# Understanding why Einstein found Maxwell's electrodynamics not relativistic

Dale
Mentor
The electric field is the electromotive force which would be on a unit charge at a given location if there is a charge present.
Not necessarily. The electromotive force can also be due to a magnetic field, as in the case under consideration. The mere presence of an elecromotive force does not provide enough information to determine if it is an electric or a magnetic field or both. In the case under consideration it is due purely to the magnetic field, there is no electric field.

See #14 above.
The notation is archaic, and I don't know what all the variables are, so I cannot comment much on that equation. However, only three variables could possibly be the components of the electric field, and there are more than three variables. So clearly, the mere fact of an electromotive force does not imply an electric field by that equation either.

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Not necessarily. The electromotive force can also be due to a magnetic field, as in the case under consideration. The mere presence of an elecromotive force does not provide enough information to determine if it is an electric or a magnetic field or both. In the case under consideration it is due purely to the magnetic field, there is no electric field.
I don't think I made myself clear. I'll try again.

1. The electric field is the electromotive force at a given location that hypothetically *would* act if a charge were placed there. [The italics are from BCrowell's response a few posts back.] The point is that the electric field is the map of the electromotive force, irrespective of the agent(s) that bring about the force.

2. Experiment shows that there is an electromotive force when a magnet is stationary and a conductor is moving. The electromotive force is acting on the charges in the conductor. Considering a single charge in the conductor, experiment shows that there is an electromotive force at a given location when a charge is present and moving.

3. Therefore, experiment shows that there is an electric field when the magnet is stationary and there is a conductor moving. [There is a charge; there is electromotive force on the charge; by definition there is an electric field.]

The three statements above are presented as proof that there is an electric field present when the magnet is stationary. In short: It has been said that there is no electric field present when the magnet is stationary. I respond, give me a magnet and a conductor and I will demonstrate that there is an electric field.

The notation is archaic, and I don't know what all the variables are, so I cannot comment much on that equation. However, only three variables could possibly be the components of the electric field, and there are more than three variables. So clearly, the mere fact of an electromotive force does not imply an electric field by that equation either.
Addressing the text in bold: The electric field is defined as the map of electromotive force. By definition, then, the mere fact of an electromotive force is the very same thing as an electric field.

Consider the case when the magnet is moving and the conductor is stationary. The assertion is that there is an electric field. On what basis is the assertion made? On the basis of an observed electromotive force. In both cases, electromotive force is observed. Why is the observed electromotive force taken as proof of an electric field in the one case, but not the other?

Dale
Mentor
I don't think I made myself clear. I'll try again.
You were clear, just wrong.

1. The electric field is the electromotive force at a given location that hypothetically *would* act if a charge were placed there. [The italics are from BCrowell's response a few posts back.] The point is that the electric field is the map of the electromotive force, irrespective of the agent(s) that bring about the force.
Close, but not quite. The electric field is proportional to the force that would act if a stationary charge were located there.

2. Experiment shows that there is an electromotive force when a magnet is stationary and a conductor is moving. The electromotive force is acting on the charges in the conductor. Considering a single charge in the conductor, experiment shows that there is an electromotive force at a given location when a charge is present and moving.
Yes.

3. Therefore, experiment shows that there is an electric field when the magnet is stationary and there is a conductor moving. [There is a charge; there is electromotive force on the charge; by definition there is an electric field.]
No, there is an electromotive force on moving charges, which is not the definition of an electric field.

The three statements above are presented as proof that there is an electric field present when the magnet is stationary. In short: It has been said that there is no electric field present when the magnet is stationary. I respond, give me a magnet and a conductor and I will demonstrate that there is an electric field.
Not if you move the conductor.

Addressing the text in bold: The electric field is defined as the map of electromotive force. By definition, then, the mere fact of an electromotive force is the very same thing as an electric field.
The electric field is a map of the electromotive force on stationary charges, so the mere fact of an electromotive force on a moving charge is not the very same thing as an electric field.

Consider the case when the magnet is moving and the conductor is stationary. The assertion is that there is an electric field. On what basis is the assertion made? On the basis of an observed electromotive force. In both cases, electromotive force is observed. Why is the observed electromotive force taken as proof of an electric field in the one case, but not the other?
Because the charges are moving in one case but not the other.

Dale
Mentor
In modern notation the electromotive force, aka the Lorentz force, is given by:
$F = q (E + v \times B)$

Note that this force consists of a term arising from an E field and a velocity dependent term arising from a B field. If you want to claim that an emf is only due to an E field, then you must get the B field term to drop out, which happens if v=0.

I generally think of this equation as defining the E and B fields.

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vanhees71
Gold Member
2019 Award
I think one has to elaborate this a bit since it's important to make clear what's meant by "force" here. Force is a pretty difficult concept if it comes to relativity. It's better to derive everything from the action principle. As will become clear, the "Lorentz force", as defined in the previous posting is not a manifestly Lorentz covariant quantity and must be handled with some care.

Let's take, as the most simple example, a single classical charged particle moving in a given (external) electromagnetic field. The latter is given by the four-vector potential $A^{\mu}$. The relativistically covariant description of the field itself is given by the Faraday tensor,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
In conventional 1+3-dimensional writing we have
$$\vec{E}=\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
where $$\vec{A}=(A^1,A^2,A^3).$$

The Lagrangian for the particle's motion in the given em. field reads
$$L=-m c \sqrt{c^2-\vec{v}^2}-\frac{q}{c} A^{\mu} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} t}.$$
This is of course not covariant, but the action is
$$A[x]=\int \mathrm{d} t L.$$
This can be written as a functional of the world line $x^{\mu}(\lambda)$, where $\lambda$ is an arbitrary scalar parameter
$$A[x]=\int \mathrm{d} \lambda \left (-m c \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}-\frac{q}{c} \dot{x}_{\mu} A^{\mu} \right ).$$
Here the dot means a derivative with respect to the "world parameter" $\lambda$. The equations of motion follow from Hamilton's principle. To derive them we note that
$$\frac{\partial L}{\partial \dot{x}^{\mu}}=-m c \frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}}-\frac{q}{c} A_{\mu}, \quad \frac{\partial L}{\partial x^{\mu}}=-\frac{q}{c} \dot{x}^{\nu} \partial_{\mu} A_{\nu}.$$
To simplify the further calculation we note that we now can use the proper time of the particle $\mathrm{d} \tau = 1/c \sqrt{\mathrm{d} x_{\mu} \mathrm{d} x^{\mu}}$ as the "world parameter". This leads to $\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}$, and the equations of motion finally read
$$-m \ddot{x}_{\mu} -\frac{q}{c} \dot{x}^{\nu} \partial_{\nu} A_{\mu} = - \frac{q}{c} \dot{x}^{\nu} \partial_{\mu} A_{\nu}$$
or
$$m \ddot{x}_{\mu}=\frac{q}{c} F_{\mu \nu} \dot{x}^{\nu}:=K_{\mu}.$$
This is the equation of motion for a particle in an electromagnetic field in manifestly covariant form. Both sides of the equation are obviously four vectors. The right-hand side is what's usually called the Minkowski-force vector.

As expected, because we have derived the equation of motion from a world-parameter-independent action, it fulfills the necessary constraint
$$\dot{x}^{\mu} \ddot{x}_{\mu}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} \tau} (\dot{x}_{mu} \dot{x}^{\mu})=0.$$
Thus the constraint
$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}$$
is compatabile with the equation of motion.

Thus only three of the four equations of motion are independent. It's thus sufficient to only consider $\vec{x}$ as independent degrees of freedom. Written in terms of the proper time the equation of motion for these components read, rewritten in three-dimensional notation,
$$m \frac{\mathrm{d}^2 \vec{x}}{\mathrm{d} \tau^2}=q \left (\vec{E} \frac{\mathrm{d} t}{\mathrm{d} \tau}+\frac{1}{c} \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \times \vec{B} \right )=\vec{K}.$$
Of course, we can further rewrite this in terms of the coordinate time of an arbitrary inertial observer, using
$$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2/c^2}=\mathrm{d} t/\gamma,$$
where we have set
$$\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \quad \text{and} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
This gives after some algebra the equation of motion in the non-covariant form
$$m \frac{\mathrm{d}}{\mathrm{d} t} \left (\gamma \vec{v} \right ) = q \left (\vec{E}+\frac{1}{c} \vec{v} \times \vec{B} \right)=:\vec{F}.$$
From the derivation it's clear that $\vec{F}$ is not the spatial part of a four-vector. That's the case only for
$$\vec{K}=q \left (\vec{E} \frac{\partial t}{\partial \tau} +\frac{1}{c} \gamma \vec{v} \times \vec{B} \right )=q \gamma \left (\vec{E} +\frac{1}{c} \vec{v} \times \vec{B} \right )=q \gamma \vec{F}.$$
It's of course correct that the physical meaning of the electromagnetic field $(\vec{E},\vec{B})$ is defined by its action in terms of the "Lorentz force", but one should keep in mind that the Lorentz force itself is a non-covariant quantity, while the electromagnetic field is, because it's nothing but the components of the Faraday tensor, which is a 2nd-rank antisymmetric tensor in Minkowski space-time.

In modern notation the electromotive force, aka the Lorentz force, is given by:
$F = q (E + v × B)$

Note that this force consists of a term arising from an E field and a velocity dependent term arising from a B field. If you want to claim that an emf is only due to an E field, then you must get the B field term to drop out, which happens if v=0.

I generally think of this equation as defining the E and B fields.
Okay. I did not realize that the electric field applies to a stationary charge.

In the Maxwell equation for emf, the second and third terms will be included in E. These are the forces on a stationary charge from a change in the magnetic field and from the potential gradient. The first term is v × B.

Even with the knowledge that the electric field applies to a stationary charge, I am still at a loss to understand Einstein's statement, "In the [moving] conductor, however, we find an electromotive force, to which in itself there is no corresponding energy."

Clearly, the energy for the emf on the charges in the moving conductor comes from the magnetic field through which they are moving: v × B.

If Einstein has a legitimate quarrel on that point, it is not with Maxwell. Maxwell attributes the cause of the emf to the magnetic field, regardless of whether the charge is moving through the field or the field is changing in strength (as when a pole moves):
This, then, is a force acting on a body caused by its motion through the
electromagnetic field, or by changes occurring in that field itself; and the effect
of the force is either to produce a current and heat the body, or to decompose
the body, or, when it can do neither, to put the body in a state of electric
polarization...

Dale
Mentor
Even with the knowledge that the electric field applies to a stationary charge, I am still at a loss to understand Einstein's statement, "In the [moving] conductor, however, we find an electromotive force, to which in itself there is no corresponding energy."

Clearly, the energy for the emf on the charges in the moving conductor comes from the magnetic field through which they are moving: v × B.
This too is corrrect, although rather cumbersome in my opinion.

The integral of the work done by a force from a static E field about a closed loop is always 0, so the force is said to have a corresponding energy and scalar potential. The integral of the work done by a force from a static B field is not always 0, so the force is not said to have a corresponding energy or scalar potential.

Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional. So the fact that the B field does have energy "in itself" does not contradict the fact that the force from the B field does not.

This too is corrrect, although rather cumbersome in my opinion.

The integral of the work done by a force from a static E field about a closed loop is always 0, so the force is said to have a corresponding energy and scalar potential. The integral of the work done by a force from a static B field is not always 0, so the force is not said to have a corresponding energy or scalar potential.
Einstein says that the force from the B field is the same as the force from the E field, given the same relative velocity and distance. If the forces and paths are the same, how can the integrals of force over path be different? (The appropriate answer to this may be, "Go read this book.")

Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional. So the fact that the B field does have energy "in itself" does not contradict the fact that the force from the B field does not.
Without contradicting you, I will note that the first term of Maxwell's equation for emf (the term which deals with a moving charge) consists entirely of the addition and subtraction of terms of this form:
(magnetic intensity along a coordinate axis) * (velocity along a coordinate axis)

On inspection, it appears as though the force is proportional to the strength of the magnetic field. (Again, the appropriate answer to this may be, "Go read this book.")

At any rate, given that the two emfs are the same, as Einstein says they are, it seems to me that Einstein's complaint is along the lines of a legal technicality, and of no practical consequence.

I have been wondering why he did not directly raise the issue that Maxwell requires the velocities of the magnet and conductor to be reckoned relative to the ether, rather than relative to each other. Perhaps the issue he raises (the emf on the moving conductor not having energy in itself) resolves to the larger issue of velocities being reckoned relative to the ether, instead of relative to the bodies under test. I note that in the body of the paper he transforms the coordinates (including time) of the moving conductor to the frame in which the conductor is stationary, taking into account only the relative velocity of the conductor and magnet. It would not be possible to do this (and get the correct answer) if the force is transmitted by the ether, and both the magnet and the conductor are moving relative to the ether.

Dale
Mentor
Einstein says that the force from the B field is the same as the force from the E field, given the same relative velocity and distance. If the forces and paths are the same, how can the integrals of force over path be different? (The appropriate answer to this may be, "Go read this book.")
Because the force from the B field also depends on the velocity, which can be different, even over the same path.

EDIT: note, I was talking about an arbitrary closed loop I.e. For defining a conserved energy. Not a straight line at constant speed.

Without contradicting you, I will note that the first term of Maxwell's equation for emf (the term which deals with a moving charge) consists entirely of the addition and subtraction of terms of this form:
(magnetic intensity along a coordinate axis) * (velocity along a coordinate axis)

On inspection, it appears as though the force is proportional to the strength of the magnetic field. (Again, the appropriate answer to this may be, "Go read this book.")
Yes. And you are correct, that doesn't contradict me.

At any rate, given that the two emfs are the same, as Einstein says they are, it seems to me that Einstein's complaint is along the lines of a legal technicality, and of no practical consequence.
That is essentially his point. The equations are non relativistic, but the phenomena are not.

Look, the mathematical fact is that Maxwells equations are non (Galilean) relativistic. The historical fact is that scientists of Einstein's day were aware of that. What else is relevant here?

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Yes. And you are correct, that doesn't contradict me.
But I'm not sure why I'm not contradicting you. You said, "Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional." I countered, "On inspection, it appears as though the force is proportional to the strength of the magnetic field." I expected you to tell me I am wrong. Instead you agreed with me.

The B field is the map of the strength of the magnetic field in space. (That is, the map at any instant. Variations in the magnetic field over time are accounted for in the second term of the equation.) That being so, the subject of your statement (the B field) is the same as the subject of my counter (the strength of the magnetic field). I don't see how your statement and my counter can both be true.

That is essentially his point. The equations are non relativistic, but the phenomena are not. Look, the mathematical fact is that Maxwells equations are non (Galilean) relativistic. The historical fact is that scientists of Einstein's day were aware of that. What else is relevant here?
I wanted to understand the particular example Einstein gives in the introduction to his 1905 paper. The biggest part of my difficulty was not understanding that "energy in itself" is a technical term, as you have explained. It sounded to me as though Einstein was asserting that Maxwell's theory does not assign a cause to the rise of the emf in the moving conductor.

I also had some difficulty in understanding why the Maxwell equation for emf is not relativistic, in the sense of depending only on the relative velocity of the magnet and conductor. The equation is relativistic in that sense, so long as the "stationary" body is at rest relative to the ether. But of course, that stipulation opens the whole can of relativistic worms, doesn't it?

Dale
Mentor
But I'm not sure why I'm not contradicting you. You said, "Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional." I countered, "On inspection, it appears as though the force is proportional to the strength of the magnetic field." I expected you to tell me I am wrong. Instead you agreed with me.
A vector has a magnitude (strength) and a direction. So one vector (force) can be proportional to the strength (your claim) of another vector (B field) while not being proportional (my claim) to the other vector (B field) if they point in different directions. A closer inspection would show that they don't point in the same direction.

However, this is a minor point. The more important point is the one about work over closed paths.

Dale
Mentor
This is easy enough to check. The Lorentz force is given by $f=q(E+v \times B)$. When we do a Galilean boost by a velocity u then q is unchanged as are E, and B, but v transforms to v+u. So in the boosted frame $f=q(E+(v+u) \times B)$ which is not the same as in the original frame. So the electromotive force is definitely not invariant wrt a Galilean transform.
So, if we return to this post and use the correct pair of Galilean transforms we get the following (in units where c=1):

Magnetic limit:
$E' = E + u \times B$
$B' = B$
$f'=q(E' + v \times B') = q(E + u \times B + (v + u) \times B) \neq f$

Electric limit:
$E' = E$
$B' = B - u \times E$
$f'=q(E' + v \times B') = q(E + (v + u) \times (B - u \times E)) \neq f$

So the Lorentz force is not Galilean invariant under either limit. For Einstein's scenario, clearly the magnetic limit is the appropriate one. For the special case of u=-v which he described f'=f even though it is not, in general, invariant.

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