While deriving potential energy stored in space due to two stationary opposite charges we end up with negative value of energy which upon on dividing by '$c$ square' provides us negative value of mass. What is the significance of this mass other than reducing the total mass of system of charges. ($c$ stands for speed of light in vacuum).
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$\begingroup$ If the mass is stored in space then what is happening with the space containing negative mass? $\endgroup$– Vasu GoyalCommented Jan 3, 2017 at 15:22
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2$\begingroup$ Possible duplicates: physics.stackexchange.com/q/18925/2451 , physics.stackexchange.com/q/34115/2451 , physics.stackexchange.com/q/44934/2451 , and links therein. $\endgroup$– Qmechanic ♦Commented Jan 3, 2017 at 15:22
1 Answer
Let's call the two particles a proton and electron for convenience.
Suppose we start with the proton and electron at rest at some distance $r$ apart. Then to separate the particles to infinity we have to do some work on them. This work is, of course, just (minus) the potential energy $U(r)$.
If we now measure the total mass of the separated particles we find it is just the mass of the proton plus the mass of the electron:
$$ M_\infty = m_p + m_e $$
So far so good, but hang on we had to put in an energy $U(r)$ to separate the two particles and that corresponds to a mass $m = U/c^2$. So if the final mass is $M_\infty$ that must mean that the initial mass was lower:
$$\begin{align} M_r &= M_\infty - U/c^2 \\ &= m_p + m_e - U/c^2 \end{align}$$
And this is quite correct. What you've discovered is that the mass of a bound state is always less that the total mass of its component parts. The difference is the binding energy.
But this doesn't mean there is a negative mass hiding somewhere. It just means that for composite systems mass is a more complicated concept than you thought. The mass of a composite system is not located at any specific point or points - it is a property of the system as a whole.
If you want to pursue this further there are aleady a lot of questions and answers dealing with it.