It is generally agreed upon that electromagnetic waves from an emitter do not have to connect to a receiver, but how can we be sure this is a fact? The problem is that we can never observe non-received EM-Waves, because if we observe them the instrument of observation becomes a receiver.

Electromagnetic waves have changing electric and magnetic fields and are both electric and magnetic. Electric current connects like from an anode to a cathode. Magnetic fields illustrated by flux lines connect from one magnetic pole to another, and no non-connecting flux lines are observed.

So electric currents connect and magnetic fields connect, so why doesn’t the electromagnetic wave also always connect to a receiver? A receiver which could be a plasma particle, a planet, a star and anything else which can absorb EM-radiation.

There is one big problem. If a photon has to be emitted in the direction of a future receiver, the photon must know where a future receiver will be. So this conflicts with our view on causality, or a cause creating an effect. And as the emitter doesn’t know where the receiver will be some time in the future, it can't emit an EM-wave against it.

But how can we know that the causality principle is always valid without exceptions? There seems to be reasons for questioning the universal validity of the causality principle:

  • Information does not have a mass and may then not be restricted by the speed of light, so the causality principle may not always hold for massless particles/waves.

  • When something travels with the speed of light, it will experience that distance become zero. If there is no distance, there is a full connection and a continuous electromagnetic wave between the emitter and receiver. Again, using the photon as a reference frame is not something relativistic physicists seem to like.

  • Maxwell's electromagnetic wave equation has a simple and an advanced solution. The advanced solution is usually discarded because the effect happens before the cause. But in Wheeler–Feynman absorber theory the advanced solution is used because it works. See this link for more information: http://en.wikipedia.org/wiki/Wheeler%E2%80%93Feynman_absorber_theory

  • The field of quantum mechanics is discussing many different causality problems. Like the observation of a particle might decide where the particle will be in time and space. Relevant to this discussion is the question of what triggers the atom to emit light:

Over the last hundred years, physicists have discovered systems that change from one state to another without any apparent physical “trigger.” These systems are described by quantum mechanics.

The simplest such system is the hydrogen atom. It’s just an electron bound to a proton. Two particles – that’s about as simple as you can get. According to QM, the electron can occupy one of a discrete set of energy levels. The electron can be excited to a higher energy level by absorbing a photon…

When the electron drops from a higher energy level to a lower level, it emits a photon: a quantum of light…

Quantum mechanics describes this process beautifully, but it only predicts the average time the electron will stay in the higher energy level. It doesn’t give any clue as to the specific time the electron will drop to the lower level. More precisely, the transition rate (the probability of a transition per unit time) is constant: it doesn’t matter how long it has been since the atom was excited, the transition rate stays the same…

When you first encounter this, you can’t quite wrap your brain around it. Surely there must be some internal mechanism, some kind of clock, that ticks along and finally “goes off,” causing the transition!

But no such mechanism has ever been found. QM has had an unexcelled record of accurate predictions, without any need for such a mechanism…” -George Mason University physicist, Robert Oerter

So is the excited atom a random generator or is it something external that triggers the release of a photon? It seems like it’s something external, and this external trigger might be the unphysical connection to a future receiver described by the advanced solution to Maxwell’s equation of electromagnetic radiation.

So it seems to me like we currently can’t be sure if a photon is always emitted against a receiver, or it is emitted randomly in any direction into space. But this question might be one of the most important questions ever asked, because if an electromagnetic wave is always connected to a receiver the implications are vast. It could shed light on the discussion of many topics. It might change our view on time and space. It might not only be the past pushing the present forward, but the future pulling on the present, making a syntropy which will create order out of chaos, and describe the marvelous universe we live in. Even the view of the present itself as a sharp line between the past and the future could be questioned. Time itself might not be totally linear, and the future may change the past. To avoid paradoxes with time travel we have to allow a number of parallel universes, as suggested by American physicist Hugh Everett who formulated the idea of their existence to explain the theory that every possible outcome of every choice we have actually does happen.

But before we can fully dive into all these fascinating questions, we have to solve this question:

Does an electromagnetic wave always have to connect to a receiver?

This hypothetical question might seem purely philosophical, but it is not. And it might even be confirmed by observations. We can’t directly observe non-received photons, but we might indirectly observe the existence or nonexistence of these photons. Any answer or suggestions are most welcome.

  • $\begingroup$ It is hard to imagine our universe without charged particles. It has as many emitters as absorbers (receivers). $\endgroup$ Feb 14, 2013 at 14:50
  • $\begingroup$ "According to the results of the double slit experiment, if experimenters do something to learn which slit the photon goes through, they change the outcome of the experiment and the behavior of the photon. If the experimenters know which slit it goes through, the photon will behave as a particle. If they do not know which slit it goes through, the photon will behave as if it were a wave when it is given an opportunity to interfere with itself." en.wikipedia.org/wiki/Wheeler%27s_delayed_choice_experiment $\endgroup$
    – Enos Oye
    Feb 25, 2013 at 14:14
  • $\begingroup$ I always found it amazing that if one claims that photon must have real spin or electron must have extension, physics calls him naive, and yet the same physics plays with time working back, etc. Apparently, definitions of "naive" have changed over the years. (Unfortunately, you are late with your concept. Holger Bech Nielsen and Masao Ninomiya claim that Higgs boson works from the future and malignantly caused (will have caused?) LHC to brake down to prevent its own discovery: nytimes.com/2009/10/13/science/space/…. You are in a good company, Enos Oye) $\endgroup$ Jun 20, 2014 at 7:18
  • 1
    $\begingroup$ @brightmagus you know that particle was found in 2012, right? $\endgroup$ Jul 15, 2014 at 19:38
  • 12
    $\begingroup$ Isn't this just, "If a tree falls in a forest..." but with bigger words? $\endgroup$
    – Señor O
    Jan 23, 2015 at 21:35

6 Answers 6


Richard Feynman's PhD thesis was about just this topic, if I am understanding your question rightly. Here is an earlier question about Feynman's thesis that addresses some of the fascinating issues involved with this.

At the suggestion of his thesis adviser John Wheeler, Feynman explained photon emission as a two-way interaction in which the regular photon is emitted and follows the "retarded" solutions to Maxwell's equations. "Meanwhile" (in some rather abstract sense of the word indeed) a target atom or particle in the distant future emits its own photon, but a very special one that travels backwards in time -- a type of solution to Maxwell's equations that had been recognized since Maxwell's time but had been ignored. These solutions were called the "advanced" solutions. This advanced photon travels back in time and "just happens" to arrive at the source at the exact instant when the regular photon is emitted, causing the emitting atom to be kicked backwards a tiny bit.

Amazingly, Wheeler and Feynman were able to write a series of papers showing that despite how mind-boggling this scenario sounded, it did not result in violations of causality, and it did provide a highly effective model of electron-photon interactions. From this start, and with some important changes, Feynman eventually produced his Feynman-diagram explanation of quantum electrodynamics, or QED. The curious time relationship continue in Feynman's QED, where for example a positron or anti-electron simply become an ordinary electron traveling backwards in time.

Staying fully consistent with his own ideas, Feynman himself described photon interactions as always having an emission and a reception event, no matter how far apart those events occur in ordinary time. In his view, if you shone a flashlight into deep space, the photons could not even be emitted until they found their "partner" advanced photon emission events somewhere in the distant future. The proof of it is in the very slight push back on your hand that happens when you shine the light, that kick coming from the advanced photons arriving from that distant point in the future and nudging the electrons in your flashlight filament.

  • $\begingroup$ You nailed my question very well. I will look into Feynman/Wheelers work, at first glance it seems like the theory works very well. Currently I don't see how backwards time travel with a photon do not violate causality, but I will do my homework. And if I got it right this is just weird: When I look at a distant star, my eyes emit photons that travel backwards in time and nudges atoms in the star and make them emit light back to my eyes. Strange concept. How do we know it is a photon who is traveling backwards in time and not some other triggering energy/informational parcel? $\endgroup$
    – Enos Oye
    Feb 25, 2013 at 8:58
  • 3
    $\begingroup$ It is just weird, delightfully so. Feynman also elsewhere speaks of "instantaneous" photon interactions from the perspective of the photon. A bit overly anthropomorphic, sure, but it also nicely captures the oddity of photons as particles that travel at c and so are not subject to normal time from their "perspective". A "must get" book, if you don't already have it, is Laurie Brown's Feynman's Thesis. Also relevant is this. $\endgroup$ Feb 25, 2013 at 17:07
  • 2
    $\begingroup$ The "instantaneous" part -- the photon perspective -- addresses your question, albeit a bit indirectly: The photon "sees" itself as a single integral unit and interaction. It is only when we try to parse that single interaction from the perspective of classical space and time that it breaks down into what looks like (to us) a forward in time retarded photon and a backwards-in-time advanced photon. $\endgroup$ Feb 25, 2013 at 17:10
  • 1
    $\begingroup$ I have to go with Feynman 100% on this one. Due to his Lagrangian viewpoint, Feynman flatly did not see time as working the same way most of us do. When folks take the interpretation that a photon can just radiate to "somewhere", all they are really doing is taking a local perspective of time where they don't care what happens after that. But QED, which is heavily based on Feynman's thesis ideas, remains one of the most accurately predictive physical theories of all time. So I think Feynman glommed onto something very deep there, and that includes photons needing two partners to exist. $\endgroup$ Feb 26, 2013 at 12:03
  • 3
    $\begingroup$ Surly you must be joking mr. Bollinger! Because the implications of your answer is just mind blowing! We cause stars to emit light many millions of years ago, creating a change, cascading and amplifying into the future. So just by existing we change the whole universe for every choice we make. And as we apparently still exist, the change we create in the past must create an infinite number of parallel realities. I can’t fully accept this answer to this fundamental question without further proof, so we have to question it further, and the quest continues. $\endgroup$
    – Enos Oye
    Feb 27, 2013 at 9:14

You clarify in the comments to @FredericBrünner 's answer:

The question is Can a photon get emitted without a receiver?

Yes. An atom in an excited state will emit a photon into space, vacuum, whatever

And it seems like photons that don't hit a receiver never can be measured,

Wrong. If you set up an experiment with atoms at an excited state you know that a photon has been released by finding it at the ground state. That is a definite measurement.

so its hard to test if they are there.

If you want to test for the existence of photons you have to have something that can interact with them, yes. It is not hard.

But the energy input of a light bulb in space will produce a certain amount of photons,

Whether in space or not this is true. The sun is a huge light bulb in space

and if we have a connected receiver we would expect a rise in measured radiation,

our eyes connect with sunlight and they do measure the electromagnetic radiation . Different detectors are needed if the radiation is absorbed and turned into heat.

if all photons must be connected to a receiver.

No. This is a wrong premise. The flux of light/em-waves from the sun can be calculated accurately and we know it disperses the same photons per unit area at the same distance from it whether there exists an absorbing or reflecting body or not.

Its a observational experiment which might fully dismiss or confirm the hypothetical question.

certainly the hypothesis that a photon has to have a receptor is dismissed from the experiment with the sun.

  • 1
    $\begingroup$ Of course I agree that photons are emitted and I understand that we can measure the energy state of an atom and prove that a photon has been emitted. I also understand that the observed concentration of photons is the same at the same distance from the emitter. But the instrument of observation serve as a receiver, so it seems like its almost impossible to directly observe photons that don't hit a receiver. Then, how can we know for sure that they are there? They are probably there, but if a photon needs a receiver to be emitted the non-interacting photons are not there. $\endgroup$
    – Enos Oye
    Feb 15, 2013 at 9:07
  • $\begingroup$ That is where mathematics comes in, both classically and quantum mechanically. When a rocket is directed against a building you only know a rocket will hit the building by mathematics. Nevertheless, if you are in the building and a rocket is coming and your detecting system projects it there, you do not stay there to verify it. the light from the sun follows a mathematical formula, and it is tested continually with any new satellite sent up, let alone the existing planets and their reflections. The light falls on the objects where the calculations say it should. It then it is just $\endgroup$
    – anna v
    Feb 15, 2013 at 9:27
  • $\begingroup$ a philosophical point you make here. Mathematics allows us not to do the experiments placing detectors all over the place because it is a shorthand of a lot of measurements that led to the distillation of the mathematical formula, so one need not put a detector to know that there will be something there to detect. $\endgroup$
    – anna v
    Feb 15, 2013 at 9:29
  • 1
    $\begingroup$ In the field of quantum mechanics, the observation of a particle can decide where the particle will be. This seems strange to us because it has some problems with causality or the timing of cause and effect, but it explains the observations. If we have particles with mass like from an explosion I agree that they will be more or less randomly spread in all directions, and we don't have to observe all the particles to understand this. But the photon is a mass less electromagnetic wave with the speed of light in vacuum, and might have a different behavior, like a wave connection to a receiver. $\endgroup$
    – Enos Oye
    Feb 15, 2013 at 10:19
  • $\begingroup$ If photons have to have a receiver, it will to us seem like the effect happens before the cause, so it conflicts with our view of causality. But the advanced solution to Maxwells equations and the electromagnetic wave equation has just such a solution where the effect happens before the cause (See the link in my post). So can this syntropi which seems unbelievable to us be real? Magnetic fields connects, and electric currents like a spark connects, so do the electromagnetic wave also have to have a connection to a receiver? I don't know, and the hypothetical question still seems unsolved. $\endgroup$
    – Enos Oye
    Feb 15, 2013 at 10:20

Do you mean that the emitter, the electrically charged particle, is the only particle in the universe? If that is how you mean your question, then here is a possible answer.

Since the question is about a very hypothetical situation, one must start with a hypothetical scenario, and then build from that position.

Let us imagine there is an electron in space all by itself. The question is:

Can this completely isolated electron emit a photon?

We assume the laws of physics hold in this case as normal.

Some of the facts we do know about the electron

(i) According to classical electrodynamics an electrically charged particle radiates electromagnetic waves only when it is subjected to acceleration, or for some reason it lowers its energy.

(ii) From quantum mechanics point of view the electron cannot be in a state of absolute rest, because then its momentum will increase in unpredictable ways by quantum fluctuations of the vacuum.

(iii) If the electron is moving with constant momentum, then according to the uncertainty principle its position will be totally undetermined, i.e. the electron will be spread all over the space available to it.

(iv) The vacuum has a Lorentz invariant structure, which requires the presence of a positron. This is a result of Dirac’s theory.


According to (i): the electron will not be able to emit a photon. The emission of a photon by an atom, as mentioned in another answer, assumes the electron has absorbed some amount of energy at an earlier time, so it will have to re-emit it, as there is a lower energy level below it. Anyway, in this case the electron is not an isolated particle in an “empty” space as hypothesised.

According to (ii): the electron will accelerate and therefore will emit photons, and it is even possible it will reabsorb them (self energy diagrams).

According to (iii): the energy of the electron would be well defined and constant, hence it would not be able to emit any energy, so no photon emission. If the electron kept emitting photons from that state, it would soon lose all its energy and would end up a massless electrically charged particle!!

According to (iv): the electron cannot be on its own, without the positron. This is necessary by Lorentz invariance of the vacuum. So the electron will exchange photons with the positron, and might even suffer pair annihilation.

Since Lorentz invariance is an inherent property of nature, in my opinion, scenario (iv) is the most likely than any of the others.

  • $\begingroup$ Your answer/question is fascinating. But I did not mean that the charged particle is the only particle in the universe, I meant all particles that can emit photons. Terry here understood my question, maybe even better than myself. He explains that Feynman has researched this topic. Good thing I did not know that in advance because then I wouldn't have gotten the full joy out of it. So following the links I ended up with another theory of Feynman, the hypothetical One-electron universe. The syntropi seems to create fascinating synchronicity;-) en.wikipedia.org/wiki/One-electron_universe $\endgroup$
    – Enos Oye
    Feb 22, 2013 at 10:01
  • $\begingroup$ I don't think it is possible to have a one electron universe, because the electron is not neutral and it has mass. It is then not in charge balance and are restricted by the speed of light and has momentum. But a universe made up by a mass less prime radiant going at infinite speed way above C, going everywhere at the same time and interacting with itself, creating gravity, particles, magnetism, electricity and the universe itself, could be a possible neat solution. Then everything is one thing, and all is one, and everything is connected. And if all is connected, the photon is no exception. $\endgroup$
    – Enos Oye
    Feb 22, 2013 at 10:37
  • $\begingroup$ @EnosOye I am glad to see that your conclusion agrees with mine (see bold faced text at the bottom of my answer.) I am not sure I would extrapolate that conclusion, to include all the interesting effects you are mentioning. The known universe is in a way interconnected this way via the laws of quantum mechanics. It is an "undivided universe" as David Bohm said! $\endgroup$
    – JKL
    Feb 22, 2013 at 18:40
  • $\begingroup$ We agree on that there are a specific number of building blocks with equal properties that make up our universe. Some break these further down to strings and the string theory. But to create equal particles like electrons, protons or quarks...we need a specific string with set properties to be the prime building block. And if this string has no mass and is infinitely fast and can interact with itself to create more and more complex structures, it could be just one string creating the whole universe. So how much energy does it take to accelerate a mass less string to infinite speed? Zero? $\endgroup$
    – Enos Oye
    Feb 25, 2013 at 9:35
  • $\begingroup$ Can anything without mass, like energy or information, have a speed above the speed of light? This link is interesting: nature.com/news/2008/080813/full/news.2008.1038.html Are there really a speed limit for mass less particles? This is also relevant to Terry's answer. $\endgroup$
    – Enos Oye
    Feb 25, 2013 at 10:03

First of all, the "point of view" of a photon is not well defined. One can not use a Lorentz transformation in order to get to the rest frame of a photon.

Furthermore, a photon is a physical entity of its own which can exist independently of any receiver. In principle, it can go on "forever" without ever being absorbed by something.

Regarding the tree in the forest: Suppose I throw a baseball as far as I can into empty space. Even if I never hear about it again and even if nobody catches it, it is nevertheless real.

  • $\begingroup$ I think I see the problem with the photons frame of reference, if the distance between the emitter A and receiver B becomes zero, A and B merge into one point. Relative to this point of reference we have instantaneously transfer of energy and information. If the photon is slowed down when transferring through the thin plasma of space, then we might have a relative distance between A and B. $\endgroup$
    – Enos Oye
    Feb 14, 2013 at 12:02
  • 1
    $\begingroup$ Yes, it is. To better understand its nature, you should realize that within quantum field theory, photons are described as field quanta of the electromagnetic field. If you haven't studied the mathematics behind this statement so far, you should take it like this: a photon is no "continuous wave"; it can be envisioned as a localized particle moving through spacetime. $\endgroup$ Feb 14, 2013 at 12:18
  • 1
    $\begingroup$ If you neither send them out nor detect them, you don't know that the photons are there. $\endgroup$ Feb 14, 2013 at 12:32
  • 1
    $\begingroup$ I might see another possible observational method of photons in space. Photons can accelerate electrons by Wakefield acceleration without being absorbed. So if we can observe the accelerated electrons or the magnetic field they create, we may indirectly observe photons. This link may have nothing to do with this, but it could and is interesting reading anyway: science.nasa.gov/science-news/science-at-nasa/2008/30oct_ftes $\endgroup$
    – Enos Oye
    Feb 15, 2013 at 12:24
  • 1
    $\begingroup$ I don't see how I didn't answer the question, but here is my answer again: Yes, it can. $\endgroup$ Feb 17, 2013 at 4:57

The Cosmic Microwave Background includes photons that will not be absorbed before the universe inflates to the point where there is nothing to hit ever again. If parts of the last scattering were somehow barred from releasing that energy, I think we would notice.

The photon carries energy. Particles do that. How is it any different from the electron, which is allowed to exist without complaint? If a flashlight (or photon rocket) emits particles bearing momentum, it recoils without any regard to what happens to the exhaust later elsewhere. I wonder if the problem is getting confused with non-radiative energy-bearing fields and so-called "virtual photons"? A charged object won't react without another charged object exchanging virtual photons with it.


yes this is how things are... on the other hand the QED makes it more clear (having Feynman started from the absorber theory to get into QED). However, in this regard, I can suggest two experiments that indicate that this is actually the correct direction of interpretation: the Purcell effect (I have a paper entitled <>) and <>

  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 6, 2022 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.