## Relativistic Kinetic Energy

The previous relationship between work and kinetic energy are based on **Newton’s laws of motion**. When we generalize these laws according to the principle of relativity, we need a corresponding generalization of the equation for **kinetic energy**. If an object’s speed is close to the speed of light, it is necessary to use **relativistic mechanics** to calculate its **kinetic energy**.

In** classical mechanics**, kinetic energy and momentum are expressed as:

Derivation of its **relativistic relationships** is based on the **relativistic** **energy-momentum relation:**

It can be derived, the **relativistic kinetic energy** and the relativistic momentum are:

The first term (**ɣmc ^{2}**) of the

**relativistic kinetic energy**increases with the speed v of the particle. The second term (

**mc**) is constant; it is called the

^{2}**rest energy**(rest mass) of the particle, and represents a form of energy that a particle has even when at

**zero velocity**. As velocity of an object approaches the speed of light, the

**kinetic energy approaches infinity**. It is caused by the

**Lorentz factor**, which approaches infinity for

**v → c**. Therefore the speed of light cannot be reached by any massive particles.

The first term (ɣmc^{2}) is known as the** total energy E** of the particle, because it equals the rest energy plus the kinetic energy:

E = K + mc^{2}

For a particle at rest, i.e. K is zero, so the total energy is its rest energy:

E = mc^{2}

This is one of the striking results of **Einstein’s theory of relativity** is that **mass and energy are equivalent and convertible** one into the other. **Equivalence** of the mass and energy is described by Einstein’s famous formula **E = mc**** ^{2}**. This result have been experimentally confirmed countless times in nuclear and elementary particle physics. For example, see Positron-electron Pair Production or Conservation of Energy in Nuclear Reactions.

See also: Relativistic Mass

## Example: Proton’s kinetic energy

A proton (**m = 1.67 x 10 ^{-27} kg**) travels at a speed

**v = 0.9900c = 2.968 x 10**. What is its

^{8}m/s**kinetic energy**?

According to a classical calculation, which is not correct, we would obtain:

K = 1/2mv^{2} = ½ x (1.67 x 10^{-27} kg) x (2.968 x 10^{8}m/s)^{2} = **7.355 x 10 ^{-11} J**

With relativistic correction the **relativistic kinetic energy** is equal to:

**K = (ɣ – 1)mc ^{2}**

where the Lorentz factor

ɣ = 7.089

therefore

K = 6.089 x (1.67 x 10^{-27} kg) x (2.9979 x 10^{8}m/s)^{2} = **9.139 x 10 ^{-10} J = 5.701 GeV**

This is about** 12 times higher** energy as in the classical calculation. According to this relationship, an acceleration of a proton beam to 5.7 GeV requires energies that are in the order different.

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