# Exotic matter

In physics, exotic matter is matter that somehow deviates from normal matter and has "exotic" properties. A broader definition of exotic matter is any kind of non-baryonic matter—that is not made of baryons, the subatomic particles (such as protons and neutrons) of which ordinary matter is composed.[1] Exotic mass has been considered a colloquial term for matters such as dark matter, negative mass, or complex mass.[2][3][4][5][6][7]

## Types

There are several proposed types of exotic matter:

## Negative mass

Negative mass would possess some strange properties, such as accelerating in the direction opposite of applied force. Despite being inconsistent with the expected behavior of "normal" matter, negative mass is mathematically consistent and introduces no violation of conservation of momentum or energy. It is used in certain speculative theories, such as on the construction of artificial wormholes and the Alcubierre drive. The closest known real representative of such exotic matter is the region of pseudo-negative-pressure density produced by the Casimir effect.

According to mass–energy equivalence, mass ${\displaystyle m}$ is in proportion to energy ${\displaystyle E}$ and the coefficient of proportionality is ${\displaystyle c^{2}}$. Actually, ${\displaystyle m}$ is still equivalent to ${\displaystyle E}$ although the coefficient is another constant[8] such as ${\displaystyle -c^{2}}$.[9] In this case, it is unnecessary to introduce a negative energy because the mass can be negative although the energy is positive. That is to say,

${\displaystyle E=-mc^{2}>0}$
${\displaystyle m=-{\frac {E}{c^{2}}}<0}$

Under the circumstances，

${\displaystyle dE=Fds={\frac {dp}{dt}}ds={\frac {ds}{dt}}dp=vdp=vd(mv)}$
${\displaystyle -c^{2}dm=vd(mv)}$
${\displaystyle -c^{2}(2m)dm=2mvd(mv)}$
${\displaystyle -c^{2}d(m^{2})=d(m^{2}v^{2})}$
${\displaystyle -m^{2}c^{2}=m^{2}v^{2}+C}$

When ${\displaystyle v=0}$,

${\displaystyle C=-m_{0}^{2}c^{2}}$

Consequently,

${\displaystyle -m^{2}c^{2}=m^{2}v^{2}-m_{0}^{2}c^{2}}$
${\displaystyle m={m_{0} \over {\sqrt {1+\displaystyle {v^{2} \over c^{2}}}}}}$

where ${\displaystyle m_{0}<0}$ is invariant mass and invariant energy equals ${\displaystyle E_{0}=-m_{0}c^{2}>0}$. The squared mass is still positive and the particle can be stable.

Since