[NOTES/EM-12001]-Lorentz transformations

1. Lorentz transformation and four vectors

\begin{FileText}{ Let \(K\) and \(K ^\prime\) be two Lorentz frames such that \(K^\prime\) moves with velocity \(v\) w.r.t. the frame \(K\) and at initial time \(t=t^\prime=0\) the coordinate axes of the two frame coincide. The space time coordinates of an event as observed in the two frames will be related by Lorentz transformation \begin{eqnarray} t^\prime = \frac{t-vx/c^2}{\sqrt{(1-(v/c)^2}}, \qquad x^\prime= \frac{x-vt}{\sqrt{1-(v/c)^2}},\quad y^\prime=y, \quad z^\prime=z. \label{E1} \end{eqnarray} We will use units in which \(c=1\). We use the four vector notation \(x^\mu=(x^0,x^1,x^2,x^3)=(t, \vec{x})\) to denote the coordinates of an event. The greek indices \(\mu,\nu, \lambda \) etc will take values 0,1,2,3 and latin indices \(i,j,k\) etc will take values 1,2,3. A {\it contravariant four vector} \(V^\mu\) is an object whose components transform like \(x^\mu\) under a Lorentz transformation. Examples of other four vectors are energy-momentum and velocity four vectors. The energy-momentum four vector \(p^\mu\) has components \begin{equation} p^\mu = (E, \vec{p}) = \Big( \frac{mc^2}{\sqrt{1-v^2}}, \frac{m\vec{v}}{\sqrt{1-v^2}}\Big). \end{equation} The velocity four vector is \(v^\mu= p^\mu/m\) and we have \begin{equation} v^\mu= ( \frac{1}{\sqrt{1-v^2}}, \frac{\vec{v}}{\sqrt{1-v^2}}\Big) . \end{equation} % We introduce the metric tensor \(g^{\mu\nu}\) by \begin{equation} g^{00}=1, g^{11}=g^{22}=g^{33}=g^{44}=-1 \end{equation} and all other, the off diagonal, components are zero, \(g^{mn}=0\) if \(m\ne n\). Its inverse \(g_{\mu\nu}\) is defined by \begin{equation} g^{\mu\nu} g_{\nu\rho}= \delta^{\mu}_\rho. \end{equation} and is also diagonal \begin{equation} g_{00}=1, g_{11}=g_{22}=g_{33}=-1\\ \end{equation} and we have \(g_{mn}=0\) if \(m\ne n\). The metric tensor, written in a matrix form, is \begin{equation}\label{EQ07} g^{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 &-1 & 0 & 0\\ 0 & 0 &-1 & 0\\ 0 & 0 & 0& -1 \end{pmatrix} \end{equation} The matrix for \(g_{\mu\nu}\) will be the inverse of the above matrix and hence the expression is identical with \eqRef{EQ07}. A general Lorentz transformation will be written as \begin{equation} x^{\mu\prime} = \sum_\nu {\Lambda^\mu} _\nu x^\nu . \end{equation} For a scalar field, the derivatives w.r.t to \(x^\mu\) will be written as \begin{equation} \Big(\pp[\phi]{x^0}, \pp[\phi]{x^1}, \pp[\phi]{x^2}, \pp[\phi]{x^3} \Big) = \partial_\mu \phi = \Big(\pp[\phi]{x^0}, \nabla \phi \Big) \end{equation} The partial derivatives \( \pp{x^{\mu}} = \Big( \pp{x^0}, \nabla \Big)\). are seen to transform differently and one has \begin{eqnarray} {\partial_\mu}^\prime = \pp{x^{\mu\prime}} = \sum_\nu \pp[x^\nu]{x^{\mu\prime}} \pp{x^\nu} = \sum_\nu{\Lambda^\nu} _\mu \partial_\nu \end{eqnarray} A four component object which transforms like derivatives will be called a {\it covariant four vector}. Form now onwards Einstein summation convention will be used and a summation over all values of a repeated index will be understood. % The metric tensor will be used to switch from contravariant to covariant vectors and from covariant to contravariant forms. So for example \(a_\mu\), defined by \begin{equation} a_\mu = g_{\mu\nu} a^\nu, \qquad a^\mu= g^{\mu\nu} a_\nu, \end{equation} will be a covariant vector. Also we will have \begin{equation} a^\mu = g^{\mu\nu} a_\nu. \end{equation} For example, we will have \begin{equation} \partial^\mu \phi(x)= g^{\mu\nu}\partial_\nu \phi(x) = \Big(\pp[\phi(x)]{t}, -\nabla \phi\Big). \end{equation} % Let \(a^\mu=(a^0, \vec{a})\) and \(b^\mu=(b^0, \vec{b})\) be two four vectors. Define \(a\cdot b\) by \begin{equation} a\cdot b = a^0b^0 - \vec{a} \cdot \vec{b}. \end{equation} Then \(a\cdot b\) is invariant under Lorentz transformations, {\it i.e.} it has the same value in all Lorentz frames. The scalar product can be written as \begin{eqnarray} a\cdot b &=& a^\mu b_\mu = a_\mu b^\mu \\ &=& g_{\mu\nu} a^\mu b^{\nu} = g^{\mu\nu}a_\mu b_\mu. \end{eqnarray} For velocity four vector and four momentum vector, we have \[v\cdot v=1 \text{ and } p\cdot p= m^2.\] Also we have \begin{eqnarray} \partial^\mu\partial_\mu &=& \partial^\mu \partial_\mu = -\square^2, \\ \text{where } \square &=& \DD{t} - \sum_{k=1}^3 \PP{x_k}\\ &=& \DD{t} - \nabla^2. \end{eqnarray} The operator \(\square\) is called de Alembertian operator. Note that some books use notation \(\square^2\) for de Alembertian operator.

2.EM theory quantities in four vector notation

The current density \( \vec{j}\) and charge density \(\rho\) form components of four vector current density \(j^\mu=(\rho, \vec{j})\). Similarly the scalar and vector potential form the time and space components of four vector potential \(A_\mu= (\phi, \vec{A})\). Then we have \begin{eqnarray} \left(\DD{t} - \nabla^2 \right) \vec{A} &=& \mu_0 \vec{j}\\ \left(\DD{t} - \nabla^2 \right) \phi &=& \frac{\rho}{\epsilon_0}. \end{eqnarray} Note that \(\mu_0=1/(\epsilon_0c^2)\) and we have taken \(c=1\). The above equations can be combined into a single four vector equation: \begin{equation} \square A_\mu = \frac{1}{\epsilon_0} j_\mu, \qquad \mu=0,1,2,3. \end{equation} The equation of continuity \begin{equation} \dd[\rho]{t} + \nabla \cdot \vec{j} =0 \end{equation} takes the form \begin{equation} \dd[\rho]{x^0} + \nabla \vec{j}=0; \qquad \partial_\mu j^\mu = 0. \end{equation} The transformation properties of the current and charge density is the same as four vector. For a frame \(K^\prime\) moving w.r.t frame \(K\) along the \(X^1\) axis with velocity \(v\), we have \begin{equation} \rho^\prime = \frac{\rho- v j_x}{\sqrt{1-v^2}},\quad j^\prime_x = \frac{j_x-v\rho}{\sqrt{1-v^2}}, \quad j_y^\prime= j_y, \quad j_z^\prime = j_z. \end{equation} Similarly, we have \begin{equation} \phi^\prime = \frac{\phi- v A_x}{\sqrt{1-v^2}}, \quad A^\prime_x = \frac{A_x-v\phi}{\sqrt{1-v^2}},\quad A_y^\prime= A_y, \quad A_z^\prime =A_z. \label{E25} \end{equation} for the vector potential.

3.Potentials of a Moving Charge

Consider a charge moving with velocity \(v\) as seen from a frame \(K\). Let \(K^\prime\) be the moving frame in which the charged particle is seen to be rest. Therefore the potentials in the moving frame at a point \(P\) will be given by \begin{equation} \phi^\prime(P) = \frac{1}{4\pi\epsilon_0 r^\prime}, \quad\quad\vec{A}^\prime=0. \end{equation} The potentials at the point \(P\) in the frame \(K\) will be obtained by transformation, inverse to that in \eqref{E25} \begin{equation} \rho(P)=\frac{1}{4\pi\epsilon_0}\frac{1}{r^\prime} \end{equation} For a charge at rest at the origin in a frame \(K\) the potentials are \begin{eqnarray} \phi(P)= \frac{\phi^\prime(P)+ v A_x^\prime(P)}{\sqrt{1-v^2}}, &\quad& A^\prime_x(P) = \frac{A_x^\prime(P)+v\phi^\prime(P)}{\sqrt{1-v^2}}\\[2mm] A_y^\prime(P)= A_y(P), &&\quad A_z^\prime(P) =A_z(P). \label{E28}. \end{eqnarray} Remembering that the coordinates of a space-time point \(P\) in the two frames are related by \EqRef{E1}, we get \begin{eqnarray} \phi(\vec{r},t)&=& \frac{1}{\sqrt{1-v^2}} \frac{q}{4 \pi\epsilon_0 } \Big(\sqrt{x^{,\prime 2}+y^{\,\prime 2}+z^{\,\prime 2}}\Big)^{-1}\\ \phi(\vec{r},t) &=& \frac{1}{\sqrt{1-v^2}} \frac{q}{4\pi\epsilon_0} \left\{\frac{(x +vt)^2}{(1-v^2)}+y^2+z^2\right\}^{-1/2}\\ \end{eqnarray} and \begin{equation} \vec{A}(\vec{r},t) = \frac{\vec{v}}{\sqrt{1-v^2}} \frac{q}{4\pi\epsilon_0} \left\{\frac{(x +vt)^2}{(1-v^2)}+y^2+z^2\right\}^{-1/2}\\ \end{equation}

References

• Sec 25-4 ELectrodynamics in four dimensional notation Sec 26-1 The four-potential of a moving charge R. P. Feynman, Robert B. Leighton and Mathew Sands Lectures on Physics, vol-II, B.I. Publications (1964)
• Sec 12.3.4 Electrodynamics in tensor notation David Griffiths, Introduction to Electrodynamics, 3rd EEE edn, Prentice Hall of India Pvt Ltd New Delhi, (2002).
• Sec 22.7 Covariant Electrodynamics  Zangwill, Modern ELectrodynamics, Cambridge University Press, Cambridge, (2012)