Potential Energy of Quantum Field?

Discussion in 'Physics & Math' started by John J. Bannan, Jul 29, 2008.

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  1. John J. Bannan Registered Senior Member

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    What is the difference between energy and potential energy in a quantum field? If I put an apple on top of a ladder, we say the apple has potential energy due to gravity which turns into kinetic energy when the apple falls. If a quantum field contains potential energy, then what force is being applied to the quantum field to give it potential energy?
     
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  3. Bishadi Banned Banned

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    electromagnetic lines or perpetrations to space.

    in orientation; the entanglement of the 2 bodies

    i.e... the apple on the latter has a force (weight) existing identified as potential energy, yet simply entangled energy between the 2 bodies (apple/earth).

    isolate that apple in a magnetic field and find that 'gravity' potential is isolated...(diamagnetism)


    drop the apple and lines are crossed, increasing the potential with additional kinetic energy, and why the speed increases

    or to see BEC (Bose Einstein condensate), a sort of reverse form to recognize the potential change (in an opposite flow... an entropy buster); when observing how BEC is reached

    http://jilawww.colorado.edu/bec/Publications/JLTP_Lewandowski2003.pdf
     
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  5. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Your line of thought is backwrads. You should think of forces as deriving from potentials---when you studied Maxwell's electromagnetism, you learned that all properties/forces in an electromagnetic field can be derived from the scalar potential V and the vector potential \(\vec{A}\).

    One can derive the force exerted by the field on the body in the standard way, that you learned (presumably) when you studied electromagnetism.
     
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  7. Bishadi Banned Banned

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    which is the error to the time scaling of velocity; or simply velocity is not the potential, it is the energy captured during.

    ie.... in a particle accelerator, the increased mass is thought as you suggest but in reality it is the from the entangled environment.

    so you are correct my friend, this polock is all messed up as energy is upon mass, rather than a potential difference between them.

    Ever notice if you have 2 balls on the ground, they rest but then to have one higher than another, it had to increase its potential from somewhere.

    that is how i look at it.

    max assisted in the relation of the perpendicular planes of electric and magnetic fields but the velocity of light is incorrect, the exchanges are t<0 if a perfect vacuum could exist between 2 points

    friend i love a thinker and if you like, let's look over some new items and allow yourself to absorb what you find 'good or bad'.....
     
  8. Guest254 Valued Senior Member

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    On a pinicity (is that a word?) aside, this isn't true in general, right - i.e if there is non-trivial cohomology lurking, then the existence of a global potential isn't cool. Or have I mis-understood?
     
  9. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    I'm affraid I don't know some of these fancy words, like "non-trivial cohomology"---are we talking instanton/boundary effects, or putting a QFT on some weird background? It seems to me that as long as you have some sort of translation invariance (i.e. 4-d Poincare invariance) that ensures your laws of physics are as good in this little piece of space as they are in that little piece of space (i.e. you conserve momentum), you're probably ok. Of course, I am the first one to admit that I am probably wrong.
     
  10. Guest254 Valued Senior Member

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    Well I assumed when you mentioned Maxwell's equations, potentials etc you were suggesting that curl-free => some potential. In the language of the exterior calculus, this is the statement "closed => exact". This implication depends on the topology of your domain: i.e on a domain X with non-trivial cohomology, there exists differential forms that are closed on all of X, but that are NOT exact. You can then relate this back to curls, grads etc.

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  11. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Ahhh right. So in forms I'd write

    \(dA = F\)

    as a definition of the fields, and

    \( dF = 0\)

    as a statment of Maxwell's equations. Your statment is that there are forms (say, H) which satisfy

    \(dH = 0\), but cannot be written as \(dX = H\) for some X?

    How is this gauge invariant? Or is this a stupid question?
     
  12. Guest254 Valued Senior Member

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    Yes to everything you said, then with regards the last bit:

    Surely you still have the freedom: \(H \rightarrow H + \mathrm{d}A\)? (i.e \(\mathrm{d}^2=0\) is a fact no matter the topology of your domain - it's fundamental to the notion of "cohomology" - google "exact seqeuce").
     
  13. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Ahh right. Stupid question.

    ----Edit

    Well, I always understood gauge invariance as \(A \rightarrow A + d\alpha\). A is the thing which has all the gauge degrees of freedom, not F. And A is not related to H at all.

    I'm a bit confused, but now I'm also intrigued...
     
  14. Guest254 Valued Senior Member

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    Well I don't know about that! I've seen a few questions on this forum that have set the bar pretty low!
     
  15. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Check edit

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    Probably still a stupid question...
     
  16. Guest254 Valued Senior Member

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    It's the same thing (I guess). Sure, there is a freedom (in your notation now) in \(F\rightarrow F + \mathrm{d}\beta \), but this would just be a re-definition of \(A\). I.e:

    \(\mathrm{d}(A - \beta) = F\)
     
  17. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Ok... I'll buy it.

    So then, in the absence of some weird topology, forces are derivable from potentials.

    Agreed?
     
  18. Guest254 Valued Senior Member

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    Aye-aye cp'n. (the Ok... didn't indicate you were skeptical did it!?!)
     
  19. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Well I was skeptical untill I picked up my pen

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    As an aside, I AM wearing my pirate shirt today...
     
    Last edited: Jul 29, 2008
  20. temur man of no words Registered Senior Member

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    With non-trivial cohomology, what is the remnant of the relation "closed -> exact"? I can imagine that you can write any closed form as the sum of an exact form plus something that depends on the cohomology class. Is there something like this? For a treatment of Maxwell's equations on nontrivial cohomology what should I look for?
     
  21. Guest254 Valued Senior Member

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    First question: see de Rham cohomology (google will give loads).
    Second question: I have absolutely no clue I'm afraid! But there are lots of physics+topology books about, so I'm sure one of them will cover that this kind of thing
     
  22. Guest254 Valued Senior Member

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    Perhaps it's clearer if I said \(F\rightarrow F + \mathrm{d}\beta\) doesn't leave the equations invariant - rather than "you need to redefine A".
     
  23. temur man of no words Registered Senior Member

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    Thanks!

    So by a result in de Rham cohomology theory, in a nontrivial cohomology, you can write

    \(F=dA+\gamma\),

    where \(\gamma\) is a harmonic form: \(\Delta\gamma=0\). Probably you can start with this and build a general theory of potentials (now not only \(A\), but \(\gamma\) is also a kind of potential) in EM.

    http://en.wikipedia.org/wiki/De_Rham_cohomology
     
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