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Alternating Current and Simple Harmonic Motion

  1. Aug 26, 2013 #1
    Hello,
    I was being taught AC in High School, It was good but the way they taught us DC, things like drift velocity, no of electrons per unit volume etc, it was easy to visualize electrons rushing in a conductor. I tried to visualise AC(which was not taught to us) and I came to a conclusion that as in case of SHM or more specifically spring mass system the force direction periodically changes, the EMF in the conductor carrying AC also changes periodically hence the force on every electron is also changing direction periodically. So the electrons are having SHM in the conductor similar to the particles oscillating in organ pipes' first harmonics. Am I correct? IF not please describe me what is microscopically going in AC.
     
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  3. Aug 26, 2013 #2

    sophiecentaur

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    "Electrons rushing" anywhere is really not a good model for many instances of electric current. (the CRT being a rare exception). In AC, they hardly get a chance to go anywhere before reversing their direction and some of them will be travelling much further than that in their random thermal motion.

    There is nothing in the term 'AC' that implies SHM. It just Alternates - or changes sign regularly. Sinusoidal AC is generated when you rotate a single loop of wire in a uniform Magnetic field and practical Alternators will approximate to this. But look at the Voltage wavefrom on any Mains outlet (with an oscilloscope) and it will be far from sinusoidal, being bumpy and mis-shapen and it will also change as loads are added or taken away. It is very hard to distinguish, visually, a good sinewave from a distorted one, in any case.
    What you get from a cheap and cheerful Inverter will be more like a squarewave than anything yet it is still 'AC".
     
  4. Aug 26, 2013 #3
    by electrons rushing I meant they along with thermal motion are having a drift velocity. I am just asking about the visualization of AC. Also I was talking of pure AC as that is what is taught to us. If that SHM thing is wrong then how to think about AC microscopically?
     
  5. Aug 26, 2013 #4

    sophiecentaur

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    Don't think about it microscopically any more than you would think of thermodynamics or Boyle's Law microscopically. Once you have derived the Kinetic Theory by considering a single molecule, you stay Macroscopic. 'Electricity' is a Macroscopic phenomenon. You are not 'compromising' or getting it wrong by thinking macroscopically.
    BTW, The nearest to a good SHM form of AC would be what you get in an LC resonant circuit, with no driving source and left to itself.
     
  6. Aug 26, 2013 #5

    vanhees71

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    To visualize currents as flow of electrons can help a lot when it comes to more complicated issues, particularly in connection with moving bodies and relativity. E.g., the homopolar generator is treated in very many textbooks in a what that's quite difficult to understand. For a simple and correct treatment, see the Feynman Lecture. This problem is, however, also related to the fact that very often Faraday's Law in integral form is incompletely stated. But that doesn't answer the question.

    So let's see, how to understand the question on AC from the point of view of the simplified classical microscopic theory. In a conducting material (metals) you have a certain amount of quasi-free conduction electrons with density [itex]n[/itex]. They are, however, subject to friction due to collisions with impurities in the crystal lattice etc. So an effective equation of motion for low-frequency low-intensity electromagnetic fields is
    [tex]m \dot{\vec{v}}+m \gamma \vec{v}=-e \vec{E}(t).[/tex]
    Here, we have assumed that the ciruit is small compared to the typical wavelenght of the field so that we can assume the the electric field is homogenous along the wire and that the velocity of the electrons is very small compared to the speed of light so that you can neglect the magnetic component of the Lorentz force and use the Newtonian expression for momentum instead of the relativistic one.

    To interpret this equation further, we first look for the Green's function of the linear differential operator on the left-hand side. Since we want to describe causal signal propagation, it should be the retarded Green's function, i.e., we look for a solution of
    [tex]\dot{G}(t) + \gamma G(t)=\delta(t), \quad G(t) =g(t) \Theta(t),[/tex]
    where [itex]\Theta[/itex] is the Heaviside unit-step function. Plugging in this ansatz and using [itex]\dot{\Theta}(t)=\delta(t)[/itex] gives
    [tex][\dot{g}(t) + \gamma g(t)] \Theta(t)+g(0) \delta(t)=\delta(t).[/tex]
    This means that we need [itex]g(0)=1[/itex] and
    [tex]\dot{g}+\gamma g=0 \; \Rightarrow\; g(t)=\exp(-\gamma t).[/tex]
    Thus we have
    [tex]G(t)=\Theta(t) \exp(-\gamma t).[/tex]
    The solution of the original equation then is
    [tex]\vec{v}(t)=-\frac{e}{m} \int_{\mathbb{R}} \mathrm{d} t' G(t-t') \vec{E}(t).[/tex]
    The current density is
    [tex]\vec{j}=-e n \vec{v}=\frac{e^2 n}{m} \int_{\mathbb{R}} \mathrm{d} t' G(t-t') \vec{E}(t).[/tex]
    Let's apply this to a constant electric field first. Then we get
    [tex]\vec{j}=\frac{e^2 n \vec{E}}{m} \int_{-\infty}^t \mathrm{d} t' \exp[-\gamma (t-t')]=\frac{e^2 n E}{\gamma}=\sigma \vec{E}.[/tex]
    This gives a nice microscopic approximation for conductivity [itex]\sigma[/itex].

    For AC that's not very different, i.e., you just have to evaluate
    [tex]\vec{j}=\frac{e^2 n \vec{E}_0}{m} \int_{-\infty}^t \mathrm{d} t' \exp[-\gamma(t-t')]\cos(\omega t)=\frac{e^3 n \vec{E}_0}{m} \frac{\gamma \cos(\omega t)+\omega \sin(\omega t)}{\omega^2+\gamma^2}.[/tex]
    This shows that there is a phase shift between current and electric field due to damping. What we calculate here is of course the stationary state of the circuit, because we switch on the electric field at [itex]t \rightarrow \infty[/itex]. As you see it's indeed as expected a harmonic motion of the electrons in the external AC field.

    In Fourier space the AC case is even simpler. Defining the Fourier transform of a quantitiy in the form
    [tex]\tilde{f}(\omega)=\int_{\mathbb{R}} \mathrm{d} t \exp(\mathrm{i} \omega t) f(t) \; \Leftrightarrow \; f(t)=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{2 \pi} \exp(-\mathrm{i} \omega t) \tilde{f}(\omega),[/tex]
    you have
    [tex]F(t) =[f*g](t)=\int_{\mathbb{R}} \mathrm{d t'} f(t-t') g(t') \; \Leftrightarrow \; \tilde{F}(t)=\tilde{f}(\omega) \tilde{g}(\omega),[/tex]
    which is known as the convolution theorem.

    Now
    [tex]\tilde{G}(\omega)=\int_{\mathbb{R}} \mathrm{d} \omega \exp(\mathrm{i} \omega t) \Theta(t) \exp(-\gamma t)=\frac{1}{\gamma - \mathrm{i} \omega}.[/tex]
    Thus in frequency space we have
    [tex]\tilde{\vec{j}}(\omega)=\frac{n e^2}{m} \frac{1}{\gamma-\mathrm{i} \omega} \tilde{\vec{E}}(\omega) \; \Rightarrow \sigma(\omega)=\frac{n e^2}{m (\gamma-\mathrm{i} \omega)}.[/tex]
    In frequency space you can express Ohm's Law for AC as for DC but with a complex [itex]\omega[/itex]-dependent conductivity.
     
  7. Aug 26, 2013 #6

    sophiecentaur

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    I'm not sure that is a genuine "microscopic" approach. It is just using a 'representative' electron and assuming that they are all the same. In actual fact, they are all doing different things and the analysis sidesteps the statistics involved. It is, in effect, using a macroscopic (bulk) model and could replace the electronic charge with 'unit charge' without making any difference to the argument.
     
  8. Aug 27, 2013 #7
    Thank you vanhees71 for your explanation but you see I am in highschool I don't understand the things you quoted above. I think I can understand it If you could give any example. Classical Physics is Intuitive and just tell me how electron behave when applied AC to a conductor. Also I think Classical Physics at its elementary level could be understood by not using mathematics also. Thanks
     
  9. Aug 27, 2013 #8

    vanhees71

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    @sophiecentaur: That's true. I left out thermal motion. A full transport-theoretical treatment is more complicated but interesting. If you are interested in this, a very good book on this is

    S.R. de Groot, L.G. Suttorp, Foundations of Electrodynamics, North-Holland Pubilishing Comp. (1972)

    @aayushgsa: the only thing I did was to solve the equation of motion for an electron that is subject to a friction force and the electric field. I don't think that you can understand physics without mathematics and calculus. It's just the language to express how nature behaves.

    The use of the Green's function is of course too difficult at high school. Thus let's solve the equation of motion with more elementary techniques and for motion in one dimension. The equation is
    [tex]m \dot{v}=-m \gamma v-e E(t),[/tex]
    where [itex]E(t)][/itex] is the component of the electric field in direction of the motion. We assume that the other components of the field are vanishing.

    Now suppose you have two solutions [itex]v_1(t)[/itex] and [itex]v_2(t)[/itex] of the above differential equation. Because it's an equation that is linear in [itex]v[/itex] for [itex]w=v_1-v_2[/itex]
    [tex]m \dot{w}=-m \gamma w \; \Rightarrow \; \dot{w}=-\gamma w[/tex]
    This equation is easy to solve. You just look for a function, that's derivative is proportional to the function itself. The only function with this property is the exponential function. Thus the solution is
    [tex]w(t)=w_0 \exp(-\gamma t),[/tex]
    where [itex]w_0[/itex] is a constant.

    Now you only need one solution for the full equation. The trick is to make the ansatz of the "variation of the constant", i.e., you write
    [tex]v(t)=u(t) \exp(-\gamma t)[/tex]
    Then you get
    [tex]\dot{v}(t)=[\dot{u}(t)-\gamma u(t)] \exp(-\gamma t) = -\gamma u(t) \exp(-\gamma t)-\frac{e}{m} E(t).[/tex]
    Simplifying this equation gives
    [tex]\dot{u}(t) \exp(-\gamma t)=-\frac{e}{m} E(t).[/tex]
    Thus the solution for [itex]u[/itex] is
    [tex]u(t)=-\frac{e}{m} \int_{t_0}^t \mathrm{d} t' \exp(\gamma t') E(t')+u_0.[/tex]
    and thus
    [tex]v(t)=u(t) \exp(-\gamma t)=-\frac{e}{m} \int_{t_0}^t \mathrm{d} t' \exp[-\gamma(t-t')] E(t')+u_0 \exp(-\gamma t).[/tex]
    If you now think of the external field to be switched on at [itex]t_0 \rightarrow -\infty[/itex] with the electron at rest at [itex]t=t_0[/itex] you get the solution, I've given with help of the Green's function in the previous posting.
     
  10. Aug 27, 2013 #9

    sophiecentaur

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    You are also leaving out any possible non linearity by assuming you can treat all electrons as if they were, in fact, all at the same velocity. (The distribution is ignored and I don't think it can be, except in a true metal (?)) As I said, you could just as easily replace 'electron' with 'unit charge carrier' or 'coulomb'. Nothing wrong with this at all, for getting a result, but it can lead to misinterpretation, I think.
    aayushgsa has made the comment that
    but it really is not necessarily the case. Some aspects of Classical Physics may be 'familiar' enough to people with a given level of knowledge but 'intuition' can be just as misleading in classical affairs as in QM. The simplification of charge flow by assuming a lot of identical electron movements is an example of what I mean. Going further back, "nature abhors a vacuum" is an intuitive statement but we now know that it is not a good description of the Physics involved.

    The title of the thread is to do with AC and SHM, which implies a macroscopic view so why try to use a half way house sort of model which uses the electron to 'help' with the understanding of what's happening? Intuition can easily run out in the microscopic world. The Particle Explanation is very popular in Education these days but I have seen too many cases of Teachers and Students struggling to explain perfectly straightforward / familiar situations in terms of particles (even convection, for God's sake) when the macroscopic description is clearly the appropriate one.
    aayushgsa wants, very reasonably, to use intuition where he/she can. Intuition works in the macroscopic world but we all use macroscopic models to explain the microscopic world - pulling ourselves up by our own intuitive bootstraps.
     
  11. Aug 27, 2013 #10

    vanhees71

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    Of course, for strong fields, this linear-response ansatz is not valid anymore, but you can show from very basic principles (QED at finite temperature or even non-equilibrium many-body QED) that for weak disturbances from equilibrium there is a general linear-response approximation. This leads to the usual equations of macroscopic electrodynamics. In almost any textbook one goes even further and treating the motion of the charge carriers (treated as fluids/plasmas) non-relativistically, which leads to the usual simplified constitutive relations
    [tex]\begin{split}
    \tilde{\vec{j}}(\omega,\vec{x}) \simeq \sigma(\omega) \tilde{\vec{E}}(\omega,\vec{x}), \\
    \tilde{\vec{D}}(\omega,\vec{x}) \simeq \epsilon(\omega,\vec{x}) \tilde{\vec{E}}(\omega,\vec{x}),\\
    \tilde{\vec{B}}(\omega,\vec{x}) \simeq \mu(\omega,\vec{x}) \tilde{\vec{H}}(\omega,\vec{x}).
    \end{split}
    [/tex]
    This approximation also assumes that the wavelength of the fields is small compared to intermolecular distances, so that spatial dispersion can be neglected. This approximation is quite valid for most applications in optics with visible lights and usual everyday matter. It's not good for plasmas, but you still can go far with the non-relativistic approximation as long as the matter involved is at rest.

    For moving bodies there are situations where even when all velocities are small compared to the speed of light, one has to use the relativistic treatment. An example is the homopolar generator (see Feynman Lectures, vol. II, where this problem is treated in the spirit of the handwaving arguments in my posting below).
     
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