31.1.2 The Transformation of Electric and Magnetic Fields

Imagine that Alec has measured the electric field E⃗A and the magnetic field B⃗A in reference frame A. Our investigations thus far give us no reason to think that Brittney’s measurements of the fields will differ from Alec’s. After all, it seems like the fields are just “there,” waiting to be measured.

To find out if this is true, Alec establishes a region of space with a uniform magnetic field B⃗A but no electric field (E⃗A=0⃗). Then, as shown in FIGURE 31.3, he shoots a positive charge q through the magnetic field. At an instant when q is moving horizontally with velocity v⃗ CA, Alec observes that the particle experiences force F⃗A=qv⃗ CA×B⃗A. The direction of the force is straight up.

FIGURE 31.3 A charged particle moves through a magnetic field in reference frame A and experiences a magnetic force.

The figure illustrates a positive charge particle moving in a uniform magnetic field in a reference frame.
FIGURE 31.3 Full Alternative Text

Suppose that Brittney, in frame B, runs alongside the charge with the same velocity: v ⃗BA=v ⃗CA. To her, in frame B, the charge is at rest. Nonetheless, because both experimenters must agree about forces, Brittney must observe the same upward force on the charge that Alec observed. But there is no magnetic force on a stationary charge, so how can this be?

Because Brittney sees a stationary charge being acted on by an upward force, her only possible conclusion is that there is an upward-pointing electric field. After all, the electric field was initially defined in terms of the force experienced by a stationary charge. If the electric field in frame B is E⃗B, then the force on the charge is F⃗B=qE⃗B. But we know that F⃗B=F⃗A, and Alec has already measured F⃗A=qv ⃗CA×B⃗A=qv ⃗BA×B⃗A. Thus we’re led to the conclusion that

E⃗B=v ⃗BA×B⃗A(31.3)

As Brittney runs past Alec, she finds that at least part of Alec’s magnetic field has become an electric field! Whether a field is seen as “electric” or “magnetic” depends on the motion of the reference frame relative to the sources of the field.

FIGURE 31.4 shows the situation from Brittney’s perspective. There is a force on charge q, the same force that Alec measured in Figure 31.3, but Brittney attributes this force to an electric field rather than a magnetic field. (Brittney needs a moving charge to measure magnetic forces, so we’ll need a different experiment to see whether or not there’s a magnetic field in frame B.)

FIGURE 31.4 In frame B, the charge experiences an electric force.

The figure illustrates a charged particle at rest placed in a uniform electric field in a reference frame.
FIGURE 31.4 Full Alternative Text

More generally, suppose that an experimenter in reference frame A creates both an electric field E⃗A and a magnetic field B⃗A. A charge moving in A with velocity v ⃗CA experiences the force F⃗A=q(E⃗A+v ⃗CA×B⃗A) shown in FIGURE 31.5a. The charge is at rest in a reference frame B that moves with velocity v ⃗BA=v ⃗CA so the force in B can be due only to an electric field: F⃗B=qE⃗B. Equating the forces, because experimenters in all inertial reference frames agree about forces, we find that

E⃗B= E⃗A+v ⃗BA×B⃗A(31.4)

FIGURE 31.5 A charge in reference frame A experiences electric and magnetic forces. The charge experiences the same force in frame B, but it is due only to an electric field.

The figure consists of two parts labeled "a" and b that illustrates the electric and magnetic field of a charged particle with respect to two frames "A" and B.
FIGURE 31.5 Full Alternative Text

Equation 31.4 transforms the electric and magnetic fields measured in reference frame A into the electric field measured in a frame B that moves relative to A with velocity v ⃗BA. FIGURE 31.5b shows the outcome. Although we used a charge as a probe to find Equation 31.4, the equation is strictly about fields in different reference frames; it makes no mention of charges.

EXAMPLE 31.1 Transforming the electric field

A laboratory experimenter has created the parallel electric and magnetic fields E⃗=10,000 î V/m and B⃗=0.10 î T. A proton is shot into these fields with velocity v⃗=1.0×105 ĵ m/s. What is the electric field in the proton’s reference frame?

MODEL Let the laboratory be reference frame A and a frame moving with the proton be reference frame B. The relative velocity is v⃗ BA=1.0×105 ĵ m/s.

VISUALIZE FIGURE 31.6 shows the geometry. The laboratory fields, now labeled A, are parallel to the x-axis while v ⃗BA is in the y-direction. Thus v⃗ BA×B⃗A points in the negative z-direction.

FIGURE 31.6 Finding electric field E⃗B.

The figure illustrates a frame in which a proton moves in the direction of parallel electric and magnetic field.

SOLVE v⃗BA and B⃗A are perpendicular, so the magnitude of v⃗BA×B⃗A is (1.0×105 m/s) × (0.10 T)(sin 90°)=10,000 V/m. Thus the electric field in frame B, the proton’s frame, is

E⃗B=E⃗A+v⃗ BA×B⃗A=(10,000 î−10,000 k^) V/m=(14,000 V/m, 45° below the x−axis)

REVIEW The force on the proton is the same in both reference frames. But in the proton’s reference frame that force is due entirely to an electric field tilted 45° below the x-axis.

To find a transformation equation for the magnetic field, FIGURE 31.7a shows charge q at rest in reference frame A. Alec measures the fields of a stationary point charge:

E⃗A=14πϵ0 qr2 r^   B⃗A=0⃗

What are the fields at this point in space as measured by Brittney in frame B? We can use Equation 31.4 to find E⃗B. Because B⃗A=0⃗, the electric field in frame B is

E⃗B=E⃗A=14πϵ0 qr2 r^(31.5)

In other words, Coulomb’s law is still valid in a frame in which the point charge is moving.

FIGURE 31.7 A charge at rest in frame A is moving in frame B.

The figure consists of two parts labeled "a" and b that illustrates the nature of a charged particle with respect to two frames in an x y plane.
FIGURE 31.7 Full Alternative Text

But Brittney also measures a magnetic field B⃗B, because, as seen in FIGURE 31.7b, charge q is moving in reference frame B. The magnetic field of a moving point charge is given by the Biot-Savart law:

B⃗B=μ 04π qr2 v⃗ CB×r^=−μ 04π qr2 v⃗ BA×r^(31.6)

where we used the fact that the charge’s velocity in frame B is v ⃗CB=−v ⃗BA.

It will be useful to rewrite Equation 31.6 as

B⃗B=−μ 04π qr2 v⃗ BA×r^=−ϵ0 μ 0v⃗ BA×(14πϵ0 qr2 r^)

The expression in parentheses is simply E⃗A, the electric field in frame A, so we have

B⃗B=−ϵ0 μ 0v⃗ BA×E⃗A(31.7)

Thus we find the remarkable idea that the Biot-Savart law for the magnetic field of a moving point charge is nothing other than the Coulomb electric field of a stationary point charge transformed into a moving reference frame.

We will assert without proof that if the experimenters in frame A create a magnetic field B⃗A in addition to the electric field E⃗A, then the magnetic field E⃗B is

B⃗B=B⃗A−ϵ0 μ 0v⃗ BA×E⃗A(31.8)

This is a general transformation matching Equation 31.4 for the electric field E⃗B.

Notice something interesting. The constant μ 0 has units of T m/A; those of ϵ0 are C2/N m2. By definition, 1 T=1 N/A m and 1 A=1 C/s. Consequently, the units of ϵ0 μ 0 turn out to be s2/m2. In other words, the quantity 1/ϵ0 μ 0, with units of m/s, is a speed. But what speed? The constants are well known from measurements of static electric and magnetic fields, so we can compute

1ϵ0 μ 0=1(8.85×10−12 C2/N m2)(1.26×10−6 T m/A)=3.00×108 m/s

Of all the possible values you might get from evaluating 1/ϵ0 μ 0, what are the chances it would turn out to be c, the speed of light? It is not a random coincidence. In Section 31.5 we’ll show that electric and magnetic fields can exist as a traveling wave, and that the wave speed is predicted by the theory to be none other than

vem=c=1ϵ0 μ 0(31.9)

For now, we’ll go ahead and write ϵ0 μ 0=1/c2. With this, our Galilean field transformation equations are

E⃗B=E⃗A+v⃗BA×B⃗AB⃗B=B⃗A−1c2v⃗BA×E⃗A(31.10)

where v⃗ BA is the velocity of reference frame B relative to frame A and where, to reiterate, the fields are measured at the same point in space by experimenters at rest in each reference frame.

NOTE   We’ll see shortly that these equations are valid only if vBA≪c.

We can no longer believe that electric and magnetic fields have a separate, independent existence. Changing from one reference frame to another mixes and rearranges the fields. Different experimenters watching an event will agree on the outcome, such as the deflection of a charged particle, but they will ascribe it to different combinations of fields. Our conclusion is that there is a single electromagnetic field that presents different faces, in terms of E⃗ and B⃗, to different viewers.

EXAMPLE 31.2 Two views of a magnet

The 1.0 T field of a large laboratory magnet points straight up. A rocket flies past the laboratory, parallel to the ground, at 1000 m/s. What are the fields between the magnet’s pole tips as measured—very quickly!—by scientists on the rocket?

MODEL Let the laboratory be reference frame A and a frame moving with the rocket be reference frame B.

VISUALIZE FIGURE 31.8 shows the magnet and establishes the coordinate systems. The relative velocity is v⃗ BA=1000 î m/s.

FIGURE 31.8 The rocket and the magnet.

The figure illustrates a magnet and a moving rocket placed in two frames labeled "A" and B respectively placed in x y coordinate plane.

SOLVE The fields in the laboratory reference frame are E⃗A=0⃗ and B⃗A=1.0 ĵ T. Transforming the fields to the rocket’s reference frame gives first, for the electric field,

E⃗B=E⃗A+v⃗ BA×B⃗A=v⃗ BA×B⃗A

From the right-hand rule, v⃗ BA×B⃗A is out of the figure, in the z-direction. v⃗ BA and B⃗A are perpendicular, so

E⃗B=vBA BA k^=1000 k^ V/m

Similarly, for the magnetic field,

B⃗B=B⃗A−1c2 v⃗ BA×E⃗A=E⃗A=1.0 ĵ T

Thus the rocket scientists measure

E⃗B=1000 k^ V/m and B⃗B=1.0 ĵ T