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September 1

Greenhouse carbon dioxide level

According to the article Carbon Dioxide, "Plants can grow up to 50 percent faster in concentrations of 1,000 ppm CO2 when compared with ambient conditions, though this assumes no change in climate and no limitation on other nutrients."

And from the same article, "Carbon dioxide content in fresh air ... varies between 0.036% (360 ppm) and 0.039% (390 ppm), depending on the location." And in the Toxicity section: "In concentrations up to 1% (10,000 ppm), it will make some people feel drowsy".

With this as background, is it practical and worthwhile to raise the CO2 level by lighting a small woodstove in a greenhouse? (Other things being equal. I.E., assuming that the level is not raised so high as to be toxic to the greenhouse workers, that the carbon monoxide level is not raised, and that the greenhouse is not overheated.) Thanks, C7nel (talk) 01:36, 1 September 2014 (UTC)[reply]

Sounds dangerous. Using a stove in an unventilated area can produce deadly carbon monoxide and other combustion products. If you really wanted to raise CO₂ levels, I'd suggest buying a tank of it. You could still get too much carbon dioxide in the air, but at least that would eliminate the other dangers. Small tanks would be safer, since even if it all leaked out at once, it wouldn't be that much. StuRat (talk) 01:57, 1 September 2014 (UTC)[reply]

From my experiments the limiting factor in plant growth is light. Not CO2, water or other nutrients. Obviously if those are lacking the plant will grow slower, but once those needs are met the limit is light. I have not tried using a solar concentrator to see if that helps. Ariel. (talk) 02:13, 1 September 2014 (UTC)[reply]

I doubt it is just light or just CO2, and this paper at least agrees that ~99% of various plants grew more when exposed to more CO2 alone Fig 1 http://www.science.poorter.eu/1993_poorter_vegetatio.pdf . One obvious mechanism is that if there is more CO2 then the stomata don't need as much air circulation so the plant loses less water via its leaves, which is one of the other limitations on plant growth. Greglocock (talk) 02:08, 2 September 2014 (UTC)[reply]

Yea, marijuana grow houses seem to install major grow lights to make it grow faster. I am guessing that the cost is prohibitive for less valuable crops. StuRat (talk) 02:30, 1 September 2014 (UTC)[reply]

No, I'm not growing marijuana :) Just regular plants in a window box. Ariel. (talk) 07:24, 1 September 2014 (UTC)[reply]

In Victorian times, some rich people in England maintained "pine houses" – heated greenhouses in which they raised pineapples to serve at table. Some heated their pine houses with a coal boiler (in an nearby outhouse) and hot-water pipes, others used heaps of vegetable matter (grass cuttings, potato peelings, etc.) decaying under the benches of the pine house. The latter was found more effective in promoting growth of the pineapple plants, and people wrote of the benefits of "live heat" relative to "dead heat". Later they realised that heat is just heat, but the decaying heaps were promoting pineapple growth by increasing the CO₂ level in the pine houses. Maproom (talk) 06:54, 1 September 2014 (UTC)[reply]

Pineapple is special in that it uses the efficient CAM photosynthesis - perhaps it can use the extra CO2, if so than corn and sugar cane should also be able to since they use C4 carbon fixation. Ariel. (talk) 07:24, 1 September 2014 (UTC)[reply]

Some plants grow much better with added carbon doxide, others don't. And many of those that do tend to become a bit woody or tasteless. So you'd need to be careful to choose what plant you wanted to grow in the greenhouse. Dmcq (talk) 12:26, 9 September 2014 (UTC)[reply]

Is there such a thing as absolute voltage (and thus true electrical neutral)?

This is based on the next-to-last question above, but deserves its own heading: someone said that voltage somehow is only relative as a result of it being a gauge theory. But ~~my impression was that a device such as an electroscope truly measures, if only approximately, a zero voltage when the little gold leaves hang down next to each other.~~ (sorry, on second thought that part was dumb) Talk:Voltage is full of debate like this, with some saying that neutral voltage can be defined as the voltage "at infinity". Given the extreme electrical charges in space I have a hard time picturing that as practical, but is it theoretically valid? Last but not least there was my supposition above that the masses of electrons and positrons (or associated with their presence in some way) ought to vary depending on where they are in an electric field -- it seems like Stackexchange leans toward the idea that a charged battery contains more mass than a discharged battery, and reddit the same of capacitors. So I'm thinking a third way to define zero voltage is where positrons and electrons have precisely the same mass. Note: even if they're valid, I don't know all three methods agree. So let's try to settle this, and hopefully find reliable sources to update Voltage to dispel the world's confusion: is there an absolute zero voltage, and do we know what it is relative to the range encountered on the ground and in the air of our planet? Wnt (talk) 13:19, 1 September 2014 (UTC)[reply]

This is a very common cause of confusion. I tried to explain the answer as clearly as I could in the "Voltage" section of the membrane potential article, and I don't think I can do any better than refer to that. But let me quote from the third paragraph there: In mathematical terms, the definition of voltage begins with the concept of an electric field $E$ , a vector field assigning a magnitude and direction to each point in space. In many situations, the electric field is a conservative field, which means that it can be expressed as the gradient of a scalar function $V$ , that is, $E = -\nabla V$ . This scalar field $V$ is referred to as the voltage distribution. Note that the definition allows for an arbitrary constant of integration—this is why absolute values of voltage are not meaningful. Looie496 (talk) 15:17, 1 September 2014 (UTC)[reply]

Perhaps it is time to review constant of integration.

With due respect, Wnt, the best way I can answer your question without writing any equations is this explanation. Please understand that this is my best and sincere effort to represent your question conceptually (and not in any way an attempt to be condescending, although if you can understand this explanation, you might get a good laugh). You ask whether there is absolute voltage. This is logically equivalent to the nonce question, "what is the length of a piece of string?" Then you refine this concept by expressing confidence that we can find a "true" electrical neutral. This is logically equivalent to saying, "what if we tied one end of the string down at the position 'zero'?"

Does this help? Voltage is the path integral of the electric field. It is computed by adding up a quantity along a path integral - sort of like how we measure the length of string. There are many clever methods to compute this value, but it always depends on where the two ends of the string are. We can measure the length of any specific string, but we cannot answer the question in general - at least, not by providing a numerical value. When we define the two points - i.e. we measure voltage between two points - it has a definite value, because it is a definite integral. When we do not define the endpoints, we have an indefinite integral, to which we may add a constant of integration. Nimur (talk) 16:11, 1 September 2014 (UTC)[reply]

Staying with the "piece of string" analogy, it might be useful to consider gravitational potential energy. The gravitational potential energy of a mass m in a uniform gravitational field is mgh. This quantity is meaningless unless we know what h is; until we define the point we're measuring the height from. On the electroscope issue, no deflection of the leaf means there's no voltage between the leaf and the case of the electroscope, so there's a definite but arbitary reference potential. Tevildo (talk) 16:47, 1 September 2014 (UTC)[reply]

On consideration I suppose I was making a basic error thinking about the electroscope -- even though the leaves obviously move apart, this isn't due to their voltage but due to their charge; i.e. it is conceivable they could be inside a very large spherical metal electrode, wired directly to it, but because the charge repels itself to the outside of the electrode, the charge goes away and the leaves fall together.

Nonetheless, the mgh analogy seems like a bad one. Relative to one source, matter really does have a zero potential at infinity, and it really does emit energy to reflect the loss of relativistic-ish mass as it falls in; in the case of a black hole, as I understand it, even most of the rest mass consumed can end up being radiated as energy. And there's a specific phenomenon, the event horizon, that forms a neat ruler line marking the precise absolute bottom of the gravity well. The energy of a gravitational potential is -G mM/r, the Schwarzschild radius is r = 2 GM/c^2, so the potential there is - c^2 m /2 ... I forget why there's a /2 here, but I remember seeing claims that actually the amount of energy was equal, so there might be a relativistic correction I've missed. The point is though, it's an absolute potential as evidenced by these clear ruler markings at the top and the bottom of it. No? Wnt (talk) 17:09, 1 September 2014 (UTC)[reply]

You're conflating "really convenient reference point" with "absolute reference point." Nimur (talk) 17:17, 1 September 2014 (UTC)[reply]

Also I should mention the photon sphere at 3 GM/c^2, i.e. where the absolute gravitational potential is precisely -1/3 of the mass of any given particle. (Hmmm, wonder if there's anything at 4...) Wnt (talk) 17:22, 1 September 2014 (UTC)[reply]

Wnt, you are trying to explore some very complicated equations, while you are demonstrating to us that you are uncomfortable with elementary mathematics. For your own sake, put the advanced topics in mathematical physics to the side, for a little while, and review the elementary concepts of analytical mathematics, because you cannot meaningfully understand mathematical physics if your comprehension of calculus is broken. Would you like a recommendation on some good books to help you refresh those concepts? Or, if you've never learned them, would you like some recommendations for introductory texts?

Nimur (talk) 17:25, 1 September 2014 (UTC)[reply]

I understand the calculus of doing an integral; what I don't understand is why the two of you use that as proof of something. Sure, position is the integral of velocity ... so there's no such thing as absolute position ... so you can put the equator wherever you want it, it doesn't mean anything. It is too easy to let the math obscure a point. Just because you can calculate some quantity doesn't mean you've explained the full system.

P.S. apparently there's yet another "ruler marking" in the gravity potential at 9/4 G M / c^2, where the redshift reaches two, that controls the possible compactness of stars. [1] (I haven't yet even looked at this) Wnt (talk) 17:30, 1 September 2014 (UTC)[reply]

But that's exactly the point. Voltage is the antiderivative of the electrical field in exactly the same way that position is the antiderivative of velocity. Incidentally, I think your explanation of the electroscope is a bit off the mark, perhaps the article is not very well written. The electroscope is actually an instrument for detecting an electrical field. It does not directly detect voltage. A zero reading for an electroscope tells you that the voltage is locally constant, not that it is zero. Looie496 (talk) 17:41, 1 September 2014 (UTC)[reply]

Wnt, we aren't presenting "proof." We are presenting definition of terms. If you aren't familiar with this distinction, you are again demonstrating that you are unprepared for analytical exercises in advanced mathematical physics. (Contrast: voltage is defined; while its path independence is proved, subject to specific definitions). This proof would be a good one for you to study, and ask questions about it if you get stuck!

We want to help you understand these things. I volunteer my time here because I like helping people understand things! But we have to be pragmatic: I can see that you are presently on completely the wrong side of a massive conceptual gulf, and the way to bridge this gap is by reviewing mathematical concepts. Every hour you spend trying to apply your present methods to understand black holes is taking you deeper into the ravine of self-induced confusion. Nimur (talk) 17:36, 1 September 2014 (UTC)[reply]

I may indeed be on the wrong track, but I don't think it is philosophically wrong to try to understand a confusing point by asking thought-experiments. For example... suppose you have a plasma of electrons and positrons in a chamber, all with the same rest mass, and now you apply a strong electric field to it. Most of the electrons go to + and most of the positrons go to - of course, but the electrons that were furthest from the + pick up the most speed (more relativistic mass) reflecting their higher electrical potential when they were near the - electrode. But now suppose you have some positronium "atoms" amid the mix; being neutral, they are little affected. Now my thinking here was that the potential energy of the electrons being closer to the negative electrode is equivalent to some amount of mass, which might be measured. But ... I'm not sure a measurement actually would back this up. When I picture those positronium atoms spinning and going into and out of excited modes, I suppose the electron wouldn't start to remain at the center in one of these because it has "increased mass" from being near the negative electrode, while the positron takes the same role in those at the other end. The mass is real, we have an article on electric potential energy that explains how to calculate it, but understanding just where it is and what it affects isn't as obvious. Wnt (talk) 18:10, 1 September 2014 (UTC)[reply]

Should be interested in the speed and speed accelerations in natural magnetism as physical-chemical phenomena which had creating an inert masses, and in particular should be interesting in the electromagnetic induction.--Alex Sazonov (talk) 03:32, 2 September 2014 (UTC)[reply]

If in physics is been assumed that the speed of light have not a limits, why does a photons do not had a masses?--Alex Sazonov (talk) 11:53, 9 September 2014 (UTC)[reply]

Obviously, that the electrons always had been a masses, because they always been occur only in substances, but photons never had been a masses, although a light is been a substance.--Alex Sazonov (talk) 13:19, 6 September 2014 (UTC)[reply]

What does always the most important for elementary particles had been presence of a mass or had been presence of a speed acceleration?--Alex Sazonov (talk) 18:11, 7 September 2014 (UTC)[reply]

In classical physics of Isaac Newton the more most important always had been a speed and speed acceleration, but not a mass!--Alex Sazonov (talk) 14:11, 8 September 2014 (UTC)[reply]

Obviously, that electromagnetism is always been powerfully than all another, because the elementary particles of electromagnetism always had been a masses!--Alex Sazonov (talk) 15:03, 9 September 2014 (UTC)[reply]

Obviously, that in the lasers photons are never had been a masses!--Alex Sazonov (talk) 02:47, 7 September 2014 (UTC)[reply]

If the photon had be a mass, so the photon would be a full electron.--Alex Sazonov (talk) 16:11, 7 September 2014 (UTC)[reply]

It is been known that light is been capable of transferring a substances (lasers spraying of cover), so that why the photons can not had a mass?--Alex Sazonov (talk) 09:43, 8 September 2014 (UTC)[reply]

Does it possible the fact in nature that photons had been a mass?--Alex Sazonov (talk) 15:03, 9 September 2014 (UTC)[reply]

Potential energy gravitates, which means that we can in principle measure the absolute quantity of it, and even its spatial distribution, by measuring spacetime curvature.

Potential energy is the energy of the field, and there's only one field in total, not one per charged particle. Thus it doesn't make sense to talk about "the potential energy of a particle". A charged battery or capacitor does have more mass than a discharged one, but that mass doesn't belong to any of the particles. In a capacitor, it's mostly located in the dielectric.

Voltage is only well defined when the total potential energy can be written in the form (U₀ +) q·f(x) where q and x are the charge and position of some object. Within its domain of applicability, only voltage differences matter. There's probably nothing deeper to say about voltage because at a deeper level it no longer makes sense as a concept. -- BenRG (talk) 21:56, 1 September 2014 (UTC)[reply]

@BenRG: This was a very good answer, and similar to what I found at [2]. It would be interesting to further explore when and how "voltage no longer makes sense as a concept". And our article on electric field says that the energy density of the field is 'proportional to the square of the amplitude', which perplexes me first, well, because they link amplitude referring to sinusoidal waves, but here we're speaking of a static field (I assume that's a mistake?), and mostly because if I take two protons and jam them side by side, the amplitude of the field should be double the two individual fields, so the energy should only be 4x, but classically I could sink an infinite amount of energy into that push, and even IRL a ridiculously large amount. Stackexchange gives me a formula but alas not the definitions or explanation:

\int |\nabla \phi |^{2}dV=-\int \phi \nabla ^{2}\phi =\int \phi \rho

and says that the self-potential of a point charge is divergent (which I suppose would explain why the 4x is a big deal?) but I'm still not quite seeing it. I ought to look further here, but I might as well poke head up and see if someone tells me I'm on the wrong track. Wnt (talk) 14:39, 2 September 2014 (UTC)[reply]

Saying that "voltage" doesn't make sense in general is like saying that "up" doesn't make sense in outer space. It's more a matter of word definition than physics.

\int |\nabla \phi |^{2}dV=-\int \phi \nabla ^{2}\phi

is Integration by parts#Higher dimensions together with the assumption that φ vanishes on the boundary of the region of integration (which is probably at infinity).

-\int \phi \nabla ^{2}\phi =\int \phi \rho

uses Poisson's equation in weird units (normally there would be a factor of 4π or 1/ε₀ in front of ρ—I suspect they just forgot it).

"Amplitude" in the article means vector norm/magnitude. I changed it.

Superimposed point charges should have twice the energy of the same charges separated to infinity because (E+E)²/(E²+E²) = 2. But E = +∞, so "twice" is infinitely more. The usual way of dealing with the infinities is to impose some sort of cutoff in the integral, justified because Maxwell's equations must break down at short distances. See regularization (physics). In this case, there's a trick you can use: write

{\frac {1}{2}}\int |E_{1}(\mathbf {x} )+E_{2}(\mathbf {x} )|^{2}={\frac {1}{2}}\int |E_{1}(\mathbf {x} )|^{2}+\int E_{1}(\mathbf {x} )\cdot E_{2}(\mathbf {x} )+{\frac {1}{2}}\int |E_{2}(\mathbf {x} )|^{2}

and then just discard the first and third terms because they're independent of the separation of the particles. The middle term should integrate to

q^{2}/|\mathbf {x} _{1}-\mathbf {x} _{2}|

. -- BenRG (talk) 17:30, 2 September 2014 (UTC)[reply]

This makes sense. If it takes an infinite amount of energy to push two point charges together, then it must take an infinite amount of energy to push one point charge together. But the point charge presumably can be replaced by a radially symmetric distribution, approximatable as a sphere of charge with no electrical force (so no field) inside by the shell theorem, and acting like a point charge outside. Which leaves me to integrate for r from x to infinity, where x is some small value. The volume is 4 pi r^2 * dr ; the magnitude squared = ( e / 4 pi eps0 r^2 )^2 ... I think. So that's the integral of e^2 dr / (4 pi eps0^2 r^2), I think. Which gets me -e^2 / (4 pi eps0^2 r) for r=(some small value) - (r at infinity), i.e. only the small value counts.

Now to check sanity by units... eps0 (I mean, vacuum permittivity) is ε0 = 8.854 187 817... x 10−12 [F/m] where 1 F = 1 s4·A2·m−2·kg−1 so that's 8.854 187 817... x 10−12 s4·A2·m−3·kg−1. The charge on the electron e is −1.602176565(35)×10−19 C where 1 C = 1 A s. So (might as well multiply the values too) I'm getting, oh, 3.204 x 10-38 A^2 s^2 / 9.8509 x 10-22 s4·A2·m−3·kg−1 = 3.253 x 10-17 m^3 kg /s^2 / r. And a joule, incredibly, is actually m^2 kg /s^2, which means I couldn't possibly have fouled up this calculation. 🙈 🙉 🙊 :) (no, really do speak evil of it when you find the bug)

As a sanity check, the energy actually in the field around an electron can't really be more than the mass of the electron itself, because wherever the electron goes it goes... right? The mass-energy of an electron is 9.10938291(40)×10−31 kg * ( 299792458 m/s) ^2 = 8.187 x 10-14 m^2 kg / s^2, which means that r can't be less than, oh, a millimeter. Arrrgh. I oughtn't post what's probably just a really stupid math error, but might as well check in and see if it turns out that there's something I ought to know about this calculation. Wnt (talk) 21:30, 2 September 2014 (UTC)[reply]

I think what you're trying to calculate is the classical electron radius, but you must have made a mistake since the answer (according to the article) is on the order of 10⁻¹⁵ m. This is still much larger than the best experimental upper bound on the electron size, which is 10⁻²² m, I think ([3]). -- BenRG (talk) 06:42, 4 September 2014 (UTC)[reply]

@BenRG: Thanks for the roadmap! The bug is proving annoying to find. The calculation I used started with Coulomb's law

|{\boldsymbol {E}}|={1 \over 4\pi \varepsilon _{0}}{|q| \over r^{2}}

.

This means that the square of the magnitude has the square of

\varepsilon _{0}

in the denominator. But somehow the squaring has to go away - I ended up doing that in the unit analysis above, but not in the calculation. And I'm having a heck of a time trying to think where I would get a spare

\varepsilon _{0}

from.

The place I want to go with this calculation is the more complex case of an electron inside a sphere with a very strong positive charge. The "net mass" of this electron ought to be negative as I said above because it could have emitted more than its mass in kinetic energy to get to this point. Inside the sphere, however, its field has a positive mass (indeed, the sphere of charge has no effect at all). Outside, however, the sum of all the charge present means that the square of a very large amount of push from positive charges is being calculated, so that reducing it by the charge of one electron can indeed take down the mass of the field by a substantial amount. Which leaves the numerical question of whether the mass of the field shifts as you move the electron around inside the sphere in a pattern that mimics having a negative mass bound to the electron. Then there's the physics question of whether the system "knows that" so that the electron would fall upward in response to gravity, i.e. whether the pull of gravity on the real mass of the field can force the electron to go upward so that more of the field is closer to the ground.

That said, I have to concede, while fun to figure out how the mass works, I realize now this doesn't actually get me a route to measure zero voltage. I'm afraid I was making the same mistake as with the electroscope -- just because there is no net charge on something doesn't mean there is no voltage on it. The sphere I test with could be grounded to a much larger sphere surrounding it, and be neutral yet be at the same potential as that is, and I have no idea what that is. Wnt (talk) 13:03, 4 September 2014 (UTC)[reply]

@Wnt: The field energy is

{\frac {\epsilon _{0}}{2}}\int E^{2}

in SI units—I think that's your missing factor of ε₀.

The notion of a "buoyant force" on electromagnetic field mass-energy in regions of destructive interference is an interesting one. There might actually be such an effect, but I suspect it's undetectably tiny if it exists, and it should depend on the local electric field (since that's what determines the local field mass density), not on the electric potential. -- BenRG (talk) 16:19, 5 September 2014 (UTC)[reply]

Inside your car there is a zero voltage point at the negative terminal of the battery; inside your house there is a zero voltage point where mains wiring is grounded (often to a water supply pipe). A zero voltage point for the whole universe is hard to find, hence the OP's question. The Bohr model of the atom describes it as a tiny orbital system of charged particles and the potentials of the particles are calculated in terms of the work needed to bring a test charge "from infinity". A more realistic wording is "from a point so distant that the field strengths of the atom's particles are negligible, and the atom being modelled is alone". The Bohr model, and the better refined Atomic orbital model, do not require an absolute zero voltage to exist anywhere; they are isolated atom models that work the same way whether the atom of, say copper, is deep in the ground or in a high-voltage cable. The article Introduction to gauge theory explains here how our understanding of electricity and magnetism by Maxwell's equations having gauge symmetry is consistent with every practical voltmeter needing two probes. A single-probe voltmeter that could indicate absolute volts is as elusive as the Magnetic monopole. We would conclude that no absolute zero voltage exists but for the implication of mainstream Cosmology that speculates A) the singular source of the Big Bang had no potential gradient i.e. it was all at the same voltage which we may call the absolute voltage reference, and B) whatever the Ultimate fate of the universe there cannot remain any voltage difference (because that could drive an electric motor, meaning that the story isn't over) so we may also call that the absolute voltage reference. There is no mainstream cosmology that claims the voltages in A) and B) would be different, from which we conclude that there is an absolute zero voltage reference but it presently exists only as the theoretical average voltage that matter would attain if all insulators including vacuum became conductors. 84.209.89.214 (talk) 13:23, 2 September 2014 (UTC)[reply]

Actually that's not really convincing as an argument. Unless there is proton decay - and even electron decay - one expects the universe to retain differences in electrical potential. And the existence of a literal singularity before the Big Bang is something at least I don't believe in. Wnt (talk) 14:43, 2 September 2014 (UTC)[reply]

Wnt, I know this has been said already, but I have to say it again: by definition, voltage is the potential difference between two points. By definition. By analogy, distance is the position difference between two points. What's the absolute distance of New York? What's the absolute distance of your house? Do you understand why the question makes no sense? --Bowlhover (talk) 17:37, 2 September 2014 (UTC)[reply]

Come on, not the same thing at all. In certain models of the Universe it makes perfect sense to talk about the absolute potential at a point — it's the potential difference between that point and the "point at infinity". For example, if space is asymptotically flat, and electrically neutral on the average, then I'm pretty sure this is well-defined (I don't know GR well enough to be sure there are no gotchas hiding in there somewhere). The "absolute distance of your house" relative to the point at infinity would always be infinity, but this is not true for electrical potential. --Trovatore (talk) 20:01, 2 September 2014 (UTC)[reply]

Trovatore, you are also conflating a "really convenient reference-point" with an "absolute reference-point."

A large number of mathematical formulae can be dramatically simplified if their definite integral is written with one limit set to infinity. This is called an improper integral. You're a mathematically-inclined individual! You already know this. This choice for the limit of integration can make the computation easier, especially for many of the potential functions we find in real-world problems.

However, solving the improper integral doesn't change the definition of potential-difference: it just asks us to consider the conceptual idea that a potential difference converges when considered across large distances. Nothing about that observation changes anything. The choice to use "a point that is very far away" as the reference-point is still totally arbitrary. The fact that the integral does converge is an important detail: it means that the potential energy functions we normally encounter are non-pathological. They frequently have radial symmetry. And so on. These observations are important to the physics! But they don't make the definitions change.

Nimur (talk) 21:18, 2 September 2014 (UTC)[reply]

Well, OK, now this has become more of a philosophical argument than a scientific or mathematical one. If potentials are well-defined relative to the point at infinity, but that does not qualify as an "absolute reference point", then what, if anything, would qualify? In the hypothesized case (flat, electrically neutral universe), it's the same for all observers, or at least all observers in our universe, right? How much more "absolute" than that do you want to get? I think we're flirting with Scholasticism here. --Trovatore (talk) 21:25, 2 September 2014 (UTC)[reply]

You bring up an interesting point; but one that only holds for those non-pathological, well-behaved functions we talk about in basic physics, like the coulomb potential. If we want to build a mathematical framework that is useful for more general problems - both real and theoretical potential functions - we need to stick to our definitions very carefully. I'm not just defending the definition for the sake of being argumentative!

For example, the Yukawa potential generalizes the Coulomb potential. In the general form, its improper integral does not always converge. Or consider the magnetic vector potential, or any other field that may represent a non-conservative potential. As you can see, there are potentials - real and theoretical - for which two observers at infinity could disagree about magnitude!

If we were to stretch the definition of the word "absolute," or if we were to assume that "a point at infinite distance" is the universal reference point, then we would not be able to mathematically describe many real observations. We would not be able to mathematically explore theoretical potential fields that represented anisotropic or inhomogeneous interactions. We need to stick to the mathematical definitions so that we can consider real, testable hypotheses for a more general class of problems than simple electrostatics. We could, with equally valid abuse of notation, stretch the meaning of "absolute" to define an inertial reference frame for which the Lorentz factor is equal to unity. It's "absolute" with respect to a specific problem-setup! Obviously you see why this is not "absolute" at all, for an entire category of interesting physics problems!

You asked "what, if anything, would qualify as absolute?" ... Nothing. Nothing qualifies as absolute. This isn't simply a regurgitation of "physics establishment" dogma: it's a real thing that we have to be aware of when we frame our problems with mathematical models.

Nimur (talk) 21:57, 2 September 2014 (UTC)[reply]

Well, so if "nothing qualifies as absolute" , my question is, is this a contingent or a priori truth? That is, is there any possible world in which there would be absolutes? If not, can you explain what work the word "absolute" is doing, and why we don't simply redefine it to be more useful, by referring to some property that is possibly (even if not actually) instantiated? On the other hand, if it's a contingent truth, then can you explain in what counterfactual situation you would call something absolute? That would help me understand what you mean by the word. --Trovatore (talk) 23:46, 2 September 2014 (UTC)[reply]

Sure: there are a lot of abstract, non-physically-realizable absolutes. Most of them are pretty abstract. For example, the total ordering of the set of all integers is an absolute: two is absolutely larger than one; there is no "relative" scheme in which two is smaller than one or equal to one. This absolute relationship follows from our definitions of what integers are.

The cardinality of the reals is absolutely larger than the cardinality of the integers. There is no conceivable hypothetical universe in which this is untrue. This is an absolute relationship between two entities that directly follows from the definition of those entities.

I think it's no coincidence that I'm readily able to find mathematical abstractions that form absolute relationships, but that I'm unable to contrive good examples of any physical manifestations of absolute relationships.

Nimur (talk) 23:58, 2 September 2014 (UTC)[reply]

(ec) Sure, you can define the reference point to be infinity, and then call the resulting potential difference an "absolute voltage". I don't really care what term you call it. It remains true that 1) voltage is, by definition, a potential difference, 2) infinity is no better or worse than any other reference point, and 3) if I define voltage to be 10 volts plus whatever you think it is, no physical prediction about the world would change. --Bowlhover (talk) 22:01, 2 September 2014 (UTC)[reply]

Well, it's "better" than any other reference point in at least one way; namely, in the hypothesized situation, it can be defined independently of any artifact or of the observer. Yes, you can also say that about "take the potential difference to infinity, and then add 10V"; it's "better" than that one in the sense of being more naturally motivated. --Trovatore (talk) 23:46, 2 September 2014 (UTC)[reply]

To reset the philosophy here, my question wasn't meant to be merely "is voltage as the term is conventionally used provided with an absolute zero value", though that has been answered and it is interesting that it is only defined in relative terms. It's also not meant to be one of whether we can define a purely arbitrary zero, a ceremonial platinum-iridium Leyden jar that they can keep in that place in France where they have the standard kilogram. Rather what I really want to get at is whether there is a universal standard reproducible procedure whereby we can work out a zero voltage point that would be the same whoever referenced data by it. If such a thing exists, then voltage is not purely relative, at least in my opinion.

Now, for gravitational potential, I think it's pretty clear there is such a point. Given access to any black hole in the cosmos, you could measure the exact position of its event horizon, then measure the exact velocity of a probe dropped from your position as it approaches it. Given access to any neutron star you could use a probe to measure where the photon sphere is located, and measure the distance dropped to it. These things are impractical, not impossible. By contrast, I'm not sure if measuring gravity "at infinity" is possible because whichever way you look there are more stars. (N.B. I am assuming that the event horizon of a black hole in a cosmic void would be smaller than one at the center of a galaxy made from the same mass; if that's wrong please enlighten me!) For electrical potential, I still don't know any method other than "at infinity" exists, and we know how extreme the voltages in space can be even quite far from the sun. Wnt (talk) 22:06, 2 September 2014 (UTC)[reply]

Ah, black hole event horizons are a little more subtle than that—they're not fixed dotted lines in space. Different observers of a black hole will see the event horizon as being in different locations. (Fun fact: to an observer approaching a black hole, the event horizon will always be beneath him.) TenOfAllTrades(talk) 14:28, 3 September 2014 (UTC)[reply]

@TenOfAllTrades: (and @Wnt:) the location of the black hole event horizon is an objective property of the spacetime. You may be thinking of Rindler or de Sitter event horizons, which are observer-dependent (and physically meaningless). -- BenRG (talk) 06:42, 4 September 2014 (UTC)[reply]

@TenOfAllTrades: Wait a minute. Wait a minute! How can the event horizon be subjective like that? Either you can shine light on arbitrary distant objects outside the hole or you can't, right? Either you cannot avoid striking the singularity in a finite period of time or you can't, right? If the event horizon recedes at all as you approach it, what that means is that from a point a short distance outside the "real" event horizon, you can pause and relay out messages from inside the hole, no? I know our article presently says this but this is contrary to all I've read about black holes. Wnt (talk) 00:21, 4 September 2014 (UTC)[reply]

Relativity sure is weird, isn't it? Curvature of space does some tricky things to length and time for observers in different reference frames. TenOfAllTrades(talk) 00:54, 4 September 2014 (UTC)[reply]

It's more weird than that. Multiple observers in different reference frames will even view events as occurring in different orders. For any two events, A and B, separated by any distance, which to one observer occur simultaneously, there is always a frame of reference where an observer will see A happen before B, and a different reference frame where A happens AFTER B. Three viewpoints, three orders. And since there is no privileged vantage point, all three (or none of them) is actually correct. See relativity of simultaneity. Reality, unfortunately, doesn't need one's belief to keep on working. It will just keep doing what it does, without needing us to feel comfortable with it. --Jayron32 01:13, 4 September 2014 (UTC)[reply]

Alright, I see now from various places especially the talk page that "While this seems to allow an infalling observer to relay information from objects outside their perceived horizon but inside the distant observer's perceived horizon, in practice the horizon recedes by an amount small enough that by the time the infalling observer receives any signal from farther into the hole, they've already crossed what the distant observer perceived to be the horizon, and this reception event (and any retransmission) can't be seen by the distant observer." I was thinking of the event horizon in terms of the fixed Schwartzschild radius from which light cannot escape to distant observers, but the actual definition concerns where events can be observed by someone at any point, i.e. further in. And as such, I may indeed be begging the question if the calibration of where the "real event horizon" (i.e. the Schwartzschild radius) is is dependent on from what 'distant objects' you're making the determination, since those are the gravitational potential zero. I wonder what the relationship of the event horizon's altered position is relative to the potential energy position of the observer? Wnt (talk) 04:11, 4 September 2014 (UTC)[reply]

Errr, on second thought, I spoke too soon about understanding this. If I drop a probe into a black hole surrounded by an accretion disk to study it, that probe shouldn't return any video of the material in the disk that is inside the Schwartzschild radius, not even "the instant before" it crosses the event horizon (though due to redshift it would take a long time for that light to escape anyway). If the event horizon is receding from it I don't know what it's seeing, except I imagine that the redshifted light from things that passed before it would provide something to see, that should be from outside the Schwartzschild radius. Wnt (talk) 13:20, 4 September 2014 (UTC)[reply]

Wnt may be interested in this paper. Count Iblis (talk) 02:13, 4 September 2014 (UTC)[reply]