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If we state that, on one hand, energy is conserved because :

$$\Delta PotentialEnergy+\Delta KineticEnergy=0 \tag{1}$$

And we state on the other hand that:

$$Energy=mc^2 \tag{2}$$

Don't we run into a contradiction? As I understand, $E=mc^2$ doesn't work for potential energy (potential energy doesn't show up as mass). Therefore potential energy isn't strictly speaking energy at least in $E=mc^2$'s context. Therefore we can't say that energy is conserved but rather that the sum of kinetic energy and potential energy is conserved.

Manu de Hanoi
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  • See Binding energy. It is a kind of potential energy and it is released in form of mass defect. –  Apr 02 '20 at 15:30
  • @FakeMod how does it relate to the question ? – Manu de Hanoi Apr 02 '20 at 15:39
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    "$E=mc^2$ doesn't work for potential energy (potential energy doesn't show up as mass). Therefore potential energy isn't strictly speaking energy." I am trying to give you examples which contradict this assertion of yours. –  Apr 02 '20 at 15:41
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    As far as I am aware, potential energy does show up as mass. Can you give a specific example which does not? – Guy Inchbald Apr 02 '20 at 15:47
  • @GuyInchbald I cant give an example – Manu de Hanoi Apr 02 '20 at 16:13
  • Note that just in a classical mechanics sense, when we say "energy is conserved" we in fact do mean "sum of kinetic and potential energy". – BioPhysicist Apr 02 '20 at 16:25
  • @GuyInchbald how would you distribute the mass pertaining to potential energy? Mass belong to an object, potential to a system. Let's take 2 planets close to each other, now we move them apart (adding potential) which planet increases in mass ? Is it both ? Now we hold one planet while we let the other fall on it. Is the planet we're holding loosing mass because the other one is falling ? – Manu de Hanoi Apr 02 '20 at 16:44
  • @ManudeHanoi In $E=mc^2$ the $m$ is the rest mass. It doesn't change. Just like others have already told you, the idea of relativistic mass isn't used anymore, as it brings along its own issues. – BioPhysicist Apr 02 '20 at 16:51
  • potential energy doesn't show up as mass Potential energy does show up as mass. For example, the negative electrostatic potential energy of a proton and electron explains why the mass of a hydrogen atom is less than the sum of the masses of a proton and an electron. – G. Smith Apr 02 '20 at 17:27
  • @G.Smith Ah, I guess I was misinterpreting the OP. I thought they were asking about changes in mass due to relative motion. – BioPhysicist Apr 02 '20 at 17:30
  • In the case of a bound system potential and kinetic energy show up as mass. – my2cts Apr 02 '20 at 17:30
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    @AaronStevens It isn’t clear to me what the OP is asking. I was responding to what the OP had written, not to what you wrote. – G. Smith Apr 02 '20 at 17:31
  • @ManudeHanoi you ask, "how would you distribute the mass pertaining to potential energy?" and suggest, "Mass belong to an object." No. Mass can also belong to something more nebulous like an energy field; energy bends spacetime just as if it were mass according to $e=mc^2$ and it makes no difference if the energy is potential. Moving two planets apart requires positive work from a 3rd source, which exactly equals the reduction in negative potential energy of the gravitational field such that the total system energy and mass do not change. – Guy Inchbald Apr 02 '20 at 17:44
  • @GuyInchbald if your system is the 2 planets and you add energy to the system by pulling the planets apart, then according to your idea that potential energy registers as mass , the 2 planets must increase mass somehow – Manu de Hanoi Apr 02 '20 at 18:18
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    Each planet stays the same, but the 2-planet system effectively gains mass via its gravitational potential, because the equations of Relativity are nonlinear. See https://physics.stackexchange.com/questions/66359/does-potential-energy-in-gravitationall-field-increase-mass – Guy Inchbald Apr 02 '20 at 20:05

2 Answers2

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The formula $E=mc^2$ gives the rest energy of an isolated system. By definition it includes only internal kinetic and potential energy.

my2cts
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  • thanks, how do you define internal potential energy ? – Manu de Hanoi Apr 02 '20 at 18:20
  • if your system is made of 2 planets and you pull them apart thereby increasing the potential energy oft the system, how will each planet increase mass ? Now we hold one planet while we let the other fall on it. Is the planet we're holding loosing mass because the other one is falling ? In other words, potential energy isnt localised but mass is, so where does mass go ? – Manu de Hanoi Apr 02 '20 at 18:28
  • @ManudeHanoi If your system is made of 2 planets and you pull them apart, you increase the total mass of the two-planet system, not the mass of the individual planets. Some energy is stored in the gravitational field between the two planets, and that energy is factored into the total rest energy (i.e. the total mass) of the two-planet system. When you pull the planets apart, the energy contained in the gravitational field increases, and so the total rest energy (and therefore the mass) of the two-planet system increases. – probably_someone Apr 02 '20 at 19:48
  • The concept of mass is specifically useful when the internal degrees of the system are not or negligibly affected by whatever you do to it. So it is less useful to speak of the mass of a two planet system. – my2cts Apr 02 '20 at 19:55
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In the formula $E=mc^2$, $E$ is the rest energy of the object or system and $m$ is its rest mass. The use of the letter $E$ is misleading because it implies it is the total energy of the object, which is in fact the sum of the rest, kinetic and potential energies. This total energy is conserved, even when energy is transferred between rest energy and other energy stores, e.g. in matter-antimatter annihilation.

bemjanim
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  • I believe E=mc2 also registers kinetic energy because kinetic energy will increase the mass , am I wrong ? – Manu de Hanoi Apr 02 '20 at 15:38
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    @ManudeHanoi - Relativistic mass is a terribly outdated concept. The only relevant mass for any object is its rest mass (aka invariant mass) and it does NOT change with speed. $E = m c^2$ holds only at rest. The correct formula for moving objects is $E = \gamma mc^2 = \sqrt{ m^2 c^4 + p^2 c^2}$ where $p = \gamma m v$. – Prahar Apr 02 '20 at 15:42
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    @Manu de Hanoi it depends whether you take m to be the rest mass or relativistic mass. Taking it to be the relativistic mass, $mc^2$ is the sum of the rest energy and the kinetic energy. If taken to be the rest mass, the correct formula is $E=\gamma mc^2 +V$, where $\gamma$ is the Lorentz factor and V is the potential energy. Relativistic mass is a fairly outdated concept so I'd stick with using rest mass. – bemjanim Apr 02 '20 at 15:44
  • to prahar : according to your equation, potential energy doesnt register into E, so my OP stands To bemjanim I googled for your equation but couldnt find references, do you have any ? – Manu de Hanoi Apr 02 '20 at 16:08
  • @Manu de Hanoi the full formula would have a four vector potential rather than a scalar potential as energy transforms as a four vector. The appropriate formula can be found here: https://physics.stackexchange.com/questions/69080/potential-energy-in-special-relativity. See also https://arxiv.org/abs/physics/9803023 – bemjanim Apr 02 '20 at 16:50
  • @benjanim if you look at the comments to my question you'll see many pple say that potential energy registers as mass but I dont see that in your equation – Manu de Hanoi Apr 02 '20 at 18:32
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    Rest mass can be defined as the magnitude of the four momentum, so to find the rest mass of a system sum the four momenta and find the magnitude of the resultant. This includes the contribution from the potential momentum, so the potential energy contributes to the mass. – bemjanim Apr 02 '20 at 19:43
  • @my2cts dont want to – bemjanim Apr 02 '20 at 22:05