What is a Field?

Discussion in 'Physics & Math' started by Willem, Apr 29, 2019.

  1. NotEinstein Valued Senior Member

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    So electromagnetism doesn't involve time (the fourth dimension)?

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  3. NotEinstein Valued Senior Member

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    No, you should have worked on it before you posted it. Remember, this is the science-section of the forum.
     
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  5. QuarkHead Remedial Math Student Valued Senior Member

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    I do not see why a field has to be a "property" of anything (other than itself).

    After all, a field, in the sense that physicists use the term is just the assignment of a value (scalar, vector, tensor etc) to each point in space. I paraphrase Einstein: to a physicist, space without a metric is
    unthinkable. And since we have no reason a priori to assume the metric is the same at every point in space, it is convenient to think of it as a metric field.

    As it happens, it is a tensor field. Of possible interest (but probably not) in 4-dimensional space with the semi-Reimann metric, the curvature field is the second derivative of the metric field and therefore from elementary calculus is zero (flat spacetime) when the metric is constant, and not otherwise (curved spacetime)
     
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  7. exchemist Valued Senior Member

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    Oh hello QH I was hoping someone like you would show up on this thread. I take your point that perhaps I should have said "value" rather than "property".

    However the fields I am struggling with are those employed by QFT: as I understand it, this treats QM entities ("particles") as disturbances in the relevant fields. Perhaps all that means, then, to take your point on board, is that there are non-zero values, of some defined kind (I won't say property) , in the areas around where the entity is located.
     
  8. Write4U Valued Senior Member

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    This may be pertinent to the question;
     
  9. arfa brane call me arf Valued Senior Member

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    The most understandable explanation I've seen of what a field is in physics is pretty much what QH posted, with the addition of a particle (the field's "quantum") which essentially captures the d.o.f. of the field with spin.

    So a scalar field has a spin 0 "particle" associated at every point, a vector field has a spin-1 particle; in general a spin n particle is associated to a field with 2n + 1 d.o.f., for the propagation of quantized energy (through a spacetime, of course).
     
  10. Confused2 Registered Senior Member

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    Spin? Who invited that?
     
  11. arfa brane call me arf Valued Senior Member

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  12. exchemist Valued Senior Member

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    Not very helpful, I'm afraid.

    What's d.o.f?

    And what is meant by a "particle" being "associated at every point", given that the purpose of the theory is to model the behaviour of actual particles, which obviously do not exist at every point?

    And why do you only mention bosons?
     
    Last edited: May 1, 2019
  13. Q-reeus Banned Valued Senior Member

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    That is inconsistent. A uniform acceleration field (constant 'g') has different values for temporal and spatial metric components as a function of elevation wrt 'g' field, but has by your own (correct) definition, zero curvature. Hence a non-zero 1st derivative of metric coefficients is inconsistent with flat spacetime. Obvious example where temporal metric varies linearly - redshift as a function of elevation in a notionally uniform (thus zero 2nd derivative of metric) g field.
     
  14. arfa brane call me arf Valued Senior Member

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    d.o.f = degrees of freedom

    For a quantum field (a bosonic field is one of these), "particles" are excitations of the field itself. So every point can be such an excitation, which does not mean every point "is" a particle. Superfluids have particles which obey Bose-Einstein statistics where every point in a given volume is not a "single particle".
     
  15. exchemist Valued Senior Member

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    Thanks, that's a bit clearer.

    But now you start to illustrate my difficulty (as a mere chemist who did not study QFT) with this. You and QH originally describe a field as simply something that assigns a "value" (scalar, vector, etc) to every point in space. That idea is easy to grasp, as clearly these "values" can be, say, zero, or some nominal datum level, everywhere except where there is a particle. So it just looks like a 3D graph, with a flat "line" everywhere except where there is something of interest. The impression is of the field as something abstract: a mathematical representation of how these "values" are distributed in space.

    However now, you start to speak, as others I have read also do, of an "excitation" of the field. Now that has a very different connotation, implying there is an insubstantial, but real, physical "something" there all the time, everywhere, which can be "excited" by an influence.

    That is an assertion about nature, rather than just a mathematical representation of a distribution of values.

    P.S. And the fermions? Are they not also represented by a vector field, or do we get into spinors etc., and that is why you have not included them in the discussion?
     
  16. Willem Banned Banned

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    No, just with the proto-particles defined in "Constructing Time from a Axiom" time already exist.
     
  17. Willem Banned Banned

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    No, but it has the 5th dimension in place of time.

    Maxwell's equations, suitably modified for 5th D replacing 4th D.
     
  18. NotEinstein Valued Senior Member

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    So those proto-particles generate the fourth dimension, and the force of gravity. Without those proto-particles, the universe would be one dimension "smaller" and have no gravity?

    Why is this thread in the science-section of the forum?

    So according to you, the universe has (at least) two time dimensions: one associated with gravity, and a second one associated with electromagnetism. Now please, pray tell, explain to me how I can distinguish those two?

    And what happens with the fourth dimension in electromagnetism when you do that? And what happens with this fifth dimension when you consider gravity?
     
  19. QuarkHead Remedial Math Student Valued Senior Member

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    Yes, hold on to that thought
    You are assuming what you are trying to ascertain

    I don't get it - loads of things whose reality we don't for a second question, have no tangible substance.

    OK. The following may be wildly wrong about QFT, so I do not advise any reader to take it too seriously.

    Assume first a quantum system, and further assume it is not particularly localized. Now since the state of a quantum system is given by a state vector, we may assume a vector field of state vectors'

    And every time we have a vector of any complexion, then we may apply an operator to it. Let us choose the Hamiltonian, this being the sum of positional energy and momentum.

    Now the so-called ground state of a system is the lowest available eigenvalue for our operator. Let us call this eigenvalue as zero. Then if at any point the state vector there has non-zero eigenvalue, we may assume that the position and momentum operators are likewise non-zero.

    In QFT, something - anything - with non-zero momentum and position is called a particle.

    I repeat, I just made that up.

    JamesR will probably ban me for faking intelligence!
     
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  20. Willem Banned Banned

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    The fourth dimension does not create gravity, mass does, spacetime just reacts to mass and allow gravitons (that sees warped spacetime) to transmit a force.

    Yes.

    I guess so.

    Yes.

    Answers to the other 3 questions is a long way off my current work.
     
  21. NotEinstein Valued Senior Member

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    This cannot be correct: in the context of curved spacetime (i.e. GR), there is no gravitational force.

    That's not how the universe works: its dimensions are not determined by the presence or absence of particles.

    In that case, please provide evidence for the existence of these proto-particles. This must be simple to do for you, because gravity is obviously real.

    Ah, so you've come to a conclusion, but you haven't yet realized it doesn't make much sense. I'll wait for you to catch up, then.
     
  22. origin Heading towards oblivion Valued Senior Member

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    Electromagnetic time - time from a clock that runs on batteries or that you plug in.
    Gravity time - that is the time you can get from a hour glass.

    This science stuff is easy!
     
  23. NotEinstein Valued Senior Member

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    Don't forget there's 6 more dimensions to explore! I guess the weak force clock is obviously: an atomic clock, but what about a "quark clock" of some kind? Sounds like I need funding!

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