Why don't electron-positron collisions release infinite energy?What happens where an electron is annihilated by a spontaneously generated positron-electron pair?Will the absorption of high energy gamma rays end up heating the absorping material?Positron and other particlesWhat happens to K.E. in matter antimatter annihilation?Potential energy function for high energy continuum?Black body radiation at atomic levelParticle production at angle relative to beam axisChemical bond and virtual particlesMuons produced in electromagnetic particle cascadesRelease of energy during electron jumps

Why dont electromagnetic waves interact with each other?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

Modeling an IPv4 Address

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

Why are electrically insulating heatsinks so rare? Is it just cost?

How to test if a transaction is standard without spending real money?

Minkowski space

How to find program name(s) of an installed package?

How much RAM could one put in a typical 80386 setup?

In Japanese, what’s the difference between “Tonari ni” (となりに) and “Tsugi” (つぎ)? When would you use one over the other?

Can I make popcorn with any corn?

Problem of parity - Can we draw a closed path made up of 20 line segments...

Accidentally leaked the solution to an assignment, what to do now? (I'm the prof)

How does one intimidate enemies without having the capacity for violence?

How to say job offer in Mandarin/Cantonese?

Fully-Firstable Anagram Sets

Arthur Somervell: 1000 Exercises - Meaning of this notation

"You are your self first supporter", a more proper way to say it

Why can't I see bouncing of a switch on an oscilloscope?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

How to write a macro that is braces sensitive?

Why Is Death Allowed In the Matrix?

If I cast Expeditious Retreat, can I Dash as a bonus action on the same turn?

Prove that NP is closed under karp reduction?



Why don't electron-positron collisions release infinite energy?


What happens where an electron is annihilated by a spontaneously generated positron-electron pair?Will the absorption of high energy gamma rays end up heating the absorping material?Positron and other particlesWhat happens to K.E. in matter antimatter annihilation?Potential energy function for high energy continuum?Black body radiation at atomic levelParticle production at angle relative to beam axisChemical bond and virtual particlesMuons produced in electromagnetic particle cascadesRelease of energy during electron jumps













21












$begingroup$


Questions of the form:




An electron and a positron collide with E MeV of energy, what is the frequency of the photons released.




quite often come up in my A Level course (for often fairly arbitrary E). But this got me thinking. There is energy stored in the separation of an electron and a positron, which, as they get closer and closer together, should all be converted into kinetic energy. As the potential is of the form $frac1r$, this implies that at arbitrarily small distances and arbitrarily high amount of energy is given off. Given that both electrons and positrons are typically regarded as point particles, in order for them to collide, they would have to be arbitrarily close together, which would imply that over the course of their collision they should have released arbitrarily high amounts of energy, in the form of kinetic energy. As this would imply photons of arbitrarily high frequency given off, I assume that I must have missed out some piece of physics somewhere, but I am uncertain where. Ideas I have had so far include:



  • Energy should be given off, anyway, by an accelerating electron, in the form of light, according to classical EM, although I don't know how this changes from classical to quantum ideas of EM - we certainly can't have all the energy given off in a continuous stream, because we need quantised photons, so does the electron itself experience quantised energy levels as it accelerates inwards (my only issue with treating the electron in such a quantised way is that, to my mind, it'd be equivalent of treating it mathematically as a hydrogen-like atom, where the probability of the electron colliding with the positron is still extremely low, and unlike electron capture, there'd be no weak force interaction to mediate this 'electron-positron atom').

  • The actual mechanism for the decay occurs at a non-zero separation distance, perhaps photons pass between the two particles to mediate the decay at non-infinitesimal distances.

  • At relativistic speeds our classical model of electrodynamics breaks down. Now, I know this to be true - considering the fact that magnetism is basically the relativistic component of electrodynamics. However, given the fact that magnetism is the only relativistic force which'd be involved, I don't see how it'd act to counteract this infinite release of energy - so is there another force which I'm forgetting?

These are just ideas I've come up with whilst thinking about the problem, and I don't know if any of them have any physical significance in this problem, so any advice is appreciated!










share|cite|improve this question









$endgroup$
















    21












    $begingroup$


    Questions of the form:




    An electron and a positron collide with E MeV of energy, what is the frequency of the photons released.




    quite often come up in my A Level course (for often fairly arbitrary E). But this got me thinking. There is energy stored in the separation of an electron and a positron, which, as they get closer and closer together, should all be converted into kinetic energy. As the potential is of the form $frac1r$, this implies that at arbitrarily small distances and arbitrarily high amount of energy is given off. Given that both electrons and positrons are typically regarded as point particles, in order for them to collide, they would have to be arbitrarily close together, which would imply that over the course of their collision they should have released arbitrarily high amounts of energy, in the form of kinetic energy. As this would imply photons of arbitrarily high frequency given off, I assume that I must have missed out some piece of physics somewhere, but I am uncertain where. Ideas I have had so far include:



    • Energy should be given off, anyway, by an accelerating electron, in the form of light, according to classical EM, although I don't know how this changes from classical to quantum ideas of EM - we certainly can't have all the energy given off in a continuous stream, because we need quantised photons, so does the electron itself experience quantised energy levels as it accelerates inwards (my only issue with treating the electron in such a quantised way is that, to my mind, it'd be equivalent of treating it mathematically as a hydrogen-like atom, where the probability of the electron colliding with the positron is still extremely low, and unlike electron capture, there'd be no weak force interaction to mediate this 'electron-positron atom').

    • The actual mechanism for the decay occurs at a non-zero separation distance, perhaps photons pass between the two particles to mediate the decay at non-infinitesimal distances.

    • At relativistic speeds our classical model of electrodynamics breaks down. Now, I know this to be true - considering the fact that magnetism is basically the relativistic component of electrodynamics. However, given the fact that magnetism is the only relativistic force which'd be involved, I don't see how it'd act to counteract this infinite release of energy - so is there another force which I'm forgetting?

    These are just ideas I've come up with whilst thinking about the problem, and I don't know if any of them have any physical significance in this problem, so any advice is appreciated!










    share|cite|improve this question









    $endgroup$














      21












      21








      21


      2



      $begingroup$


      Questions of the form:




      An electron and a positron collide with E MeV of energy, what is the frequency of the photons released.




      quite often come up in my A Level course (for often fairly arbitrary E). But this got me thinking. There is energy stored in the separation of an electron and a positron, which, as they get closer and closer together, should all be converted into kinetic energy. As the potential is of the form $frac1r$, this implies that at arbitrarily small distances and arbitrarily high amount of energy is given off. Given that both electrons and positrons are typically regarded as point particles, in order for them to collide, they would have to be arbitrarily close together, which would imply that over the course of their collision they should have released arbitrarily high amounts of energy, in the form of kinetic energy. As this would imply photons of arbitrarily high frequency given off, I assume that I must have missed out some piece of physics somewhere, but I am uncertain where. Ideas I have had so far include:



      • Energy should be given off, anyway, by an accelerating electron, in the form of light, according to classical EM, although I don't know how this changes from classical to quantum ideas of EM - we certainly can't have all the energy given off in a continuous stream, because we need quantised photons, so does the electron itself experience quantised energy levels as it accelerates inwards (my only issue with treating the electron in such a quantised way is that, to my mind, it'd be equivalent of treating it mathematically as a hydrogen-like atom, where the probability of the electron colliding with the positron is still extremely low, and unlike electron capture, there'd be no weak force interaction to mediate this 'electron-positron atom').

      • The actual mechanism for the decay occurs at a non-zero separation distance, perhaps photons pass between the two particles to mediate the decay at non-infinitesimal distances.

      • At relativistic speeds our classical model of electrodynamics breaks down. Now, I know this to be true - considering the fact that magnetism is basically the relativistic component of electrodynamics. However, given the fact that magnetism is the only relativistic force which'd be involved, I don't see how it'd act to counteract this infinite release of energy - so is there another force which I'm forgetting?

      These are just ideas I've come up with whilst thinking about the problem, and I don't know if any of them have any physical significance in this problem, so any advice is appreciated!










      share|cite|improve this question









      $endgroup$




      Questions of the form:




      An electron and a positron collide with E MeV of energy, what is the frequency of the photons released.




      quite often come up in my A Level course (for often fairly arbitrary E). But this got me thinking. There is energy stored in the separation of an electron and a positron, which, as they get closer and closer together, should all be converted into kinetic energy. As the potential is of the form $frac1r$, this implies that at arbitrarily small distances and arbitrarily high amount of energy is given off. Given that both electrons and positrons are typically regarded as point particles, in order for them to collide, they would have to be arbitrarily close together, which would imply that over the course of their collision they should have released arbitrarily high amounts of energy, in the form of kinetic energy. As this would imply photons of arbitrarily high frequency given off, I assume that I must have missed out some piece of physics somewhere, but I am uncertain where. Ideas I have had so far include:



      • Energy should be given off, anyway, by an accelerating electron, in the form of light, according to classical EM, although I don't know how this changes from classical to quantum ideas of EM - we certainly can't have all the energy given off in a continuous stream, because we need quantised photons, so does the electron itself experience quantised energy levels as it accelerates inwards (my only issue with treating the electron in such a quantised way is that, to my mind, it'd be equivalent of treating it mathematically as a hydrogen-like atom, where the probability of the electron colliding with the positron is still extremely low, and unlike electron capture, there'd be no weak force interaction to mediate this 'electron-positron atom').

      • The actual mechanism for the decay occurs at a non-zero separation distance, perhaps photons pass between the two particles to mediate the decay at non-infinitesimal distances.

      • At relativistic speeds our classical model of electrodynamics breaks down. Now, I know this to be true - considering the fact that magnetism is basically the relativistic component of electrodynamics. However, given the fact that magnetism is the only relativistic force which'd be involved, I don't see how it'd act to counteract this infinite release of energy - so is there another force which I'm forgetting?

      These are just ideas I've come up with whilst thinking about the problem, and I don't know if any of them have any physical significance in this problem, so any advice is appreciated!







      quantum-mechanics particle-physics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 15 hours ago









      DoublyNegativeDoublyNegative

      424211




      424211




















          2 Answers
          2






          active

          oldest

          votes


















          22












          $begingroup$

          This is a great question! It can be answered on many different levels.



          You are absolutely right that if we stick to the level of classical high school physics, something doesn't make sense here. However, we can get an approximately correct picture by "pasting" together a classical and a quantum description. To do this, let's think of when the classical picture breaks down.



          In relativistic quantum field theory, particles can only be localized on the scale of their Compton wavelength
          $$lambda = frachbarmc.$$
          This means that the classical picture of point particles must break down as we approach this separation distance. Now, the electric potential energy released at this point is
          $$E = frace^2r = frace^2 m chbar$$
          in cgs units. Here one of the most important constants in physics has appeared, the fine structure constant which characterizes the strength of electromagnetism,
          $$alpha = frace^2hbar c approx frac1137.$$
          The energy released up to this point is
          $$E approx alpha m c^2$$
          which is not infinite, but rather only a small fraction of the total energy.



          Past this radius we should use quantum mechanics, which renders the $1/r$ potential totally unapplicable -- not only do the electrons not have definite positions, but the electromagnetic field doesn't even have a definite value. Actually thinking about the full quantum state of the system at this point is so hairy that not even graduate-level textbooks do it; they usually black-box the process and only think about the final results, just like your high school course is doing. Using the full theory of quantum electrodynamics, one can show the most probable final outcome is to have two energetic photons come out. In high school you just assume this happens and use conservation laws to describe the photons long after the process is over.



          For separations much greater than $lambda$, the classical picture should be applicable, and we can think of part of the energy as being released as classical electromagnetic radiation, which occurs generally when charges accelerate. (At the quantum level, the number of photons released is infinite, indicating a so-called infrared divergence, but they are individually very low in energy, and their total energy is perfectly finite.) As you expected, this energy is lost before the black-boxed quantum process starts, so the answers in your school books are actually off by around $1/137$. But this is a small enough number we don't worry much about it.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
            $endgroup$
            – DoublyNegative
            15 hours ago










          • $begingroup$
            @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 1




            $begingroup$
            @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            Ah, ok, that makes sense, thanks!
            $endgroup$
            – DoublyNegative
            14 hours ago


















          4












          $begingroup$

          Positronium is what you're describing in your first idea.



          Until/unless a believable model of the electron is developed, pretty much the best we can do is say that the rest mass of the electron-positron system is the only energy available to be converted to radiation in an annihilation event.



          This experiment will reveal a lot more about what happens in very close encounters between electrons and positrons.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 2




            $begingroup$
            Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
            $endgroup$
            – S. McGrew
            14 hours ago











          Your Answer





          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "151"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470922%2fwhy-dont-electron-positron-collisions-release-infinite-energy%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          22












          $begingroup$

          This is a great question! It can be answered on many different levels.



          You are absolutely right that if we stick to the level of classical high school physics, something doesn't make sense here. However, we can get an approximately correct picture by "pasting" together a classical and a quantum description. To do this, let's think of when the classical picture breaks down.



          In relativistic quantum field theory, particles can only be localized on the scale of their Compton wavelength
          $$lambda = frachbarmc.$$
          This means that the classical picture of point particles must break down as we approach this separation distance. Now, the electric potential energy released at this point is
          $$E = frace^2r = frace^2 m chbar$$
          in cgs units. Here one of the most important constants in physics has appeared, the fine structure constant which characterizes the strength of electromagnetism,
          $$alpha = frace^2hbar c approx frac1137.$$
          The energy released up to this point is
          $$E approx alpha m c^2$$
          which is not infinite, but rather only a small fraction of the total energy.



          Past this radius we should use quantum mechanics, which renders the $1/r$ potential totally unapplicable -- not only do the electrons not have definite positions, but the electromagnetic field doesn't even have a definite value. Actually thinking about the full quantum state of the system at this point is so hairy that not even graduate-level textbooks do it; they usually black-box the process and only think about the final results, just like your high school course is doing. Using the full theory of quantum electrodynamics, one can show the most probable final outcome is to have two energetic photons come out. In high school you just assume this happens and use conservation laws to describe the photons long after the process is over.



          For separations much greater than $lambda$, the classical picture should be applicable, and we can think of part of the energy as being released as classical electromagnetic radiation, which occurs generally when charges accelerate. (At the quantum level, the number of photons released is infinite, indicating a so-called infrared divergence, but they are individually very low in energy, and their total energy is perfectly finite.) As you expected, this energy is lost before the black-boxed quantum process starts, so the answers in your school books are actually off by around $1/137$. But this is a small enough number we don't worry much about it.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
            $endgroup$
            – DoublyNegative
            15 hours ago










          • $begingroup$
            @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 1




            $begingroup$
            @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            Ah, ok, that makes sense, thanks!
            $endgroup$
            – DoublyNegative
            14 hours ago















          22












          $begingroup$

          This is a great question! It can be answered on many different levels.



          You are absolutely right that if we stick to the level of classical high school physics, something doesn't make sense here. However, we can get an approximately correct picture by "pasting" together a classical and a quantum description. To do this, let's think of when the classical picture breaks down.



          In relativistic quantum field theory, particles can only be localized on the scale of their Compton wavelength
          $$lambda = frachbarmc.$$
          This means that the classical picture of point particles must break down as we approach this separation distance. Now, the electric potential energy released at this point is
          $$E = frace^2r = frace^2 m chbar$$
          in cgs units. Here one of the most important constants in physics has appeared, the fine structure constant which characterizes the strength of electromagnetism,
          $$alpha = frace^2hbar c approx frac1137.$$
          The energy released up to this point is
          $$E approx alpha m c^2$$
          which is not infinite, but rather only a small fraction of the total energy.



          Past this radius we should use quantum mechanics, which renders the $1/r$ potential totally unapplicable -- not only do the electrons not have definite positions, but the electromagnetic field doesn't even have a definite value. Actually thinking about the full quantum state of the system at this point is so hairy that not even graduate-level textbooks do it; they usually black-box the process and only think about the final results, just like your high school course is doing. Using the full theory of quantum electrodynamics, one can show the most probable final outcome is to have two energetic photons come out. In high school you just assume this happens and use conservation laws to describe the photons long after the process is over.



          For separations much greater than $lambda$, the classical picture should be applicable, and we can think of part of the energy as being released as classical electromagnetic radiation, which occurs generally when charges accelerate. (At the quantum level, the number of photons released is infinite, indicating a so-called infrared divergence, but they are individually very low in energy, and their total energy is perfectly finite.) As you expected, this energy is lost before the black-boxed quantum process starts, so the answers in your school books are actually off by around $1/137$. But this is a small enough number we don't worry much about it.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
            $endgroup$
            – DoublyNegative
            15 hours ago










          • $begingroup$
            @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 1




            $begingroup$
            @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            Ah, ok, that makes sense, thanks!
            $endgroup$
            – DoublyNegative
            14 hours ago













          22












          22








          22





          $begingroup$

          This is a great question! It can be answered on many different levels.



          You are absolutely right that if we stick to the level of classical high school physics, something doesn't make sense here. However, we can get an approximately correct picture by "pasting" together a classical and a quantum description. To do this, let's think of when the classical picture breaks down.



          In relativistic quantum field theory, particles can only be localized on the scale of their Compton wavelength
          $$lambda = frachbarmc.$$
          This means that the classical picture of point particles must break down as we approach this separation distance. Now, the electric potential energy released at this point is
          $$E = frace^2r = frace^2 m chbar$$
          in cgs units. Here one of the most important constants in physics has appeared, the fine structure constant which characterizes the strength of electromagnetism,
          $$alpha = frace^2hbar c approx frac1137.$$
          The energy released up to this point is
          $$E approx alpha m c^2$$
          which is not infinite, but rather only a small fraction of the total energy.



          Past this radius we should use quantum mechanics, which renders the $1/r$ potential totally unapplicable -- not only do the electrons not have definite positions, but the electromagnetic field doesn't even have a definite value. Actually thinking about the full quantum state of the system at this point is so hairy that not even graduate-level textbooks do it; they usually black-box the process and only think about the final results, just like your high school course is doing. Using the full theory of quantum electrodynamics, one can show the most probable final outcome is to have two energetic photons come out. In high school you just assume this happens and use conservation laws to describe the photons long after the process is over.



          For separations much greater than $lambda$, the classical picture should be applicable, and we can think of part of the energy as being released as classical electromagnetic radiation, which occurs generally when charges accelerate. (At the quantum level, the number of photons released is infinite, indicating a so-called infrared divergence, but they are individually very low in energy, and their total energy is perfectly finite.) As you expected, this energy is lost before the black-boxed quantum process starts, so the answers in your school books are actually off by around $1/137$. But this is a small enough number we don't worry much about it.






          share|cite|improve this answer









          $endgroup$



          This is a great question! It can be answered on many different levels.



          You are absolutely right that if we stick to the level of classical high school physics, something doesn't make sense here. However, we can get an approximately correct picture by "pasting" together a classical and a quantum description. To do this, let's think of when the classical picture breaks down.



          In relativistic quantum field theory, particles can only be localized on the scale of their Compton wavelength
          $$lambda = frachbarmc.$$
          This means that the classical picture of point particles must break down as we approach this separation distance. Now, the electric potential energy released at this point is
          $$E = frace^2r = frace^2 m chbar$$
          in cgs units. Here one of the most important constants in physics has appeared, the fine structure constant which characterizes the strength of electromagnetism,
          $$alpha = frace^2hbar c approx frac1137.$$
          The energy released up to this point is
          $$E approx alpha m c^2$$
          which is not infinite, but rather only a small fraction of the total energy.



          Past this radius we should use quantum mechanics, which renders the $1/r$ potential totally unapplicable -- not only do the electrons not have definite positions, but the electromagnetic field doesn't even have a definite value. Actually thinking about the full quantum state of the system at this point is so hairy that not even graduate-level textbooks do it; they usually black-box the process and only think about the final results, just like your high school course is doing. Using the full theory of quantum electrodynamics, one can show the most probable final outcome is to have two energetic photons come out. In high school you just assume this happens and use conservation laws to describe the photons long after the process is over.



          For separations much greater than $lambda$, the classical picture should be applicable, and we can think of part of the energy as being released as classical electromagnetic radiation, which occurs generally when charges accelerate. (At the quantum level, the number of photons released is infinite, indicating a so-called infrared divergence, but they are individually very low in energy, and their total energy is perfectly finite.) As you expected, this energy is lost before the black-boxed quantum process starts, so the answers in your school books are actually off by around $1/137$. But this is a small enough number we don't worry much about it.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 15 hours ago









          knzhouknzhou

          46.6k11125224




          46.6k11125224











          • $begingroup$
            So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
            $endgroup$
            – DoublyNegative
            15 hours ago










          • $begingroup$
            @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 1




            $begingroup$
            @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            Ah, ok, that makes sense, thanks!
            $endgroup$
            – DoublyNegative
            14 hours ago
















          • $begingroup$
            So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
            $endgroup$
            – DoublyNegative
            15 hours ago










          • $begingroup$
            @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 1




            $begingroup$
            @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
            $endgroup$
            – knzhou
            14 hours ago










          • $begingroup$
            Ah, ok, that makes sense, thanks!
            $endgroup$
            – DoublyNegative
            14 hours ago















          $begingroup$
          So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
          $endgroup$
          – DoublyNegative
          15 hours ago




          $begingroup$
          So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?)
          $endgroup$
          – DoublyNegative
          15 hours ago












          $begingroup$
          @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
          $endgroup$
          – knzhou
          14 hours ago




          $begingroup$
          @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field.
          $endgroup$
          – knzhou
          14 hours ago












          $begingroup$
          I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
          $endgroup$
          – DoublyNegative
          14 hours ago




          $begingroup$
          I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions?
          $endgroup$
          – DoublyNegative
          14 hours ago




          1




          1




          $begingroup$
          @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
          $endgroup$
          – knzhou
          14 hours ago




          $begingroup$
          @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other.
          $endgroup$
          – knzhou
          14 hours ago












          $begingroup$
          Ah, ok, that makes sense, thanks!
          $endgroup$
          – DoublyNegative
          14 hours ago




          $begingroup$
          Ah, ok, that makes sense, thanks!
          $endgroup$
          – DoublyNegative
          14 hours ago











          4












          $begingroup$

          Positronium is what you're describing in your first idea.



          Until/unless a believable model of the electron is developed, pretty much the best we can do is say that the rest mass of the electron-positron system is the only energy available to be converted to radiation in an annihilation event.



          This experiment will reveal a lot more about what happens in very close encounters between electrons and positrons.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 2




            $begingroup$
            Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
            $endgroup$
            – S. McGrew
            14 hours ago















          4












          $begingroup$

          Positronium is what you're describing in your first idea.



          Until/unless a believable model of the electron is developed, pretty much the best we can do is say that the rest mass of the electron-positron system is the only energy available to be converted to radiation in an annihilation event.



          This experiment will reveal a lot more about what happens in very close encounters between electrons and positrons.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 2




            $begingroup$
            Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
            $endgroup$
            – S. McGrew
            14 hours ago













          4












          4








          4





          $begingroup$

          Positronium is what you're describing in your first idea.



          Until/unless a believable model of the electron is developed, pretty much the best we can do is say that the rest mass of the electron-positron system is the only energy available to be converted to radiation in an annihilation event.



          This experiment will reveal a lot more about what happens in very close encounters between electrons and positrons.






          share|cite|improve this answer









          $endgroup$



          Positronium is what you're describing in your first idea.



          Until/unless a believable model of the electron is developed, pretty much the best we can do is say that the rest mass of the electron-positron system is the only energy available to be converted to radiation in an annihilation event.



          This experiment will reveal a lot more about what happens in very close encounters between electrons and positrons.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 15 hours ago









          S. McGrewS. McGrew

          9,06021236




          9,06021236







          • 1




            $begingroup$
            This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 2




            $begingroup$
            Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
            $endgroup$
            – S. McGrew
            14 hours ago












          • 1




            $begingroup$
            This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
            $endgroup$
            – DoublyNegative
            14 hours ago






          • 2




            $begingroup$
            Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
            $endgroup$
            – S. McGrew
            14 hours ago







          1




          1




          $begingroup$
          This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
          $endgroup$
          – DoublyNegative
          14 hours ago




          $begingroup$
          This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then?
          $endgroup$
          – DoublyNegative
          14 hours ago




          2




          2




          $begingroup$
          Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
          $endgroup$
          – S. McGrew
          14 hours ago




          $begingroup$
          Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass.
          $endgroup$
          – S. McGrew
          14 hours ago

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Physics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470922%2fwhy-dont-electron-positron-collisions-release-infinite-energy%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Reverse int within the 32-bit signed integer range: [−2^31, 2^31 − 1]Combining two 32-bit integers into one 64-bit integerDetermine if an int is within rangeLossy packing 32 bit integer to 16 bitComputing the square root of a 64-bit integerKeeping integer addition within boundsSafe multiplication of two 64-bit signed integersLeetcode 10: Regular Expression MatchingSigned integer-to-ascii x86_64 assembler macroReverse the digits of an Integer“Add two numbers given in reverse order from a linked list”

          Category:Fedor von Bock Media in category "Fedor von Bock"Navigation menuUpload mediaISNI: 0000 0000 5511 3417VIAF ID: 24712551GND ID: 119294796Library of Congress authority ID: n96068363BnF ID: 12534305fSUDOC authorities ID: 034604189Open Library ID: OL338253ANKCR AUT ID: jn19990000869National Library of Israel ID: 000514068National Thesaurus for Author Names ID: 341574317ReasonatorScholiaStatistics

          Kiel Indholdsfortegnelse Historie | Transport og færgeforbindelser | Sejlsport og anden sport | Kultur | Kendte personer fra Kiel | Noter | Litteratur | Eksterne henvisninger | Navigationsmenuwww.kiel.de54°19′31″N 10°8′26″Ø / 54.32528°N 10.14056°Ø / 54.32528; 10.14056Oberbürgermeister Dr. Ulf Kämpferwww.statistik-nord.deDen danske Stats StatistikKiels hjemmesiderrrWorldCat312794080n790547494030481-4