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Einstein mass- energy relation states $E=mc^2$. It means if energy of a paricle increases then mass also increases or vice-versa. My question is that what is the actual meaning of the statement "mass increases"? Is really the mass of the particle increasing or what?

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Originally Einstein derived his relationship $\Delta E=\Delta mc^2$ for macroscopic bodies capable of absorbing and radiating electromagnetic waves. So he meant literally the body mass change. This formula is not applicable to "elementary" particles incapable to absorb/emit other species. –  Vladimir Kalitvianski Nov 10 '12 at 19:14
    
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The rest mass of an object is, by definition, independent of the energy. But all other forms of mass are indeed increasing with the energy, as $E=mc^2$. With the relativistic interpretation of the kinetic energy, the total mass is $$ m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$ Here, $m_0$ is the mass measured at rest, i.e. the rest mass. The corrected, total mass goes to infinity if $v\to c$ and it holds for the following interpretations of the mass:

  • the inertial mass, i.e. the resistance towards the acceleration, increases. For example, the protons at the LHC have mass about 4,000 times larger than the rest mass (the energy is 4 TeV), and that's the reason why it's so hard to accelerate them above their speed of 99.9999% of the speed of light and e.g. surpass the speed of light. It's impossible to surpass it because the object is increasingly heavy, as measured by the inertial mass

  • the conserved mass. If you believe that the total mass of all things is conserved, it's true but only if you interpret the "total mass" as the "total energy over $c^2$". In this conserved quantity, the fast objects near the speed of light indeed contribute much more than their rest mass. If you considered the total rest mass of objects, it wouldn't be conserved

  • the gravitational mass that enters Newton's force $Gm_1m_2/r^2$. If an object is moving back and forth, by a speed close to the speed of light, it produces a stronger gravitational field than the same object at rest. For example, if you fill a box with mirrors by lots of photons that carry some huge energy and therefore "total mass" $m=E/c^2$, they will increase the gravitational field of the box even though their rest mass is zero. Be careful, in general relativity, the pressure from the photons (or something else) creates a gravitational field (its independent component curved in a different way), too.

Despite this Yes Yes Yes answer to the question whether the total mass indeed increases, Crazy Buddy is totally right that especially particle physicists tend to reserve the term "mass" for the "rest mass" and they always prefer the word "energy" for the "total mass" times $c^2$.

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The correct expression is

$$E=\frac{mc^2}{\sqrt{1 - {v^2}/{c^2}}}$$

where $v$ is the velocity of the particle. The mass $m$ of the particle is a constant and cannot increase [*]. The energy of a particle increases with its velocity. For a particle at rest $v=0$ and $E=mc^2$.

[*] This is the modern concept of mass used in physics. For instance the mass of an electron is $m_\mathrm{e} = 9.10938188 \times 10^{-31}$ kg. Some answers here mention the old "relativistic mass" concept $m_\mathrm{rel} \equiv {m}/{\sqrt{1-v^2/c^2}}$ and mention a supposed "rest mass". The term "rest mass" is a misnomer, for instance $m_\mathrm{e}$ is not the mass of an electron at rest and this term is still more misleading for particles such as photons (which are never at rest). For additional criticism see the recent Okun paper Mass versus relativistic and rest masses

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That's not a relation, that's more like the equivalence between mass and energy. The main concept under this formula is that mass is a property of any energy and energy is the property of any mass, connected with a constant - $c^2$.

It means that if you have a body at rest with a mass of $m$, it would have energy $E=mc^2$.

For example, the rest mass of electron is $m_e=0.511\mathrm{MeV}/c^2$. And when electron and positron annihilates the photons with total energy of $E = 1.022\mathrm{MeV}$ is radiated.

So, about the question: yes, heavier elementary particles would have more energy under annihilation. But you can't make electron with bigger mass. And, btw!, this energy is hard to utilize due to energy conservation law.

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I've pointed out to rest state of mass and provide a small real-life example –  m0nhawk Nov 10 '12 at 17:59
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Yes. You're somewhat (not whole part) right. But, The actual terminology of mass-energy equivalence is that, It gives the energy associated with a "massive" particle or the total energy content obtainable from the system.

But Nowadays, relativistic mass $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ isn't considered as a good term by Physicists. Because, they're eager & more satisfied with the true rest mass $m_0$ and rest energy $m_0c^2$ they actually measure (with their goggles) rather than the apparent relativistic mass.

Moreover, there are satisfactory explanations that the rest mass does not vary with an object's inertial motion relative to the observer's frame of reference.

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