# E=MC^2 - Do I have this right?

1. Mar 3, 2005

OK first let me explain, I have NO formal physics background, not even high school, so please don't flame me if these are dumb questions

1. What unit of measure is E? If it were distance we could say miles or kilometers.

2. I understand that this formula is for mass at rest. If the mass were moving at 50% light speed, would the answer to E=MC^2 be half of what it would be for the same mass at rest?

3. In nuclear fusion I understand the energy is based on applying this formula to the mass left over when 2 atoms fuse. So looking at a Hydrogen atom, if .00794 atomic mass were converted into energy during the fusion, does that mean the energy gained from this fusion is Energy = .00794 x c^2? I don't know if that is the right number but I needed to apply one for the purpose of this question.

Thanks

2. Mar 3, 2005

### dextercioby

Energy,and it is measured in Joules.

Not exactly in that form.$$E=m_{0}c^{2}$$ is for mass at rest.Notice the fact that the subscript "0" indicates that thing...

No.It would be more.It's really easy to compute the energy,once you know the connection between rest mass $m_{0}$,movement mass $m$ (or "M",as you denoted it) and the velocity $v=|\vec{v}|$.Just arithmetics.

EXACTLY...Einstein's formula tells us how much enegy is involved in nuclear reactions...Pay attention with the units,though.That 0.00794 a.m.u.needs to be converted to Kg and then the final result (the energy) would be in Joules,energy's unit.

Daniel.

3. Mar 3, 2005

### James R

1. Any unit of energy will do. The standard SI unit is the Joule. If m is in kilograms and c is in metres per second, then E works out in Joules.

2. The complete formula is:

$$E = \frac{mc^2}{\sqrt{1-(v/c)^2}}$$

where $v$ is the speed of the object.

If you plug in $v=c/2$, you get:

$$E = \frac{mc^2}{\sqrt{1-(1/2)^2}} = \frac{2}{\sqrt{3}}mc^2 \approx (1.15)mc^2$$

When an object is moving, the energy is greater than $mc^2$. In fact, $mc^2$ is the rest energy only. The "extra" energy is kinetic energy.

3. Essentially, you are right, but you need the mass in kilograms rather than atomic mass units. The conversion factor is:

$1$ amu = $1.66 \times 10^{-24}[/tex] kilograms. Hope this helps. 4. Mar 3, 2005 ### TsunamiJoe also, not to sound stupid, but what exactly is E=mc^2 solving, the energy at rest? and if that is the case would that be more commonly known as potential energy? 5. Mar 4, 2005 ### dextercioby No.In the form where that "m" is the rest mass,it is just the rest energy of the particle.If that "m" is not the rest mass,it is the TOTAL ENERGY OF THE PARTICLE,rest+kinetic... Daniel. 6. Mar 4, 2005 ### SpaceTiger Staff Emeritus Physicist - Joules Astrophysicist - Ergs Particle Physicist - eV Nutritionist - Calories Electrician - Kilowatt-hours King of England - Foot-pounds Oil Tycoon - Barrels Enron Executive - Dollars Mars Spacecraft Operator - Joules, no....foot-pounds, wait, no... 7. Mar 4, 2005 ### pmb_phy The unit is Jopules. This a derived unit and as such can be exressed in term of basic units, i.e. [joules] = [N = Newton's][L = distance] where [joules] = [kg][m2]/[s2] If you intend m to mean rest mass then your equation leaves something to be desired. The actuall expression is [itex]E = \gamma mc^2$. You should have written E0 = mc2. The expression for energy for a particle in motion is

$$E = \gamma m c^2$$

recall that

$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

Its rather simple to get the energy you're speaking of. Merely replace v by v/2.

Pete

8. Mar 4, 2005

### Moose352

He meant to write c/2, not v/2.

9. Mar 5, 2005