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I Why do fields carry momentum?

  1. Dec 2, 2016 #1
    So why, in a classical sense, do we need to come up with the idea that fields carry momentum?

    I mean, charge 1 pushes charge 2 and charge 2 pushes back via newtons third law.

    Of course, we can say that charge 1 creates a field that carrys the momentum, which then affects charge 2 etc.
    But doesn't this seem to go against ocams razor?
  2. jcsd
  3. Dec 2, 2016 #2


    Staff: Mentor

    The forces are not equal and opposite in general. Newton's third law can be violated for two charges.
  4. Dec 2, 2016 #3
    I think you are talking about a situation like the T interection between two charges.

    But still, I don't see why a field is needed to address this since whatever a field does it's only because it is emanated from the charge. The charge is the source of the abstract mathematical entity that is the field.
  5. Dec 2, 2016 #4


    Staff: Mentor

    OK, so then where would you put the missing momentum? It isn't in either of the charges, so where else should it be?
  6. Dec 2, 2016 #5
    I understand, but then wouldn't that just be making something up to fix the problem? Kind of like aether.
  7. Dec 2, 2016 #6


    Staff: Mentor

    Except that the fields explain many measurements. The aether doesn't explain any.
  8. Dec 2, 2016 #7
    Ok so for example the electric force on a force q is just q*E created by another charge. Maybe one can say that it's from the field created from the other charge, but not the other charge itself. But then it's just semantics at that point.

    The field may make up for the missing momentum, but isn't the field generated from the charges themselves? Can't we say that the two charges interact in some manner of pushing or pulling happen in some direction, maybe not NTL, such that the momentum is conserved without invoking the idea of a field?
  9. Dec 2, 2016 #8


    Staff: Mentor

    So then what is the formula for the momentum of a charge?

    Not that I am aware of, particularly not for fields that radiate off to infinity and don't interact with any charge in the future. I mean, in the end to calculate the momentum I think you need a computation that is just as complicated as the field equations, even if you don't call them fields.
  10. Dec 2, 2016 #9

    Wouldn't the formula just be p=mv? Where v is the velocity at a particular time that is reached somewhere within the acceleration produced by the electric force from the other charge.

    How would you detect a field that radiates off to infinity without using any test charges?

    You can detect the warping of spacetime using massless photons, showing that the curvature space is indeed a real thing. I haven't heard of detecting the electrostatic field, for example, using chargeless particles.
    Last edited: Dec 2, 2016
  11. Dec 2, 2016 #10


    Staff: Mentor

    No, that is the whole point. If you use only p=mv for the charges then you get that momentum is NOT conserved for a system of two charges. If you want to get rid of the field then you cannot use it to account for that missing mechanical momentum. Therefore your expression for the momentum of a charge must change to something that accounts for the missing momentum.

    By the energy and momentum that it carries out of the system.
  12. Dec 2, 2016 #11
    I think I understand. So p=m*v needs an adjustment term to account for the missing momentum. Kinda analogous to kinetic energy needing adjustment term for the rotational motion in addition to the already familiar translational motion.

    So does that mean that if we observe the system, we can calculate the missing momentum, and therefore that must be the momentum that radiates off to infinity?
    If that is the only way, then doesn't that mean that if we are at infinity, we cannot detect a field from a source afar without using test charges?
  13. Dec 2, 2016 #12


    Staff: Mentor

    Yes, and I cannot see such a term being any simpler than the fields.

    The laws of physics are symmetric under spatial translations, so being at a different location doesn't change the physics, even if it is very far away. If a pair of charges is missing momentum it doesn't matter how near or far you are.
  14. Dec 2, 2016 #13
    Say that there are two charges in the universe and that we have a lab here where we can measure the missing momentum. Well, in a very far away lab for some alien species, they would have no way of detecting the charge. Those observers would have no knowledge of the field. The laws of physics may hold everywhere, but they have no experimental way of knowing the laws.

    But this is contrasted with the case that in a universe with two masses. A lab infinitly far away can detect the gravitational field( i.e warped spacetime) using massless photons.

    It's as if there is a fundamental asymmetry between masses and charges that allow those situations to occur.
  15. Dec 2, 2016 #14


    Staff: Mentor

    None of that argument makes sense to me.
    Why not?

    Why not? And really, so what?

    Why not?

    EM radiation is far easier to detect than gravitational radiation.
    Last edited: Dec 2, 2016
  16. Dec 2, 2016 #15
    Oh I completely forgot. Yes accelerating charges generate electromagnetic radiation. The only way this can happen is if the there are electric and magnetic field lines/loops that pinch off and radiate outwards.
  17. Dec 3, 2016 #16


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    The entire physics is just making up stuff to match observation.
  18. Dec 3, 2016 #17


    User Avatar

    Staff: Mentor

    Including observations that haven't yet been made; matching these is what predicting experimental results is all about.
  19. Dec 3, 2016 #18


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    I'd like to add another aspect in answering the excellent question in the OP. This question, in principle leads you to the answer why we have to introduce the field concept into physics at all.

    The reason is that physics has taught us that the world is described by relativistic rather than Newtonian physics, and this implies that there cannot be "actions at a distance", where indeed you'd not need a field. The paradigmatic example is Newton's theory of gravity. It says that two point masses attract each other with an inverse distance-squared law, and that this holds instantaneously, i.e., if you move mass 1 then it will instantaneously change the force on mass 2 due to gravity. According to relativistic physics, this is impossible since there cannot be causal effects (motion of mass 1) instantaneously propagating to a distant place (change of force on mass 2). So the way out, found by pure imagination and experience with experiments concerning electromagnetism by Faraday, is the field concept, according to which the interactions are mediated by fields, i.e., the force on mass 2 due to the presence of mass 1 is not instantaneous but it's a local effect, i.e., there's a gravitational field associated with any mass spreading over the entire space (mass 1 as source of its gravitational field) and mass 2 feels a force due to the presence of the field of mass 1.

    Now, what has this to do with the question concerning momentum? That's also a deep concept of modern physics, namely the symmetry principles, brought into the physicists's thinking by Einstein in 1905 and into nearly final form by Emmy Noether in 1918. The idea is to address the question, which symmetries the physical laws obey, i.e., to look for transformations that don't change the form of the equations describing the dynamics of all kinds of stuff. The first thing that comes to mind is space, where symmetry has a direct geometrical meaning, and indeed both Newtonian space and special relativistic space for an inertial observer is of very high symmetry. It's the Euclidean space, and it is symmetric under translations (i.e., no point looks any different than any other) and rotations around any point (i.e., there's also no somehow preferred direction in space). Now Noether has shown that if the dynamical laws obey a continuous symmetry like the translations, there must be some conserved quantity associated to that symmetry transformation (and the math tells you precisely which quantities are these conserved ones). In the case of the spatial translations it's momentum.

    Now, since special relativity, obeys translation invariance, also in special relativity one must be able to define a momentum that is conserved due to the symmetry of space under translations. Now you see another dilemma with the action-at-a-distance models: When mass 1 changes its momentum and there is just some delayed reaction of mass 2 interacting with mass 1, for some time the momentum-conservation law would be violated since particle 2 couldn't make up for the change in particle 1's momentum instantaneously, but here also the field concept helps immediately. Since the field is there everywhere, it can take up the momentum change and carry itself momentum, such that momentum conservation is always fulfilled, and that's how things work (at least in special relativity): The field is not just somehow statically associated with their sources but themselves dynamical entities that can carry energy, momentum, and angular momentum, and they obey themselves dynamical laws (equations of motion) like the Maxwell equations of the electromagnetic field.
  20. Dec 4, 2016 #19
    I've been doing some research. Appearently, charges are just spikes in intensity of the EM field. It's the field that is more fundamental in QFT. But this view isn't classical.

    So, It's just like an ice cube that is partially melted. The solid part of the ice is the particle(solidified field); the liquid part is the spread out part of field. They are essentially the same substance. One part of it is just in a different state than the other part, but they are the same object.
  21. Dec 4, 2016 #20
    Ah, so the idea is that the charges are fundamentally connected with the field that it creates. So in the electrostatic case, I can think the field as an extention of the charge but also seperate at the same time. Kind of like a planet with a soild core being the charge and the gaseous atmosphere being the connected field, where it gets weaker and less soild the further one moves out. And in the electrodynamic case, EM radiation is kind of like one of those intro momentum problems where the masses fragment, except one of the fragment is solid and the other is spread out/wavy.
  22. Dec 4, 2016 #21


    Staff: Mentor

    Yes. This is correct. However, since this thread is in the classical physics section and since the first sentence in the first post specified that you were asking about classical fields then you really should not get into a quantum discussion here. Instead, open a new thread in the QM section.

    Ugh, no!!! This is terrible. Discard this image entirely from your mind. Go to the QM section and learn if you wish, but be forewarned, there is no classical analogy to be had. A quantum particle is its own kind of thing and it needs to be understood in terms of its actual quantum behavior, not a sloppy classical analogy.

    Again, no!!!
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