Can I simply claim that, according to the mass-energy equation $E=mc^2$, whenever the energy of an physical object (not necessarily a microcosmic one) changes, its mass also change?

Okay, I noticed that when the external field change, the potential energy of the object will change but its mass remain the same. But with what kind of energy change can we observe a mass change?

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    $\begingroup$ What do you mean by "energy of an physical object"? I think that's to unspecific to give an answer. The mass of an object does not change when its potential energy in a gravitational field is changed for example. $\endgroup$ – Chris Feb 16 '15 at 17:00
  • $\begingroup$ "Does the Inertia of a Body Depend upon its Energy-Content?" Yes. $\endgroup$ – Robin Ekman Feb 16 '15 at 17:00
  • $\begingroup$ @Robin Ekman :Does inertia of a body depend on its energy content? But the right sentence should be ,''Does inertia of a body depends on its rest energy''? $\endgroup$ – Paul Feb 28 '15 at 2:55
  • $\begingroup$ @Paul "Does the Inertia of a Body Depend upon its Energy-Content?" is the title of a famous paper by a famous physicist. $\endgroup$ – Robin Ekman Feb 28 '15 at 3:28
  • $\begingroup$ Yes,I know by Einstein. But he really meant what I said. $\endgroup$ – Paul Feb 28 '15 at 3:30

Certainly not. You have extracted only part of the full equation, which actually looks Pythagorean in structure:

$E^2 = (mc^2)^2 + (pc)^2$

This relates the energy of an object to its mass and momentum. Its more famous cousin, $E = m c^2$ is simply the limit where $p=0$, or the energy of an object in a reference frame in which it is at rest. On the other hand, if you use a Taylor series of the full equation to approximate E in the limit where $p$ is non-zero but $pc$ is small compared to $mc^2$, after a little rearranging (a worthwhile exercise) you will get an equation with another familiar component, $E = mc^2 + \frac{1}{2}mv^2$, where the second term is the Newtonian kinetic energy of a particle at low speeds with respect to $c$. And, if you consider yet another limit of the full one and consider the case where a particle has no mass, you get $E = pc$, which is the relationship between the energy and momentum of a photon.

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