Browse & Filter

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For a self coupled scalar theory with interaction Lagrangian density given by

\begin{equation} \mathscr {L}_{\text{int}}= \frac{\lambda}{3!}\phi(x)^3 \end{equation}

Compute and write your answers as integrals of expressions involving propagator in position space.
\begin{eqnarray} (a)&&  \int d^4y_1\int d^4y_2\int d^4y_3 \matrixelement{0}{T(\phi(x_1)\phi(x_2) \mathscr{ L}(y_1) \mathscr {L}(y_2)\mathscr {L}(y_3)}{0} \\ (b) && \int d^4y_1 \int d^4y_2\matrixelement{0}{T(\phi(x_1)\phi(x_2)
\mathscr{L}(y_1) \mathscr{L}(y_2)}{0}. \end{eqnarray}

Draw position space Feynman diagram in each case.

 $\newcommand{\matrixelement}[3]{\langle#1|#2|#3\rangle}\newcommand{\dd}[2][]{\frac{d#1}{d#2}}${}$\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$
$\newcommand{\ket}[1]{|#1\rangle}$ {} $\newcommand{\bra}[1]{\langle #1|}$\(\newcommand{\PP}[2][]{\frac{\partial^2 #1}{\partial^2 #2}}\)

The Feynman propagator \(\Delta_F\) for a real scalar field is defined as the vacuum expectation value of time ordered product of fields: \[ \Delta_F(x-y) = \matrixelement{0}{T\big(\phi(x)\phi(y)\big)}{0}.\] We will now show that the Feynman propagator satisfies the equation \[ (\Box + M^2) \Delta_F(x-y)= - \delta^{(4)}(x-y).\] } Since \(\phi(x)\) obeys the Klein Gordon equation, it is obvious that \(\Delta^{(\pm)}(x-y)\) satisfy the Klein Gordon equation \begin{equation} (\Box + M^2)\Delta^{(\pm)}=0 . \end{equation} The time ordered product can be written as \begin{eqnarray} T(\phi(x)\phi(y)) &=&(\phi(x)\phi(y))\theta(x_0-y_0) + (\phi(y)\phi(x))\theta(y_0-x_0) \end{eqnarray} Now we compute action of the operator \(\displaystyle\Box =\frac{\partial^2}{x^{0\,2}}- \frac{\partial^2}{\partial x^{k\,2}}\) on the time ordered product. Note that the space derivatives in \(\displaystyle\Box = \PP{{x^0}}- \PP{{x^k}}\) will act only on the fields but the time derivative will act on theta functions also. So let us compute the time derivatives using \(\displaystyle\dd[\theta(x_0-y_0)]{x_0}= \delta(x_0-y_0)\). We will get \begin{eqnarray} { \pp{x_0}T(\phi(x)\phi(y))\nonumber }\\\nonumber &=&\Big\{\pp[\phi(x)]{x_0}\phi(y)\theta(x_0 - y_0) + (\phi(x)\phi(y))\pp{x_0}\theta(x_0 - y_0)\Big\} - \\ && \qquad \Big\{\phi(y)\pp[\phi(x)]{x_0}\theta(y_0 - x_0) - (\phi(y)\phi(x))\pp{x_0}\theta(y_0 - x_0)\Big\}\\\nonumber &=& \dot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\dot{\phi}(x)\theta(y_0 - x_0) + \\ && \qquad \label{EQ19} \big[\phi(x), \phi(y)\big] \delta(x_0 - y_0)\\ &=& \big(\partial_0{\phi}(x)\big)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\big(\partial_0{\phi}(x)\big)\theta(y_0 - x_0) \label{EQ20} \end{eqnarray} The last term in \eqRef{EQ19} becomes equal time commutator of the fields and hence it is zero. Differentiating \eqRef{EQ20} once again w.r.t. \(x_0\) we will get \begin{eqnarray}\nonumber { \frac{\partial^2}{\partial x^{02}}T(\phi(x)\phi(y)) }\\\nonumber &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) +\\\nonumber &&\qquad \dot{\phi}(x)\,\phi(y)\delta(x_0 - y_0) - \phi(y)\,\dot{\phi}(x)\delta(y_0 - x_0)\\\nonumber &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) \\\nonumber && \qquad + \big[\dot{\phi}(x)\,,\,\phi(y) \big] \delta(x_0-y_0) . \label{EQ21} \end{eqnarray} Substituting the value of equal time commutator \[\big[\dot{\phi}(x)\,,\,\phi(y)\big]\Big|_{x_0=y_0} = -\delta^{(3)}(\vec{x}-\vec{y}).\] we get \begin{eqnarray}\nonumber {\frac{\partial^2}{\partial x^{02}}\big\{T(\phi(x)\phi(y))\big\} }\\ &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) - \delta^{(4)}(x-y). \label{EQ26} \end{eqnarray} Hence \begin{eqnarray}\nonumber {\big(\partial_0^2-\partial_k^2 + M^2\big)T(\phi(x)\phi(y))}\\\nonumber &=& \big(\Box + M^2\big)\phi(x)\,\phi(y)\theta(x_0 - y_0)\\ \nonumber &&- \phi(y)\big(\Box + M^2\big)\phi(x)\theta(y_0- x_0) - \delta^{(4)}(x-y) \label{EQ27} \\ &=& - \delta^{(4)}(x-y) . \end{eqnarray} Remembering that the field satisfies Klein Gordon equation and taking vacuum expectation value, we get the desired answer \begin{equation} \big(\Box + M^2\big) \Delta_F(x) = - \delta^{(4)}(x-y) . \end{equation}

A conical pendulum moves in a circular path of radius \(a\) and string making an angle \(\alpha\) with vertical.

• Working in the inertial frame, draw a diagram showing the forces acting on the pendulum and prove that the angular frequency \(\omega\) is given by \[ \omega= \sqrt{\frac{g}{L\cos\alpha}}\]
• Draw the all forces acting on the pendulum as seen from the rotating frame in which the pendulum is at rest. Is the resultant of all forces is zero or not? If it is not zero, how do you explain that the pendulum is at rest in the rotating frame? Give a complete answer.
• Next consider a third frame rotating with angular velocity \(2\omega\), the direction of the angular velocity is the same as that of the pendulum. What is the motion of the pendulum as seen by an observer in this frame? Does the sum of all forces vanish now or not? Explain observed behaviour of the pendulum using the equation of motion as applicable in this frame.

%FigNoNumX{10,-65}{40}{0}{ConicalPendulum}

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A conical pendulum moves in a circular path of radius \(a\) and string making an angle \(\alpha\) with vertical.

• Working in the inertial frame, draw a diagram showing the forces acting on the pendulum and prove that the angular frequency \(\omega\) is given by \[ \omega= \sqrt{\frac{g}{L\cos\alpha}}\]
• Draw the all forces acting on the pendulum as seen from the rotating frame in which the pendulum is at rest. Is the resultant of all forces is zero or not? If it is not zero, how do you explain that the pendulum is at rest in the rotating frame? Give a complete answer. [10+10]

In equilibrium a pendulum suspended from an inextensible spring in an accelerating car will not hang vertically. The string will make an angle \(\theta\) with vertical see \Figref{me-fig-08002-Q}.\\ \FigBelow{30,20}{50}{70}{me-fig-08002-Q}{Pendulum in accelerating car}\\ Assume that \(g\) is known and the \(\theta\) is measured by an observer sitting in the car. Then without assuming the gravitational and inertial masses to be equal, which of the following quantity can be related to the angle \(\theta\) ?

• value of acceleration.
• value of inertial mass.
• value of gravitational mass.
• the ratio of gravitational and inertial masses.

Give a short explanation of your answer.

A 400-ton train runs south at a speed of \(u\) at a latitude of \(\lambda\) north. Find the expressions of

• the horizontal force on the tracks?
• the direction of the force?

Compute the numerical values of the forces assuming \(u=100 \text{km/h}, \lambda=60^o\)

A number of balloons are attached to a circular disk with string. Some balloons are filled with air and some balloons are filled with helium. The disk is hung freely from ceiling of a room and is disk is rotated about its center. Assuming that the disk remains horizontal while rotating, describe what happens to the balloons?

• All the balloons move away from the center of the disk.
• All the balloons move towards the center of the disk.
• Air balloons move outwards away from the center and helium balloon move towards the center.
• Air balloons move inwards towards the center and helium balloon move outwards the away from the center.

Sketch diagrams to show the balloons
(i) when the disk is hanging horizontally and is stationary.
(ii) when the disk is rotating Give a short argument in support of your answer.