Equation for relative Kinetic energy

Question

Relative Kinetic energy is given by

K.E = ($\gamma$-1)$m_0$c²; where $m_0$ is rest mass

but can it also be given by this

K.E= $\frac{1}{2}\gamma m_0v²$; where v is velocity of particle

can it?

Answer 1

In special relativity, the total energy of an object is given by; $$E = \sqrt{ (pc)^2 + (m_0 c^2)^2} = \sqrt{(\gamma m_0 v c)^2 +(m_0 c^2)^2}$$ where $p$ is the relativistic momentum $(\gamma m v)$.

The kinetic energy is the total energy minus the rest energy, so:

$$E_K = E - m_0 c^2 = \sqrt{(\gamma m_0 v c)^2 +(m_0 c^2)^2} - m_0 c^2$$

If you rearrange this equation you get : $$ E_K^2 + 2 E_K m_0 c^2 - (\gamma m_0 v c)^2 =0 \ .$$ This is a quadratic equation and the positive root is $$E_k = m_0 c^2 (\gamma -1) $$ in agreement with your first equation and as documented by Wikipedia.

If you take the Taylor expansion of this expression, the first three terms are: $$0 + \frac 1 2 m_0 v^2 + \frac 3 8 \frac{m_0 v^4} {c^2} $$

When v is small compared to c the terms with powers of v greater than 2 become negligible and the equation approximates to the Newtonian expectation $ 1/2 m v^2$. The Newtonian equation is not strictly correct and is superseded by the relativistic equation, but for velocities encountered in everyday mechanics the Newtonian equation is a reasonable and convenient approximation.

Your second equation $\gamma (1/2 \ m_0 v^2)$ is not equal to either the relativistic or the Newtonian equation for kinetic energy. For example, if we assume $$1/2 \gamma m v^2 = m_0 c^2 (\gamma -1) $$ this equivalence eventually reduces to $$\frac{v^2}{c^2} = 0$$ which can only be true if v=0.

How come it reduces to $v^2/c^2=0$ ?

$$1/2 \gamma \ m v^2 = m_0 c^2 (\gamma -1) $$

$$ \gamma \ v^2/c^2 = 2 (\gamma -1) $$

Substitute x for v^2/c^2

$$ \frac{x}{\sqrt{1-x} } = \frac{2 (1 - \sqrt{1-x})}{\sqrt{1-x}} $$

$$ {x} = 2 (1 - \sqrt{1-x}) $$

$$ 2 \sqrt{1-x} = 2-x $$

$$ 4 (1-x) = (2-x)^2 $$

$$ -x^2 = 0$$

$$ x = 0$$

Answer 2

No.

Relativistic kinetic energy is difference of relativistic energy and rest energy of object. To check that your proposal does not hold,- simply try to equate these expressions, like :

$$ \tag 1 (\gamma-1)m_0c^2 = \frac 12 \gamma m_0v^2 $$

Then after a couple of equation simplification steps (leaving this for you as an exercise), you will get that your gamma factor has form :

$$\tag 2 \gamma_{fake} = \frac {1}{1- {v^2}/({2c^2})}$$

This fake factor differs from true Lorentz factor $\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$ mainly by 2 aspects :

• Square root in denominator is lost
• in your case, maximum possible body speed is $\sqrt 2 ~c$ which is greater than speed limit $c$.

Hence you can't assume new form of relativistic kinetic energy expression. It's already well-defined.