, including the one you have accepted. $\endgroup$
    – ProfRob
    Commented Jan 28, 2018 at 0:34
  • $\begingroup$ Ah, okay thank you. I been sitting for most of the day figuring ways to do this question. Must say learnt a few things along the way, and thank you for the link above took a while to read but, very informative and useful. $\endgroup$ Commented Jan 28, 2018 at 2:08
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If you take a look at how the magnitude system is derived, it usually starts with an equation similar to the following

$$100^{(m_1 - m_2)/5} = \frac{F_2}{F_1}$$

where $m$ is the apparent magnitude and $F$ is the flux. The subscripts denote two different objects. This equation derives directly from the fact that, by definition, a magnitude 1 star is 100 times brighter than a magnitude 6 star.

Now if you look at the equation above, you'll see the right hand side is a ratio of fluxes, and necessarily it is unitless. What this actually means is that you can derive the magnitude system with any measure of "brightness" be it a flux which has units of $\mathrm{J\ s^{-1}\ m^{-2}}$, a luminosity which has units of $\mathrm{J\ s^{-1}}$ or power which also has units of $\mathrm{J\ s^{-1}}$. The only constraint is that you have to consider the same "type of brightness" (for want of a better term) for both sources so your ratio remains unitless. That is, you can't use flux for one star and power for another. They both have to be flux or they both have to be power. In your case, you're given the total energy output, which you can assume happens over the same time scale since they're both just explosions.

To be honest though, your problem doesn't really rely on that knowledge. In fact, your problem becomes exceedingly simple because it says to assume that $m_1=m_2$ which means the ratio of the flux from both sources is unity by the equation above. You just have to say $F_1 = F_2$, relate flux to luminosity and distance, and solve for your unknown.

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  • $\begingroup$ Power does not have units of J. The quantities n the magnitude equation do need to be powers or fluxes and be normalised by a time quantity. The only way the question can be done is to convert the energy release into some measure of power by dividing by a timescale. $\endgroup$
    – ProfRob
    Commented Jan 25, 2018 at 23:52
  • $\begingroup$ @RobJeffries Yes you are correct on both accounts. I was being too quick and sloppy with my answer. In the back of my mind, I considered the comparison between the two explosions comparable because they happen over identical timescales as the problem is constructed, I just wasn't good at explaining I was thinking that. $\endgroup$
    – zephyr
    Commented Jan 26, 2018 at 0:00
  • $\begingroup$ I still disagree with your answer. Yes, you can assume they both happen on the same timescale, but there is no reason or justification to do so. They are not both just explosions. The energy release in the form of light has a very distinct time behaviour in the case of a supernova and is nothing like the time profile (by orders of magnitude) of a nuclear explosion. $\endgroup$
    – ProfRob
    Commented Jan 26, 2018 at 7:21
  • $\begingroup$ @RobJeffries That's certainly true, but I think you're approaching this question at too high a level. It's clearly not meant for that level of dissection. It may be a bad question, but that is, from my experience, how intro physics questions are designed. $\endgroup$
    – zephyr
    Commented Jan 26, 2018 at 14:00
  • $\begingroup$ Realising there is a difference between energy and flux and that magnitudes represent a flux is NOT high level. I will put forward my response as the answer. $\endgroup$
    – ProfRob
    Commented Jan 26, 2018 at 14:35

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