# Tag Info

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The reaction $$\gamma\to e^+ e^-$$ is actually forbidden on kinematic grounds - one can show that momentum and energy cannot be simultaneously conserved if there is not something else to take "excess" momentum, e.g. a heavy nucleus nearby - pair production from gamma rays happens in metals, for example, but not in vacuum. Of course, given enough energy, it ...

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Build up four-momentum from the mass and the four-velocity: $$m \vec u = m \gamma (\vec e_t + \vec v)$$ But four-momentum can always be decomposed according to energy and momentum: $$m \vec u = E \vec e_t + \vec p$$ Equate the $t$-components. Done.

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Our Universe is a larger version of a galactic polar jet. 'Was the universe born spinning?' physicsworld [dot] com/cws/article/news/46688 "The universe was born spinning and continues to do so around a preferred axis" Our Universe spins around a preferred axis because it is a larger version of a galactic polar jet. 'Mysterious Cosmic 'Dark Flow' Tracked ...

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In cosmology, when you apply the Einstein equation to the whole [homogeneous and isotropic] universe, you get the Friedman-Lemaître-Robertson-Walker (FLRW) equations. These equations dictate how the size of the universe evolves depending on its material content. The different components are characterized by their equations of state, the ratio between ...

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I agree with atsby, and would like to say that Dark Matter and Dark Energy are two fundamental concepts of which we know very little about. However I doubt that Dark Matter can be converted to Dark Energy the same way we can convert normal matter to energy, let alone convert Dark Matter to Dark Energy at all. For starters, Dark Matter is the substance ...

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Mass and energy are the same thing. If you have a sealed internally reflective box of photons, it will have inertia and a exert a gravitational pull on any other mass/energy that may be around. http://usersguidetotheuniverse.com/?p=2865 That being said, no properties of dark matter or dark energy are known. These terms are used to refer to matter and ...

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This is a misconception. Theoretically, Dark Energy is a form of energy which accelerates the expansion of the universe, whereas Dark Matter is thought to be made up of weakly interacting particles (which account for most of the matter in the universe). Dark Matter and Dark Energy have completely different properties, and their names don't imply that they ...

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Dark matter are not related in that way. Dark energy is the energy responsible for the accelerating expansion of the universe. Dark matter is a source of gravitation which we couldn't see directly.

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The mass of a free neutron is 939.566 MeV/c$^2$ (almost 1 GeV/c$^2$, so that's probably where your instructor got the "1" value), and the mass of a free proton is 938.272 MeV/c$^2$. A free neutron will decay into a free proton, free electron ($\beta^-$), and an anti-neutrino, $\bar{\nu}$. The mass of the electron is 0.511 MeV/c$^2$, and of the ...

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There is no conservation of mass. There is conservation of mass/energy. Proton mass does not equal neutron mass.

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I'm not going to talk about the misunderstanding you have, but about the actual question you asked.The mass energy equivalence equation is $$E^2=(mc^2)^2 + (pc)^2$$ ($p$ is the momentum of the body). If we use specialized units such that $c$ (the speed of light) becomes dimensionless, we can have the same units for mass and energy. (This can be done, for ...

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Explain why the mass of a tree cannot be converted directly into energy. That's a tricky one, because it could turn out that it is possible to turn matter alone into energy. Floris hinted at this with radioactive decay, but there are potentially other methods such as melting hadrons in a quark-gluon plasma (QGP), see for example this report. The interesting ...

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When a radioactive element decays, part of its mass is converted to energy - no obvious need for antimatter anywhere. Instead, the energy is released because the binding energy of the sum of the fragments might be higher than that of the parent nucleus. However, to fully convert matter to energy you do need the antiparticle. Otherwise, you run into ...

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Energy is never created nor destroyed, and to say "X is converted into energy" is just meaningless. We don't convert things distinct from energy into energy, all we ever do is convert one form of energy into another. The badly posed question from your book probably intends to ask why we cannot convert the mass energy that any chunk of matter contains as per ...

0

See explanations on Quarks and Gell-Mann's Quark Model.

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It is because they only lose there energy when accelerating. A moving electron or proton with a constant velocity won't emit EM radiation. The electronmagnetic radiation only takes away the aceleration, slowly the proton or electron down to a constant velocity. The reason they have charges is much more complicated. For the proton example it is because ...

2

First a note not to forget that $E=m\,c^2$ is only a special case of the more general formula $\frac{E^2}{c^2} - p^2 = m^2 c^2$ for the Lorentz-invariant square "length" of the momentum four-vector in terms of the rest mass $m$. That matter out of the way, there are two methods springing to mind: Method 1: A neat way to do this is to imagine light in a ...

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A simple derivation that is accessible to lay people who can only do primary school level math, starts from the fact that a pulse of electromagnetic radiation with energy $\mathbf{E}$ has a momentum of $\dfrac{\mathbf{E}}{\mathit{c}}$. In addition, one assumes conservation of momentum. We do a thought experiment involving a closed box containing two objects ...

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The answer really depends on how much you assume. You could start with the relativistic action, $$S=-mc\int d\tau$$ where $c^2d\tau^2=dx^\mu dx_\mu$, such that the Lagrangian is $$L=-mc\sqrt{1-(v/c)^2}$$ whence the momentum follows from $p=\partial L/\partial v$ and the total energy is $E=pv-L$--this leads directly to \begin{align} E&=\gamma ...

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The energy-momentum 4-vector is simply mass times the velocity 4-vector. This gives the correct momentum in the non-relativistic limit. In the rest frame it correctly gives the (three)-momentum as being zero. And gives $E = mc^2$ for the time-component of the vector. This is clearly not going to make too much sense if you don't know relativity, but that's a ...

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