The equivalence between mass and energy in the expression *E*^{2}−(*p**c*)^{2 }= *m*^{2}*c*^{4} gives the rest mass of a particle.

${\mathit{E}}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{4}}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}$

${{\mathbf{\left(}}{\mathit{p}}{\mathit{c}}{\mathbf{\right)}}}^{{\mathbf{2}}}{\mathbf{=}}\frac{m2v2c2}{1-\frac{v2}{c2}}$

Taking the condition, v = 1.0c

Relativistic Energy and Momentum

**Learning Goal:**

To learn to calculate energy and momentum for relativistic particles and, from the relativistic equations, to find relations between a particle's energy and its momentum through its mass.

The relativistic momentum p and energy *E* of a particle with mass *m* moving with velocity v are given by

and

What is the rest mass *m* of a particle traveling with the speed of light in the laboratory frame.

Express your answer in MeV/c^{2} to one decimal place.

**Hints**

**Hint 1.** Equations for momentum and energy

Compare the equations for energy and momentum if *v*=1.0*c*. What do you find when you plug these into

*E*^{2}−(*p**c*)^{2 }= *m*^{2}*c*^{4}?

Frequently Asked Questions

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