PDA

View Full Version : Reversing E=MC^2

Indagare
2011-Oct-05, 12:37 PM
Can someone help me with the simple math here?

I've heard that E=M.
And that c is usually set to 1.
So, would the reverse of E=Mc^2 be M=Ec^2 or would it be M=E/c^2?

Ivan Viehoff
2011-Oct-05, 01:19 PM
I've heard that E=M.
And that c is usually set to 1.
So, would the reverse of E=Mc^2 be M=Ec^2 or would it be M=E/c^2?
Cart before the horse there. If you are working in units such that c=1, then E=m.
m=E/c^2 is a correct rearrangement of the formula. It isn't really appropriate to call it "the reverse".

grapes
2011-Oct-05, 01:33 PM
Can someone help me with the simple math here?

I've heard that E=M.
And that c is usually set to 1.
So, would the reverse of E=Mc^2 be M=Ec^2 or would it be M=E/c^2?M=E/c^2, of course.

If c = 1, it doesn't make any difference, as M and E would be measured in the same units.

The actual formula is E = m c^2 \frac{1}{\sqrt{1-v^2/c^2}}, but the series expansion of that is

E = m c^2 \left( 1 + \frac{1}{2}\left(\frac{v}{c}\right)^2 + \frac{3}{8}\left(\frac{v}{c}\right)^4 + \frac{5}{16}\left(\frac{v}{c}\right)^6 + ... \right)

And obviously the first term is the famous formula, the second is kinetic energy. The following terms are too small to mess with... :)

ETA: JOOC, I checked and it looks like the third term is 10% of the total "kinetic energy" at a velocity only 78% the speed of light.

Indagare
2011-Oct-05, 01:46 PM
Cart before the horse there. If you are working in units such that c=1, then E=m.
m=E/c^2 is a correct rearrangement of the formula. It isn't really appropriate to call it "the reverse".

Ah, okay. Would "solving for M" be more appropriate? My first love is English, so my math skills and terms tend to be a bit rusty.

M=E/c^2, of course.

If c = 1, it doesn't make any difference, as M and E would be measured in the same units.

The actual formula is E = m c^2 \frac{1}{\sqrt{1-v^2/c^2}}, but the series expansion of that is

E = m c^2 \left( 1 + \frac{1}{2}\left(\frac{v}{c}\right)^2 + \frac{3}{8}\left(\frac{v}{c}\right)^4 + \frac{5}{16}\left(\frac{v}{c}\right)^6 + ... \right)

And obviously the first term is the famous formula, the second is kinetic energy. The following terms are too small to mess with... :)

Okay. How would all the rest of it factor in for determining how much matter you can get from energy? Or am I asking it the wrong way?

grapes
2011-Oct-05, 01:51 PM
Okay. How would all the rest of it factor in for determining how much matter you can get from energy? Or am I asking it the wrong way?It's the first formula that is important, not the series. Of course, the less energy you spend in contributing to the velocity the better, so your best conversion would be for zero velocity. And, you're back to the famous formula, M = E/c^2.