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Total energy is the sum of rest energy = and relativistic kinetic energy: = = + Invariant mass is mass measured in a center-of-momentum frame. For bodies or systems with zero momentum, it simplifies to the mass–energy equation E 0 = m 0 c 2 {\displaystyle E_{0}=m_{0}c^{2}} , where total energy in this case is equal to rest energy.
For example, for a speed of 10 km/s (22,000 mph) the correction to the non-relativistic kinetic energy is 0.0417 J/kg (on a non-relativistic kinetic energy of 50 MJ/kg) and for a speed of 100 km/s it is 417 J/kg (on a non-relativistic kinetic energy of 5 GJ/kg). The relativistic relation between kinetic energy and momentum is given by
Due to kinetic energy and binding energy, this quantity is different from the sum of the rest masses of the particles of which the system is composed. Rest mass is not a conserved quantity in special relativity, unlike the situation in Newtonian physics.
This implies the kinetic energy, in both Newtonian mechanics and relativity, is 'frame dependent', so that the amount of relativistic energy that an object is measured to have depends on the observer. The relativistic mass of an object is given by the relativistic energy divided by c 2. [10]
Derivation of the relativistic Doppler shift If an object emits a beam of light or radiation, the frequency, wavelength, and energy of that light or radiation will look different to a moving observer than to one at rest with respect to the emitter.
The above derivation makes use of the vector calculus identity: ... (total energy) can be viewed as the sum of the relativistic energy (kinetic+rest), ...
The relativistic mass-energy relation: = + where again E = total energy, p = total 3-momentum of the particle, m = invariant mass, and c = speed of light, can similarly yield the Klein–Gordon equation: ^ = ^ + ^ = ^ + where ^ is the momentum operator.
The left side represents the square of the momentum operator divided by twice the mass, which is the non-relativistic kinetic energy. Because relativity treats space and time as a whole, a relativistic generalization of this equation requires that space and time derivatives must enter symmetrically as they do in the Maxwell equations that ...