4 ms·
Not really, c^2 has the unit of velocity squared, not acceleration: [c^2] = [(v^2/t^2)] != [(v/t^2)] = [a] We can also choose natural units where c==1, which
by maze-le 4y ago
Not really, c^2 has the unit of velocity squared, not acceleration:
[c^2] = [(v^2/t^2)] != [(v/t^2)] = [a]
We can also choose natural units where c==1, which leads directly to e==m
- boomboomsubban 4y agoI took "accelerated" to mean at a certain speed, like in the sentence "we accelerated to 1 meter per second," not acceleration itself.
- javaunsafe2019 4y agoStill wrong. It’s a constant here that plays a role in the relation between mass and energy. That’s all. the speed of light is always the same. The mass was not accelerated to it.
- boomboomsubban 4y agoNo, e=mc^2 is the equation for something at rest. The full equation involves the momentum as well, see https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati... It's why I said "broadly what the equation shows." It's hard to tell at rest, but it is still technically including the momentum.
- javaunsafe2019 4y agoOk I understand now what you mean and agree with your point. I guess I over simplified my take.