Major doubt on relativistic energy

Question

We know of the equation that the kinetic energy of a body is:

$$K=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2 \ \ ,$$

where $m_0c^2$ is the rest mass energy. If $v$ is very small, then:

$$\frac{v^2}{c^2} \to 0 \ \ ,$$

then we get:

$$K=m_0c^2-m_0c^2=0 \ \ .$$

Does that mean a moving body has no kinetic energy? How is it possible? We surely know the kinetic energy of that body should be $\frac{1}{2}mv^2$ according to newtonian mechanics.

Also, I think I have a major misconception regarding $E=mc^2$. I think of $E$ as the total energy of the body of mass $m$ (this is what our teacher taught us). But then does this mean the potential + kinetic energy of a body($E$) is always $mc^2$? Even if the body is stationary, how does it still have energy $mc^2$? If this is a great misconception, please enlighten me since our teachers taught us wrongly and they can't even answer my questions. All they do is use the formulae to solve problems.

Answer 1

If $v$ is very small,then $\frac{v^2}{c^2}\to 0$, then we get $K=m_0c^2−m_0c^2=0$.

No, you approximated too coarse. Doing it more carefully using the approximation $\frac{1}{\sqrt{1-\beta^2}}\approx 1 + \frac{1}{2}\beta^2$ for $\beta\ll 1$ you will get

$$\begin{align} K&=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}} - m_0c^2 \\ &\approx m_0c^2\left(1+\frac{v^2}{2c^2}\right) - m_0c^2 \\ &= m_0c^2 + \frac{1}{2}m_0v^2 - m_0c^2 \\ &= \frac{1}{2}m_0v^2 \end{align}$$ which is just the non-relativistic expression for the kinetic energy.

Answer 2

1. Your formula is wrong.
2. You just set $v=0$ instead of considering $v \ll c$ or $v/c \ll 1$, if $v=0$ the kinetic energy is 0. For small $v^2/c^2$ one knows the approximation ${1/\sqrt{1-v^2/c^2)} \approx 1+v^2/2c^2}$. Use this and you find the classic kinetic energy. If the body does not move $m=m_0$ and it has this energy, which can be partly changed to other forms of energy for example in a nuclear reactor.

Answer 3

If $v$ is very small,then $\frac{v^2}{c^2} \to 0$,then we get $K=m_0c^2-m_0c^2=0$. Does that mean a moving body has no kinetic energy? How is it possible?

No, it means that as a body stops moving (limit as $v \to 0$) the kinetic energy goes to zero. $v \to 0$ is not the appropriate limit for "small" velocities, it is the appropriate limit for stopping. When you stop your KE does go to zero.

If you want to see what the kinetic energy is for small but non-zero velocities then you do a Taylor series expansion about $v=0$. $$K=\frac{m_0 c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2 = \frac{1}{2} m_0 v^2 + \frac{3}{8}\frac{m_0}{c^2}v^4+ O(v)^6$$ So the relativistic KE is equal to the Newtonian KE to 4th order.

The formula $E=mc^2$ applies for a body at rest. The general formula for a body not at rest is $m^2 c^2 = E^2/c^2 - p^2$. As you can see the formula simplifies to the usual $E=mc^2$ for the special case of $p=0$, and it simplifies to a photon's $E=pc$ for the special case of $m=0$

Answer 4

As others have pointed out, $v$ being very small is not the same as $v = 0$. If $v$ is very small then the kinetic energy (KE = $\frac {1}{2}mv^2$) is indeed very small compared to the energy of the rest mass, $E = mc^2$. That's why a kilogram of uranium in an atom bomb can generate a lot more energy than a kilogram mass kinetic projectile.

Which comes to the question as to how a stationary object $m_0$ can have energy.

Simply put, although an object might be stationary on the outside, there is a lot happening on the inside. The atoms are vibrating due to heat, the electrons in the atoms are in orbits around the nucleus and the electromagnetic field holds them in place. The quarks in the nucleus are similarly not fixed and are held in place by the strong nuclear force. Particles also interact with the Higgs field. All of these things constitute the energy of the object, even though it is not externally in motion.

This energy is the rest mass, $m_0$. Your first equation applies the Lorenz transformation to the rest mass as a moving object. Basically, there we are saying, what does all of this energy look like if viewed from a moving reference frame? It has more energy.

We can make a Taylor series expansion (a mathematical trick) to your formula and it will give us ${m_0}c^2 + \frac{1}{2}{m_0}v^2 $ plus some smaller terms which we can ignore at low velocities.