Kinetic Energy: Relativistic kinetic energy of rigid bodies

Relativistic kinetic energy of rigid bodies

In special relativity, we must change the expression for linear momentum.
Using m for rest mass, v and v for the object's velocity and speed respectively, and c for the speed of light in vacuum, we assume for linear momentum that $\mathbf{p}=m\gamma \mathbf{v}$ , where $\gamma = 1/\sqrt{1-v^2/c^2}$ .
Integrating by parts gives

$E_k = \int \mathbf{v} \cdot d \mathbf{p}= \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) = m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} = m \gamma v^2 - \frac{m}{2} \int \gamma d (v^2)$

Remembering that $\gamma = (1 - v^2/c^2)^{-1/2}\!$ , we get:

$\begin{align} E_k &= m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d (1 - v^2/c^2) \\ &= m \gamma v^2 + m c^2 (1 - v^2/c^2)^{1/2} - E_0 \end{align}$

where E₀ serves as an integration constant. Thus:

$\begin{align} E_k &= m \gamma (v^2 + c^2 (1 - v^2/c^2)) - E_0 \\ &= m \gamma (v^2 + c^2 - v^2) - E_0 \\ &= m \gamma c^2 - E_0 \end{align}$

The constant of integration E₀ is found by observing that, when $\mathbf{v }= 0 , \ \gamma = 1\!$ and $E_k = 0 \!$ , giving

$E_0 = m c^2 \,$

and giving the usual formula:

$E_k = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - v^2/c^2}} - m c^2$

If a body's speed is a significant fraction of the speed of light, it is necessary to use relativistic mechanics (the theory of relativity as developed by Albert Einstein) to calculate its kinetic energy.
For a relativistic object the momentum p is equal to:

$p = \frac{m v}{\sqrt{1 - (v/c)^2}}$ .

Thus the work expended accelerating an object from rest to a relativistic speed is:

$E_k = \frac{m c^2}{\sqrt{1 - (v/c)^2}} - m c^2$ .

The equation shows that the energy of an object approaches infinity as the velocity v approaches the speed of light c, thus it is impossible to accelerate an object across this boundary.
The mathematical by-product of this calculation is the mass-energy equivalence formula—the body at rest must have energy content equal to:

$E_\text{rest} = E_0 = m c^2 \!$

At a low speed (v<<c), the relativistic kinetic energy may be approximated well by the classical kinetic energy. This is done by binomial approximation. Indeed, taking Taylor expansion for the reciprocal square root and keeping first two terms we get:

$E_k \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2\right) - m c^2 = \frac{1}{2} m v^2$

Kinetic Energy

Thursday, 14 July 2011

Relativistic kinetic energy of rigid bodies

Relativistic kinetic energy of rigid bodies

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