# E =mc{squared} How Did He Arrive At This ?

1. Apr 3, 2005

E =mc{squared}
How Did He Arrive At This ?
What Was The Guy Thinking When He Cooked Up This Mess .

2. Apr 3, 2005

### whozum

He guessed. :)

Try the search you'll find alot of stuff on this.

3. Apr 3, 2005

### SpaceTiger

Staff Emeritus
I think it's fair to say that most ground-breaking theories are little more than educated guesses.

4. Apr 3, 2005

### moose

Don't you think it's kind of a coincidence that the formula is that small, and it just uses our units O.O perfectly, like meters, second, joule, kilograms......
am I missing something here?

5. Apr 3, 2005

### SpaceTiger

Staff Emeritus
Units are a human invention, so their consistency has no physical meaning. The real key thing here is that the physical concept of energy is related to that of mass, things which had been considered separately prior to Einstein.

6. Apr 4, 2005

### eNathan

Does this mean units of the "meter" and "kilogram" were invented to have a relationship too?

7. Apr 4, 2005

### dextercioby

Nope.Physics (menaing experiments and theories) "cooked" up relations between quantities which automatically mean relations between units.

Daniel.

8. Apr 4, 2005

### SpaceTiger

Staff Emeritus
Not sure what you mean. The units of meters and kilograms are arbitrary scalings to two different kinds of quantities. Nobody writes an equation with one side having units of meters and the other with units of kilograms. That would imply "inconsistent" units.

9. Apr 4, 2005

### SpaceTiger

Staff Emeritus
A small amendment to my previous statements. There ought to be some correlation between our choice of units for different quantities, but only a very rough one (noticable on a logarithmic scale). In other words, the units will usually be chosen so that "1 unit of x" will be typical on the scales of interest. For example, mks units are mostly chosen to be relevant on a human scale, while astronomers will often express things in terms of much larger units (like solar masses and AU).

10. Apr 4, 2005

### eNathan

yerz

We are talking about e=mc^2 here, what this means is that the energy that could come from mass (in joules) is equavilant to the product of the mass (in KG) and the speed of light, c (in meters/second).

So what I am saying is this. What if we re-arranged the equation to this.

e=mc^2 where "m" is in pounds, and "c" is in miles per hour, and "e" is still in joules. Would this equation work? no, the units have a relationship. The American Standard Units are terrible :rofl:

11. Apr 4, 2005

### chroot

Staff Emeritus
Of course it would still work:

The units (meters, kilograms, etc.) are meaningless. The dimensions (length, mass, etc.) are quite significant, however you can choose any units you wish to represent those dimensions.

- Warren

12. Apr 4, 2005

### dextercioby

Yes,but it would not be that simple

$$1 \ J=1\ Kg\cdot 1ms^{-2}\cdot 1m=\frac{1}{0.453}\mbox{pounds}\cdot \frac{1}{1609}mi\cdot \left(\frac{1}{3600}hr\right)^{-2}\cdot \frac{1}{1609} mi$$

It's not pretty anymore.

Daniel.

13. Apr 4, 2005

### Staff: Mentor

I wouldn't quite say "arbitrary". Unlike the English system, SI was designed specifically to be easy to use. I don't know the specifics of how it was done, but it is very convenient that (for example) water has a mass of 1g/cc and a heat capacity of 1cal/g*C.

14. Apr 4, 2005

### dextercioby

Well,Russ,trust me,SI-cgs (or cgs-gauss for electrodynamics) is as difficult as the Anglo-Saxon system,at least for someone like me who's worked with SI-mKs all his short life...

Daniel.

15. Apr 4, 2005

### Integral

Staff Emeritus
Why are we getting into unit bashing? It really doesn't make any difference which units are used as long as you are consistent.

Just for the record, Einstein did not just "guess" this relationship. It is a derived result of aplying the Lorentz transforms to the kinematics equations of physics.

16. Apr 4, 2005

### DaveC426913

Consider a slightly different example (because that's the one I know).

$$F = \frac{m_1.m_2}{r^2}$$
This is the formula that shows the relationship to between two masses and gravity. It tells us that the gravitational attraction is directly proportional to the masses of the two bodies and inversely proportional to the square of their distances. Note that there are no units, and that you cannot use this formula to calculate actual values. you can only show relationships (double the mass to double the force, halve the distance to quadruple the force).

To use it to provide actual figures, we need to provide some units. We will measure the mass in grams and the radius in meters. But now our formula needs a constant: G.

$$F = G.\frac{m_1.m_2}{r^2}$$

The constant G is equal to $$6.672.10^-11 N m^2 kg^-2$$. Note the units.

If you plug all these together into the formula, including some mass and distance units, you will end up with a number and a unit of measurement:m.a - as in F=m.a. Hey! That's a unit of force!

Note that you could plug inches and stones into the equation, but to do that, your G constant would have to be converted to those units too.

The law itself is universal. The application of that law, is man-made.

Last edited: Apr 4, 2005
17. Apr 4, 2005

### dextercioby

You can't do that,it's illogical.You know that in the LHS you must have a force,since that's what you're measuring (along with mass & distance),so the units for G are obtained as a consequence,not as a premise...

Daniel.

18. Apr 4, 2005

### DaveC426913

When all is washed and dried, does that make a difference?

I was merely pointing out the difference between proportionality and equality.

E=mc^2 is a proportionality.

Or am I completely wrong?

19. Apr 4, 2005

### whozum

I was joking.

20. Apr 4, 2005

### whozum

E = mc^2 is a proportionality and an equality.

For every unit mass m multiplied by the square of the speed of light in any unit, will result in energy released with units composed of those m and c are made of. Its redundant, I dont know why you would want to toy with the units, but the relationship holds anway.

Changing the units will just require a scalar conversion factor, such as G in gravitation or K in electrics to get an answer in joules.

Last edited: Apr 4, 2005