4 ms·

I mean that the theory splits into two parts: the part that tells you the fields given the charges, and the part that tells you the charges given the fields. Fo

by jules 8y ago

I mean that the theory splits into two parts: the part that tells you the fields given the charges, and the part that tells you the charges given the fields. For instance, it doesn't tell you what two charges with given initial velocities will do. (not even if you also give an initial electromagnetic field)

Certhas 8y ago
Well, the Lorentz force does though. And that follows from energy conservation and the Maxwell equations, e.g.: https://physics.stackexchange.com/a/77028 https://physics.stackexchange.com/a/77028 So that's not really terribly deep or even true.
- jules 8y ago
  So what are the equations for those two charges?
  - Certhas 8y ago
    For any number of particles, the equation for each of them is this one here: https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism#Charged_particle https://en.wikipedia.org/wiki/Covariant_formulation_of_class... Edit: Coupled differential equations. If I have an equation for x in terms of y, and one for y in terms of x, then in total I have a set of equations for x and y. e.g.: dx/dt = y dy/dt = -x then the solution is x = C e^(i t) y = i C e^(i t) with C a constant determined by the initial conditions. Nothing mysterious.
    - jules 8y ago
      I understand differential equations ;-) That page describes Maxwell's equations and the Lorentz force law. That a naive approach cannot work can be seen by considering the following example. Take one charge initially at rest with huge mass, and another light charge orbiting around it. According to Lorentz law it will orbit in a circle for the right initial conditions. However, then according to Maxwell's equations it will radiate electromagnetic waves, violating energy conservation. The field of the accelerated charge will affect the charge itself, but this is not easy to take into account, because that field is infinite at the location of the charge.
      - DoctorOetker 8y ago
        define radiate? if you calculate the poynting vector you see energy is not really leaving the system in this case, although there certainly is electromagnetic oscillation/rotation/circulation, so there is no energy violation. 1) If we consider the ground state of a system this behaves as expected quantum mechanically: there is motion in the ground state but no energy leaves the system! Why hold Maxwell equations to a perhaps higher or perhaps falser standard than we hold quantum mechanics?? 2) But if a system is not in its ground state we do expect (by the observation that excited molecules emit light) this oscillation/rotation of the electromagnetic field to radiate away energy that leaves the system. (Actually the simplest, most widely taught formulations of quantum mechanics, also don't predict decay of an excited energy state: The eigenstates of energy are valid solutions, and the solutions rotate in the complex plane indefinitely, so here it is both deterministic solutions to Maxwell Equations and Introductory QM predicting incorrectly no decay, with the important difference that Introductory QM can already calculate the energy spectrum, and hence absorption and emission spectrum, glossing over the fact that the transition is not in fact yet predicted) 3) As I said, I am willing to agree to calling Maxwell's equatins schizophrenic, but not because of their coupled equations nature, but because of its indeterministic nature (again a similarity with quantum mechanics!). If we subdivide the worldline of each charge's motion into individual infinitessimmal but continuous motion events, then we can spend a total of one Lienard Wiechert potentials (say 1.0 delayed + 0 advanced, the other way around, 0.7 delayed and 0.3 advanced, or -10.4 delayed and +11.4 advanced, etc!) on each such event individually, i.e. we can have this division that sums to 1 depend on both particle and time, as long as the total applied force in the past and future respect are compatible with the trajectory of the particle... this is quite schizophrenic indeed! I since long suspect Maxwells Equations to contain quantum mechanical behaviour that has not been explored yet. EDIT: There are roughly speaking 2 kinds of physicists/students: 1) those who either deny, ignore, gloss over this indeterminacy problem, or point at a pseudo-contradiction, and ignore they are holding Maxwell equations to a higher standard than Introductory QM and "just move on to QM, btw shut up and calculate!" 2) those who recognize explicitly or implicitly this schizophrenia (of the second indeterminacy kind, not the coupledness kind) and have attempted various approaches or formulations to rigorously enumerate/solve for all mathematically valid (self-consistent) histories/trajectories/solutions of Maxwell's equations. Einstein was famously trying to resolve this issue (among others) for the rest of his life, Feynmann worried about it (you can read this between the lines in his Absorber Theory), and so on. I believe many of those approaches and attempts will turn out to be different but equivalent perspectives, once the problem of solving for all self-consistent histories of ME's with given initial conditions has been solved... I obviously belong to group 2) and given your dissatisfaction, I presume you too belong to group 2). However I believe that the issue is not the maxwell equation's themselves, but our lack of mathematics to solve for all self-consistent histories... Those who do not express dissatisfaction at the lack of complete enumeration of solutions to this insidious/schizophrenic/indeterminacy of ME, or just move on to QM belong to camp 1). There is no shame in that as long as they don't ridicule/demotivate those of us trying to bridge (semi-)classical physics with quantum mechanics... and make the analogies Dyson points out between Maxwell Equations and QM (the first and second layer, etc) more complete. Perhaps this problem has already been resolved between Maxwell's publication of the ME's, and today. Perhaps the solution is present in the literature, but with very low uptake because most people are in camp 1). Perhaps the person who solved the issue was too modest in describing the significance of what he has concluded, and we will read about it in 70 years...
    - aj7 8y ago
      As a depressed MIT freshman I modeled my mood as a pair of coupled equations. I got growing oscillations!

by jules 8y ago

Certhas 8y ago

Well, the Lorentz force does though. And that follows from energy conservation and the Maxwell equations, e.g.: https://physics.stackexchange.com/a/77028 https://physics.stackexchange.com/a/77028 So that's not really terribly deep or even true.

jules 8y ago
So what are the equations for those two charges?
- Certhas 8y ago
  For any number of particles, the equation for each of them is this one here: https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism#Charged_particle https://en.wikipedia.org/wiki/Covariant_formulation_of_class... Edit: Coupled differential equations. If I have an equation for x in terms of y, and one for y in terms of x, then in total I have a set of equations for x and y. e.g.: dx/dt = y dy/dt = -x then the solution is x = C e^(i t) y = i C e^(i t) with C a constant determined by the initial conditions. Nothing mysterious.
  - jules 8y ago
    I understand differential equations ;-) That page describes Maxwell's equations and the Lorentz force law. That a naive approach cannot work can be seen by considering the following example. Take one charge initially at rest with huge mass, and another light charge orbiting around it. According to Lorentz law it will orbit in a circle for the right initial conditions. However, then according to Maxwell's equations it will radiate electromagnetic waves, violating energy conservation. The field of the accelerated charge will affect the charge itself, but this is not easy to take into account, because that field is infinite at the location of the charge.
    - DoctorOetker 8y ago
      define radiate? if you calculate the poynting vector you see energy is not really leaving the system in this case, although there certainly is electromagnetic oscillation/rotation/circulation, so there is no energy violation. 1) If we consider the ground state of a system this behaves as expected quantum mechanically: there is motion in the ground state but no energy leaves the system! Why hold Maxwell equations to a perhaps higher or perhaps falser standard than we hold quantum mechanics?? 2) But if a system is not in its ground state we do expect (by the observation that excited molecules emit light) this oscillation/rotation of the electromagnetic field to radiate away energy that leaves the system. (Actually the simplest, most widely taught formulations of quantum mechanics, also don't predict decay of an excited energy state: The eigenstates of energy are valid solutions, and the solutions rotate in the complex plane indefinitely, so here it is both deterministic solutions to Maxwell Equations and Introductory QM predicting incorrectly no decay, with the important difference that Introductory QM can already calculate the energy spectrum, and hence absorption and emission spectrum, glossing over the fact that the transition is not in fact yet predicted) 3) As I said, I am willing to agree to calling Maxwell's equatins schizophrenic, but not because of their coupled equations nature, but because of its indeterministic nature (again a similarity with quantum mechanics!). If we subdivide the worldline of each charge's motion into individual infinitessimmal but continuous motion events, then we can spend a total of one Lienard Wiechert potentials (say 1.0 delayed + 0 advanced, the other way around, 0.7 delayed and 0.3 advanced, or -10.4 delayed and +11.4 advanced, etc!) on each such event individually, i.e. we can have this division that sums to 1 depend on both particle and time, as long as the total applied force in the past and future respect are compatible with the trajectory of the particle... this is quite schizophrenic indeed! I since long suspect Maxwells Equations to contain quantum mechanical behaviour that has not been explored yet. EDIT: There are roughly speaking 2 kinds of physicists/students: 1) those who either deny, ignore, gloss over this indeterminacy problem, or point at a pseudo-contradiction, and ignore they are holding Maxwell equations to a higher standard than Introductory QM and "just move on to QM, btw shut up and calculate!" 2) those who recognize explicitly or implicitly this schizophrenia (of the second indeterminacy kind, not the coupledness kind) and have attempted various approaches or formulations to rigorously enumerate/solve for all mathematically valid (self-consistent) histories/trajectories/solutions of Maxwell's equations. Einstein was famously trying to resolve this issue (among others) for the rest of his life, Feynmann worried about it (you can read this between the lines in his Absorber Theory), and so on. I believe many of those approaches and attempts will turn out to be different but equivalent perspectives, once the problem of solving for all self-consistent histories of ME's with given initial conditions has been solved... I obviously belong to group 2) and given your dissatisfaction, I presume you too belong to group 2). However I believe that the issue is not the maxwell equation's themselves, but our lack of mathematics to solve for all self-consistent histories... Those who do not express dissatisfaction at the lack of complete enumeration of solutions to this insidious/schizophrenic/indeterminacy of ME, or just move on to QM belong to camp 1). There is no shame in that as long as they don't ridicule/demotivate those of us trying to bridge (semi-)classical physics with quantum mechanics... and make the analogies Dyson points out between Maxwell Equations and QM (the first and second layer, etc) more complete. Perhaps this problem has already been resolved between Maxwell's publication of the ME's, and today. Perhaps the solution is present in the literature, but with very low uptake because most people are in camp 1). Perhaps the person who solved the issue was too modest in describing the significance of what he has concluded, and we will read about it in 70 years...
  - aj7 8y ago
    As a depressed MIT freshman I modeled my mood as a pair of coupled equations. I got growing oscillations!