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My gut feeling tells me things should have energy because of their charge, like they have energy because of their mass. You are not the first one who had this idea. There was indeed a concept calles "electromagnetic mass", where the electrostatic energy $E_{em}$ of a charged particle at rest would be $$E_{em}=\frac{e^{2}}{2\cdot r}$$ (where $e$ is the ...


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This question is quite broad, but I'll try my best. When making the jump from special relativity, we have two good rules of thumb: Wherever there is a Minkowski metric in SR, put a general metric in GR. Wherever there is a partial derivative in SR, put a covariant derivative in GR. Take, for instance, mass-energy equivalence. In SR, we have ...


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At hyperrelativistic speeds, a massive object has the bulk of its energy in kenetic form so the invariant mass is small residue. It acts like a very low mass particle at meerly relativistic speeds, or a almost massless neutrino at any speed. A 7Tev proton is travelling at $0.999999991c$ and its mass is only contributing on the order of one one hundredth of ...


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Your question is so confused that it's hard to give a meaningful answer... if matter were to travel the speed of light, would it become energy? Anything that has a rest mass $m_0$ has an energy associated with that mass of $E=m_0 c^2$. If this mass is also moving with some momentum $p$, it has a kinetic energy associated with the motion. The total ...


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$c$ is not only an invariant speed, $c$ is also a physical constant that factors in many well known formula, e.g., the electromagnetic fine structure constant $$\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}$$ In the case of the famous $$E = mc^2$$ the particle with mass $m$ has zero speed (in this frame of reference). If the particle has a speed $v$ in ...


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$E=mc^2$ is actually not the full equation, although it is clearly the most famous form of it. The full equation is $E^2 = p^2c^2 + m^2c^4.$ This formula says that a particle's energy squared is equal to the sum of the momentum squared and the mass squared (with factors of the speed of light thrown in to make it dimensionally correct). The simple ...


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Certainly not. You have extracted only part of the full equation, which actually looks Pythagorean in structure: $E^2 = (mc^2)^2 + (pc)^2$ This relates the energy of an object to its mass and momentum. Its more famous cousin, $E = m c^2$ is simply the limit where $p=0$, or the energy of an object in a reference frame in which it is at rest. On the other ...


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If the velocity is slower than the speed of light (which is always the case) the newtonian limit for small velocities is $$E_{kinetic}=\frac{m\cdot v^2}{2}$$ and in relativity $$E_{kinetic}=\frac{m\cdot c^2}{\sqrt{1-v^2/c^2}}-m\cdot c^2$$ The rest energy is always $$m\cdot c^2$$ no matter what the velocity is.


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One of the insights of special relativity is that space and time are not absolute or distinct; instead of thinking of space and time, we should think of space-time. One consequence is that we should, in principle, measure space and time with the same unit. We haven't done that; we've measured time in seconds and space (distance) in metres. The factor of ...



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