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In the non-relativistic case we still have that $E=mc^2$. Since energy is a conserved quantity, any mass differences resulting from a process must be compensated for by energy differences (units are often used with $c=1$).

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1) $E$ being a conserved quantity is a principle of physics; your theory must by construction preserve whatever it defined to be $E$. The Taylor expansion you mentioned is a plausibility argument to convince you that $mU$ is indeed the correct relativistic generalization of $E$ in newtonian mechanics. 2) You are correct about Noether's theorem. In a ...

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We then defined $E$ via $E/c=P^0$ and claimed that $E$ is a conserved energy quantity based on the first 2 terms of its Taylor expansion. Taylor expanding $E$ around $v=0$ gives that $E = mc^2 + \frac{1}{2}mv^2 + \mathcal O(v^4)$, which means that at low velocities (ignoring the constant term, because constant shifts in energy are irrelevant in Newtonian ...

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When you extend Newtonian mechanics to special relativity you use the principle that the equations of SR must reduce to the Newtonian in the non-relativistic limit. By this principle you can find suitable new definitions of relativistic energy and momentum, as you describe. So, why are relativistic enery and momentum conserved? There are two slightly ...

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You are better to take conservation of energy and momentum as a fundamental empirical principle that underpins everything else in physics. You can prove them from Newton's laws, but that only covers classical mechanics. Actually conservation of energy and momentum is a fundamental principle contained in Einstein's equation in general relativity, and they can ...

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The correct equation is $E=\frac{1}{\sqrt{1-v^2/c^2}}m_0c^2$, or, if you like ($\vec{p}$ is the three-momentum)- $$E^2=|\vec{p}|^2c^2+m_0^2c^4$$. This shows you how the velocity(momentum) of the object comes into the picture when calculating energy. More the $p$, more the $E$, as you'd intuitively expect. The kinetic energy is $$K.E=E-m_0c^2$$ and clearly $... 1 The mass energy relation states that mass and energy are equivalent and interconvertible. The law which governs the conversion of mass into energy is$E=mc^2$as you stated. But it seems that you have misunderstood the context of this equation. Think it in this way. The energy$E$is the maximum possible energy the mass can give due to mass energy ... 2 Yes,$p^2 := p \cdot p := \sum_{i=1}^n p_i^2$with$\cdot$the scalar product as you correctly guessed, and$n$the number of dimensions in Euclidean space. Note that there might be a minus sign for the space components in Minkowski space, which is the usual signature for special relativity. 2 Chemical reactions can be described fully in terms of electrons, (given) nuclei and sometimes photons. 2 The formula$E=mc^2$gives the rest energy of an isolated system. By definition it includes only internal kinetic and potential energy. 1 In the formula$E=mc^2$,$E$is the rest energy of the object or system and$m$is its rest mass. The use of the letter$E$is misleading because it implies it is the total energy of the object, which is in fact the sum of the rest, kinetic and potential energies. This total energy is conserved, even when energy is transferred between rest energy and other ... 1 Your first formula,$\Delta=(m(A,Z)-A)c^2$is the binding energy, which is equal to$\mu c^2$,$\mu$being the mass defect. Here,$m(A,Z)$is the observed mass of the nucleus, while$Zm_H+Nm_n\neq Zm+(A-Z)m=Am$is the sum of masses of the nucleons making up the nucleus (and we assume that the proton and the neutron have equal mass$m$. For a stable nucleus, ... 1 As for most questions, a more succinct answer is needed, rather than a more detailed one, so here it is: Stored energy is equivalent to mass. Thus we can never store more energy than mass, as the OP already suspected. This is just a very fundamental fact about nature. To understand and accept it, one can study some Special Relativity. To make some more ... 1 I will try to clarify. To simply the formulae, I will put the speed of light equal to unity,$c=1$, so that if time is seconds, distance is in light seconds and something traveling at half the speed of light has$v=1/2$. Energy-momentum can be written as a 4-vector$(E, \mathbf p)$. The magnitude of energy-momentum is mass,$m$, and obeys the relationship$$... 0 Well, first of all, it should be cleared that as velocity increases mass does not increase. So, when an object(in this case the flywheel) gains more kinetic energy, its mass does not increase, rather its inertia increases. that is to say, it becomes harder to accelerate the flywheel as the kinetic energy goes up. So when you add energy to a flywheel of mass ... 4 Relativistic energy is$E = \sqrt{p^2c^2 + m^2 c^4}$, so while stationary objects with no moving parts have an upper bound of$E=mc^2$(by, for example, matter-antimatter interaction), the energy of a moving object is only capped by the momentum,$p$. Relativistic momentum,$p$, is$\frac{m v c}{\sqrt{c^2-v^2}}$. This has no upper bound. Once an object ... 1 While @G. Smith's answer is correct and straightforward (with$\gamma = 1/\sqrt{(1 - \beta^2)}$), it might be of interest to interpret the form of this equation. Using the substitutions in terms of hyperbolic-trigonometric functions of rapidity$\theta$:$\beta=\tanh\theta$,$\gamma=\cosh\theta$, and$\cosh^2\theta - \sinh^2\theta\equiv 1$(so$E=m\cosh\...

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It follows straightforwardly from $E=\gamma m$ and $p=\gamma\beta m$.

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I'm not a physicist, but I will attempt to answer. Mostly, I will restate a lot of your question and attempt to show why antimatter is the most efficient. As you know, a 1 kg mix of matter and antimatter can react to create energy of $e=mc^2$ where $m=$ 1 kg. The principle of mass-energy equivalence says that the energy released also weighs 1 kg. Now let'...

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Binding energy tends to lower the total energy of the system, making it more stable. Thus, a nucleus has less energy than the particles it's made up of, and since the nucleus arrangement has a lower total energy, it's a more stable configuration of the system. Thus, binding energy is how much the nucleus is below the energy where particles could break free ...

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