E=mc2

[budcamp]

e=mc2

What numerical value is used for C?
What number am I to use the square?

It seems to me a simlpe question, but when I started to work out a problem: it dawned on me I had no idea of how to square the speed of light.

[Daniel]

Depends on your units,
but mostly it is meters/sec (in the sane part of the world :D )
which equals around 292000

[budcamp]

If you square klm per sec or klm per hour you get a diferent answer.

e=m*c2

if m= 10 kilos, what will e equal?

[neubjr]

I think that daniels answer was in kilometers per second. the speed of light is ~3 x 10^8th meters / second (2.92 x 10^8).

E = 10kg * (2.92x10^8)^2 m/s
E= 8.53x10^17 J

[Daniel]

ooppsss... yes km/sec :o

[budcamp]

Why did you use klm/sec rather than klm/hour? or even klms/year. Is klm/sec the generally accepted time period for all equations concering the speed of light? I'm not trying to be arguementative, I just don't know the answer.

[Daniel]

Well you have two major units systems

M.K.S - Meter/kilogram/second

C.G.S - Centimeter/Gram/Second

(did i say I hate units :D )

[budcamp]

Thanks for the reply Daniel, but I really didn't understand it. Could you be more specific?

[Daniel]

Well, in the metric system you use mainly 2 sets of unit system - the one I gave above.
As you can see, one of them refers to distance as CMeters and the other as Meters.

So while you can use KM and anyother unit, in physics etc... you will either use meters and centimeters.
For time you will always use seconds. Again it can vary but in physics it is mostly seconds.

Actually the KM/sec is also pretty widespread but more for popular science.

[budcamp]

I am obviously not expressing myself very well.

There are many ways to express the speed of light. All of these represent the speed of light.

miles per hour=670,000,000
miles per sec= 186,000
miles Per year= 6,300,000,000,000

We could also make them metric.

when "E=MC^2 it makes quite a difference which one is used.

E=10kilos*c^2

Which nunber do I square
186,000 670,000,000 6,300,000,000,000 or some other number. they are all the speed of light.

There must have been some agreement as to this already, but I don't know what it is.

[Skulled]

That's not that hard, just take the equation and write down the units of all the quantities involved. They must equate, the units in which, of the speed of light, the equation is exact, is the most common or appropriate one used.

For example, if you use Joules for Energy, Kg for mass, only one value of the speed of light will match the equation. On the other hand if you use Joules for Energy and Grams for mass, you will have to use another value.

So the point you have to concentrate is what are the units you want the other quantities to be in, once you have this , getting the coorect value for the speed of light shouldn't be a problem.

Hope this helps.

[Steve Langford]

For c, you can use any valid units for velocity. For m, you can use any valid units for mass. Your choice of velocity and mass units determines the units of energy that appear in the result. Most choices of units for mass and velocity lead to unusual energy units, but they work.

Since energy units are messy, we usually choose the energy unit we want and use the mass and velocity units that give this energy unit. In the SI system, the natural energy unit is the Joule. In terms of mass, length, and time units, the Joule is one kg m^2/s^2. If you express the mass in kg and the velocity in m/s, the energy comes out in Joules.

In the old English pound-force system (where the pound is treated as a force and not a mass), the natural unit of energy is the foot-pound. One foot-pound equals one slug ft^2/s^2. (The slug is a unit of mass. One slug weights 32.17 pounds.) When you express the mass in units of slugs, and velocity in units of feet/second, the energy will be in foot-pounds.

The equation is valid no matter what units you use for mass and velocity. If you express the mass in units of grams and the velocity in units of miles per hour, the resulting energy will be in units of gram mile^2/hour^2. Most of us are not familiar with these energy units, and therefore scientists and engineers are discouraged from using them. Nevertheless, the energy units are legitimate energy units, and the equation is true using these units.

You pay your money and get your choice.

[timeshifter]

e=mc2

if you can keep up with me here,

energy = mass * speed of light 2


i'm fairly sure that is what everything stands for.

[steve]

Lots of energy from a bit. About 90 terawatts from each gram of matter converted to energy.

Ever read about a muon generator? Makes a few nanograms of negatively charged muons, tosses them at a few Ugm. of protons, they attract electrostatically and you get E=MCexp2.

Quite a nice matter/anti-matter engine, frankly. Been used in particle physics for years, and the yearly scale up in muons is an interesting fact.

Maybe a rather better source of energy in space?

But where do we get those dilithium crystals to make it work efficiently? Or is there a smaller particle than the muon & which is less massive and therefore more efficiently made?

[budcamp]

There have been some great explanations so far. I have learned alot from them. DON'T STOP NOW! Keep those cards and letters coming folks!

budcamp

[timeshifter]

Well, he told us to keep going, I just want to be nice to someone. Except I don't have any fun remarks, except that the speed of light is about 300,000 km/h. Fast? I thought so too.

[Tac-Tics]

Keep in mind that you can square UNITS like anything else.

(1 meter)^ 2 = (1 square meter) (area measurement)

(1 second)^2 = (1 squared second) (useless measurement)

The latter is useless alone, but its inverse is significant.

(1 second)^ -2 = (1 time per second every second)

Example:

acceleration = m / s ^2 = meters per second (speed gained) every second

Energy is a represented in weird units called Joules. A joule is an acceleration applied to a mass (a force) THROUGH a given distance or kg * m^2 / s^2.

Looking at just the units:

E = MC^2

E = energy is in joules or kg * m^2/s^2

M = mass is in kilograms or kg

C = velocity of light is in units m/s
C^2 = C * C is in units m^2/s^2

So MC^2 is in kg * m^2/s^2 or Joules, the unit for energy.

[timeshifter]

So, we have determined that we can square any measurement known to man. The use of these squared measures has yet to be determined. But, the sybol of squared is also used to signify area. Such as, something is twenty-five feet squared. That does not mean you square the twenty-five feet. That is just telling you that you are measuring area, not a linear measure.

So, the equation in question:

E=mC^2

Is that squaring C or is that telling us we are working with area?

If we are talking about squaring C, we can change the equation to the following:

E=m*89,875,517,873,681,764 m/s

A very large number indeed, but it would remove all questions about "are we squaring C or is it area?"

Keep in mind though, that would only work if we are squaring C alone.

[timeshifter]

So, we have determined that we can square any measurement known to man. The use of these squared measures has yet to be determined. But, the sybol of squared is also used to signify area. Such as, something is twenty-five feet squared. That does not mean you square the twenty-five feet. That is just telling you that you are measuring area, not a linear measure.

So, the equation in question:

E=mC^2

Is that squaring C or is that telling us we are working with area?

If we are talking about squaring C, we can change the equation to the following:

E=m*89,875,517,873,681,764 m/s

A very large number indeed, but it would remove all questions about "are we squaring C or is it area?"

Keep in mind though, that would only work if we are squaring C alone.

[timeshifter]

So, we have determined that we can square any measurement known to man. The use of these squared measures has yet to be determined. But, the sybol of squared is also used to signify area. Such as, something is twenty-five feet squared. That does not mean you square the twenty-five feet. That is just telling you that you are measuring area, not a linear measure.

So, the equation in question:

E=mC^2

Is that squaring C or is that telling us we are working with area?

If we are talking about squaring C, we can change the equation to the following:

E=m*89,875,517,873,681,764 m/s

A very large number indeed, but it would remove all questions about "are we squaring C or is it area?"

Keep in mind though, that would only work if we are squaring C alone.

[neubjr]

The equation is:

E=m*C*C or written in exponential notation E=m*(C^2)

if you look at the equation in dimensional terms it would be:

[joules] = [kg]*[m/s]*[m/s]

[timeshifter]

So, in English, we are still dealing with that very large number, right?
Now we are multiplying the speed of light squared by the mass of the object in kilograms. This will result in the amount of energy made.

So, why can't we surpass C?

[neubjr]

If you want to look into why you cannot actually surpass the speed of light (c) you have to take into account the relativistic effects.

First of all for a little back ground. The whole equation is actually:

E^2 = m^2*c^4 + p^2*c^2

but forget that for now...


If you don't know much of relativity it is a little hard to explain. But basically the fact that you cannot break the speed of light has to do with special relativity. In relativity there is this factor of gamma which comes into play in all of the relationships.

Gamma = [1-(v/c)^2]^(-1/2) where v is the speed of the object and c is the speed of light. So as v approaches c gamme becomes very very large. Basically this means that the closer you come to reaching the speed of light, the more energy you have to put into the system.

E = mc^2 in relativity (forgetting the p^2*c^2 term) becomes:

E = (gamma) * m*c^2

so you can see the energy explodes as you approach c. Does that make sense?

Joshua.

[budcamp]

Timeshifter:

Squareing a number does give it area.

With right triangles the hight^2 + the base^2 = the hippopotamus^2!!

a^2 + b^2 = c^2

If you draw out a right triangle and then make each side of the triangle one side of a square. You can use the area of (square a) and the area of (square b) and you willl find they equal the area of (square c).

Almost like magic isn't it?

[timeshifter]

Hmm... Length squared plus height squared equals hippopotamus squared?!? I didn't know you could square an animal, but anyway...

Thanks for the analogy, I thought it a really amusing one.

[budcamp]

I'm glad you detected the humor, but the fact remains; if you draw out a right triangle and make each side of the triangle one side of a square. The area of the square for the hight, added to the area for the square of the base, equals the area of the square of the (h). If you have a grudge against large gray animals, you can call (h) the slanty line, or anything else that you please. But, whatever you call it, you will be dealing with area.

[budcamp]

My question to begin with has to do with the problem of level of the speed of light needs to be squared.

If you square 60 minutes you end up with 3600 minutes or 60 hours.

If you square one hour you have one hour! Same amount of time being squared but with vastly diferent answers.

[timeshifter]

Well, in that context, we have a slight problem. Here is my opinion..

I think you would use the basic unit of measurement in each category if you are working with equations.

Time: seconds
Length: depends on distance, mm, m, or km
Weight: g or kg
Speed: Km/h
Volume: cm^3

You see my point? Very true, if you take 60 minutes squared, you get sixty hours. If you take 1 hour squared, you get one hour.
And, if you take sixty seconds squared, you get six minutes. But, this is all working on the assumption that time is not just linear, which it is. When you think about it in that regard, things change a little, for time is linear.

Now, what makes us able to square time is the fact that it is tied in with space.

In that regard, it becomes something like....
1 second ^2 = 1 second cm^3

Maybe i am way off, maybe not, but I just thought i should bring up linearity to the string.

[timeshifter]

Well, in that context, we have a slight problem. Here is my opinion..

I think you would use the basic unit of measurement in each category if you are working with equations.

Time: seconds
Length: depends on distance, mm, m, or km
Weight: g or kg
Speed: Km/h
Volume: cm^3

You see my point? Very true, if you take 60 minutes squared, you get sixty hours. If you take 1 hour squared, you get one hour.
And, if you take sixty seconds squared, you get six minutes. But, this is all working on the assumption that time is not just linear, which it is. When you think about it in that regard, things change a little, for time is linear.

Now, what makes us able to square time is the fact that it is tied in with space.

In that regard, it becomes something like....
1 second ^2 = 1 second cm^3

Maybe i am way off, maybe not, but I just thought i should bring up linearity to the string.

[timeshifter]

Well, in that context, we have a slight problem. Here is my opinion..

I think you would use the basic unit of measurement in each category if you are working with equations.

Time: seconds
Length: depends on distance, mm, m, or km
Weight: g or kg
Speed: Km/h
Volume: cm^3

You see my point? Very true, if you take 60 minutes squared, you get sixty hours. If you take 1 hour squared, you get one hour.
And, if you take sixty seconds squared, you get six minutes. But, this is all working on the assumption that time is not just linear, which it is. When you think about it in that regard, things change a little, for time is linear.

Now, what makes us able to square time is the fact that it is tied in with space.

In that regard, it becomes something like....
1 second ^2 = 1 second cm^3

Maybe i am way off, maybe not, but I just thought i should bring up linearity to the string.

[PhysBrain]

Sorry, but I have to stop you here. Your discussion is becoming very muddled and you are not making any progress towards resolving your misconceptions. Units are just special algebraic variables, and you must treat them appropriately. Some units have geometrical or physical interpretations, but these should not be confused with the way they are manipulated in algebraic expressions. By the same token, some combinations of units have no real interpretation, but can still be manipulated in exactly the same way in algebraic expressions.

Take for example: (10kg)*(10kg)=(100kg^2). There is no such physical thing as a square kilogram, and trying to interpret it as a mass-area is not helpful... it does not exist. For the purposes of this expression, you just treat the units like unknown algebraic variables. That means that you can only add together quantities that have identical units:

a*c + b*c = (a+b)*c
where you can think of c as your unit. And when you multiply quantities together, you get a completely new type of quantity:

(a*c + b*d)*c = a*c^2 + b*d*c

where you could think of c, d, c^2 and d*c as units, each one of them denoting a new kind of quantity which is unique from the others.

All of this is basic algebra and dimensional analysis (which is just basic algebra applied to units and unit conversions). Please consult any introductory physics text for more examples of how to work with units.

[budcamp]

I would like to thank all of you who provided me with a solution to this question. While I am not now an expert, I am able to understand how the units system works.

Thanks all!

[raptor]

e represents energy in ergs (a unit of energy)=m represents mass in grams * and c represents the speed of light in centimeters per second squared.
Because light travels at 30 billion centimeters per second, the value of c2 is 900 billion billion centimeters per second.

See:
http://digilander.libero.it/brunoosimo/Einstein.doc
for more info.

[timeshifter]

Not quite. If you weren't squaring a velocity, that is what you would come up with. However, you are squaring a velocity (299,792 km/s). When you square a velocity, you come up with an acceleration, hence why I believe it is supposed to be E=(mc)(mc), or E=(mc)^2. Otherwise, we wind up with Energy=Mass * 299,792 km/s/s. Something about that method just doesn't strike me as being correct, although I'm not very well versed in General Relativity, so I wouldn't know.

[PhysBrain]

When you square a velocity, you come up with an acceleration, hence why I believe it is supposed to be E=(mc)(mc), or E=(mc)^2. Otherwise, we wind up with Energy=Mass * 299,792 km/s/s.

You will not get a force from this (mass * accelleration) because velocity squared (m^2/s^2) is not the same as accelleration (m/s^2). You will instead get units of energy. Recall that work and energy have the same units (the total energy of an object is its capacity to do work, or the maximum amount of work the object is capable of performing), and that one definition of work is (force*distance). In terms of units, this expression gives (N*m) or (kg*m^2/s^2), or simply (J), where J is a Joule (unit of energy), and N is a Newton (unit of force). As you can see, this expression gives the same units as E=mc^2. Keep in mind that the units I have chosen here are not unique, but the quantities they represent (i.e. mass, length, time, force, energy) are unique and must be conserved.

[timeshifter]

So, no offense intended, but could you please just rewrite the equation the way it should be read after all the fancy interpretations? If it's not possible, then I don't think I know anything about physics at all.

[alana]

So, no offense intended, but could you please just rewrite the equation the way it should be read after all the fancy interpretations? If it's not possible, then I don't think I know anything about physics at all.

Timeshifter,

Does the following help?

[period indicates multiplication]

E = mc2
E (energy in ergs) = m (mass in gm) c^2 (vel2 in cm2 / sec2)

To show that the units are correct, we want to show

erg = grams . cm2 / sec2

Defn:
energy(work) = force . distance
Units:
erg = dyne . cm

Defn:
force = mass . acceleration
Units:
dyne = gram . cm / sec2

So
erg = dyne . cm
= (grams . cm / sec2) . cm
= grams . cm2 / sec2

which is what we wanted to show.

alana

[fisherabode]

I understand that the units are arbitrary but their relation to each other is not. But the equation does give you the relationship between the units. Thus if you choos miles hours and grams for the equation, what tells you the units to use for energy. do you find it by experiment?

[timeshifter]

Something just struck me that I learned a couple of weeks ago in my Physics class, and from what my teacher said, Einstein's equation is just a relativistic version of the force equation, F=ma. I'm wondering now if my prior interpretation of the equation might have been flawed...

[Capable Hands]

In thids equation
E= Total energy of a body
M=mass of a body
c=speed of light

The speed of light is a constant fixed by Maxwell's Theory of Electromagnetism. It has a value of 300 hundred million meters per second or 300 thousand kilometers per second. If you would like to see a proof let me know and I will see what I can do.

[alana]

. . . what my teacher said, Einstein's equation is just a relativistic version of the force equation, F=ma....

Well, if you take the equation F=ma and multiply both sides by distance, you get energy on the left side and mass times velocity squared on the right - just like
E=mc2.

I think of the equation as showing that mass is a form of energy, with c2 being a units conversion factor.

alana

[timeshifter]

If what we are saying is true, then it would have to be either a unit conversion statement or just a constant that we can base everything off of. That in itself would basically prove what I just came up with.

[crazymarie2002]

The original question was what is C, the speed of light, this could be almost any number from one up. EG one light year a year gives a value of one. This could give e=mc2 a value of 1. A mass of one ton multiplies by one squared is one. Not a very big bang for a nuclear explosion, but if you use something else, say furlongs per fortnight, then you get a figure of 325,140,480 furlongs per fortnight, which if you square and multiply by one gram gives you a much bigger bang for your bucks. So the only answer to the question is that C is whatever Einstein thought it was. Enjoy..

[timeshifter]

Please allow me to apologize for my stupidity in saying that a velocity squared is an accelleration. School has been burning up my brain, and coming here and trying to make intelligent replies after all of that is kinda tough, if you know what I mean.

Now that I'm a little more focused, let's try and figure out what c equals. Does something along the lines of 299,792 km/s sound reasonable?

[skitsefranic]

ITs to my understanding that energy= MASSx velocity2. for example in a thermal nuclear explosion, mass being the weight of the core and velocity being the rate of expansion measured in a unit of distance( meters, feet). The answer will also be in a similar unit.

[budcamp]

It's energy= MASSx the velocity of light Squared.

Bud

[raptor]

I just read in a yr 10 science textbook that:

E(Joules)=M(kg)*(C(m/s))squared.


PS: I like how the site's been done up-I haven't been here for a while.

[MyFinalAnswer]

Hi, I'd like to clear up a few issues regarding Einstein's famous equation, E = mc².

Section 1: Where did E = mc² come from? (Requires: Calculus)
Let's see... Energy is some silly abstract idea which helps physicists keep a tally to help them solve their problems. You may recall that they have defined energy to by the product of force and distance:
E = ∫F(x) · dx
Ok, we all know that force (another abstract idea) is given by the product of the mass of an object and the acceleration its experiencing. Since the acceleration is given by the derivative of the velocity (with respect to time), we can write
F = ma = m(dv/dt).
We can plug in the second equation into the first (by substitution), and rearrange a little, to arrive at a third equation:
E = ∫m(dv/dt)·dx = ∫m(dx/dt)·dv = ∫mv·dv
Now, if we were to integrate right here, we would arrive at the classical mechanical formula for kinetic energy, E = ½mv².
However, this would not take relativity into account. We must use another relationship of special relativity
m = m0/√(1-v²/c²).
Ok, where the heck did this come from, and what does it mean. Ok, we all know that, according to speical relativity, as you travel faster and faster, time slows down. However, two other things occur at the same time (the reason why this is not well known is a mystery). The travelling vehicle shinks lengthwise (the length contraction effect), and the travelling vehicle gets more massive (not bigger, it just starts to weigh more)! Its the later effect that's being described by the equation above. In words the equation says, "the new mass of the object is equal to the rest mass of the object divided by the sqaure root of stuff."
Ok, enough of that, we simply plug this ugly equation into our equation that we have been working on, and arrive at
E = ∫m0v/√(1-v²/c²)·dv.
Ok, to finish things up, we integrate the right side of the equation (with the limits being 0 ―>v) Integration by trigonometric substitution will work best here (or a TI-89, or Mathematica will do nicely).
E = c²m√(1-v²/c²)
Since m√(1-v²/c²) = m0, we arrive at Einstein's famous equation,
E=mc².
(Note, the 0 on m has been dropped for aesthetic purposes)

[Daxid]

> What numerical value is used for C?
> What number am I to use the square?

The answer simply depends on which units you want to end up with.

If you want the answer in Joules(kgm²/s²), then you want to multiply kg*m/s*m/s -- (mass in kg, c in m/s)

If you want the ansewr in ergs(gcm²/s²), then you want to multiply g*cm/s*cm/s -- (mass in g, c in cm/s)

[Veroman]

> What numerical value is used for C?
> What number am I to use the square?

The answer simply depends on which units you want to end up with.

If you want the answer in Joules(kgm²/s²), then you want to multiply kg*m/s*m/s -- (mass in kg, c in m/s)

If you want the ansewr in ergs(gcm²/s²), then you want to multiply g*cm/s*cm/s -- (mass in g, c in cm/s)

Veroman

March 13th, 2005, 18:05 PM

This note I’ve done maybe helps to anyone else, it is an addendum to Daxid note dated August 28th, 2004, 8:38 AM. Thanks.

Einstein’s Equation

E=mc²

What numerical value is used for c?
What number am I to use the square?

The answer simply depends on which units you want to end up with.

If you want the answer in Joules(kgm²/s²), then you want to multiply kg*m/s*m/s –
m = mass in Kilograms
c = velocity in metres/second

thus c² equals (3 x 10exp8)² = 9 x 10exp16 = 90.000.000.000.000.000

which means that 1 kg mass makes 90 thousand billions Joules energy

If you want the answer in ergs (gcm²/s²), then you want to multiply g*cm/s*cm/s –
m = mass in grams
c = velocity in centimetres/second

thus c² equals (3 x 10exp10)² = 9 x 10exp20 = 900.000.000.000.000.000.000

which means that 1 gram mass makes 9 hundred trillions Ergs energy

Hope this can be helpful to someone else

[jj.mv]

Hi, my name is juan and sems to me very interesting that u are in that field....

That ecuation show us how the mass and the light are the same thing in the same time , that why einstain told : tha matter only tranforms u cant creat matter becouse the matter exist alone, the universe off all that stars that u see and question so much is very live as us, becose we are that same mass in a incredeble dinamic imposible to stop, we go to the speed of light becauses we are that light in the midle of that incredeble space.

To compare the magnitud of the universe with the feelings, you can see the love for example, what is that, define love, ther are a million definitions of love, like in the space millions of galaxies or stars, it doesnt matter the way u name it.

[Max]

Hi, my name is juan and sems to me very interesting that u are in that field....

That ecuation show us how the mass and the light are the same thing in the same time , that why einstain told : tha matter only tranforms u cant creat matter becouse the matter exist alone, the universe off all that stars that u see and question so much is very live as us, becose we are that same mass in a incredeble dinamic imposible to stop, we go to the speed of light becauses we are that light in the midle of that incredeble space.

To compare the magnitud of the universe with the feelings, you can see the love for example, what is that, define love, ther are a million definitions of love, like in the space millions of galaxies or stars, it doesnt matter the way u name it.

I couldn't understand a thing you wrote

[Max]

For all worried about the units try using the metric si system

Time - s (Seconds)
Mass - kg (KiloGram)
Distance - m (Meters)
Force - N (Newtons)
Energy - J (joules)

For instance In the equation E=mC^2
E(j)=m(kg)C(m/s)^2

[JediJosh]

Awww

But what about us astronomers wanting to use CGS units?

:-(

[bhootnath]

with reference to the discussion regarding 'c' units

John Dobson makes the following statement
(does it make sense?)


quote

It was from the geometry that Einstein saw that what we call rest mass, that which is responsible for the heaviness of things and for their resistance to being shaken, is really just energy. Einstein's famous equation is E = mc2. Probably most of you have seen that equation. It says that for a particle at rest, its mass is equal to its energy. Those of you who read Einstein know that there is no 'c' in that equation. The c2 is just in case your units of space and time donÕt match. If you've chosen to measure space in an arbitrary unit and time in another arbitrary unit, and if you have not taken the trouble to connect the two units, then, for your system you have to put in the c2. If you're going to measure space in centimeters, then time must not be measured in seconds. It must be measured in jiffies. A jiffy is the length of time it takes light to go one centimeter. Astronomers are rather broad minded people, and they have noticed that the universe is quite a bit too big to be measured conveniently in centimeters, and quite a bit too old to be measured conveniently in seconds; so they measure the time in years and the distance in light-years, and the units correspond. That 'c' in the equation is the speed of light in your system of units, and if youÕve chosen years and light-years then the speed of light in your system is one. And if you square it, it's still one, and the equation doesnÕt change. The equation simply says that energy and mass are the same thing.

unquote

tks n brgds

[bhootnath]

regarding my post, I had a question

is Mr dobson talking sense
i.e
can we write e=m, if c is in light years (1 light year) what should be the units of energy and mass

tks n brgds

[PhysBrain]

Depends on your units of mass. For example, if you use kilograms for mass, then

(kg)(ly/y)(ly/y) = kg ly^2 / y^2

Or if you use solar masses (i.e. the approximate mass of our sun - it's constantly decreasing b.t.w.)

SM ly^2 / y^2

However, I am not aware of any simplification of these units into anything more convenient like Joules (kg m^2 / s^2).

With regards to Dr. Dobson's comments, the chief advantage of using this notation is that you could choose the mass of whatever object you are studying as the unit mass and what you get is an equivalence. If you are studying the sun, then your units of mass would be the solar mass, and your units of energy would be the solar watt (for lack of a better term). In this case, one solar watt would be energy equivalent of a one solar mass object.

Ok, so what is the energy output of the sun in solar watts? I'll leave that as an exercise for the reader, although with these units, the answer is fairly straight forward. The energy output of the sun is the fraction of its own mass that it is loosing per unit time (1 year) due to mass energy conversion in the fusion process and convective losses due the solar wind. It's interesting to note that the sun is loosing mass-energy as the result of both radiative transport (primarily energy) and convective transport (primarily mass) phenomena.

[snikmij]

Try turning the suns mass into joules using E=MC^2

A whopping 1.790 x 10^47 Joules.

Or, in a slightly different way, a 100kg meteorite hitting a surface at a speed of, say, 50 metres per second would equal 250,000 Joules.

The atomic bomb dropped on Hiroshima was 13 million joules as a comparison.

Regards