Question

Obtain the field (41.5), (41.6) of a uniformly moving charge \(q\left[=\left(4 \pi \varepsilon_{0}\right)^{-1} Q\right]\) by the following alternative method: Assume that the field in the rest frame \(S^{\prime}\) of the charge is given by $$ \mathbf{e}^{\prime}=\left(Q / r^{\prime 3}\right)\left(x^{\prime}, y^{\prime}, z^{\prime}\right), \quad \mathbf{b}^{\prime}=0, \quad r^{\prime 2}=x^{\prime 2}+y^{\prime 2}+z^{\prime 2} $$ then transform this field to the usual second frame \(\mathrm{S}\) at \(t=0\). [Hint: obtain \(\mathbf{b}=\mathbf{u} \times \mathbf{e} / c^{2}\) from \((39.2)\); from the inverse of \((39.1)\) obtain \(\mathbf{e}=\left(Q \gamma / r^{3}\right)(x, y, z) ;\) finally use (41.7). ]

Short answer

The expressions for the electric field \(\mathbf{e}\) and magnetic field \(\mathbf{b}\) are given by: \(\mathbf{e} = \left(\frac{Q}{\gamma^2 r^3}\right)(\gamma x, y, z)\) and \(\mathbf{b} = \frac{\mathbf{u} \times \mathbf{e}}{c^{2}}\), where \(\gamma\) is the Lorentz factor, \(\mathbf{u}\) is the constant velocity, and \(r\) is the distance from the charge.

Step-by-step solution

Obtain magnetic field \(\mathbf{b}\) in the rest frame \(S^{\prime}\)

First, we need to find the magnetic field \(\mathbf{b}\) using equation (39.2). Since the charge is moving uniformly, its velocity \(\mathbf{u}\) remains constant. Equation (39.2) relates the magnetic field \(\mathbf{b}\) with the electric field \(\mathbf{e}\) and velocity \(\mathbf{u}\) as: $$ \mathbf{b}=\frac{\mathbf{u} \times \mathbf{e}}{c^{2}} $$ Given that the magnetic field in the rest frame \(S^{\prime}\) is zero (\(\mathbf{b}^{\prime}=0\)), we can write the equation as: $$ \mathbf{b}=\frac{\mathbf{u} \times \mathbf{e}}{c^{2}} $$

Obtain electric field \(\mathbf{e}\) in the rest frame \(S^{\prime}\)

Next, we need to obtain the electric field \(\mathbf{e}\) in the rest frame \(S^{\prime}\). From the inverse of equation (39.1), we can find the relationship between the electric field in both frames as: $$ \mathbf{e} = \left(\frac{Q\gamma}{r^3}\right)(x, y, z) $$ Where \(\gamma\) is the Lorentz factor, which is defined as: $$ \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} $$

Transform the fields to the second frame \(S\) at \(t=0\)

Finally, we need to transform the obtained electric field \(\mathbf{e}\) and magnetic field \(\mathbf{b}\) to the second frame \(S\) at \(t=0\). Using equation (41.7), we can write the transformation as: $$ t=0 \implies x^{\prime}=\gamma x \text{ and } \mathbf{r}^{\prime}=\gamma \mathbf{r} $$ Substituting \(x^{\prime}=\gamma x\) and \(\mathbf{r}^{\prime}=\gamma \mathbf{r}\) into the expressions for \(\mathbf{e}\) and \(\mathbf{b}\) (from steps 1 and 2), we get the fields in the second frame \(S\) as: $$ \mathbf{e} = \left(\frac{Q}{\gamma^2 r^3}\right)(\gamma x, y, z) \text{ and } \mathbf{b} = \frac{\mathbf{u} \times \mathbf{e}}{c^{2}} $$ Now we've obtained the electric field \(\mathbf{e}\) and magnetic field \(\mathbf{b}\) of the uniformly moving charge \(q\) in the second frame \(S\) at \(t=0\).

Key concepts

Electric Field Transformation

When dealing with a charge moving uniformly, the electric field appears differently when observed from different reference frames. This is known as electric field transformation.

In the charge's rest frame, denoted as \(S'\), the electric field \(\mathbf{e}'\) is given by:

• \(\mathbf{e}' = \left(Q / r'^{3}\right)(x', y', z')\)

Here, \(r'\) represents the distance from the charge. In the observer’s frame, or the second frame \(S\), the electric field \(\mathbf{e}\) is transformed using the Lorentz transformation properties. When we apply this transformation, the equation becomes:

\[ \mathbf{e} = \left(\frac{Q \gamma}{r^{3}}\right)(x, y, z) \]

Where \(\gamma\) is the Lorentz factor. This transformation shows how the electric field reshapes by stretching or squeezing due to relativistic effects when a charge moves relative to the observer.

Magnetic Field Transformation

The magnetic field experienced by a moving charge also changes when viewed from different reference frames. In the rest frame \(S'\) of a charge at rest, the magnetic field \(\mathbf{b}'\) is typically zero, meaning there is no magnetic effect. However, things change in the observer's frame \(S\).

For a uniformly moving charge, the magnetic field \(\mathbf{b}\) in frame \(S\) can be derived using the relation:

• \(\mathbf{b} = \frac{\mathbf{u} \times \mathbf{e}}{c^{2}}\)

Where \(\mathbf{u}\) represents the velocity of the moving charge, \(\mathbf{e}\) is the electric field, and \(c\) is the speed of light. This relation signifies how the magnetic field arises due to the motion of the charge across an existing electric field, a direct consequence of electromagnetic field transformation in relativity.

Lorentz Transformation

A cornerstone of relativity, Lorentz transformation is crucial for understanding how physical quantities interchange between different frames of reference. It describes how measurements of space and time by two observers are related to each other.

For an electric and magnetic field, this transformation is vital because:

• The factor \(\gamma\) is introduced, defined as \(\gamma = \frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}}\), accounting for time dilation and length contraction in moving frames.
• Spatial coordinates and time are blended together, leading to changes in observed fields.

Applying the Lorentz transformations allows us to convert the rest frame field values to those in an arbitrary observer's frame, revealing how electromagnetic phenomena remain consistent across different viewpoints.

Uniform Motion of Charge

Understanding the behavior of a uniformly moving charge is essential in electromagnetism, especially within the context of relativity. A charge moving at a constant velocity indicates unchanging position and velocity at any given time.

This uniform motion simplifies the analysis of field transformations because:

• The velocity \(\mathbf{u}\) remains constant, eliminating any acceleration components in calculations.
• Accelerated charges produce radiation, whereas uniform motion does not, leading to consistent field transformations.

Thus, studying the uniformly moving charge lays a foundational understanding of how electromagnetic fields behave under special relativity and serves as the basis for more complex scenarios where motion is not constant.