# Can energy be defined as the most fundamental particle which exists in different forms as protons, electrons etc

What we see all the different forms of matter around is just a form of energy. Why can't this energy be the most fundamental particle. Given dust can turn into a star and then emit all sorts of radiation loosing energy over a period of time either collapse to worst case of black hole or burst as a supernova. The dust and gas would again collapse again at some point of time and start the cycle all over again. Why can't energy be then the fundamental particle?

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Have a look at the standard model of particle physics en.wikipedia.org/wiki/Standard_Model . all the entries in the table are "particles" the fundamental building blocks of nature. They interact with each other according to the energies of the system and the dynamical equations. All of them have energy, both as rest mass, and possibly kinetic, if in motion. –  anna v Feb 15 at 7:26

Students have a tendancy to think of the energy as a thing in it's own right - possibly as a result of watching too many episodes of Star Trek. As various people have commented, energy is a property of a system not an object in it's own right.

You can see this very easily. Take two objects of mass M coming towards each other at speed $v$. What is the kinetic energy of the system? If you're in the centre of mass frame you see two objects both moving at speed $v$, so the total kinetic energy is $mv^2$. But suppose you're sitting on one of the objects. Now you see your object has zero KE because it's stationary while the other object is moving at $2v$, so the total KE is $2mv^2$, which is different from the previous result.

So energy is frame dependant unlike for example the number of particles, which is frame independant (we'll gloss over the Unruh effect here!).

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Very clean demonstration –  Michael Brown Feb 15 at 8:00
One object of mass $m$ moving at speed $v$ has KE $0.5mv^2$. Two such objects have KE $0.5mv^2 + 0.5mv^2 = mv^2$. In the second frame KE = $0.5m(2v^2) = 2mv^2$. –  John Rennie Feb 15 at 8:14
Think you mean $\frac{1}{2}m(2v)^2$ in the second frame –  Michael Brown Feb 15 at 8:19
@MichaelBrown: oops, yes, typo! –  John Rennie Feb 15 at 8:24