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How do the Maxwell equations transform under a time reversal?

  1. Dec 13, 2013 #1

    Let me ask you the silliest question of the year. I am looking at the Maxwell equations in their standard form. No 4-dim potential A, no Faraday tensor F, no mentioning of special relativity - just the standard form from a college-level textbook.

    I know that the eqns are NOT form-invariant under the pure Galilean transformation (addition of a constant velocity). That's fine.

    Now, how about the other subgroups of the full Galilean group?

    That the equations are form-invariant under time-independent spatial rotations, spatial translations and time dilations is most obvious.

    Form-invariance under inversion of the spatial axes will also become clear if we agree that B is not a vector but a pseudo-vector.

    What remains is the inversion of time. From looking at the equations with curl E and curl B, I see two options:

    1. One would be to agree that B changes its direction to opposite under a time inversion. This option is not easy to justify, because B = curl A is not expected to "feel" a time reversal.

    2. Another option would be to agree that an inversion of time yields a simultaneous change of the sign of charge and of the E field. My guess is that this option is right and that this has something to do with the CPT business.

    How would a physicist of the late XIX century resolve this issue, a guy who knew nothing about special relativity or CPT, and who operated with only elementary concepts?

    Many thanks!!

  2. jcsd
  3. Dec 13, 2013 #2


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    Interesting question Michael! I can't immediately comment on the subtleties but I have read through the following paper and while it deals with the issue using the generally covariant Maxwell equations, the discussion and references within should guide you along: http://philsci-archive.pitt.edu/1406/1/Albert-TRI.pdf
  4. Dec 13, 2013 #3
    Thanks for the informative reference. I shall try to digest it, though it does not seem a simple piece of reading.

    I see that the author of that article had to resort to pretty advanced concepts, and I am trying to express this in simple words. From the author's equations (22 - 23), I see that, indeed, an inversion of time necessitates a simultaneous change of the sign of charge and of the E field. (Please correct me if I am wrong!)

    Now, let's see what happens to the Lorentz force F = q [ E + v x B ] . In the brackets, both E and v change their signs, with B staying intact. But then the charge q changes its sign too, so the force F remains unchanged.

    This is very counter-intuitive. But on the other hand, if we recall that the force is equal to the mass multiplied by the SECOND derivative w.r.t. time - than all seems to work nicely.

    Do you agree with my sketchy logic?

    Many thanks,

  5. Dec 14, 2013 #4


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    I also think within classical electromagnetic theory, i.e., Maxwell theory, it is easier to argue from the basics of the operational definition of the fields involved, starting from purely mechanical arguments. I'll use Heaviside-Lorentz units with [itex]c=1[/itex] and argue with special-relativistic space-time arguments, because the Maxwell theory is the paradimatic example of a relativistic classical field theory. However, I'll stick to the non-covariant three-vector formalism, because that's more intuitive than the covariant formulation, particularly when it comes to time reversal ;-)).

    Let's start with the definition of momentum of a point particle, which is most easily inferred from the definition of a four-vector that behaves like energy and momentum in the Galileian limit. A particle's motion is described as a worldline in four-dimensional Minkowski space. The natural parameter is the proper time of the particle, defined by
    [tex]\mathrm{d} \tau=\sqrt{\mathrm{d} x_{\mu} \mathrm{d} x^{\mu}}=\mathrm{d} t \sqrt{1-\vec{v}^2},[/tex]
    where [itex]t[/itex] is the coordinate time in an inertial frame of reference, and [itex]x^{\mu}[/itex] is the time-position four vector, and [itex]\vec{v}[/itex] the usual (non-covariant) velocity [tex]\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t.[/tex]

    With the invariant mass [itex]m[/itex] of the particle, we have
    [tex]p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{m}{\sqrt{1-\vec{v}^2}} \begin{pmatrix} 1 \\ \vec{v} \end{pmatrix}.[/tex]
    Note that the temporal component is the kinetic energy of the particle, including it's rest energy [itex]E_0 = m[/itex], and [itex]\vec{p}[/itex] is the three-momentum of the particle (as can be seen by going to the non-relativistic limit [itex]\vec{v}^2 \ll 1[/itex].

    The electromagnetic field is introduced together with the notion of charge [itex]q[/itex] of the point particle by its action on such a point particle, i.e., the Lorentz-force Law:
    [tex]\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=q (\vec{E}+\vec{v} \times \vec{B}).[/tex]
    Now to see, how the various quantities transform under time-reversal, we must assume that the theory is time-reversal invariant and see, whether this assumption really allows us to define a time-reversal operation on the various physical quantities involved that at the same time is also a symmetry of the Maxwell equation, the field equations, which tell us, how the electromagnetic fields are to be determined from its sources. As is well known, this is a delicate issue of self-consistency, and it's not even totally clear, whether we are able to formulate a fully consistent classical theory of interacting charges and the electromagnetic fields (for sure that's not possible for the naive point-particle picture, I use here). Thus we shall be content with the standard approximation of fields, created by a charge-current distribution on the one hand and the motion of "test charges" in these fields on the other (if necessary one can correct for radiation damping in leading order perturbation theory as detailed in many textbooks, e.g., Jackson). For more information on this difficult problem, see also

    F. Rohrlich, Classical Charged Particles, World Scientific

    So let's discuss the time-reversal operation on the quantities involved so far. By definition, the time reversal operation is defined on the components of the four vector wrt. an inertial frame as
    [tex]t \rightarrow t'=-t, \quad \vec{x} \rightarrow \vec{x}'=\vec{x}.[/tex]
    This immediately implies for the four-momentum vector
    [tex]p^0=E \rightarrow p'^0=E'=E, \quad \vec{p} \rightarrow \vec{p}'=-\vec{p}.[/tex]
    Now consider the left-hand side of the Lorentz-Force Law. Assuming that mass, as an intrinsic parameter is invariant under time reversal, which makes sense, since the mass of a classical particle should be a positive scalar real quantity, then the left-hand side is even under time reversal, because
    [tex]m \frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=m \frac{\mathrm{d} \vec{p}'}{\mathrm{d} t'}.[/tex]
    That means that the right-hand side must also stay unchanged under time reversal. Assuming again that [itex]q[/itex] is a scalar under time reversal, we must conclude that
    [tex]\vec{E} \rightarrow \vec{E}'=\vec{E}, \quad \vec{B} \rightarrow \vec{B}'=-\vec{B}.[/tex]
    Now we have to check, whether the Maxwell equations are also invariant under time reversal.

    Let's start with the equations, not involving time derivatives. These are
    [tex]\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \cdot \vec{B}=0.[/tex]
    Since [tex]\vec{\nabla}'=\vec{\nabla}[/tex] under time reversal, and since the electric charge is a scalar, which implies that also the charge density is a scalar, the first equation is invariant under time reversal. The magnetic field flips sign under time reversal, but that doesn't matter in the above Gaussian Law, because there are no magnetic charges (if one assumes the existence of magnetic charges, we have to assume that these flip sign under time reversal as then does also the magnetic-charge density, but we'll not discuss this generatlization of Maxwell theory further).

    Now we come to the laws involving time derivatives. Let's start with Faraday's Law:
    [tex]\partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0.[/tex]
    This is also invariant under time reversal, because obviously [itex]\partial_t \rightarrow \partial_{t'}=-\partial_t[/itex]. Since [itex]\vec{B}[/itex] is odd and [itex]\vec{E}[/itex] is even, and there are only spatial derivatives of the latter, the equation is indeed invariant under time reversal.

    Last but not least we must investigate the Maxwell-Ampere Law
    [tex]-\partial_t \vec{E}+\vec{\nabla} \times \vec{B}=\vec{j}.[/tex]
    The left-hand side flips sign under time-reversal. This must be also true for the right-hand side, which becomes immediately clear when we consider [itex]\vec{j}[/itex] as given by
    [tex]\vec{j}=\vec{v} \rho.[/tex]
    As argued above, [itex]\rho[/itex] is even under time reversal. Since the fluid-flow velocity [itex]\vec{v}[/itex] is odd, also [itex]\vec{j}[/itex] thus flips sign as it must be.

    Considering also magnetization we have to add a current,
    [tex]\vec{j}_m=\vec{\nabla} \times \vec{m},[/tex]
    which must also flip sign under time reversal, and thus [itex]\vec{m}[/itex] also must flip sign.

    With these definitions of the behavior of the various field quantities under time reversal we such have time-reversal invariance of Maxwell's theory.

    This is an interesting example for the derivation of transformation laws under a certain transformation from an assumed symmetry of a physical theory. Only, if you can show that such transformations cannot be defined consistently with the equations of motion, you can prove that the theory is not invariant under the transformation, and thus it has not the corresponding symmetry.

    You can also easily show by similar arguments that electromagnetic Maxwell theory is invariant under space reflections (parity). Of course these symmetries, which have been argued in terms of microscopic Maxwell theory, also hold for macroscopic Maxwell theory which can be derived from the microscopic theory using arguments of kinetic theory.
  6. Dec 14, 2013 #5
    If you time reverse a moving particle it will move backwards which changes the sign of the magnetic field with no effect on the electric field. It's as simple as that.
  7. Dec 14, 2013 #6

    Jano L.

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    Inversion of time is a formal operation, it does not necessitate any other operation. However, sole inversion of time does not give the equations of the same form. But if in addition magnetic field is reversed as well, the resulting equations have the same form, so when we say electromagnetic theory is time-reversal invariant, we mean that both time and magnetic field are to be inverted in the operation "time reversal". What this really means is that if all velocities and magnetic fields were reversed at some instant, the system (positions of charged particles and electric fields) would revisit its past states in inverted order (velocities and magnetic fields will also revisit past magnitudes, but with opposite signs).

    Here you've got it wrong. Charge and electric field do not change sign, but the magnetic field does.
  8. Dec 16, 2013 #7
    Dear vanhees71, dauto and Jano L.,

    Thank you very much for your help !!

    Much food for thought, really
  9. Dec 2, 2014 #8
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