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Photons, Antimatter- My unsolved Questions

  1. Jun 19, 2013 #1
    Since this is my first post on this forum, I would like to start by saying:

    Hello everyone! :smile:


    Now let's get straight down to business:

    Over the past few months, I have been doing quite a bit of research in terms of physics, and have had the following questions pop up. I've asked some of the professors at my college about them, and they had trouble answering me. I mean, they "answered" them, but in this sort-of weird, staggering tone of voice, as if they were unsure themselves. Not knowing where to turn, I have decided to try and get some kind of consensus of answers (or an answer that makes sense). And I guess a forum would be the best place to start. Okay so, let's begin...

    __________________________________________________________________________________

    • 1) The photon is considered to be a mass-less particle (ie: having zero mass), correct? Okay, so then using Einstein's equation:
      E= mc2 , we can conclude that the energy of a photon is 0* speed of light2= 0. Now, correct if I'm wrong, but don't things like light and electromagnetic fields, which are solely comprised of these photons, have energy greater than zero (again, unless I'm mistaken, and have picked up a fallacy somewhere)? If this is indeed the case, isn't there a mistake somewhere in this logic?

      Think about it, I mean how can even a huge amount of these photons give rise to something that has more than zero energy? If photons themselves have an energy of zero, the net energy of a system comprised of them is still: 0+0+0+..... +0= 0. Something has to be wrong here. Either I must be missing something or our current understanding of photons is inaccurate. That or maybe the human understanding of what zero actually is, is well... incorrect. Otherwise, this seems to be a paradox.


    • 2) Antimatter + matter= both "die", and create energy + photons. That's a really simplified version of what actually happens, but let it be the case for now. Question is: Doesn't this violate the law of conservation of energy? In a sense, are we not destroying matter, and creating energy? Should this perhaps be changed to something like: "The 'net energy' in a system cannot be created/ destroyed." This is similar to simple math, when you take away something, you have to add it again to make sure you haven't changed the equation/ expression (ie: x + y - y= x).


    • 3) Stemming off from question #2, there is an old saying that goes something like: "To be like God, one must be able to create from nothing, everything." It seems that theoretically, such a thing is indeed possible. Is it ludicrous to suggest that we could grab something like a photon or a system of 0 energy, and transform it to produce a certain amount of matter/ antimatter? This way, not violating the law of 'net energy' conservation, and yet producing matter.

      OR better yet, just like in mathematics, the idea of carrying out a multiplication of negative one to a number and transforming it, is it ludicrous to suggest that such an operation could be done to real matter in order to transform it into antimatter?


    • 4) The existence of antimatter provides us with a very kind of scary possibility. If the universe is comprised equally of antimatter and matter, doesn't that mean that the universe has in fact, zero energy? This also goes back to my first question, because once all this matter and antimatter collides with itself, nothing but photons are left, and again, zero energy. This gives new meaning to the phrase "even nothing, is something." Except in this case, nothing, is everything, haha. (think about how crazily that would change philosophy).



    I guess I'm not only looking for answers, but also for a general discussion, and thoughts about such ideas/ concepts. Please feel free to correct me (I'm still in college, and pretty dumb, so go easy on me, haha). All advice, criticism, corrections on ANYTHING (even on something that looks slightly/ completely wrong to you) are all welcome. But please, no flaming, hacking, trolling, spitting on the side-walk. And I guess that's that; so without further ado, Discuss!

    Thank you for reading,
    -Cat_Ion
     
  2. jcsd
  3. Jun 19, 2013 #2
    I'll take the first question. Others will probably follow up.

    The rest mass of a photon is taken to be 0, but photons never rest. They are always in motion and have only one speed: c. The energy of a photon is related to the wave length of the radiation and is given by the formula: [itex]E=hc/\lambda[/itex] where h is Planck's constant and lambda is wave length. It's clear that the energy of a photon is in inversely related to wave length. As the wave length increases the energy decreases. As the energy approaches zero (equivalent to zero mass), the wave length approaches infinity. The is an asymptotic limit that presumably is never reached.
     
    Last edited: Jun 19, 2013
  4. Jun 19, 2013 #3

    Bandersnatch

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    Hello, Cat_ion, welcome to PF!

    It seems to me that all your doubts have basis in one misunderstanding stated in 1).
    [itex]E=mc^2[/itex] is only applicable to particles possessing rest mass, and not moving. For moving particles, one has to use the more general expression: [itex]E^2=m^2c^4+p^2c^2[/itex].

    If a particle is not moving, then p=0, and the second part of the equation dissapears.

    In the case of photons, the particle does not posses rest mass, and is always moving, so the above equation reduces to [itex]E=pc[/itex].
     
  5. Jun 19, 2013 #4

    mfb

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    E= mc2 is a formula for the rest energy of a particle. A photon does not have a rest energy, as it cannot be at rest.
    Photons have energy, but they have no mass.

    Mass is a type of energy, it is just converted to another type (radiation).

    No experiment ever found a conversion from matter to antimatter, and there are well-defined conservation laws corresponding to this. It is expected that those conservation laws are not exact at very high-energetic processes, but even then you cannot just press a button and convert matter to antimatter. And even if you could, it would not violate energy conservation. Antimatter has positive energy.

    No significant amount of antimatter has been observed anywhere in the universe.
    No, at least not in the way you ask here.

    I guess the other questions are answered with "photons have energy", too.
     
  6. Jun 19, 2013 #5

    mathman

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    Short answer: Once you understand that photons have energy and E = mc2 is a way of relating the rest mass of particles to energy when converted.

    Example: electrons have a mass equivalent to 511 Kev of energy. When an electron and positron annihilate, the result is two photons, each of which has 511 Kev of energy.

    Anti-matter question: This is one of the open questions in physics today. Physicists believe that the universe is essentially all matter. What happened to the anti-matter (after the big bang there was an equal amount of matter and anti-matter)?
     
  7. Jun 19, 2013 #6
    Seems like you are using wrong equatiom for Energy. For a moving particle the equation gor energy is

    E = (gamma)*mc^2, which becomes mc^2, when particle is at rest(v=0).

    This equation for energy is for the total energy of the particle that is kinetic plus (energy stored as mass).

    I think this what SW VandeCarr meant to say. Your first question's answer is the answer to the rest questions.

    I think the redioactivity helps understand the mass-energy relation better. Solving numericals of alpha, beta and gamma decay helped for me.
     
  8. Jun 19, 2013 #7

    sophiecentaur

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    @cat_ion
    It may be worth while pointing out that 'Antimatter' does not have negative mass - just the opposite charge to its counterpart particle. It's exotic but not THAT exotic. Negative mass really would be something!
     
  9. Jun 19, 2013 #8
    Well, the equation I used doesn't have a mass term. The energy of a photon is inversely related to the wave length of the radiation. I don't really understand how equations with mass or momentum terms on the right can be used for massless particles. In the equation I used, the energy term is on the left and means total energy.
     
    Last edited: Jun 19, 2013
  10. Jun 19, 2013 #9
    Wow, a lot of responses in such a short amount of time! Thanks for replying everyone!


    Ah, okay. I didn't know that it was specifically for objects at rest. That makes sense; I knew I was missing something. And yes, I realize that in the case of photons it simplifies to E=pc. Only question is what does the pc stand for exactly? I read that it represents the total vector of an object in classical spacial geometry, is that correct? But, when dealing with a single particle, like a photon it refers to the magnitude of the particle's momentum. Is that also correct?


    +
    I didn't mean to say/ imply that antimatter had negative mass. Forgive me, if I had made that seem apparent; sometimes I have trouble explaining what my mind is thinking. And to be honest, actually, its not only a matter of (Pun absolutely intended) opposite charge, they also have opposite structures (in terms of quark formations and such).... now, in terms of no experiment has ever found a conversion from matter to antimatter. Well, hey, you never know what's possible. We didnt think antimatter existed at all, until we "discovered" it mathematically. Hopefully we'll be flying around with matter/ antimatter warp cores soon enough though. :smile:


    I'm not sure how this really helps, as in this case, doesn't the energy of the photon become zero as well?


    Actually, that is a very helpful example. Thanks!

    Many thanks to you. And Interesting. So what if we were to look at the relationship in reverse? As the wavelength becomes zero, the energy becomes infinite. Do I smell a machine that will shrink photon wavelengths some time in the future? :biggrin: (I realize that it's not so simple, but who knows, anything is possible)


    Also, I realize that technically you can't just mix equations, but I'm just curious:
    E= pc and E=hc/λ; Therefore, pc=hc/λ. So, basically (the way I understand it), the momentum of a photon= plank constant*speed of light/ photon wavelength. Is that right?


    Thanks once again. Looking forward to some more replies. Thanks.
    -Cat_Ion
     
  11. Jun 19, 2013 #10

    Bandersnatch

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    It's perfectly valid to "mix" these two equations like you did.
    p is the momentum, and c is the speed of light. The momentum of a photon is therefore just [itex]p=\frac{h}{\lambda}[/itex]
     
  12. Jun 19, 2013 #11

    Dale

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    Welcome to PF! I hope you enjoy it here.

    Definitely not. In a matter/anti-matter anhilation interaction momentum, energy, mass, charge, and probably other things are all conserved. For example, the most common anhilation is the electron-positron annhilation. Typically they will have a low relative velocity, so the energy of the pair is approximately twice the mass-energy: 511 keV. This gets transformed into two photons of 511 keV each, going in opposite directions so that momentum is conserved. And the total charge is 0 before and after.
     
  13. Jun 19, 2013 #12
    Cat_ion:"Also, I realize that technically you can't just mix equations, but I'm just curious:
    E= pc and E=hc/λ; Therefore, pc=hc/λ. So, basically (the way I understand it), the momentum of a photon= plank constant*speed of light/ photon wavelength. Is that right?"

    No, the equation is for total energy. Mathematically you can substitute [itex] \rho c [/itex] for total energy, but rho is a symbol for momentum which in mechanics is mass*velocity. This approach seems wrong to me. You can talk about the total energy of a photon, but how do you attribute momentum to a photon? I'm not saying you can't. I'm just not sure how for a massless particle. Perhaps someone else might explain. You can calculate the mass equivalent of energy, but as mfb said, photons have energy, but no mass. I do know that in particle physics the electron-volt is used as a measure for both mass and energy for moving particles, but I don't know if the term applies to massless particles.
     
    Last edited: Jun 19, 2013
  14. Jun 19, 2013 #13

    Nugatory

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    Photons carry momentum - intuitively they have to, because photon-electron collisions (think photoelectric effect, or ionizing radiation) change the speed and hence the momentum of the electron.

    And it's important to remember that ##p=mv## is only true when v is much smaller than c and you're talking about massive particles. If you're dealing with speeds that aren't small compared with c, or with massless particles you have to use the ##E^2=(m_0c^2)^2+(pc)^2## relationship which provides a perfectly satisfactory definition of the momentum of a photon.
    I added one very important word to the quoted text above, and with that addition mass/energy equivalence applies to both massless (that is, particles with zero rest mass like the photon) and massive particles.
     
  15. Jun 19, 2013 #14
    Thank you Nugatory. That clears things up and it works out dimensionally with the wave length equation. Also, I've been using rho instead of p for momentum. Rho is for density and p is for momentum. I'm a little rusty at this stuff.
     
  16. Jun 20, 2013 #15
    This helps. Einstien's mass energy relation is valid for any particle wether moving/not moving/massless.

    E = mass*c^2,

    For particle at rest mass= m, that is rest mass, for moving particles mass= gamma*m. And for particle at rest(v=0) gamma becomes 1 so the relation E = gamma*mc^2 is the general relation

    And this is useful for massless particles as well.Energy doesn't becomes zero as for massless particles. As massless particles can only move at speed c, so gamma becomes infinite for them. infinite*zero is not zero but a finite term!.

    Ther relation
    E^2 = (mc^2)^2 + (pc)^2 is also derived using, E = gamma*c^2, both these equarions are more or less same but E = gamma*c^2 gives a better undersranding of mass-energy relation.

    For photons, the energy is also given by E= hf, which is used the calculate the mass, momentum of moving photons.
     
  17. Jun 20, 2013 #16

    Dale

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    This definition of "mass" is the "relativistic mass". It is a concept which has been out of favor for modern science for quite some time now. When most scientists today use the unqualified term "mass" they mean "invariant mass", not "relativistic mass". So if you are using the concept of relativistic mass you should always correctly qualify it to avoid confusion.
     
  18. Jun 20, 2013 #17
    I use m0, for rest mass and m, for relativistic mass which is gamma*m0, but in thread most are using m, for rest mass. That is why used m for rest mass and gamma*m for relativistic mass.

    Why isn't it favoured? I don't find any problem with it.
     
  19. Jun 20, 2013 #18

    Dale

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    Yes, your notation is fine. I was just referencing your English description. When you write the word "mass" most scientists will think "invariant mass". So if you intend to refer to "relativistic mass" then you should write it explicitly.

    There are four issues.

    1) Relativistic mass is frame variant, and invariant quantities are almost always favored over frame variant quantities.

    2) Invariant mass is a property of an object itself, but relativistic mass is as much related to the observer as the object.

    3) Relativistic mass is just another name for total energy, so we already have a perfectly good word for it and don't need to use up the word "mass" to refer to it.

    4) Newton's 2nd law, f=ma, cannot be used with relativistic mass since it would require separate masses in the parallel and transverse direction. However f=ma is correct where m is the invariant mass and f and a are the force and acceleration four-vectors respectively.

    None of these issues mean that relativistic mass is wrong and invariant mass is right, but those are the reasons that modern physicsts use invariant mass more than relativistic mass. Using relativistic mass is not inherently wrong, but it is also not the usual meaning and so it needs to be identified when it is used.
     
  20. Jun 20, 2013 #19

    ZapperZ

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