# Tag Info

## New answers tagged mass-energy

9

The source of gravity is not mass, but stress-energy-momentum, so you are correct that the energy converted in this process already has gravity and that that gravity is only rearranged The change in the gravitational field needs time to propagate, though, and this does indeed happen at the speed of light.

3

If rest mass does not change with v then why is infinite energy required to accelerate an object to the speed of light? The momentum of a material particle, a conserved quantity, is theoretically and experimentally a non-linear function of velocity given by $$\vec p = m \frac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ which goes to infinity as $v ... 1 In relativity the rest mass is the mass of an object measured from a reference frame in which it is at rest. But this is not the mass involved in acceleration or inertial mass. Inertial mass, or the opposition of the body to the change of movement (directional or in magnitude), will grow with the speed of the body: $$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$ ... 2 For physicists it can be very annoying that our historically evolved units of measurement cause the speed of light$cto differ from unity. So physicists often apply a trick to avoid distracting conversion factors corresponding to the numerical value of (powers of) the speed of light popping up in their equations. That trick is simply to define your own ... 1 A sample unit conversion for the second half of your question: \begin{alignat}{2} 0.511\,\mathrm{MeV}/c^2 &= 0.511\,\mathrm{MeV}/c^2 \times \frac{10^6\,\mathrm{eV}}{1\,\mathrm{MeV}} \times \frac{1.60\times10^{-19}\,\mathrm{joule}}{1\,\mathrm{eV}} \\ &\quad\qquad \times \frac{1\,\mathrm{kg\cdot m^2/s^2}}{1\,\mathrm{joule}} \times \left( ... 9 Capital\mathrm{C}$, in upright font, is the symbol for the coulomb. Lowercase$c$, italicized, is the speed of light in vacuum. Thanks to Einstein's equation, we can switch between mass and energy ($\mathrm{MeV}$is a unit of energy) by using factors of$c^2$, and sometimes it's more convenient to know the energy equivalent of a particle's mass rather than ... 1 I didn't see anyone mention the practical reason to use an approximation for energy. It is that in most problems you will be computing differences in energy. In that case, for small velocities, you can not only go the approximation${m v^2\over 2}+m c^2$, but if you are also not converting mass to energy or vice-versa, you can drop the$m c^2\$ as well ...

Top 50 recent answers are included