A problem with $E = mc^2$

Question

(Rest) mass of proton = (rest) mass of neutron (approximately) = $1.67 \times 10^{-27}$ kg.

let $m$ be the mass of nitrogen atom then $m = 2.32 \times 10^{-26}$ kg.

$c = 3 \times 10^8$ m/s (approx).

Energy $E = mc^2$.

$E = 2.32 \times 10^{-26} \times 3 \times 10^8 \times 3 \times 10^8$ Joules.

$E = 20.88 \times 10^{-26+16}$ Joules.

$E = 20.88 \times 10^{-10}$ Joules.

Energy required to propel a nitrogen atom at the speed of light is very less. Is my calculation wrong?

Because even if we take a cluster of 10000 atoms of nitrogen we can make them cross the speed of light easily.

Answer 1

Your calculation is wrong because $E=mc^2$ doesn't mean that the object has velocity $c$.
To make my answer useful i will give a very brief overview of Dynamics at higher velocities which is a consequence of Special theory of relativity.
At higher speeds(of order of $c$) Newtonian mechanics is not valid.
The linear momentum is defined as: $$\vec p=m_0\gamma \vec v \tag{1}$$
$m_0$ is the rest mass which appears in Newton's second law. $\gamma$ is defined as: $\gamma= \dfrac{1}{\sqrt{1-v^2/c^2}}$.
Newton's second law in the modified form is$$\vec F=\dfrac{d\vec p}{dt}$$ where $\vec p$ is as defined in equation $1$.
Now for increasing velocity of a material particle from $0$ to $c$ we have to apply some force but as long as we apply the force the moving mass $m_0\gamma$ increase boundlessly making it impossible for a finite force applied for a finite time to make $v=c$.
Let us apply a constant Force $F$ on a material object for time $t$ and see if we can make $v=c$ or not.
Let the particle starts from rest so $\vec F$ and $\vec v$ are in same direction. So we can right: $$F=\dfrac{dp}{dt}$$
In differential form
$$\lim_{dp \to 0}dp=F\lim_{dt \to 0}dt$$
Intagrating within proper limits, given $F$ is constant throughout
$$\int_0^pdp=F\int_0^tdt$$
$$\implies p=Ft$$
Using equation $1$ with a little algebra
$$v=\dfrac{Ftc}{\sqrt{m_0^2c^2+F^2t^2}}$$
However large $t$ may become, $v$ can never reach $c$ or exceed $c$.
About $E=mc^2$. It means Total Kinetic energy of a particle moving at a speed $v$ is given by: $$E=m_0\gamma c^2$$
An important conclusion can be made from the above written equation. Any material object if is at rest in certain co-ordinate system its kinetic energy is $m_0c^2$ where $m_0$ is the rest mass which appears in the usual classical equation of Newton's second law $\vec F=m_0 \vec a$. It also means that mass is equivalent to energy, that is you can convert this mass into pure energy like heat.

Answer 2

1. The mass $m$ in the formula is NOT the rest mass $m_0$ and therefore dependent on the velocity: $$ m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma m_0 $$ This means you cannot simply take the normal mass of a nitrogen atom and put it in there, if you assume a speed $v \neq 0$.

2. The $c$ in the formula doesn't mean that the particle travels at light speed. It's a factor, nothing more. The actual particle velocity comes into play by the mass $m$ with the above relation. There is also an expression for the energy in terms of the rest mass and momentum: $$ E^2 = p^2c^2 + m_0^2c^4 $$ This might remind you of Pythagoras ;) In fact this equation is even better because it's also valid for massless particles like the photon, where $E = mc^2$ is only valid for massive objects.

3. You can easily see with the help of the first equation that if $v$ approaches $c$ the term for $m$ becomes $\infty$ and therefore $E$ becomes also $\infty$. This means that it's impossible to accelerate a particel with a rest mass $m_0 \neq 0$ to light speed.