# Nuclear Fusion question **URL:** \ **Category:** Factual Questions\ **Created:** [December 12, 2002, 6:22am UTC](https://boards.straightdope.com/t/nuclear-fusion-question/142214 "2002-12-12T06:22:34Z")\ **Posts on this page:** 2\ **Page:** 2
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [December 12, 2002, 11:51pm UTC](https://boards.straightdope.com/t/nuclear-fusion-question/142214/21 "2002-12-12T23:51:18Z")
Okay, we’re defining the mass of a system as its total energy in its zero momentum frame? If so, I’m not sure I see yet why this definition is justified, but now I understand most of your comments so far. But not this one: > [@](#): > > \*Originally posted by Ring \* > However in your example… the sum of their masses equals the total energy of the system in the zero momentum frame. I don’t see why this is true at all. If the individual particles are moving at a speed v = 0.6 (c = 1), then they each have total energy E = 1.25m[sub]e[/sub], and the total energy of the system is E = 2.5m[sub]e[/sub], not 2m[sub]e[/sub]. What am I doing wrong? ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [December 13, 2002, 12:21am UTC](https://boards.straightdope.com/t/nuclear-fusion-question/142214/22 "2002-12-13T00:21:45Z")
> [@](#): > > \*Originally posted by Achernar \* > \*\*Okay, we’re defining the mass of a system as its total energy in its zero momentum frame? If so, I’m not sure I see yet why this definition is justified, but now I understand most of your comments so far. But not this one:I don’t see why this is true at all. If the individual particles are moving at a speed v = 0.6 (c = 1), then they each have total energy E = 1.25m[sub]e[/sub], and the total energy of the system is E = 2.5m[sub]e[/sub], not 2m[sub]e[/sub]. What am I doing wrong? \*\* Nothing, I misread your post. I thought the particles were stationary. [Previous page](https://boards.straightdope.com/t/nuclear-fusion-question/142214.md?page=1)