About the wonderful connection between mathematics and physics
Different mathematical models represent a certain picture of the world - real, possible, or theoretical. For example, Euclidean geometry represents a flat universe and curved universes are represented by spherical or hyperbolic geometries. Also, multi-dimensional universes are represented for example by theories such as Klotza-Klein (in the universe without the nuclear forces), or string theory which may represent our universe.
Sometimes mathematics predicts physical phenomena, as in a certain example that I was exposed to years ago, when I studied the theory of electricity - to this day I am amazed by it. I will try to share my admiration with you.
This is a seemingly strange phenomenon, according to which in a certain process part of the energy is lost, according to the calculation, without initially understanding why.
This is a charged capacitor with an electric charge to which another uncharged capacitor is connected at the same time. After the connection, the charge that was previously all in one capacitor, is divided between the two capacitors.
There are formulas that link the capacitance, the charge, the voltage and the energy. We will not copy the formulas here, but those interested can find them in a physics textbook or on Wikipedia.
And if you calculate the energy, it turns out that after connecting the capacitors, half of the energy is lost (if the two capacitors are equal in value).
I'm not familiar with the history of science, but maybe when it was first calculated, the question was asked where the energy goes and it's possible that at the time there wasn't a satisfactory answer.
By the way, a similar phenomenon also exists in mechanics. I mean plastic collision. When such a collision occurs between two equal bodies, one is moving and one is at rest and all the momentum is transferred from the first to the second, half the energy is lost. In such a collision, the formulas are also similar to those for calculating capacitors and have the same form only that instead of the parameters charge, capacitance and voltage, the parameters momentum, mass and speed respectively appear.
It is understood that in a plastic collision the lost energy is wasted in the form of heat generated due to the collision. Energy loss is present in a plastic collision by definition (as opposed to a purely elastic collision where there is no energy loss). But in the case of capacitors, where the hell does the energy go?
Anyone know?
And the bigger question - how does mathematics know that half of the energy is supposed to be lost?
76 תגובות
In the described case, why does the voltage on the two capacitors actually decrease from the voltage that was on the charged capacitor before it was connected to the empty capacitor? Thanks.
No one said that in all conditions it is half. Half the energy is wasted when an empty capacitor is connected to a charged capacitor - only if the two capacitors are equal. Any other case on its own merits. Your new paradox does not seem like a paradox to me because if the resistor is small then the current is large and the charge transfer time is small. If the resistor is large then the current is small and the charge transfer time is large. And the energy is a function of the current and time. You want - think of examples. For this you need a formula from the Hadoah field and I'm not good at it. Instead of calculating, you can do an experiment with some resistor values.
One of the respondents here - Hanan - did an experiment. Earlier he suggested that perhaps the voltage drops to half a root for years and then no energy is lost. But then he wasn't lazy, he did an experiment and found out that the voltage really drops in half. He told us this here and said he eats the hat.
Why in all conditions is half! Just like when you lower this capacitance or it should be 10 times as well
I don't disagree with the reality, I just think that the Argy beacon should also contain an element of losses when you connect a resistor, in my opinion you will get less than half (or I'm wrong) if it doesn't matter what size resistor we use, the total energy of the two capacitors will be preserved, this is a new paradox!
You have to calculate with two or three values of resistance and we will see.
Ran - in your response you demonstrate the initial astonishment expressed in the title of the article. But the answer has already been given. If some source charges the capacitor, you as an electrical engineer can calculate all the energies in the circuit and process. If a capacitor charges another capacitor through resistance, you can easily calculate the energy wasted on it. And in the theoretical case of zero resistance, you know as an engineer and knowledgeable in radio that a sudden current pulse, such as that created by charging a receiving capacitor after a charged one, creates electromagnetic radiation - so what is the theoretical problem here?
Let's assume that the source that charges the capacitor will be connected through an electricity meter that will measure the energy, it will measure how much enters the capacitor, whether the source is another capacitor or a power supply, when the capacitor is charged to a certain voltage, its energy is the same, it doesn't matter who the charging factor is! Len, there is a fundamental problem here! There is a real theoretical problem here!
Let's define it this way. The voltage of a charged capacitor is defined as the ratio of the charge to the charge divided by the voltage. It is clear that when you double the energy of the capacitor, the voltage drops to half (if the charge remains the same).
Something is missing in explaining the process. The formula gives an answer for the voltage to the capacitance to the charge but not to the energy. It is possible to claim in the back regulator to measure the losses and also to calculate them, they are spread out less than fifty percent. What will happen if there are no losses and we get more than half the voltage? In fact, no!
That's why there is a problem here! I suggest that a high-level physicist try to give a solution. (I'm just an engineer)
A. In my opinion, it will be possible with the help of a branded supplier to transfer energy from one recipient to another.
Also regarding boards if you move the board from a distance of one meter to ten meters I don't think that the effect of the charge will be significant I think it is beautifully presented a paradox that has no simple answer and you are trying to force a solution.
Ran - your other example is of course different from the problem presented in the article. You are right in your words - if you decrease the capacitance of a charged capacitor, it adds energy and vice versa. But this should not be considered a mystery, even if the mechanical energy does not appear in the formulas. Even in a plastic collision, the energy that is lost in the form of collision heat does not appear in the formulas. In a variable capacitor you invest energy in moving the plates apart, or removing the dielectric (the insulating material between the capacitor plates whose presence increases capacitance) and this energy is added to the energy stored in the capacitor. And the opposite in the reverse process - the plates get closer by themselves, if they are allowed to, as a result of the attraction of the opposite charges, or if the dielectric is allowed, it is drawn into the capacitor on its own, and this is where the energy going down from the capacitor goes. It is clear that the energies in question are of a small order of magnitude compared to the mechanical energies (friction and acceleration) that you apply to change the capacitance of a practical capacitor.
And we will return to the matter of the article - unlike what I wrote in previous comments here, in the practical case that there is resistance in the circuit between the two capacitors, or if we connect the capacitors with a resistor, half of the energy actually goes to heating the resistor or resistance and the leakage of energy through radiation is negligible (by and large - for practical purposes - Not as I wrote that the wasted energy is divided between heating and radiation). In this respect, all the commenters who insisted on introducing the resistance to the solution of the presented problem were right, but again - in the theoretical case of zero resistance - the lost energy is all converted into radiation.
We will take another example of a charged variable capacitor if we reduce the capacitor to half the charge will be multiplied by
Double the capacity, the energy will be reduced to half. That is, it is a reversible process in both directions and works in every respect! Here there is no current, and sparks! Maybe a mechanical energy investment doesn't appear either
In no way.
66 - Your words can be interpreted not according to Haim's intention. According to him I am the only chatterbox who gave a correct answer. Haim - you wrote too fast. Not half of the energy goes to the second capacitor, but a quarter (half to radiation and the other half is distributed between the capacitors).
Haim:
Both the "solution" and I gave a correct answer.
I understand from your words that you do not include us in "chatters" and my question is why you include Aryeh Seter in chatters.
The first thing an electronics student learns is that when two identical capacitors are connected, half of the energy goes to the other capacitor and the other half becomes electromagnetic radiation. The energy on the resistance is divided by the radiation and it can be calculated simply. Of all the chatters here, only Aryeh Seter, who according to his response is an electronics or physicist, gave a correct answer.
Funny that such a simple question was dragged into such an unnecessary debate with 62 responses.
Those who do not know how energy turns into radiation should go back to Maxwell's equations.
The comment needs approval?????????
No wonder there are articles like this.
Aryaster
Charges do not pass in zero time, neither does a conductor.
Although tunneling, which is a quantum process, has zero time.
In the situation you described zero time is not possible.
And to conclude - when there is resistance then it is not a "miracle" that energy is wasted on resistance, but even when the resistance is zero, energy is wasted - and it is indeed a "miracle", until you realize that it is wasted on electromagnetic radiation and that is the idea of the article.
Josh's friend:
Friends don't have to agree but that's usually what happens.
You didn't talk about radiation, so I talked about it.
I assume that the law of conservation of energy is known to you and therefore all energy that is not lost as heat is lost as radiation.
If the total energy lost is constant (which is clear) and if, as you claim, the amount of energy lost as heat is constant, then it is clear that the amount of energy lost as radiation is constant.
just no?
That's why I asked if that was your intention.
Danny - in the theoretical case of zero resistance, it is not true that it takes time for the charge to go from receiving to receiving. The functions you are talking about describe charging or discharging a capacitor in an RC circuit. According to these functions, charging or discharging takes, for practical purposes, a time of 5RC (at which time more than 99 percent of the discharging or charging is completed). From these functions the energy loss is derived only as a result of ohmic losses (and not from radiation). But in the case of zero resistance, then also the RC product (which in unit calculations is t - time) is zero, which means that the charge passes between the capacitors (theoretically) in zero time.
Josh's friend - the simple case is zero resistance, then all the "lost" energy becomes radiation; When there is resistance, then you have to start making calculations how much energy went on radiation and how much on heating the conductor.
As my good friends said, the simple case is when there is resistance (and then when you solve the simple case of an RC or RLC circuit, you get that half of the energy went to heat, it is clear that there is a simplification of the problem since part really goes to radiation, but it is negligible). When the resistance strives to zero, the problem must be handled carefully, how exactly do you strive to zero the resistance, etc.
The only way (in my opinion) to check what happens when there is no resistance, is when you treat a conductor (there is no point in trying to just reset the normal solution since it has a lot of neglect, etc. as I wrote) and then the problem is solved by itself (it will not stay on a conductor) and it is possible that we will get Sinusoidal solutions without energy loss (and practically everything will go to radiation depending on how you solve).
Michael, I didn't understand what you said. Do friends have to agree? Did I mention the radiation?
Danish:
Do you understand what you are talking about?
It really doesn't seem that way.
Can you write the same "asymptotic curve" (what a crooked expression! I tell you this as a mathematician).
Josh's friend:
are you sure you are his friend It turns out that you do not agree with him!
You don't agree with reality either.
If the amount of energy emitted as heat is constant then the amount of energy emitted as radiation is also constant.
Is that what you think?
Arya, as you mentioned - when solving questions in mechanics, you start without friction, then add it. In this way, the simple cases are understood, and then the problem is complicated. But here precisely when there is resistance the problem is simple, and when it is removed the calculations get complicated because of the radiation. Therefore, in my opinion, it is not worthwhile to ignore the case where there is an objection. It seems to me that you can learn general things from him that may be relevant even in the complicated case.
And if you still allow me to say a few words about solving the problem with resistance: the amount of energy that has turned into heat (from the time the discharge begins to infinite time) does not depend on the resistance at all. Meaning: we can reduce the resistance as much as we want, without changing the amount of heat emitted. I don't know what you mean by "ideal conductor", but for me "we can reduce the resistance as much as we want" is close enough to ideal.
Machel
The passage of the particular load at a particular time
described by a mathematical function (asymptotic curve)
from which the energy loss is derived.
There is no wonder about it, the only wonder is why this article was written at all.
Josh's friend:
It's really not clear what you're talking about.
Josh didn't talk about water at all and anyone who talked about water besides me presented an analogy that doesn't even have a capacity equivalent.
The only "watery" analogy presented here was mine and this example does not need air or any medium around the capacitor and therefore has no sound waves.
In addition, sound waves, being the movement of molecules, are more similar in the present context to heat than to electromagnetic radiation.
Danish:
It is not clear to me what you are trying to say in response 49.
If the subject does not interest you - why are you trying to share your lack of interest with everyone.
Or did I miss in your words some attempt to answer the question where did the energy go?
Can you tell us what function you are talking about or are you just talking?
Not all comments need approval.
There are keywords that cause the response to be automatically delayed until manual confirmation
The response would have been released even without you asking because it does not violate the site's procedures.
Josh's friend - I am not underestimating the issue of resistance. I simply claim that the assumption in the question presented in the article is that there is no resistance and then the "lost" energy - all that half that was "lost" becomes radiation and nothing else. In a practical case where there is little resistance, a part of the energy that can be easily calculated becomes heat in the conductor. I do not now know how to present the calculation of the energy that becomes radiation.
And another note to Josh - you keep talking about a super conductor. I'm talking about a theoretical conductor with zero resistance where there is no current limit and it always remains zero resistance.
To the editor:
Here I got the answer.
Apparently only her contact with you does not need approval.
Be healthy
To the editor:
Do all comments here require approval?
that's new!
Michael
You can do the opposite and the opposite again. But the emphasis is on the process of transition from state to state.
The electric charge does not suddenly appear on the second capacitor.
There is no tunneling or quantum leap process here.
There is a certain amount of time it takes for the cargo to go through.
The function that describes this transition also describes the loss of energy.
Therefore, it seems to me that all the fuss is unnecessary.
Arya, I think you are too quick to underestimate the issue of resistance. The idea is that there is a mechanism for losing energy. It could be radiation, and it could be heat. The calculation is much simpler and familiar to everyone in the case of heat, so Josh prefers to focus on it. It is true that it would be interesting to see a detailed calculation of energy loss to radiation. If you can present such a calculation, I would love to see.
Michael, in the case of water it is likely that some of the energy is transferred to sound waves, so the analogy is still appropriate. Of course, the details of the physics of electromagnetic radiation are different from those of sound "radiation", but at least at the principle level there is a similarity here. Also, similar to the case of capacitors and superconductors, it is possible to perform the bucket experiment with a superfluid.
Josh - I repeat - the formulas do not take into account resistances (nor inductances), neither of the conductors, nor of the capacitors nor of the switch. In such a theoretical case, half the energy is radiated. In a practical circle, as Hanan tried, the difference is not significant. If there is resistance in the circuit, we all know how to calculate such simple RC circuits. So weak. The question asked in the article talks about a circle without objections, so why should we constantly refer to objections; It's like solving a fundamental problem in mechanics and insisting on taking friction into account without it being required or given as a given in the problem. And in general, to say that the assumption that the resistance of the conductors and capacitor plates is zero is complicated, it is, to put it mildly, nonsense. It is precisely the opposite assumption, that there is resistance, that complicates the calculations.
Danish:
Maybe we'll do the opposite?
Maybe you will see what it is similar to?
For example - in the situation you describe - what is the analogy of the capacitance of the capacitors?
You can invent some story here, but usually - when describing a capacitor using a water transmission system - the most successful description in my opinion is that of a water tank that is divided by a sealed rubber sheet.
I assume that those who know what Kebal is also understand why this parable is good, so I will spare myself the detail.
In any case - it's just a parable that nicely illustrates the voltage, the capacitance, and the movement of the charge, but it has no analogy to radiation.
This is one of the reasons why there is a difference between the electronics profession and the plumbing profession.
point:
It is not important at the moment to check if there is an analogy between the cups you described and the capacitor.
What I said is that your words do not address the question that was "where did the energy go?"
Do you think they do address this question?
block:
In response 29 I already pointed out that in response 28 you admitted that radiation is also a source of energy loss but because you did so with a weak answer and because you limited it to a superconductor without seeing the superconductor as one end in a sequence of states that are a function of the resistance (when at one end, when the resistance is large, The majority goes for heat and at the other end, when the resistance approaches zero, the majority goes for radiation) I wanted to find a way to illustrate things to fully convince you.
lion:
Perhaps you will see how this matter is different from connecting the interlaced vessels of 2 buckets of water, one full and one empty.
In both cases the water or electric charge does not suddenly appear on the empty side.
There is a process of flow that is caused by the released energy
lion:
Perhaps you will see how this matter is different from connecting the interlaced vessels of 2 buckets of water, one full and one empty.
In both cases the water or electric charge does not suddenly appear on the empty side.
There is a process of flow that is caused by the released energy.
You do not need to perform this test, in the second answer I did correct myself, and said that in a normal conductor there will also be a loss of energy through radiation and the majority will be lost to heat (as is obtained if you do the integral)
The more interesting question is what happens with a conductor I answered several answers (none of which may be correct)
The capacitors themselves are made of normal conductive material, so there is resistance (the simplest answer).
If you complicate things and say that the capacitors are also above a conductor, you can say that the conductor has a critical current above which it behaves like a normal conductor, and since the discharge current that will flow through the wire will be infinite (after all, there is no resistance), so anyway the superconductor will lose its properties (and will be a conductor without resistance) because there is A critical current above which it is a normal conductor
You didn't take into account the self-inductance the circuit will have and that also creates resistance
It is possible that there will be an oscillation between discharge and charge all the time so that the majority (and the emphasis is the majority) will go to radiation here
It is also possible that all the answers are correct
Hi Michael,
Why is the example with 2 glasses of water that connect them not similar?
I thought that the law of conservation of charge paralleled the law of conservation of mass and that the required potential equalization was the same whether it was electrical or gravitational.
So what is the fundamental difference between 2 glasses of water connected to 2 capacitors?
Josh I suggest that you sum up the work done by the resistor from zero to infinity (meaning the integral over time)
And this way you will see how much energy is "wasted" in the form of heat as a function of the resistance
OK; I am not currently thinking of a way to measure the amount of radiation energy. For this you need an array that, in my opinion, cannot be realized by the means known to me (an array like a very, very wideband antenna, which spherically embraces the capacitors, when they are connected to each other). It is possible to demonstrate without measuring - using a radio receiver that will tune to different frequencies and in all of them a click will be heard when connecting the capacitors. Besides, Josh specifically wrote in response 28 that there is indeed radiation at resistance 0 and if the connection is through a resistor then most of the energy goes to heat losses in the resistor, meaning he agrees with us.
lion:
It seems to me that you forgot in what context the test was offered.
It was offered to convince Josh that there is energy lost in radiation.
It does not seem to me that any examination of those which you have proposed is likely to convince him of this.
If neither heat nor radiation are measured, they can only confirm the claim that energy was lost, but this does not determine how it was wasted.
Michael - measuring current parameters as suggested by Hanan are fine. Such a measurement responds only to the current parameters and not to the radiation, and there is no problem integrating over the discharge time. Hanan and I know electricity well enough for you to trust us on this matter. In any case there is no need to try it. A practical resistor even with a very small value will reduce the radiation to minimal values and it is possible to show that most of the energy will be wasted in the resistor on heat. In general, things are quite clear and known and there are no magic formulas here, as the impression may have been created in the article.
Hanan:
I wrote "graphite" because I remember that as a child I would build myself improvised solders by creating an electric circuit consisting of the filling of a copper pen with a small graphite tip mounted on it (on the perforated side).
I would wrap the filling in some material that insulates heat and connect to its other end (one that does not contain graphite) an electrical wire that is connected at one end to one side of a battery. From the other side of the battery I would take out another wire.
When I wanted to solder something into an electrical circuit, I would touch the piece of tin I wanted to melt with the free end of the electrical wire and the graphite tip.
The closed circuit caused the graphite to instantly heat up and melt the tin.
In other words - the graphite is a material that converts current energy into heat that is felt.
I hoped that using such a resistor would simplify the experiment because the heat differences that would be created would be clear enough so that there would be no need to measure the calories.
Measuring the energy wasted on the resistor will not give exactly what we are looking for because it will not be clear from it whether the energy was wasted on heat or radiation.
In addition to this - it is about the unloading process and not a static situation and it is not clear to me how you will be able to follow everything that is happening in it at its various moments.
If you have a standard simulation software that also calculates radiation or heat emission, although it will not be a real experiment, it will nevertheless receive some support for one of the parties because if this software, which serves professionals, made a mistake in the calculations, people would surely overcome it.
Michael
Does the graphite represent resistance for you? If so, I can mount a normal resistor.
Regarding the experiment, I don't have the means to measure the calories that will develop on the resistor, but I can download the current or voltage curve as Scope allows me and from that calculate the energy that was wasted on the resistor.
It is much simpler to perform an electrical simulation, but it is no longer an experiment.
good day everybody.
You still haven't considered what I said about superconductors
Hanan:
As someone who has already demonstrated skill in conducting experiments - could you check for Josh if there is a difference between the amount of heat generated when connecting the capacitors with copper wire and that generated when connecting them using graphite (the material from which the filling of a pencil is made)?
I guess it's difficult because you need very serious capacitors for something to get measurably hot, but maybe you have the necessary means for that?
block:
I'm not angry - I was just puzzled by your words and tried to explain to you why.
I failed, but that's okay with me.
I really don't understand why you are upset, I answered the specific question of connecting two capacitors at the same time, read the post carefully and you will see that this is what he asked, obviously it is not about any electric circuit.
And you did not refer at all to what I answered about conductors
block:
First of all - now you already admit that part of the energy will be lost as radiation - something that contradicts your previous words.
Besides, you claim that the amount of energy that will be lost in heat does not depend on the resistance and the question arises whether this is what you claim only for the specific configuration of two capacitors connected in parallel or for any electric circuit.
How do you think radio broadcasts are produced? Why doesn't the transmitter energy turn mostly into heat?
I mean - I have no doubt that you are wrong - I'm just trying to make you understand this too.
I have already answered what happens (in my opinion) if it is a conductor (zero resistance) see answer 26
An accelerated electric charge does radiate, but still in a resistive conductor most of the energy goes to dissipation
block:
Maybe enough with the war against the laws of nature?
Are you really trying to argue that an accelerating electric charge and changing electric field does not create electromagnetic radiation?
If you prove it, you will deserve a Nobel Prize, but since it is impossible to prove what is not true, it is likely that you will not even receive an Ignobel.
What do you think would happen in a situation of zero resistance? Will the capacitors never balance?
If you do an integral over the discharge time, which is in principle from zero to infinity, the result obtained is indeed half the energy regardless of R
Regarding the question of a conductor, it can be answered as follows:
The capacitors themselves are made of normal conducting material, so there is resistance. If you complicate things and say that capacitors are superconductors too, you can say that a superconductor has a critical current above which it behaves like a normal conductor
You didn't take into account the self-inductance the circuit will have and that also creates resistance
block:
Come on!
You have already written everything here and you are starting from the beginning!
Do you disbelieve in the fact that electromagnetic radiation is created that carries with it some of the energy?
Does, according to your logic - the heat generated does not depend on the resistance at all?
Josh - it was not said that the capacitors are connected by a conductor of resistance R. Theoretically it is a perfect conductor and then, yes, amazingly, half the energy "disappears". It is not true that it can be shown that all half of the energy (even if the conductor has a certain resistance), turns into heat. If there is kinetic energy, then it is of molecules (their movement due to heat) and it is not in the language of electrons. And the "school solution" can be found in the previous comments.
When you connect the capacitors with a conductor of resistance R, a discharge occurs, and it can be shown that all half of the energy goes to heat (of the conductor with the resistance).
In the language of electrons it can be said that half of the energy goes to kinetic energy.
Hanan:
You have proven that you are a scientist at heart.
The experiment is always the final arbiter and publishing the results even when they contradict the opinion you expressed earlier is indeed the noble act expected of a scientist.
No one is wiser than that who has experience.
I did a simple test and indeed the voltage drops to half which means half the energy is gone.
So today I eat the hat.
lion:
It is not fair to claim that those who talk about opposition are wrong - for two reasons:
The first reason is that in reality some of the energy is lost due to the resistance,
The second point is that in formulating the question you said nothing about the objection and therefore you had no reason to expect people to understand that there is no objection.
The only formulas you referred to are those of the capacitance of capacitors connected in parallel.
These formulas are indifferent - both to the subject of resistance and to the subject of radiation.
All those who attribute the loss of energy to the resistance losses of the conductors (or the connection switch), both in the responses to this article and in the responses to the link brought to us by Shlomi are wrong. In fact, even those who answer the question as asked that part of the energy goes to heat losses in the resistance of the conductor and part to electromagnetic radiation - are wrong. The formulas do not take ohmic losses into account, but analyze an ideal situation where there is no resistance and then the energy loss is exactly half and this energy becomes electromagnetic radiation due to the current pulse in the transition of the charges to a capacitor. It's just like we solve problems in mechanics and ignore friction. So it is true that in the circuit of his actions there is resistance and a small part of the energy goes to the resistance of the conductors (then the radiation energy is not exactly half). Again, the exact answer to the question as asked is - electromagnetic radiation, as only a few of the respondents to the link answered. By the way, when you short a capacitor - all the energy goes (trivial), theoretically - to radiation.
Ethan - the question of what happens in a charged capacitor if its plates are (theoretically) allowed to attract each other, or if we apply a force and move them away from each other, is an interesting question in itself, but not relevant to the question we presented.
The solution of the solution is also not accurate: a frequency determined according to the properties of the capacitor and the wire, i.e. his claim that there is a resonant circuit - again the formulas speak of a theoretical circuit without resistance and without inductance, so it is not correct to claim that the electromagnetic energy is at a certain frequency (and there are also ohmic losses). According to the Fourier decomposition - a current pulse as in our case, radiates in the entire spectrum.
Hanan - the voltage drops to half and not to the root 2. You are trying to cancel the "loss" of energy in your claim, but energy does escape - in radiation.
Reut - coordination of impedances for maximum energy transfer, relevant when there is a resistive component. In the question presented here - theoretically there is no impedance (resistance + inductance or capacitance), but only capacitive feedback (reaction = inductance or capacitance) and in a network of pure feedbacks there is no energy loss.
Devil's advocate:
If you read my words carefully, you will see that I did not try to follow any special mathematical theory at all.
On the one hand, I talked about the possibility of using inappropriate models for the purpose of describing a physical phenomenon. Such usage can be completely correct mathematically but meaningless in the physical world.
On the other hand, I was talking about old mathematical teachings that are well integrated into physics and yet are not fully expressed in it. I brought as an example the term "85 dimensional space" but it was only an example.
Infinity, for example, is a legitimate, useful, and even essential mathematical term in the development of the formulas used by us in physics, yet we believe that there is nothing infinite in physics, and "infinities" obtained in calculations based on an existing theory in an attempt to describe a physical phenomenon are seen as a point of failure of the theory (that is, there is This is actually using the phenomenon I described under the heading "on the other hand" to identify a situation of an inappropriate model which I described under the heading "on the one hand").
From * Al
It is said, if I am not mistaken, about the Norwegian mathematician Lee that when he finished thinking about his bruises, he rubbed his hands together with pleasure and announced that he finally had a mathematical field that the physicists would never pollute with their research. And on this it was said "another vision for the time". And it is not certain that it is necessary to move away to other universes (if it is far away).
Aviran
The physical reality known to us does not really coincide with the one perceived by our senses (it depends on the definitions, there is no debate on the matter), it is not certain that the universe coincides with which of them, yet it is probably ours.
Michael: I claimed as you claimed. I wrote in that response - on which I actually made a comment related to Gadel's theorem (in the article to which I provided a link) - that physics is a special case of mathematics, in the sense that it is a truly contained subgroup. In other words: all physics is mathematics, but not the other way around (of course, it is possible that all mathematics is expressed physically, but not in our universe - for example, in regards to the number of dimensions, mathematics allows such a number for you to choose, while in our universe, this means the physical reality known to us or perceived by our senses is only three-dimensional).
The solution is very simple: at the moment of connection, a current passes between the capacitors and the current "spends" energy on the resistance of the wires.
To calculate this, you have to assume that the resistance of the wires is very high (aims to infinity) and then the current is weak, otherwise the current will be strong and then a significant electromagnetic field will also be created, which is also a form of energy, only that it is more difficult to calculate.
By the way, I solved this problem already in high school.
I am interested in why people keep trying to solve the problem after a solution has already been presented.
The energy will be dissipated as both heat and electromagnetic radiation.
The smaller the resistance of the conductor, the more the energy loss will tend towards the radiation.
The higher the resistance, the more energy will be lost in the form of heat.
Regarding the big question - this is simply a result of the fact that our models reflect reality well.
The formulas for calculating the capacitance of capacitors connected in parallel were not created in a vacuum. They were tested experimentally and found to be correct. So are the formulas regarding the charge distribution over a conductor.
If they are true, and if the law of conservation of energy is true, it follows that energy must be lost.
If we used incorrect formulas for the capacitance of capacitors connected in parallel, then even though the math was correct, we might not have discovered that energy must be lost.
Since we were also able to discover the fact that a charge undergoing acceleration - or a changing magnetic field produce electromagnetic radiation and also the fact that a charge flowing in a resistive medium creates heat, we can also understand where the energy went and receive additional confirmation of the correctness of the assembly.
It should be understood that if the results did not add up - this would not detract from the correctness of the mathematics, but only from the correctness of the way we chose to represent the physics within it.
Knight:
I read the comment you voted on.
At the time I didn't turn to her because I was busy arguing with Lisa, but from what I said at the beginning of the current response you can understand that I don't agree with your claim that mathematics and physics are the same.
Definately not.
Nor can it be, since mathematics embraces fields that have no realization in physics (such as 85-dimensional space or the Klein bottle).
Of course, I'm only talking here about the physics that exists in our world.
It may be that other universes have other laws and that in the end every mathematical structure has a physical representation in some universe, but this is a speculation that we cannot confirm or refute. In any case - this is the speculation represented in Tagmark's words and it is different from the interpretation you tried to give it.
Reut
Energy is never "wasted" because there is a law of conservation of energy.
Energy can only transform into another kind of energy.
It is true that we talk about wasting energy, but this term refers to wasting energy that is available for use and turning it into energy that is not available for use.
The capacitor is an energy source that transfers the energy to a capacitor identical in its properties,
The maximum energy that can be transferred from one reservoir to another occurs when the impedance of the reservoir is equal to that of the receiver of the energy (the load).
And hence half of the energy is transferred and half is wasted on the impedance of the system, in our case the internal impedance of each capacitor.
Just in general, just short between the plates and all the energy will dissipate. It's more or less the same.
Regarding this question, just assume that there is resistance between them at the moment of connection. After that you will inhale it to 0 (it will not depend on the resistor if there is conservation of energy or not). A little algebra that the current is the derivative of the charge, the charge is related to the charge on each individual capacitor, the voltage difference is related to the current (ohm) and that's enough. The heat loss per unit time through the resistor is equal to the potential energy loss (not particularly surprising, 3 lines, it's quite simple). For those who are interested, there are many mathematical questions that can be solved with the help of physics, which is very interesting. Example: Prove that every polygon has at least one perpendicular from the center of mass to a side so that it passes through the side itself (and not in continuation). Answer: Place it on the floor and if the condition is not met it will roll forever and accelerate and preserve Angia, etc. Another one: given a faun (a three-dimensional shape with wigs). For each wig I output a "normal", a vertical vector whose length is the area of the wig. Sum all the vectors, came out 0. Why? Because if not, we'll put it on the gas (pressure) and it will start spinning out of nothing. To think: I need to get from one point to another (not at the same height). Starting from speed 0, what is the shape of the curve so that I can get there the fastest? (frictionless bicycles, etc.).
There is an insane amount of such and most of them are very beautiful and original (based on constant torque, Fermat's law on Snell's law, etc.). Sorry to anyone who had to read this and it didn't really interest them.
My attempt to answer your "bigger" question can be found in the link to my message here:
https://www.hayadan.org.il/for-one-tiny-instant-physicists-may-have-broken-a-law-of-nature-0604107/#comment-265170
The universal aesthetic standards inherent in us - the fruit of the evolutionary creation - are a "pillar of fire before the camp" for those who explore the expanses and landscapes of mathematics. The pleasure involved in the excitement of a screen of elements forming together into a structure with a higher degree of "elegance" than each of its components separately, constitutes a weighty incentive in our pursuit of their invention or discoveries. The belief in a clear foundation of "elegance" in the fields of exact science serves as a guideline that helps in the discoveries of the most abstract and therefore also the most compact structures. It is these structures - precisely because of their nature - that led to the rise of the so-called "queen of the sciences", associated with mathematics.
Nice puzzle.. You should make this a daily section…
I found it:
http://www.edaboard.com/thread168490.html
And by the way: mathematics does not "know" anything.
In total, we require conservation of charge and equalization of potentials and the result is that the energy is cut in half.
Where did she go? The link explains it pretty well, although I haven't tested it myself.
And always always that the energy disappears someone/something is doing work. (In a more colloquial language it is called dissipation).
The solution presented by "the solution" is correct.
Aryeh also knows that this was my answer as part of a conversation we had before the article was published.
Hanan's solution is wrong - the result of halving is the one obtained in the calculations.
Point's answer does not address the problem at all.
Regarding the question presented by the "solution" - it seems to me a bit strange in its wording. Probably the intention is that the galaxies actually gain potential energy of gravity, and even kinetic energy and perhaps where does this energy come from should have been the question.
The answer to this is not really known, but it is called "dark energy":
http://en.wikipedia.org/wiki/Dark_energy
http://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy/
Obviously, during the connection of the capacitors, the voltage will not drop to half the voltage of the charged capacitor, otherwise the law of conservation of energy will be violated, and indeed the voltage will only drop to half the root, that is, to 0.707 of the value of the full capacitor.
It is indeed a beautiful problem that the intuition of the three children.
At the moment of connecting the capacitors, oscillations will be created at a frequency determined by the properties of the wire and capacitors, after which the energy will dissipate in the form of heat energy and electromagnetic energy.
And now for a real question?
If the universe is expanding at an accelerated rate then all galaxies are losing altitude energy?
Where does the consumed energy come from?
There is an answer to that, but we'll see you first.
The appropriate example is a glass full of water to which a second glass is attached. The water level will decrease by half and the total energy will decrease by half.
Our hypothesis is that there is potential energy in the capacitor, after all there is a force of attraction between the plates and as the distance between the plates changes, so does its energy. (both the electric and the potential related to the force acting between the plates) there is no heat loss. We haven't played with the formulas but this is the most logical explanation.
What is more intriguing is what exactly needs to be understood here.
To match with the example you gave, it can be seen as if you connected at the same time a get equal to it.
and then get a capacitor equal to the sum of the previous two and so on.
This is easy to understand if you connect a capacitor of infinite capacitance to the charged capacitor.
It's as if you shorted the plates and the energy turned into heat as it flowed.