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February 29

Gravitational potential energy

Our article gravitational energy, asserts in the first chapter:

For two pairwise interacting point particles, the gravitational potential energy $U$ is given by $U=-{\frac {GMm}{R}},$

without making any distinction between Classical mechanics and Relativistic mechanics.

I wonder if this definition leads to a vicious circle in Relativistic mechanics, since: According to the formula $E=mc^{2},$ the whole mass $m$ of a given body already contains also the body's gravitational potential energy $U,$ defined above by the whole mass $m,$ whose content contains also $U...,$ so isn't it a vicious circle?

Apparently, it's like when someone mentioning A - who is asked "what do you mean by A" - defines A by B, but when asked "what do you mean by B" - they define B by A. In our case, the whole mass $m$ depends (according to the formula $E=mc^{2})$ also on the energy $U,$ depending (according to the formula $U=-{\frac {GMm}{R}}$) also on $m,$ right? Unless I miss something here.

However, if our article (in the first chapter) did have to make a distinction between Classical mechanics and Relativistic mechanics, then how does the latter avoid the vicious circle mentioned above? HOTmag (talk) 08:42, 29 February 2024 (UTC)

See Mass in general relativity. Graeme Bartlett (talk) 10:55, 29 February 2024 (UTC)

Thank you for the link.

Having read it, let me put my original question this way:

1. Can General Relativity give a clear cut answer to any/both of the following questions?

A. Does a given body's gravitational potential energy contribute to the body's total mass?

B. Does a given body's gravitational potential energy depend on the body's total mass?

HOTmag (talk) 12:13, 29 February 2024 (UTC)

I think the answers are yes and yes. This is what makes General Relativity nonlinear and very hard. This is related to the question whether gravitational waves could exist in GR — Einstein showed that linear waves could exist in the weak-gravity limit (that's easy, and in this regime your questions are quantitatively irrelevant), but had doubts himself whether waves could actually be generated in the strong-gravity regime. This question was only settled in the 1960s. --Wrongfilter (talk) 12:17, 29 February 2024 (UTC)

If (as you say) the answers are yes and yes, i.e. a given body's gravitational potential energy, both contributes to the body's total mass, and also depends on the body's total mass, so it follows logically that the gravitational potential energy, both contributes to itself (it being a part of the total mass), and also depends on itself (for the same reason). Isn't it like Baron Münchhausen who saved himself from drowning by pulling up on his own hair? In other words, isn't it a vicious circle? All of that reminds me of the algebraic equation: x=x+1...

Indeed, sometimes defining an object by itself does not lead to any vicious circle, e.g. in the algebraic equation: 2x+1=x+2, but that's only because we can prove that this equation has a solution (actually a unique one). However, how can we be sure that the interdependence between the total mass and the gravitational potential energy leads to no contradiction, as opposed to the case of the interdependence of the sides of the equation x=x+1? In both cases, an object depends on itself !

This is my wonder from a logical viewpoint. But, let's put logic aside, and get back to physics. My practical question is: Can General relativity describe the gravitational energy, in such a way, that this gravitational energy will only depend on the mass's components other than its gravitational potential energy component? Just as we can do the same in the algebraic equation: 2x+1=x+2, by replacing it by a direct equation x=1, i.e. so that x will only depend on a pure number. For simplicity, let's focus on the two object case, assuming that the whole universe only contains two electrons alone (or any two uncharged objects alone). HOTmag (talk) 13:15, 29 February 2024 (UTC)

I've removed the equation from the title of this question, this seems to cause some problems and weird behaviour of the system. --Wrongfilter (talk) 12:19, 29 February 2024 (UTC)

Gravitational potential energy is not associated with just some of the several bodies involved; it is a quantity associated with a system of bodies in gravitational interaction. Inasmuch as one may want to consider an increase in gravitational energy as an increase in mass, it is the mass of the whole multi-body system.  --Lambiam 11:09, 1 March 2024 (UTC)

Of course, but I don't see how all of that has anything to do with how General relativity avoids the vicious circle I asked about. 15:09, 1 March 2024 (UTC) HOTmag (talk) 15:09, 1 March 2024 (UTC)

I'm still not sure whether I fully understand what question B actually means. I'll start with what I think I know for sure (this is essentially weak fields), and then indicate what I think is the case in the general, strong-field case. Taking the equation for the gravitational energy as it is, U is the energy that we need to supply to separate two bodies of mass M and m, initially at distance R from each other, i.e. move both to "infinity". This is a binding energy, it is negative, and causes a mass defect, so that the total mass of the system is less than the sum M + m. If the system (say a binary black hole) is gravitationally bound, the mass defect is ${\tfrac {U}{2c^{2}}}$ — only half of the gravitational binding energy because there is kinetic energy that keeps the system from collapsing, and the balance between gravitational and kinetic energy is given by the virial theorem. Therefore the total mass is $M_{\mathrm {tot} }=M+m-{\tfrac {U}{2c^{2}}}$. This is not a fully general relativistic treatment of the problem; I don't know what the terms "gravitational energy" and "total mass" would mean in GR; they are integrated quantities and integration in GR is a bit of a mystery to me. Having said that, the gravitational field (the metric) carries energy (gravitational waves certainly do transfer energy) so it enters the energy-momentum tensor, and the field/metric consequently appears on both sides of Einstein's equation. There are self-consistent solutions for this, which are hard to find, but they exist and the question is not a logical conundrum or "vicious circle". If you're not happy with this sort of answer (I expect your aren't), you'll have to find an actual expert or dive deep into the GR literature. --Wrongfilter (talk) 16:06, 1 March 2024 (UTC)

I don't get why the mass equivalent of the potential energy is not added to the rest masses of the components, like $M_{\text{tot}}=M+m+{\tfrac {U}{2c^{2}}}$?  --Lambiam 11:11, 2 March 2024 (UTC)

The binding energy is negative, as energy needs to be lost for the system to form a bound state, or conversely you need to supply energy to break a bound system apart. Take a comet that falls in from the Oort cloud ("from infinity"). If unperturbed, it will follow a hyperbolic orbit past the sun and vanish back to infinity. If it passes close to Jupiter, say, it may transfer some of its energy to that planet and be perturbed onto a tighter elliptical orbit. This bound orbit has a lower energy than the previous unbound orbit, hence the binding energy is negative. Similarly for a collapsing gas cloud that forms a star or a galaxy: The collapse leads to a tighter configuration only if the gas cloud loses energy, mostly through radiation (radiative cooling). This loss of gravitational energy is also what powers active galactic nuclei, quasars and such. Gravitational systems that cannot cool radiatively (e.g. dark matter) can cool through other processes like violent relaxation or in a simpler way by accelerating and losing some particles while the remainder are decelerated and become more tightly bound. In black hole mergers, this is quantitatively relevant: The models for the observed gravitational-wave events always have and end state (single black hole) of lower mass than the sum of the original black holes. Long story short, bound systems have lower total energy compared to unbound systems, hence binding energy is negative, hence the total mass is less than the sum of the masses of the constituents. --Wrongfilter (talk) 12:14, 2 March 2024 (UTC)

Using the formula $U=-{\frac {GMm}{R}},$ we have

${\frac {{\text{d}}\,U}{{\text{d}}R}}={\frac {GMm}{R^{2}}},$

which means that the energy increases and decreases with distance. The highest energy is at infinity. Isn't that consistent with a tighter orbit having a lower energy?  --Lambiam 21:43, 2 March 2024 (UTC)

Oh, I didn't see your response until now. Yes, a tighter orbit has a lower energy. But that's just what I was saying in so many words anyway. --Wrongfilter (talk) 12:32, 10 March 2024 (UTC)

Energy of one period for alternating current

In alternating current, as at the output of an amplifier for a dipole antenna, I cannot find a direct formula giving the energy over a period. Is it simply accepted that this value is given by dividing the power by the frequency? Or is there a direct formula and if so which one? Malypaet (talk) 12:10, 29 February 2024 (UTC)

Energy is power per unit time. So "energy over a period" is just power.

The formula for that isn't so much "accepted" as it is simply the definition of the terms. Energy defined to be power per unit time. (Power / unit time) * (time period) = power. Just basic dimensional analysis. PianoDan (talk) 15:32, 29 February 2024 (UTC)

I assume they mean "energy emitted over a period", i.e. power * (time period). For a monochromatic sinusoidal oscillation this would indeed be power / frequency. --Wrongfilter (talk) 15:39, 29 February 2024 (UTC)

If you use "P" for power in watt or joule/s, "E" for energy in joule, $\Delta t$ for the unit of time of 1s and T for a period in second ($T={\frac {\lambda }{c}}$), "Energy is power per unit time" is like $E={\frac {P}{\Delta t}}$ and as what I know is $P={\frac {E}{\Delta t}}$, you write $E={\frac {E}{\Delta t^{2}}}$...People have a great problem in distinguishing Energy (kinetic energy of an object or energy accounted for in an interval of time in an energy flow), and Power that is an energy rate (flow,etc...).

This is why I am looking for an equation directly giving the energy accounted for over a period ($T={\frac {\lambda }{c}}$) of a sinusoidal electric current. Energy in joule equal to W-s (watt⋅second). Malypaet (talk) 21:30, 29 February 2024 (UTC)

OK, mea culpa, I wrote that wrong. Power is energy per unit time, not the other way around. That said, it's STILL just basic dimensional analysis. (Admittedly, using the correct units does help.)

P = E / t. So E = P * t. That's all there is. PianoDan (talk) 23:26, 29 February 2024 (UTC)