$E=mc^2$ vs $(1/2)mv^2$

Question

How does the kinetic energy $E_k = \frac{1}{2} mv^2$ equation fit with Energy equation $E = mc^2$.

Becuase if you rearange it you get $\frac{1}{2} mv^2 = mc^2$, mass cancels and velocity = $\sqrt{2c^2}$ which makes velocity a constant of $423,970,560$ m/s, which doesn't make sense, or is this the maximum velocity of an object with energy. This is also seen when also mixed with Gravitational Potential energy $E_p = mgh$, giving a constant to height.

No I have not done relativity yet.

i found this while trying to explain to myself why objects lose mass the faster they go in regards to Einstein's theory of relativity, without learning it yet, but also in that regard why do the masses cancel out.

Answer 1

The total energy $E$ of a particle is the sum of its kinetic energy and its rest mass, and equals $E=\gamma mc^2$ where $\gamma$ is the Lorentz factor, $$\gamma = \dfrac1{\sqrt{1-v^2/c^2}}.$$

The relativistic kinetic energy is $K=(\gamma - 1)mc^2$. For speeds much less than that of light, this equation is approximately equal to $\dfrac12 mv^2$.

To show that, we can rewrite the kinetic energy as $$ \begin{align} K &= \gamma mc^2-mc^2\\ &= \left( 1-\left( \dfrac vc\right)^2 \right)^{-1/2}mc^2-mc^2 \end{align}$$ which by the Binomial theorem for $(1+x)^{-1/2}\approx 1 - \dfrac12 x$ can be expanded as $$K\approx \left( 1 + \dfrac12 \left(\dfrac vc\right)^2\right) mc^2-mc^2=\dfrac12mv^2$$ giving us the classical $K=\dfrac12mv^2$.

Answer 2

$E=mc^2$ is called the rest energy because it is just that: the relativistic energy of an object of mass $m$ that is at rest. Equating it to a kinetic energy does not make sense.

See How to interpret the non-relativistic limit of $E=mc^2$? or How to understand $E=mc^2$?

Answer 3

These days the symbol $m$ is taken to mean the rest mass. Go back half a century and $m_0$ was often used to mean the rest mass and $m$ was the relativistic mass:

$$ m = \gamma m_0 = \frac{m_0}{\sqrt{1 - v^2/c^2}} \tag{1} $$

Using this notation $mc^2$ is the total energy i.e. the kinetic energy plus the rest mass energy, so to get the kinetic energy we would subtract off the rest mass:

$$ KE = (\gamma - 1)m_0c^2 = \left(\frac{1}{\sqrt{1 - v^2/c^2}} - 1\right) m_0c^2 \tag{2} $$

So we have to show that in the non-relativistic limit $v \ll c$ we recover the Newtonian kinetic energy. To do this we rewrite equation (2) as (I'm going to switch back to using $m$ for the rest mass since this is the current convention):

$$ KE = ( (1 - v^2/c^2)^{-1/2} - 1 ) mc^2 \tag{3}$$

and since in the non-relativistic limit $v \ll c$ we can use a binomial expansion and approximate $\gamma$ as:

$$ (1 - v^2/c^2)^{-1/2} \approx 1 + \tfrac12 \frac{v^2}{c^2} $$

Substitute this into equation (3) and we get:

$$ KE = (1 + \tfrac12 \frac{v^2}{c^2} - 1 )mc^2 = \tfrac12 mv^2 \tag{4} $$

Answer 4

$E = mc^2$ gives the energy of an object with mass $m$ when it is at rest. In this situation, the kinetic energy is $0$.

For a particle (not necessarily at rest), the kynetic energy is therefore defined as the total energy $E$ minus the rest energy : $E_{\text{kin}} = E - mc^2$. To find the energy $E$ as a function of velocity $v$, we boost to a frame where the object is at rest and the energy is the rest mass, then boost back to the observer's frame. Because of the way $E$ transforms under Lorenz transformations (it is the $0$-th component of a $4$-vector), we get : $$E = \left(1-\frac{v^2}{c^2}\right)^{-1/2}mc^2$$

If the particle is non-relativistic, ie $|v| \ll c$, we can expand this at first order in $v^2$ to get : $$E \simeq mc^2 +\frac{1}{2}mv^2$$