# Can we create and destroy energy simultaneously at different points in a space?

Can we create an amount of energy at a point in space and destroy an equal amount of energy at another point in space, with both the processes occurring simultaneously? This will not violate energy conservation, as the total energy in the universe is constant. So is it possible?

• Is it currently possible to do them both independently of each other? Seems like if not, it would be even less possible (is that a thing?) to do them at the same time. – corsiKa Dec 7 '15 at 20:22
• @corsiKa: I don't find that argument very convincing. There are many pairs of things where you can only do one if you simultaneously do the other. (Independent electron annihilation and positron annihilation are not possible; independent magnetic north monopoles and south monopoles are not possible; independent action and reaction are not possible; etc.) – ruakh Dec 8 '15 at 7:09
• No. You are creating energy and destroying energy. Neither of those things are possible. – DeepDeadpool Dec 8 '15 at 21:48
• I know extremely little about physics, but wouldn't creating and destroying energy simultaneously allow us to move energy instantaneously over any distance? Maybe even matter? Wouldn't this be FTL travel and therefore impossible? – 11684 Dec 9 '15 at 15:53

Not if the laws of physics (particularly the laws of gravity) are as we understand them. In general relativity, there are a set of equations, called the Einstein field equations, that relate the curvature of space (roughly speaking, how much gravity there is) to how energy and momentum are distributed in space and time. To be consistent, these equations require that the energy change in any region of space is entirely due to energy flowing into/out of that region. If you try to set up the mathematical equations that say "energy disappears from this region and reappears over here without passing through the points in between", and ask what the gravity field produced by this energy distribution is, you get nonsense; the equations work out to things like 1 = 0. In technical language, general relativity requires that energy be locally conserved (i.e., conserved in every tiny region of space) rather than globally conserved (i.e., conserved in total, but allowing for the energy in small regions to be non-conserved.)

Now, it's possible that our understanding of general relativity is flawed, and maybe there are small-scale violations of this principle. The so-called Steady State theory was a theory that cosmologists came up with '40s and '50s as an alternative to the Big Bang model; and it required that matter (and therefore energy) was continually being created in the Universe. However, this model has now been discredited; and almost everyone who works on alternative models of gravity these days still requires that energy be locally conserved rather than globally conserved.

EDIT #1: For those are interested, here are the technical details. The Einstein field equations are $$G^{\mu \nu} = 8 \pi G T^{\mu \nu}$$ in units where $c = 1$. The object on the left-hand side is the Einstein tensor, which describes the curvature of spacetime; the object on the right is the stress-energy tensor, which describes how energy and momentum are distributed in space and moving through space. We can take the "four-divergence" of both sides (which is just like the regular divergence in 3D vector calculus, only with some time derivatives added in as well.) This is denoted as $$\nabla_\mu G^{\mu \nu} = 8 \pi G \nabla_\mu T^{\mu \nu}.$$ But it is always true that $\nabla_\mu G^{\mu \nu} = 0$; the way this tensor is constructed, its spacetime divergence automatically vanishes (the so-called Bianchi identity for this tensor.) This implies then that $$\nabla_\mu T^{\mu \nu} = 0.$$ which, as @SebastianRiese pointed out in his answer, means that energy (and momentum) are locally conserved rather than globally conserved. This means that if you have a theory where energy is not locally conserved ($\nabla_\mu T^{\mu \nu} \neq 0$), then you will have to modify Einstein's equations, or you get a contradiction.

EDIT #2: As far as virtual particles go, that's a trickier business. We don't yet know how gravity behaves on the quantum scale; and a full description of the situation would include the quantum-mechanical response of the gravitational field to the quantum-mechanical virtual particles. So in some sense, virtual particles are beyond the scope of this answer.

That said, it's possible to define the expectation value of the stress-energy tensor for a quantum-mechanical field, and to try to couple that to the Einstein tensor. This yields what is called the "semi-classical Einstein equation": $$G^{\mu \nu} = 8 \pi G \langle T^{\mu \nu} \rangle$$ This expectation value would include all the energy and momentum of the "virtual particle sea" that one talks about in standard QFT. The question then arises whether $\nabla_\mu \langle T^{\mu \nu} \rangle = 0$, as it must if we're going to write down the above equation. Answering this sort of question is tricky—we're now dealing with QFT in curved spacetime, after all—but it can be shown that we can always define our stress-energy operator such that its expectation value satisfies $\nabla_\mu \langle T^{\mu \nu} \rangle = 0$, along with the other nice properties we want it to have.

If you're really dedicated to the nitty-gritty of this, see Section 4.6 of Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. But be warned that QFT in curved spacetime looks very different than QFT in flat spacetime; in particular, there's no well-defined notion of "virtual particles" at all.

• Pretty sure that this is wrong or I've completely misunderstood the concept of Virtual Particles. – Aron Dec 8 '15 at 1:27
• would you care to show how the equations would produce absurd results? In particular, how would you interpret the condition(s) in question as the input to the equations? – user3209815 Dec 8 '15 at 9:46
• @Aron I think you have, yes. Virtual particles are kind of tricky to understand - but if you do accept them as being real, then they indeed do violate conservation of energy and momentum. That's more likely a problem with the concept of virtual particles, rather than conservation of energy. All in all, I think virtual particles do more harm than good - they violate too many good principles (CoE, CoM, relativities, they can have negative kinetic energy...), and all that just to retain particles where there are none (in the "we're not quite comfortable with the quantum world" sense). – Luaan Dec 8 '15 at 11:54
• @Luaan I get that virtual particles are somewhat misleading, since they can never be observed directly. But it still seems strange that Feynman diagrams with unphysical numbers are allowed to perturbate the integral. – Aron Dec 8 '15 at 14:41
• @Aron: The idea of virtual particles becomes basically useless when we're dealing with curved spacetime (i.e., gravity.) One can still define a stress-energy operator, though, and it can be shown that the expectation value of this stress-energy operator is always conserved. I've updated my answer to address this. – Michael Seifert Dec 8 '15 at 15:15

Can we create an amount of energy at a point in space and destroy an equal amount of energy at another point in space, with both the processes occurring simultaneously? This will not violate energy conservation, as the total energy in the universe is constant. So is it possible?

If you learn about Einstein's Special Relativity, you'll discover that the term "simultaneous" is all relative. Two events which may appear to be simultaneous to you will not be simultaneous according to another observer speeding by you at, say, half the speed of light. So although having the two events you speak of occur "simultaneously" as observed by you may seem to ensure the conservation of energy to you, other observers will see violations in the conservation of energy because they will see energy being destroyed, a time pause, and then energy being created. Or they will see energy suddenly being created - a violation of the conservation of energy - and then a time pause and then the same amount of energy being destroyed.

• Thank you for the answer but I do not want the experimenters at the points to observe each other.The events just occur simultaneously without any observation on each point relative to the other point. – Ananyo Bhattacharya Dec 7 '15 at 16:23
• @Ananyo - No, I'm not talking about experimenters or observers necessarily being at the two points. I'm talking about how the notion of "simultaneity" is all relative. What you think of as "simultaneous" events will not appear to me to be simultaneous if I am moving at some speed with respect to you. That's the nature of space-time according to Special Relativity. – Samuel Weir Dec 7 '15 at 16:32
• To summarize, things can only be simultaneous if they also both occur in the same place. – Ben Voigt Dec 7 '15 at 23:13
• @Taemyr if we expect some law of conservation to hold at all, then we expect it to hold in all reference frames; thus, for the purposes of original question "both the processes occurring simultaneously [...] This will not violate energy conservation", it must also happen in the same place, otherwise some (most) observers would clearly see energy being temporarily created or removed. – Peteris Dec 8 '15 at 10:27
• I think this answer should be updated to make the conclusion reached in the comments. It isn't enough to say simultaneous is relative. You need to explain that this implies that to guarantee the simultaneity, any changes must be at the same point, and that the conservation of energy requires the creation and destruction to be simultaneous. – Oddthinking Dec 8 '15 at 11:07

I think one simplifying way of thinking about this idea is in terms of causality. That is, here we see a case where two things must be intrinsically linked. The energy being destroyed and the energy being created must have some mediator and they must always at the same time.

The problem, however, is that if they must be causally related, there must be some way for the point where matter is destroyed to know when the other matter is being created. The fastest that information can travel is at the speed of light, but in this case there isn't even time for light to communicate the information, so it simply cannot happen.

As to the idea that energy could be created and destroyed simultaneously in the same location, and this be simultaneous to all observers, I don't know how to say that couldn't be true besides to say that there would be no observable change as far as I can imagine.

I hope that answer is at least a worthwhile supplement to the other correct answers.

• This is a good way to look at it. It shows you cannot create energy at one point and destroy it at another point that are timelike separated aside from the Heisenberg uncertainty relation. It doesn't deal with spacelike separation. – Ross Millikan Dec 8 '15 at 4:59

In classical theories the answer is no, in quantum theories the answer is also no, but the interpretation is more subtle (as there is a energy uncertainty relation), but basically the same arguments apply (that there is a local formulation of energy conservation).

All accepted fundamental classical theories do not only have global conservation of energy, but local conservation of energy. That means, one can derive a continuity equation $$\partial_t \rho + \nabla \cdot \vec j = 0,$$ where $\rho$ is the energy density and $\vec j$ is the energy density current.

From this equation we get global energy conservation by using the Gauss theorem (for a volume $V$): $$E = \int_V d^3r\,\rho$$ and therefore, $$\partial_t E = \int_V d^3r\,\partial_t \rho = -\int_V d^3r\,\nabla \cdot \vec j = -\int_{\partial V} d\vec\Sigma \cdot \vec j.$$ That is, the change of energy in any volume $V$ is given by the flux of energy through the volume's surface, therefore total energy is conserved if the fields vanish at infinity (or the volume is a compact manifold).

In electrodynamics such an equation is known as Poynting's theorem (with the energy density of the electro-magnetic field $u = \frac 1 2 \left(\varepsilon_0 \vec E^2 + \frac 1 {\mu_0} \vec B^2\right)$ and the Poynting vector $\vec S = \frac 1 {\mu_0} \vec E \times \vec B$): $$\partial_t u + \nabla \cdot \vec S = -\vec j \cdot \vec E,$$ where the right-hand side is the work done on mechanical systems (which conserve energy up to work done by external forces), so total energy is again conserved locally.

In general relativity this takes the form $$T^{\mu\nu}_{;\mu} = 0,$$ where $T^{\mu\nu}$ is the energy-momentum tensor. These are four equations of the form given above, describing momentum-density and energy conservation (these quantities are coupled, because $(E, \vec p)$ form a four-vector, and therefore transform to one another, there are some subtleties as the metric field carries energy as well, that is not encoded in the energy-momentum tensor, but in the covariant divergence).

At the risk of contradicting everybody else, I'd say the answer is Yes.

Other answerers use general relativity as their authority, but people who use quantum mechanics (particle physics) see the world differently. In that world, an electron continually shakes off (virtual) photons, each of which takes away or adds a little energy. Usually, they recombine with the electron quickly. However, this gives rise to a continual jitter or zitterbewegung allowed by Heisenberg uncertainty.

In the rest frame of the electron, this jitter is clearly a violation of classical conservation. In particular, since the recombination happens at a different time and place, it is a violation of locality.

When one electron scatters off of another, the photon starts at one electron and ends at the other, so you have energy disappearing at one place and appearing at another. Are these events simultaneous?

As one person mentioned, simultaneity is a bad concept in modern physics, so your thinking that the two events happen simultaneously is a misunderstanding. The answer depends on the frame of reference.

## protected by Qmechanic♦Dec 8 '15 at 1:36

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