mostly worthy to read is here, but I failed to attach file.

Assume a part of stable DC circuit with the shape of “L” rests in inertial frame S: section 1 horizontal; section 2 vertical; wire cutting face square a × a for convenience.

Another inertial frame S’ moves rightward at velocity u.

First, analysis resorts to traditional Lorentz transformation.

In frame S:

For section 1, I1 = n1qv1; for section 2, I2 = n2qv2

(Here take the moving direction of electron as the direction of current for convenience, n is the density of free electron in the wire, q is the charge carried by electron, v is the moving velocity of free electron.)

I1 = I2, n1 = n2 = n, v1 = v2 = v

In frame S’:

For section 1, △x’ = △x / γ , (γ = 1 / √1 - u²/ c² ) , △y’ = △y , △z’ = △z

V’ = △x’△y’△z’ = △x△y△z / γ = V / γ , V’ is volume in frame S’, V is volume in frame S

So, n1’ = γn1 = γn

Lorentz transformation of velocity:

v1’ = (v1 - u) / (1 - u v1/c² )

I1’ = n1’qv1’ = γnq (v1 - u) / (1 - u v1/c²)

For section 2, also V’ = V / γ

So, n2’ = γn2 = γn

Lorentz transformation of velocity:

v2’ = v2 / γ(1 - u v1/c²)

I2’ = n2’qv2’ = γnq v2 / γ(1 - u v1/c²) = nq v2 / (1 - u v1/c²)

I1’ ≠ I2’

Physics rule is not equally applicable in inertial frame S’.

Next, resort to the space contraction in all directions which is disclosed by the Mass – Space equation.

In frame S’:

For section 1, n1’ = γ³n1 = γ³n

v1’ = dx’ / dt’ = (v1 - u) / (1 - u v1/c² )

I1’ = n1’qv1’ = γ³nq (v1 - u) / (1 - u v1/c²)

For section 2, n2’ = γ³n2 = γ³n

v2’ = dz’ / dt’ = dx’ / dt’ = (v1 - u) / (1 - u v1/c² )

I2’ = n2’qv2’ = γ³nq (v1 - u) / (1 - u v1/c²)

I1’ = I2’

Physics rule is equally applicable in inertial frame S’.

Li Qiang Chen

July 17, 2019