# Are fundamental forces conservative?

I wonder whether fundamental forces are conservative.

First of all, I'm not sure if we can talk about conservative forces, since to study electromagnetism, weak and strong interactions we need QFT.

For gravity, I'd that it's not conservative, because energy conservation doesn't even apply.

• Wikipedia: "Gravity is an example of a conservative force". And "However, general relativity is non-conservative, as seen in the anomalous precession of Mercury's orbit. However, general relativity can be shown to conserve a stress-energy-momentum pseudotensor." – Keep these mind Jun 19 '13 at 22:26
• @Gugg Just classically. – jinawee Jun 19 '13 at 22:27
• The force of a magnetic field on a moving charge (the Lorentz force) is not conservative as is it not a force field. – fffred Jun 19 '13 at 22:43
• @fffred What about $F$ = $IL$ x $B$? The Lorentz force is not conservative too? And why would it be? – Pupil May 22 '14 at 4:59
• @Key Lorentz force is not conservative neither non-conservative. – jinawee May 22 '14 at 5:26

The notion of "conservative forces" is not in any way fundamental. What's fundamental is that we're able to assign a number to the state of a system, and that number is conserved.

The relatively uninteresting notion of a "conservative force" can be applied only to a force that can be expressed as a vector field that depends only on position. That means it's meaningful for Newtonian gravity and for electrostatics, but not for any other force that could be considered fundamental. Re the nuclear forces, see Do strong and weak interactions have classical force fields as their limits? .

General relativity has local conservation of energy-momentum, which is expressed by the fact that the stress-energy tensor has a zero divergence. A mass-energy scalar or energy-momentum vector isn't something that can be defined globally in GR for an arbitrary spacetime.

WP says (Gugg, where was the link from?):

However, general relativity is non-conservative, as seen in the anomalous precession of Mercury's orbit. However, general relativity can be shown to conserve a stress-energy-momentum pseudotensor.

The first sentence is wrong, because Mercury's anomalous precession can be described in terms of a test particle moving in a Schwarzschild metric. The Schwarzschild metric has a timelike Killing vector, so there is a conserved energy-momentum vector for test particles.

The second sentence is also misleading, since it doesn't make the global/local distinction. What's conserved locally isn't a pseudotensor, it's a tensor (the energy-momentum vector). Globally, there are various pseudotensors that can be defined, and the fact that they're pseudotensors rather than tensors means that they're fundamentally not well-defined quantities -- they require some specially chosen system of coordinates.

• FYI: The quoted Wikipedia passage is the last paragraph of this entry. (And this entry states: "Some people object [...] on the grounds that pseudotensors are inappropriate objects in general relativity, but the conservation law only requires the use of the 4-divergence of a pseudotensor which is, in this case, a tensor (which also vanishes). Also, most pseudotensors are sections of jet bundles, which are perfectly valid objects in GR.") – Keep these mind Jun 20 '13 at 5:38

In a 1963 lecture by Dr. Feynman, which Caltech republished, the statement is made that "All fundamental forces in nature appear to be conservative." This statement was made while developing his argument that "there are no nonconservative forces."

Here is a link to the lecture. http://www.feynmanlectures.caltech.edu/I_14.html