Yes, you're right about the upper limit being ##1##. Also, the fact that ##\sigma << 1## implies that we might as well take the upper limit to ##+\infty##. But other than a factor in front, I don't see how that changes my answer.
From this paper, I am trying to compute the coefficients in the expansion of the Gaussian wavepacket
$$\phi(x) = \frac{1}{(2\pi\sigma^2)^{\frac{1}{4}}}\exp \Big(-\frac{(x-x_{0})^{2}}{4\sigma^{2}} + ik_{0}(x-x_{0})\Big) $$ where $$\sigma << 1$$and $$k_{0} >> \frac{1}{\sigma}$$
in terms of the...
I expanded the exponential with the derivative to get:
## Z = \Bigg(1 + \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} + \frac{1}{4} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{\partial}{\partial x_{k}} A^{-1}_{kl}...
I need to prove that in a vacuum, the energy-momentum tensor is divergenceless, i.e.
$$ \partial_{\mu} T^{\mu \nu} = 0$$
where
$$ T^{\mu \nu} = \frac{1}{\mu_{0}}\Big[F^{\alpha \mu} F^{\nu}_{\alpha} - \frac{1}{4}\eta^{\mu \nu}F^{\alpha \beta}F_{\alpha \beta}\Big]$$ Here ##F_{\alpha...
Say we have a matrix L that maps vector components from an unprimed basis to a rotated primed basis according to the rule x'_{i} = L_{ij} x_{j}. x'_i is the ith component in the primed basis and x_{j} the j th component in the original unprimed basis. Now x'_{i} = \overline{e}'_i. \overline{x} =...
Riley Hobson and Bence define covariant and contravariant bases in the following fashion for a position vector $$\textbf{r}(u_1, u_2, u_3)$$:
$$\textbf{e}_i = \frac{\partial \textbf{r}}{\partial u^{i}} $$
And
$$ \textbf{e}^i = \nabla u^{i} $$
In the primed...
The ##\mu## is contravariant and the ##\beta ## covariant, right?
Shouldn't $$ \partial_{x}\partial^{x}$$ and $$ \partial_{t}\partial^{t}$$ have opposite signs, since we are working with four vectors?
I'm trying to derive the Klein Gordon equation from the Lagrangian:
$$ \mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi)^2 - \frac{1}{2}m^2 \phi^2$$
$$\partial_{\mu}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\Bigg) = \partial_{t}\Bigg(\frac{\partial \mathcal{L}}{\partial...
The equilibrium position is ##\phi_{0} = \arctan{(\frac{a}{g})}##. So if we linearize the equation, it becomes
$$\ddot{\phi} + \frac{\sqrt{a^2 + g^2}}{l} (\phi +\phi_{0}) = 0$$ if ##\phi_{0}## is small, or equivalently the acceleration is small.
Equivalently,
$$\ddot{\phi} + \frac{\sqrt{a^2 +...
We have a car accelerating at a uniform rate ## a ## and a pendulum of length ## l ## hanging from the ceiling ,inclined at an angle ## \phi ## to the vertical . I need to find ##\omega## for small oscillations. From the Lagrangian and Euler-Lagrange equations, the equation of motion is given...
Hi,
This is the statement of the problem of AP French's textbook "Newtonian Mechanics".
1. Homework Statement
The commander of a spaceship that has shut down its engines and is
coasting near a strange-appearing gas cloud notes that the ship is
following a circular path that will lead...
but the length of the rod is in the yz plane. How can there be any compression parallel to the x-axis?
I interpreted the question in the following way and ended up with the correct answer:
What is the the ratio of the frequency in the situation in which the shear force F is applied parallel...