E=mc^2 and the huge numbers it produces

[White Rabbit]

Ok, I never finished school when I was a kid so I never did much math and I did no Physics what so ever. So could someone explain something for me.

If I take Hydrogen with the atomic weight 1.00794 and plug it into the equation E=mc^2 to find out how much energy would be released if it was converted to pure energy I get the number 8.39875517873682.......... thenumbers keep going on and on. So what is this unit of enegy called that is produced in this equation?

here is my calculation.
1.0079*299 792 458^2=8.39875517873682..........

Have I got the equatioin correct?

thanks
Sandy

[renormalised]

Not quite right...what you needed to do is convert that mass (amu= atomic mass units) into grams or kilograms (or electron volts) and then plug the numbers into the equation. The answer is in an unit of measurment for energy called joules (http://en.wikipedia.org/wiki/Joule). I'll show you...

Since an atom of Hydrogen is essentially a proton (http://en.wikipedia.org/wiki/Proton) (ignore the electron, for now)

E=mc^2
= 1.673×10^−27 kg x (3 x 10^8m/s)^2
= 1.5057 x 10^-10 J

Not much, but when you add up all the reactions going on inside a star, it's an enormous amount of energy.

[Coen]

Comes down to units used.

The standards are usually mass in kilograms and velocity in m/s thus the units of m x c^2 translate to the energy unit kg . m^2 / s^2 called a Joule.

In the case of a hydrogen atom of mass 1.67262158 × 10^(-27) kilograms and the speed of light as 299 792 458 m / s you get:
(approx) 1.503x10^(-10) Joules.

To put in perspective 1 Joule is equivalent to lifting an apple straight up 1m.

The Sun, being big (relative to us), burns through about 600,000,000 tonnes of hydrogen converting it into about 596,000,000 tonnes of helium every second via a nuclear fusion process. That short fall of about 4 million tonnes is what has been converted to energy via E=mc^2 or in joules about 3.595x10^(23)

[White Rabbit]

Thanks for the replies, and as always, answers lead to more questions. I'll do a bit more research and be back with more questions.

Again, Thanks guys.

[bojan]

More perspective:
This huge power averages to ~50W per cubic metre (for the whole Sun), and about 200W per cubic metre for Sun's core.. not very efficient indeed !!

EDIT:
0.273W/m3 average, ~1W/m3 for the core.

[Coen]

To add further:
The maximum energy impacting on 1 square metre of the Earth's surface from the Sun is on the order of 1,000 watts (1kW). Thus if our solar collectors were 100% efficient then that is the maximum we can obtain if everything else was working in our favour.

[renormalised]

Actually closer to 1370W/m^2.

In so far as efficiency of energy production, the pp chain reaction at the Sun's core runs at an efficiency of 0.7%, so that means 99.3% of the fuel is not converted into energy through the fusion process.

[SkyViking]

Haven't done the calculation, but I'd expect a cubic metre of the Sun's core, which is at 15 million degrees, to emit more than 200W?

[SkyViking]

OK, I couldn't resist trying to calculate this:





So, the power density in the core must be 194 µW/kg * 150 = 29100µW/kg = 0.0291W/kg.



But, the core density is 150 g/cm3, so 1 cubic metre = 150,000kg.
This gives 0.0291W/kg * 150,000kg = 4365W/m3.

Sounds more plausible, but is it correct? I can't see where they got 0.272 W/m3 from?

[bojan]

Surprisingly, not. Despite the high temperature, this is how much escapes the Sun from each m3..

ref: http://en.wikipedia.org/wiki/Sun

Search for "power density". The actual number is 0.272 W/m3
My number comes from different source.. so somewhere there must be a typo.
But, regardless, the power generated in each cubic metre of the Sun's core is surprisingly small, compared what we would expect intuitively... which means we should abandon the "feelings" in those matters :P

[SkyViking]

Did a calculation above based on Wikipedia's numbers, but I get a much higher output?

[bojan]

yep. I saw it.. numbers do not fit.. I will try to find my original source of info later.

Perhaps the calc for power density should be done based on Solar constant (1.37kW/m2)...

OK, here is the calculation, based on Solar constant.
Their end number is right, 273mW/m3 (average across the whole Sun). However, something is not right with other numbers... .

Spreadsheet cal is attached...

[Nesti]

I wonder what the ratio is between energy in the form of particle radiation and energy in the form of gravity wave radiation, ie the ratio between physical matter against gravity waves...

Where's my slide rule?...

[bojan]

No gravity waves from Sun..
Or at least, nothing significant.

[SkyViking]

Yes, it should be possible to calculate from the solar constant.
My only comment to the spreadsheet is that the energy is only generated in the Sun's core, which is about 0.2 solar radii.
That then gives 34.18W/m3 power density in the core itself.

But still, imagine a cubic metre of super dense solar core plasma. It weighs 150 tonnes, and is at 14 million degrees. Would it only emit 34 watts - thats like a dim lightbulb??? Something's not right.

[bojan]

Everything is right here...
This is simply how much heat manages to escape from the interior of the Sun. However, it does not tell how much heat energy is still trapped inside... and this is huge amount.

[SkyViking]

Yes good point. There is energy trapped inside to counteract a gravitational collapse.

And there are also neutrinos which accounts for quite a bit of the sun's energy output. These are not part of the radiation flux as expressed by the solar constant.

But then the whole thing is simply a mix of concepts, because we cannot use the solar constant alone to calculate a W/m3 for the Sun, it is meaningless because of the above reasons.
The correct calculation must be based on the actual energy production in the core rather than only on the flux we can measure outside. The calculation I did earlier based on the Wikipedia numbers then seem methodologically correct, although I'm not sure if the actual numbers involved were OK.

[bojan]

Yep...
Wiki's numbers (except this 234mW/m3) are not right, probably a typo somewhere...

I am sure people who are into Sun theory have done a proper job (taking into account what we know about Sun and star's interiors).

[Nesti]

Hang-on, all mass creates gravity waves, a natural product of the interaction between the energy momentum and the gravitational field, right?

So even though we are not technologically advanced to detect them (personally I believe we cannot), they must surely be there and radiating...ergo eating into the Sun's mass at some [perhaps tiny] rate. Surely we must know an approximation of what's being given off too.

[bojan]

No.. Only sudden change in mass (or spacial mass distribution) will produce gravity waves, which are disturbances in gravitational field.
Nothing will radiate from mass which is static in space-time..

[Nesti]

Maybe the 'Weber Bar' design could be used for earthquake detection rather than other types of cosmic events...or perhaps just an organ pipe.

Cheers
Mark