- #1
user1139
- 71
- 8
- Homework Statement
- Assuming that the first pair of equations given below are correct, how do I show that the electric and magnetic fields obtained transform correctly under general Lorentz boost?
- Relevant Equations
- The relevant equations are provided below.
I was trying to show that the field transformation equations do hold when considering electric and magnetic fields as 4-vectors. To start off, I obtained the temporal and spatial components of ##E^{\alpha}## and ##B^{\alpha}##. The expressions are obtained from the following equations:
$$E^{\alpha}=F^{\alpha\beta}U_{\beta},\: B^{\alpha}=\frac{1}{2c}\epsilon^{\alpha\beta\mu\nu}F_{\beta\mu}U_{\nu}$$
I obtained:
\begin{align*}
E^{0}&=F^{00}U_{0}+F^{0i}U_{i}=\frac{\gamma(u)}{c}\left(\vec{E}\cdot\vec{u}\right)\\
E^{i}&=F^{i0}U_{0}+F^{ij}U_{j}=\gamma(u)\left[\vec{E}+\left(\vec{u}\times\vec{B}\right)\right]^{i}\\
B^{0}&=\frac{1}{2c}\epsilon^{0\beta\mu\nu}F_{\beta\mu}U_{\nu}=-\frac{\gamma(u)}{c}\left(\vec{B}\cdot\vec{u}\right)\\
B^{i}&=\frac{1}{2c}\epsilon^{i\beta\mu 0}F_{\beta\mu}U_{0}+\frac{1}{2c}\epsilon^{i\beta\mu j}F^{\beta\mu}U_{j}=\gamma(u)\left[\vec{B}-\frac{\vec{u}}{c^2}\times\vec{E}\right]^{i}
\end{align*}
I interpreted the above components as that of fields observed by a stationary observer. To show that the fields transform correctly I have to show that:
$$\vec E' = \gamma \left( \vec E + c\vec \beta \times \vec B\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec E \right )$$
$$\vec B' = \gamma \left( \vec B - \frac{\vec \beta}{c} \times \vec E\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec B \right )$$
i.e. I have to show that I am able to construct the RHS from the components I have found. However, I do not seem to be able to show that using the Lorentz transformation equations under general boost. The Lorentz transformation equations under general boost is given as:
$$A^{'0}=\gamma\left(A^0-\vec{\beta}\cdot\vec{A}\right)$$
$$\vec{A}'_{\parallel}=\gamma\left(\vec{A}_{\parallel}-\vec{\beta}A^0\right)$$
$$\vec{A}'_{\perp}=\vec{A}_{\perp}$$
How should I proceed?
$$E^{\alpha}=F^{\alpha\beta}U_{\beta},\: B^{\alpha}=\frac{1}{2c}\epsilon^{\alpha\beta\mu\nu}F_{\beta\mu}U_{\nu}$$
I obtained:
\begin{align*}
E^{0}&=F^{00}U_{0}+F^{0i}U_{i}=\frac{\gamma(u)}{c}\left(\vec{E}\cdot\vec{u}\right)\\
E^{i}&=F^{i0}U_{0}+F^{ij}U_{j}=\gamma(u)\left[\vec{E}+\left(\vec{u}\times\vec{B}\right)\right]^{i}\\
B^{0}&=\frac{1}{2c}\epsilon^{0\beta\mu\nu}F_{\beta\mu}U_{\nu}=-\frac{\gamma(u)}{c}\left(\vec{B}\cdot\vec{u}\right)\\
B^{i}&=\frac{1}{2c}\epsilon^{i\beta\mu 0}F_{\beta\mu}U_{0}+\frac{1}{2c}\epsilon^{i\beta\mu j}F^{\beta\mu}U_{j}=\gamma(u)\left[\vec{B}-\frac{\vec{u}}{c^2}\times\vec{E}\right]^{i}
\end{align*}
I interpreted the above components as that of fields observed by a stationary observer. To show that the fields transform correctly I have to show that:
$$\vec E' = \gamma \left( \vec E + c\vec \beta \times \vec B\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec E \right )$$
$$\vec B' = \gamma \left( \vec B - \frac{\vec \beta}{c} \times \vec E\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec B \right )$$
i.e. I have to show that I am able to construct the RHS from the components I have found. However, I do not seem to be able to show that using the Lorentz transformation equations under general boost. The Lorentz transformation equations under general boost is given as:
$$A^{'0}=\gamma\left(A^0-\vec{\beta}\cdot\vec{A}\right)$$
$$\vec{A}'_{\parallel}=\gamma\left(\vec{A}_{\parallel}-\vec{\beta}A^0\right)$$
$$\vec{A}'_{\perp}=\vec{A}_{\perp}$$
How should I proceed?