The relativistic expression of kinetic energy is one of the outcomes of Einstein’s special theory of relativity. In this article, we are discussing relativistic kinetic energy in detail with some of the numerical.
Contents:
What is Relativistic kinetic energy?
The relativistic kinetic energy is the difference between total relativistic energy (E) and rest mass energy (`E_{o}`).
`KE_{\text{rel}} = EE_{o}`
When the speed of the object reaches up to a significant fraction of the speed of light (C) then in such cases the classical (Newtonian) expression of kinetic energy (KE = `\frac{1}{2}mv^{2}`) not gives the accurate results. Therefore this expression doesn’t become valid for all inertial reference frames.
But the relativistic expression for the kinetic energy is valid for all the inertial reference frames and gives accurate results at very high speed and also at very low speed.
Hence the relativistic expression of kinetic energy obeys the 1st postulate of the special theory of relativity.
The equation of relativistic kinetic energy is expressed as follows,
`KE_{\text{rel}}` = `\frac{m_{o}C^{2}}{\sqrt{1\frac{V^{2}}{C^{2}}}}m_{o}C^{2}`
Where,
`m_{o}`= Rest mass
V = Object speed
C = Light speed
The expression can be also written as,
`KE_{\text{rel}}` = `\gamma m_{o}C^{2} – m_{o}C^{2}`
Where, `\gamma = \text{Lorentz factor} = \frac{1}{\sqrt{1\frac{V^{2}}{C^{2}}}`
Here the term `\gammam_{o}C^{2}` indicates the total relativistic energy (E) of the object and `m_{o}C^{2}` indicates the rest mass energy (Energy at rest, `E_{o}`).
Relativistic kinetic energy equation:
The equation is given by,
`KE_{\text{rel}}` = `\frac{m_{o}C^{2}}{\sqrt{1 – \frac{V^{2}}{C^{2}}}} – m_{o}C^{2}`
In terms of Lorenz factors (`\gamma`), the equation becomes,
`KE_{\text{rel}} = m_{o}C^{2} (\gamma – 1)`
The equation of relativistic kinetic energy in terms of momentum is given by,
`KE_{\text{rel}} = m_{o}C^{2} [(\sqrt{\frac{P^{2}}{m_{o}^{2}C^{2}} + 1}) – 1]`
Relativistic kinetic energy derivation:
According to the workenergy theorem, the amount of net work done by a force is equal to the change in kinetic energy of an object.
∴ `d(KE)_{\text{rel}}=dw`
`d(KE)_{\text{rel}}=F.dx`
As the force is equal to the change in momentum with respect to time, (F = dP/dt). Thus the above equation becomes,
`d(KE)_{\text{rel}} = \frac{dP}{dt}.dx`
`d(KE)_{\text{rel}} = dP.V \cdots\cdots [∵ V = \frac{dx}{dt}]`
Where P is the relativistic momentum which is given by,
P = `\frac{m_{o}}{\sqrt{1\frac{V^{2}}{C^{2}}}}.V`
Where,
`m_{o}` = Rest mass
V = Velocity of object
C = Light speed
By putting the value of momentum ‘P’ in the equation of kinetic energy, we get,
`d(KE)_{\text{rel}} = d {\frac{m_{o}}{\sqrt{1\frac{V^{2}}{C^{2}}}}.V}V`
`d(KE)_{\text{rel}} = m_{o}V.d {[\frac{1}{\sqrt{1\frac{V^{2}}{C^{2}}}}] V} [∵ m_{o} = \text{Constant}]`
By differentiation,
`d(KE)_{\text{rel}} = m_{o}V{\frac{1}{\sqrt{1\frac{V^2}{C^2}}}.dV + V.d(\frac{1}{\sqrt{1\frac{V^2}{C^2}}})}`
`d(KE)_{\text{rel}} = m_{o}V {\frac{1}{\sqrt{1\frac{V^2}{C^2}}}.dV + V(\frac{1}{2})(1\frac{V^2}{C^2})^{\frac{3}{2}}. (\frac{1}{C^2}2V).dV}`
`d(KE)_{\text{rel}} = m_{o}V {\frac{1}{\sqrt{1\frac{V^2}{C^2}}} + \frac{V^2}{C^2}(1 – \frac{V^2}{C^2})^{\frac{3}{2}}}.dV`
`d(KE)_{\text{rel}} = m_{o}V {\frac{1}{\sqrt{1\frac{V^2}{C^2}}} \times \frac{(1 – \frac{V^2}{C^2})^{\frac{3}{2}}}{(1 – \frac{V^2}{C^2})^{\frac{3}{2}}} + \frac{V^2}{C^2}(1 – \frac{V^2}{C^2})^{\frac{3}{2}}}.dV`
`d(KE)_{\text{rel}} = m_{o}V(1\frac{V^2}{C^2})^{\frac{3}{2}} {(1\frac{V^2}{C^2}) + \frac{V^2}{C^2}}.dV`
`d(KE)_{\text{rel}} = \frac{m_{o}V}{(1\frac{V^2}{C^2})^{\frac{3}{2}}}.dV`
The above equation indicates the small change in relativistic kinetic energy.
While the total relativistic kinetic energy is given by,
`KE_{\text{rel}} = \int d(KE)_{\text{rel}} = \int_{0}^{V} \frac{m_{o}V}{(1\frac{V^{2}}{C^{2}})^{\frac{3}{2}}}.dV`
`KE_{\text{rel}} = m_{o} \int_{0}^{V} \frac{V}{(1\frac{V^{2}}{C^{2}})^{\frac{3}{2}}}.dV`
Now assume, `\mathbf{\frac{V^{2}}{C^{2}}}` = a
By differentiation, `\frac{1}{C^{2}}2V.dV = da`
∴ V.dV = `\mathbf{\frac{C^{2}}{2}.da}`
Now the limits of integration become,
at V = 0, a = 0
at V = V, a = `\frac{V^{2}}{C^{2}}`
Put these values and limits in equation of `KE_{\text{rel}}`,
`KE_{\text{rel}} = m_{o}\int_{0}^{\frac{V^{2}}{C^{2}}}\frac{(\frac{C^{2}}{2}).da}{(1a)^{\frac{3}{2}}}`
`KE_{\text{rel}} = m_{o}\frac{C^{2}}{2} \int_{0}^{\frac{V^{2}}{C^{2}}} \frac{1}{(1a)^{\frac{3}{2}}}.da`
`KE_{\text{rel}} = m_{o}\frac{C^{2}}{2} \int_{0}^{\frac{V^{2}}{C^{2}}} (1 – a)^{\frac{3}{2}}.da`
`KE_{\text{rel}} = m_{o}. \frac{C^{2}}{2} [\frac{(1 – a)^{\frac{3}{2}+1}}{(\frac{3}{2}+1)(1)}]_{0}^{\frac{V^{2}}{C^{2}}}`
`KE_{\text{rel}} = m_{o}. C^{2} [(1 – a)^{\frac{1}{2}}]_{0}^{\frac{V^{2}}{C^{2}}}`
`KE_{\text{rel}} = m_{o}. C^{2} {[1 – \frac{V^{2}}{C^{2}}]^{\frac{1}{2}} – [1]^{\frac{1}{2}}}`
`KE_{\text{rel}} = m_{o}. C^{2} {[1 – \frac{V^{2}}{C^{2}}]^{\frac{1}{2}} – 1}`
`KE_{\text{rel}} = \frac{m_{o}. C^{2}}{\sqrt{1\frac{V^{2}}{C^{2}}}} – m_{o}C^{2}`
As, `\frac{1}{\sqrt{1\frac{V^{2}}{C^{2}}}} = \gamma`, Lorentz factor, Therefore the equation becomes,
`KE_{\text{rel}} = \gamma m_{o} C^{2} – m_{o}C^{2}`
`\mathbf{KE_{\text{rel}}} = \mathbf{m_{o}C^{2} (\gamma1)}`
This is the expression to calculate the relativistic kinetic energy of the object.
Relativistic kinetic energy vs Classical kinetic energy:
Sr. No.  Relativistic kinetic energy  Classical kinetic energy 

1  It is based on Einstein’s special theory of relativity.  It is based on newtons law of motion. 
2  It is used to calculate the kinetic energy when the speed of the object is a considerable fraction of light speed (C).  It is used to calculate kinetic energy when the speed of the object is much lower than light speed (C). 
3  The relativistic expression of kinetic energy is valid in all inertial reference frames.  The equation is not valid in all inertial reference frames. 
4  It is given by, `K_{\text{rel}}` = `(\gamma – 1)m_{o}C^{2}`  It is given by, K = `\frac{1}{2}m_{o}V^{2}` 
Relativistic kinetic energy and momentum relation:
The relativistic kinetic energy is given by,
`KE_{\text{rel}} = EE_{o}`
Where `E_{o} = m_{o}C^{2}`
And `E` indicates the total energy possessed by the object which is given by,
`E = \sqrt{P^{2}C^{2} + m_{o}^{2}C^{4}}`
By putting the `E` and `E_{o}` in equation of `KE_{\text{rel}}`, we get,
`KE_{\text{rel}} = \sqrt{P^{2}C^{2} + m_{o}^{2}C^{4}} – m_{o}C^{2}`
`KE_{\text{rel}} = \sqrt{m_{o}^{2}C^{4} (\frac{P^{2}C^{2}}{m_{o}^{2}C^{4}} + 1)} – m_{o}C^{2}`
`KE_{\text{rel}} = m_{o}C^{2} (\frac{P^{2}}{m_{o}^{2}C^{2}} + 1)^{\frac{1}{2}} – m_{o}C^{2}`
`KE_{\text{rel}} = m_{o}C^{2} {(\frac{P^{2}}{m_{o}^{2}C^{2}} + 1)^{\frac{1}{2}} – 1}`
This equation gives the relativistic kinetic energy in terms of relativistic momentum (P).
The above equation can be also reduced as follows,
`(\frac{KE_{\text{rel}}}{m_{o}C^{2}} + 1) = {\frac{P^{2}}{m_{o}^{2}C^{2}}+1}^{\frac{1}{2}}`
`(\frac{KE_{\text{rel}}}{m_{o}C^{2}} + 1)^{2} = \frac{P^{2}}{m_{o}^{2}C^{2}} + 1`
`\frac{KE_{\text{rel}}^{2}}{m_{o}^{2}C^{4}} + \frac{2KE_{\text{rel}}}{m_{o}C^{2}} + 1 = \frac{P^{2}}{m_{o}^{2}C^{2}} + 1`
`\frac{KE_{\text{rel}}^{2}}{m_{o}^{2}C^{4}} + \frac{2KE_{\text{rel}}}{m_{o}C^{2}} = \frac{P^{2}}{m_{o}^{2}C^{2}}`
`P = \sqrt{m_{o}^{2}C^{2}{\frac{KE_{\text{rel}}^{2}}{m_{o}^{2}C^{4}} + \frac{2KE_{\text{rel}}}{m_{o}C^{2}}}}`
`P = \sqrt{\frac{KE_{rel}^{2}}{C^{2}} + 2m_{o}KE_{rel}}`
Relativistic kinetic energy at low speed:
When the speed of the object is very less, then the expression of relativistic kinetic energy gives the same results as the expression of classical kinetic energy.
The relativistic kinetic energy is given by,
`KE_{\text{rel}} = \frac{m_{o}C^{2}}{\sqrt{1 – \frac{V^{2}}{C^{2}}}} – m_{o}C^{2}`
`KE_{\text{rel}} = m_{o}C^{2} [(1 – \frac{V^{2}}{C^{2}})^{\frac{1}{2}} 1]`
By doing binomial expansion of `(1 – \frac{V^{2}}{C^{2}})^{\frac{1}{2}}`
Note: `(1 – x)^{n} = 1 + nx + \frac{n(n+1)}{2!}x^{2} + \frac{n(n + 1)(n + 2)}{3!}x^{3} + \cdots`
`KE_{\text{rel}} = m_{o}C^{2}{[1 + (\frac{1}{2})\frac{V^{2}}{C^{2}} + \cdots] 1}`
As V<<C, thus neglecting the higher order terms,
`KE_{\text{rel}} = m_{o}C^{2} {[1 + \frac{V^{2}}{2C^{2}}] – 1}`
`KE_{\text{rel}} = m_{o}C^{2} \times \frac{V^{2}}{2C^{2}}`
`\mathbf{KE_{\text{rel}} = \frac{1}{2}m_{o}V^{2}}`
Solved examples:
1] The electron (m = 9.109 x 10⁻³¹ Kg) is moving at a velocity of 0.75 times light velocity (C), Find the kinetic energy of an electron.
Given:
`m_{o}` = 9.109 x 10⁻³¹ Kg
V = 0.75 C
C = 3 x 10⁸ m/s
Solution:
The relativistic kinetic energy is given by,
`KE_{\text{rel}} = (\frac{1}{\sqrt{1 – \frac{V^{2}}{C^{2}}}} – 1) m_{o}C^{2}`
`KE_{\text{rel}} = (\frac{1}{\sqrt{1 – \frac{(0.75C)^{2}}{C^{2}}}} – 1) 9.109 \times 10⁻³¹ \times (3 \times 10⁸)^{2}`
`\mathbf{KE_{\text{rel}}}` = 419.62 x 10⁻¹⁶ J
2] A particle moving with a velocity of 0.7 times light speed has a kinetic energy of 4.5 Kev. Find the rest mass energy of the particle.
Given:
V = 0.7 C
C = 3 x 10⁸ m/s
`KE_{\text{rel}}` = 4.5 keV = `\frac{4.5}{6.242\times10^{15}}` J = 7.209 x 10⁻¹⁶ J
Solution:
Step 1] Find rest mass(`\mathbf{m_{o}}`):
The relativistic kinetic energy of the particle is given by,
`KE_{\text{rel}} = (\frac{1}{\sqrt{1 – \frac{V^{2}}{C^{2}}}} 1)m_{o}C^{2}`
`7.209 \times 10^{16} = (\frac{1}{\sqrt{1 – \frac{(0.7C)^{2}}{C^{2}}}} 1)m_{o}(3 \times 10^{8})^{2}`
`\mathbf{m_{o}}` = 2.001 x 10⁻³² Kg
Step 2] Rest mass energy (`\mathbf{E_{o}}`):
It is given by,
`E_{o} = m_{o}C^{2}`
`E_{o}` = (2.001 x 10⁻³²)x (3 x 10⁸)²
`E_{o}` = 1.8009 x 10⁻¹⁵ J
`E_{o}` = (1.8009 x 10⁻¹⁵) x (6.242 x 10¹⁵) keV
`\mathbf{E_{o}}` = 11.24 keV
FAQs:

What happens to relativistic kinetic energy at low velocities?
At a very low speed, the relativistic kinetic energy gives the same result as a classical expression of the kinetic energy.

When to use relativistic kinetic energy?
The relativistic equation of the kinetic energy helps to find the kinetic energy of an object when the speed of the object is considerably closer to the light speed.
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