Comment on Scientists find first observational evidence linking black holes to dark energyparentComments−boomboomsubban3yThey said e is accelerated m, which is broadly what the equation shows.−margorczynski3yThe thing is E=mc^2 is the energy at rest, there is a more general equation for objects in motion−maze-le3yNot 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−boomboomsubban3yI took "accelerated" to mean at a certain speed, like in the sentence "we accelerated to 1 meter per second," not acceleration itself.−javaunsafe20193yStill 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.−boomboomsubban3yNo, 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_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.−javaunsafe20193yOk I understand now what you mean and agree with your point. I guess I over simplified my take.
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They said e is accelerated m, which is broadly what the equation shows.
The thing is E=mc^2 is the energy at rest, there is a more general equation for objects in motion
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
I took "accelerated" to mean at a certain speed, like in the sentence "we accelerated to 1 meter per second," not acceleration itself.
Still 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.
No, 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_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.
Ok I understand now what you mean and agree with your point. I guess I over simplified my take.