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As kinetic energy is the "energy of motion", it cannot exist without matter. However, the energy in an E/M wave can. I understand that E/M waves have been described in structure as electrical and magnetic waves in a sinusoidal configuration. This means that kinetic energy cannot be made of EM waves or it would be mass-independent.

So my question is, what is kinetic energy made of?

Emilio Pisanty
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kettlecrab
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  • Possible duplicates: http://physics.stackexchange.com/q/3014/2451 and links therein. – Qmechanic Feb 21 '14 at 03:37
  • @Qmechanic This question is far from being a "duplicate" of the question you linked it to; it would make more sense to say that it could be answered by it. It is a very interesting topic, I must say, but the answer I was looking for was more concrete, (like if kinetic energy is made of subatomic particles or not). – kettlecrab Feb 21 '14 at 03:51
  • @Stopforgettingmyaccounts... Kinetic energy is made from work. It's not made of anything. There's a good discussion of the difference between the phrases 'made of' and 'made from' over at the English StackExchange. – David H Feb 21 '14 at 04:03
  • @DavidH How can you have nothing (kinetic energy) made out of something? It seems like a ridiculous notion to me. If indeed kinetic energy is 'nothing', then it wouldn't be important enough to mention. I believe there must be some sort of substance, or at least energy in the vague sense, to constitute its existence. – kettlecrab Feb 21 '14 at 04:20
  • @Stopforgettingmyaccounts... Energy has no substance because energy is a property of substance. Energy is a quantitative property of physical systems that describes a system's state. As if you are trying to build numbers out of rocks every time you need to count something. Suppose you have one rock. The 'one' is merely description of how many rocks you have; the number 'one' is not itself made of rocks. – David H Feb 21 '14 at 05:23
  • @Stopforgettingmyaccounts...Indeed if numbers were made of rocks, you run into a very big problem. Say you needed to count 5 rocks. Well first you would need to build the '5' that you need in order to perform the count. So go grab five more rocks to make the '5'. In order to grab the right number of rocks, you need to count those additional 5 rocks. And to perform that count, you need to go grab five more rocks! The paradox results if you try to ascribe a material component to energy. – David H Feb 21 '14 at 05:27
  • Are you saying that energy is a just a number? If that's the case, then there would be absolutely no physical counterpart whatsoever. How would scientists even theorize the existence of an EM wave? – kettlecrab Feb 21 '14 at 05:32
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    @Stopforgettingmyaccounts... Indeed, energy has absolutely no physical component. If it did have one, then its component would have its own energy, and the energy of the material part of energy would have its own material part. And the material part of energy's energy would have energy too. So on and so forth, ad infinitum, ad nauseum. But it doesn't, which is why no one's ever succeeded in making a perpetual motion machine. – David H Feb 21 '14 at 07:40
  • "Energy is liberated matter, matter is energy waiting to happen.” ― Bill Bryson, – Sensebe Feb 21 '14 at 19:00

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Energy is an extraordinarily elusive concept, so much so that experienced physicists will use the term only with care and in well defined circumstances.

For example it might seems obvious that if I fire a pistol then the bullet has kinetic energy. But Einstein's theory of special relativity tells us that all inertial frames are equally valid, and in the bullet's frame the bullet is stationary and it's me and the pistol that are moving. So in this frame where did the kinetic energy of the bullet go and where did my and the pistol's kinetic energy come from?

If you say the bullet has a kinetic energy of $\frac{1}{2} m v^2$ what this really means is that to make the bullet move at a velocity $v$ I had to supply an energy of $\frac{1}{2} m v^2$. If I have lost an energy of $\frac{1}{2} m v^2$ then conservation of energy means the bullet has gained an energy of $\frac{1}{2} m v^2$, and this is what I measure as it's kinetic energy.

This is the way to think of kinetic energy. If some observer, like you or me, measures a kinetic energy $E$ this means we would have to put an energy $E$ into making the object move. And this carries straight over into measuring the energy carried by light. If I create a photon then this costs me energy, and that energy goes into the photon. If/when the photon hits something the energy is then transferred to whatever it hits. So the photon carries energy just like a bullet does, even though the photon has zero mass.

Now you may say this all sounds a bit contrived, because what I've just described is the total energy of the photon and not its kinetic energy, and in fact you'd be absolutely correct. This is because kinetic energy is a low energy approximation and the concept doesn't really exist in relativity. In relativity we only talk about the total energy of an object, and the total energy is given by the equation:

$$ E^2 = p^2 c^2 + m^2 c^4 $$

where $m$ is the object's mass and $p$ is its momentum. The first term, $p^2c^2$ is sort of the kinetic energy and the second term, $m^2c^4$, is sort of the rest energy, though thi separation isn't clear for fast moving particles. Anyhow, for a photon the mass, $m$, is zero so the expression simplifies to:

$$ E^2 = p^2 c^2 $$

And this could be regarded as the kinetic energy because it's the same term that is considered the kinetic energy for a massive object.

John Rennie
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