It is usually stated in various books or stories 'Energy can't be created or destroyed , but is conserved'. I guess this is within a closed system. Are there any axioms or principles in physics that would be contradicted if certain types of energy could be destroyed or changed into some form that is 'unusable' by any closed or open system? Are the conservation laws axioms? Saying the principles of energy conservation are valid approximations given what is presently known is only a partial answer to this question. I'm wondering if any principles are known now that would determine energy can not be destroyed in many situations not 'covered' by such approximations. Could conservation laws be formulated 'a priori' or without empirical evidence?

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    $\begingroup$ Repeat after me: There are no axioms in physics! $\endgroup$ – CuriousOne Dec 28 '14 at 4:42
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    $\begingroup$ Why, may I ask, not? $\endgroup$ – oink Dec 28 '14 at 4:51
  • $\begingroup$ @CuriousOne There really are. I don't know what definition of axiom you are using, but Newton's Laws in Classical Mechanics and Schroedinger's Equation in QM are both widely considered to be axioms of their respective fields. Are they supported by experimental fact? Sure. But just because $A\rightarrow B$ and $B$ does not mean that we can therefore deduce $A$. We assume that they are true and make use of them because they seem to work. That's exactly what axioms are. Granted, you could pick another starting point for QM than the SWE but that doesn't change the fact that you started somewhere. $\endgroup$ – Geoffrey Dec 28 '14 at 6:05
  • $\begingroup$ @Geoffrey: Newton's laws are approximations based on observations. We know that they are not true, but they are good enough to fly to Mars with them and get within a dozen miles of the intended landing spot. In an atom, however, they are 100% false and you can't even run your GPS on them. Schroedinger's equation describes Hydrogen OK, but it already fails to describe Helium correctly. It's not even a fundamental equation of quantum mechanics, merely one model for the non-relativistic single particle case. Neither of these is useful if you don't also know how they fail and when. $\endgroup$ – CuriousOne Dec 28 '14 at 6:13
  • $\begingroup$ @CuriousOne Perhaps we can continue this discussion in chat at a later time, but for the moment, I'll say this: whenever you make a calculation (like figuring out how to fly a rocket to Mars), you need to make some assumptions. These assumptions are the ground on which you stand, your starting point. Are these assumptions approximations to reality? Yes. Will we ever have a theory that isn't just an approximation? I doubt it. But when you tackle a real-world problem, you simplify it because the real world is scary and we don't understand it. You live in an imaginary world built on axioms. $\endgroup$ – Geoffrey Dec 28 '14 at 6:25

Noether's Theorem states that every conservation law must correspond with a symmetry; if one goes through all the math it turns out that energy conservation is related to time symmetry of physical laws (more precisely, the Lagrangian).

So we're relying on the fact that the laws of physics are symmetric with respect to time. Since we've never observed the laws of physics changing with respect to anything at all, we can pretty safely assume that energy is conserved.

Both of these are just empirically observed things, but they're very well-grounded.

I'll refer you to the wikipedia article for details, because it does it better than I can. https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:_Conservation_of_energy

(fun fact: if the laws of physics are symmetric with respect to position, too, then momentum is conserved. In X, Y, and Z directions, respectively.)

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    $\begingroup$ This is something that I see frequently, but I find it a bit disingenuous as presented. Conservation of energy doesn't merely follow from the fact that the laws of physics should have time symmetry, it must further be imposed that the laws of physics come from an action principle. This is highly nontrivial and one needs to know the laws in order to actually verify that this is true. However given Newton's laws it is easily checked that energy is conserved and one needs not invoke Noether's theorem. $\endgroup$ – JLA Dec 29 '14 at 6:27
  • $\begingroup$ @JLA Can you provide a link to stuff about the action principle thing? I've never thought of it before but it makes sense that one needs to first assume that physics actually fits the Lagrangian... how do you even determine that? (I want to know more!) $\endgroup$ – oink Jan 1 '15 at 7:33
  • $\begingroup$ You can look up the variational complex, or look at Peter Olver's book "Applicatons of Lie Groups to Differential Equations". It's similar to how you determine if a function is the gradient of a scalar actually. I've written a bit about it, but it's not the best exposition. $\endgroup$ – JLA Jan 3 '15 at 8:01

This is completely false. Energy can both be destroyed and created. For example, if two gamma rays (photons) interact with a nucleus, they form one electron and one positron, i.e. $\gamma + \gamma \Rightarrow e^- + e^+$. This is also known as pair production. In contrast, if a particle meets its antiparticle they get "converted" into a force carrier particle, such as a gluon, W/Z force carrier particle, or a photon.

The simplest reaction would be if one electron met its antiparticle, a positron. The reaction, as you might guess, ends up in two photons, more specifically in the range of gamma rays. $e^- + e^+ \Rightarrow \gamma + \gamma$. We call this particle annihilation.

But… the total energy and the energy equivalence of the mass in a given system is said as far as we know always constant. In order words, $E + \sum\limits_{k=1}^n \sqrt{{m_k}^2c^4 + \mathbf{p}_k c^2} = \text{constant}$, where $E$ is the total amount of "direct" energy (chemical, electric, radiant, etc.) in the system , $m_k$ the mass of any particle with mass, $c$ the speed of light in vacuum and $\mathbf{p}_k$ the momentum ($\mathbf{p} = \frac{mv}{\sqrt{1-{(\frac{v}{c})}^2}}$) for any particle with mass.

(For $\mathbf{p} = 0$ this long mess above simply becomes the famous $E + \sum\limits_{k=1}^n m_k c^2 = \text{constant}$ from $E = mc^2$)

Of course, we can't be sure of that either. It could be the case that energy can enter and leave from and to other Universes, or that quantum mechanics allows for energy to change within a system.

EDIT: As some already mentioned, quantum mechanics can in fact temporarily violate this principle. In Newtonian mechanics though, the law of conservations is a result of Newton's Laws of Motion. I've found Wikipedia's proof simple and easy to understand:

Suppose, for example, that two particles interact. Because of the third law, the forces between them are equal and opposite. If the particles are numbered 1 and 2, the second law states that $F_1 = \frac{\Delta \mathbf{p}_1}{\Delta t}$ and $F_2 = \frac{\Delta \mathbf{p}_2}{\Delta t}$. Therefore $\frac{\Delta \mathbf{p}_1}{\Delta t} = - \frac{\Delta \mathbf{p}_2}{\Delta t},$ or $\frac{\Delta}{\Delta t} \left(\mathbf{p}_1+ \mathbf{p}_2\right)= 0.$

If the velocities of the particles are $u_1$ and $u_2$ before the interaction, and afterwards they are $v_1$ and $v_2$, then $m_1 u_{1} + m_2 u_{2} = m_1 v_{1} + m_2 v_{2}.$

This law holds no matter how complicated the force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds up to zero, so the total change in momentum is zero. This conservation law applies to all interactions, including collisions and separations caused by explosive forces.

Source for quote above: Wikipedia

  • $\begingroup$ A change in the total mass of a system corresponds to a change in the total energy of that system. The "$m_k$" you're using is just the rest mass of each particle, but this also corresponds to some energy. And the "$E$" you're using also translates to some inertia and gravity, so it is mass. $\endgroup$ – Wood Dec 28 '14 at 5:20
  • $\begingroup$ @Wood Except that it isn't just the rest mass. The rest mass is given by $mc^2$, not $\sqrt{m^2c^2 + \mathbf{p} c^2}$ which is the total relativistic mass. $\endgroup$ – Madde Anerson Dec 28 '14 at 14:24
  • $\begingroup$ But I want to correct me here anyway. The momentum in GR is not given by $\mathbf{p} = mv$, but $\mathbf{p} = mv \cdot \gamma$, where $\gamma$ is the Lorentzfactor $\gamma = \frac{1}{\sqrt{1-{(\frac{v}{c})}^2}}$. $\endgroup$ – Madde Anerson Dec 28 '14 at 14:27
  • $\begingroup$ But that's exactly what you wrote: $E + \sum\limits_{k=1}^n \sqrt{{m_k}^2c^4 + \mathbf{p}_k c^2} = \text{constant}$. Here, $m_k$ is the rest mass. Now if you convert the relativistic mass into energy, the total energy is constant. But the important thing here is that the $E$ term also contributes to the inertia and gravity of your system. So it is mass. If you take that into account, the total mass is also conserved (and proportional to the total energy by $E=mc^2$). $\endgroup$ – Wood Dec 28 '14 at 15:37
  • $\begingroup$ @Wood - What you say is true of a bound system, but is there any meaningful way to define the "inertia" of an unbound system consisting of a number of particles flying off in different directions? $\endgroup$ – Hypnosifl Dec 29 '14 at 5:06

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