E & B under a Boost — Lecture 19

Source

Lecture: 19 — The Lorentz force & E/B mixing
Simulation: Electric & magnetic fields under a boost (lecture-demo mode)
Concept: E and B are frame-dependent facets of one field; a pure external magnetic field in the lab becomes electric and magnetic for a moving observer.
Serves: Field-transformation intuition; electric and magnetic contributions to the Lorentz force; transverse-force consistency across frames.
Inquiry: In this instantaneous snapshot, how do Bella's fields and the force on a test charge change as her velocity relative to Alice changes?
Misconception: "If Alice sees no electric field, no other inertial observer can see one." — In normalized field units, Bella measures E′_y/(cB₀) = γ_v(v/c), which reverses when v reverses.

Pedagogy principles applied

ID	Principle	How it shows up here
A1	One concept, one sentence	E and B are one field; changing observer frame mixes Alice's pure B field into Bella's E′ + B′.
H2	Structural misconceptions	Predict card tests whether a moving observer can measure an electric field when Alice measures none.
D6	Scene evolves with steps	Alice's lab frame first; later steps direct the learner to Bella's variable-velocity frame.
C1	Numbers earn their place	Signed field and force-component readouts expose cancellations that the net-force arrow alone cannot show.
E1	Animation honesty	Static single-frame diagram — field and vector lengths are proportional to the calculated quantities; Bella's speed is capped at \|v/c\| ≤ 0.9.
A3	Formal tools off by default	Field-transform equations live in the ⚛︎ The Physics revision sheet, opened on demand.

E & B under a boost

Teach me

Field view Alice · lab (static)

E field

B field

test charge q / velocity

net force on q

Guided Inquiry

1 · Alice's lab

Let’s start in Alice’s lab. A test charge q moves at u = 0.50c along +x through a uniform external field. In Alice's lab frame the electric field is zero and the magnetic field is B₀ = 1 into the page, so the force on the charge is purely magnetic, F = q u × B, pointing up for q > 0.

Now meet Bella: she moves along the same x-axis at a signed velocity v relative to Alice.

2 · Predict — Bella's electric field

Here’s the first puzzle. Alice measures no electric field anywhere in this region. What electric field do you think Bella measures as she moves along +x?

3 · Predict — and the magnetic field?

Now for the magnetic side. Bella also measures a magnetic field. Compared with Alice’s B₀ into the page, what do you think it is?

4 · Predict — a frame with no magnetic field?

Let’s push it further. Switch to Bella and sweep v/c across its full range. Do you think there’s a frame in which the magnetic field vanishes, leaving a purely electric field?

5 · Predict — the force at v/c = +0.50

One last experiment. Set Bella’s velocity to v/c = +0.50, matching the charge’s own u/c. Alice measures a force of 0.500 on the charge. What do you think Bella measures?

6 · Two views of one field

So here’s the big picture. Alice’s pure magnetic field and Bella’s mixed E′ and B′ are one field read from two frames: electric and magnetic are observer-dependent labels, not two separate things.

And what every observer agrees on is E² − c²B², here −c²B₀² — and because it is negative, no frame can remove the magnetic field.

Listen

Frame & charge

▾

Display

Observer frame

Bella velocity v/c+0.50

−0.9 (left)+0.9 (right)

Charge q+1.0

−2+2

u = 0.50c

B₀ = 1 (into page)

Readouts

▾

Alice's frame (lab)

E_y0

B_z−1.000 (into page)

|F|0.500

E² − c²B²−1.000

4-force mag.0.577

Bella's frame (moving)

E′_y+0.577

B′_z−1.155 (into page)

|F′|0.577

E′² − c²B′²−1.000

4-force mag.0.577

Field transform, boost +x

Lab frame (Alice)

Moving frame (Bella)

Force transform and invariants

The Physics

E and B under a boost · what this sim shows, the physics behind it, and every control

At a glance

A test charge crosses a uniform magnetic field. This sim reads that single instant from two inertial frames, Alice's lab and Bella's moving frame, and prints the fields and the force each observer measures. It shows how a field and a force are described by two observers in relative motion.

What you'll learn

How the fields Bella measures depend on her velocity v.
How the charge's speed and the force on it are measured in each frame.
Which quantities differ between the two frames, and which are the same for both.
How to read the two columns of the Readouts panel side by side.

The charge and Bella both move along x. The field is the same uniform B₀ everywhere.

What's on screen

Test charge q (amber disc), with its velocity as an amber arrow labelled u or u′.
B field (blue ⊗ lattice, into the page). Layer: Show B.
E field (purple arrows). Layer: Show E.
Net force on q (pink arrow, labelled |F| or |F′|). Layer: Show force.
Readouts panel: one block per frame, with E, B, the force, and two green rows (see section 07).

See itShow Alice's frame with the B and force layers on. Watch for: a blue ⊗ lattice, no purple arrows, and a pink arrow pointing up, labelled |F| = 0.500.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
Show E	Draws the electric-field arrows of the frame you are viewing.	on / off opens off
Show B	Draws the magnetic-field lattice of the frame you are viewing.	on / off opens off
Show force	Draws the net force on the charge, with its size.	on / off opens off
Observer frame	Chooses whose reading the canvas shows: Alice (lab) or Bella (moving frame). The Readouts panel always lists both.	Alice / Bella opens on Alice
Bella velocity v/c	Bella's velocity along x relative to Alice. It changes Bella's readouts and her view; Alice's stay put.	−0.9 to +0.9 opens at +0.50
Particle rest frame	Sets v equal to the charge's speed u and switches to Bella's view.	button sets v/c = +0.50
Charge q	The size and sign of the test charge. It scales the forces and never touches the fields.	−2 to +2 opens at +1.0
Readouts	Alice's and Bella's E, B, force magnitude, and two green invariant rows.	panel opens collapsed
↻ Reset	Back to Alice's view, v/c = +0.50, q = +1.0, all layers off.	top bar

Fixed: u = 0.50c in Alice's frame and B₀ = 1 into the page, in units with c = 1. The sim shows one instant, so nothing moves: the frame choice and v/c select an observer, not a motion.

In Alice's frame the force is magnetic

A charge q with velocity u feels the Lorentz force F = q(E + u × B): an electric part qE and a magnetic part q u × B.

In Alice's lab the electric field is zero, so only the magnetic part is left:

Alice's frame in the sim's symbols (c = 1).

For q = +1 the force is F_y = quB₀ = +0.500, pointing up.
It is perpendicular to u, so it bends the path and does no work on the charge.
Changing the sign of q reverses the force and leaves its size alone.

See itSet the charge to q = −1 in Alice's frame. Watch for: the pink arrow flipping to point down, with the same size, |F| = 0.500.

A boost mixes E and B

Bella moves along +x at velocity v relative to Alice. Her fields follow from Alice's, with γ_v = 1/√(1 − v²/c²):

Components along the boost are unchanged. Only the perpendicular parts mix.

The mixing terms v × B and v × E are perpendicular to v, so nothing new appears along the boost.
Alice has E = 0 and B = −B₀ẑ. Bella therefore measures E′_y = γ_v(v/c)B₀ and B′_z = −γ_vB₀ (E in units of cB₀).
At v/c = +0.50, γ_v = 1.155, so E′_y = +0.577 and B′_z = −1.155.
Reversing v reverses E′_y but not B′_z, since γ_v is the same for ±v.

Boost along x, B into the page: E′ is transverse and flips with v.

Key idea. E and B are parts of one field. A boost mixes them, so how much of it is "electric" depends on the observer.

See itSwitch to Bella at v/c = +0.50 with the E and B layers on. Watch for: purple E′ arrows pointing up between the ⊗, and E′_y = +0.577 in the readout.

See itNow reverse the boost to v/c = −0.50. Watch for: the purple arrows flipping to point down, while B′_z stays −1.155.

Bella's fields and speed change with v

In the sim's units (c = 1, B₀ = 1), the quantities Bella reads are:

E in units of cB₀, B in units of B₀, c = 1.

Both fields grow with |v| and have no limit as |v| → c (the sim stops at 0.9). Their ratio stays fixed: |E′|/(c|B′|) = |v|/c, always below 1.
The charge's speed changes too. It follows the velocity-addition rule above: u′ = 0 at v = u, u′ = +0.80c at v/c = −0.50, and |u′| never reaches c.
At v = 0 Bella is Alice: E′ = 0, B′ = −B₀, u′ = u.

See itPush Bella to v/c = +0.90. Watch for: E′_y and B′_z both larger than at +0.50, with E′_y still smaller than |B′_z|.

The force depends on the frame

The force is dp/dt, and a boost changes both dp and dt. For a force perpendicular to u, as here:

Left: the force transform. Right: two field invariants (section 07).

At v/c = +0.50: 0.500 / (1.155 × 0.75) = 0.577. At v/c = −0.50: 0.500 / (1.155 × 1.25) = 0.346. The force points along the same ±y in both frames; only its size changes.
Bella adds her own electric and magnetic parts: F′ = qE′_y − qu′B′_z. At v = u, u′ = 0, so the magnetic part vanishes and the force is purely electric: F′ = qE′_y = 0.577.

v/c q

These sliders move the sim too.

See itRide with the charge, at v = u. Watch for: the label u′ = 0 (at rest) and a pink arrow still there, |F′| = 0.577 against Alice's 0.500.

See itNow try Bella at v/c = −0.50. Watch for: |F′| = 0.346 for Bella against |F| = 0.500 for Alice.

What every observer agrees on

Some quantities have the same value in both frames. The sim prints two of them in green.

E · B = 0 in both frames: E′ points along y and B′ along z.
E² − c²B² = −c²B₀², printed as −1.000 at every v and q. A negative value means c²B² exceeds E²: the field is magnetically dominated.
No frame is purely electric. The invariant stays negative, so B never vanishes. Equivalently, |E′|/(c|B′|) = |v|/c < 1.
The four-force magnitude. Here F ⊥ u in each frame, so the time component is zero and the magnitude is γ_u|F| = 0.577|q|, the same in both panels.

The four-force dp^μ/dτ, and its magnitude in each frame.

See itTry the extreme setting v/c = −0.90 with q = −2. Watch for: E² − c²B² still reading −1.000 in both frames, and the four-force magnitude reading 1.155 in both.

Where it matters

A current-carrying wire: the same pull on a moving charge, magnetic in the lab and electric in the charge's own frame.

Wires and motors. A current-carrying wire is neutral in the lab, so it pulls a moving charge by a magnetic force. In the charge's frame the electrons and the ions are contracted by different amounts, the wire carries a net charge, and the same pull is electric.
Accelerator magnets. In a particle's own rest frame a magnet's field acts as a purely electric force on it (u′ = 0), as in the sim at v = u.
Magnetically dominated plasmas. Pulsar magnetospheres and the solar corona are commonly modelled with E · B = 0 and c²B² > E², the same class as the field in this sim.

Go deeper: the size of the wire's net charge

In the charge's frame, the wire's net line density is λ′ = −γ_u u I / c², where I is the lab current.
Its electric force is then γ_u times the lab's magnetic force, matching the force transform at v = u (here 0.577 against 0.500).
The charge density is tiny, but copper has about 10²⁹ electrons per m³. For 1 A through 1 mm², the drift speed is only about 0.07 mm/s.

Summary

In one line. A boost mixes the transverse parts of E and B and changes the three-force; E · B, E² − c²B² and the four-force magnitude are the same for every observer.

Key points

From E = 0 and B = −B₀ẑ, Bella reads E′_y = γ_v(v/c)B₀ (odd in v) and B′_z = −γ_vB₀ (even in v).
The three-force changes by F′_y = F_y/[γ_v(1 − uv/c²)], and its direction stays the same.
Here E² − c²B² = −c²B₀² < 0 in every frame, so no frame is purely electric.

Common mix-ups

"Alice sees no electric field, so no one does." E is not invariant: Bella measures E′_y = γ_v(v/c)B₀.
"If some B became E, B must get weaker." Both transverse fields grow by γ_v: B′_z = −1.155 against Alice's −1.000 at v/c = +0.50.
"Every observer measures the same force." Alice reads 0.500 and Bella 0.577 at v/c = +0.50. The invariant is the four-force magnitude, 0.577 for both.

Things to try on your own

Set q = −2 and v/c = −0.90. Compute |F′| from F′_y = F_y/[γ_v(1 − uv/c²)] and compare it with the Readouts panel.
Use Particle rest frame and divide |F′| by |F|. Compare the ratio with γ_v at v/c = +0.50.
In Bella's frame with Show E and Show B on, sweep v/c from 0 up to +0.90 and note which arrows change.