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6

You can get an exact solution for $t(p)$, although it involves a rather nasty integral that I'm not sure can be written in closed form. Here's how: The equations of motion are $$ \frac{dp}{dt} = -kx \qquad \frac{dx}{dt} = \frac{1}{m} \frac{p}{\sqrt{1 + p^2/m^2 c^2}}. $$ This second equation can be obtained by taking the equation $p = m v /\sqrt{1 - ...


4

Perhaps its a little clearer if you shorten the contents of the brackets (and lets drop the constants too): $$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$ $$ = \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} ...


3

since P is a constant and can be taken outside of the integral There is no reason whatsoever why $p$ should be a constant, unless specified so; in particular, in your exercise the task is to find a solution for isothermal transformations. For gases and fluids $p$ is a function of the volume and other variables as well, therefore the equation becomes $$ ...


2

No that is not what you must prove. It is not true that if $u^a$ is hypersurface orthogonal then $\nabla_a u_b = \nabla_{(a}u_{b)}$. In fact this is only true if $u^a$ is geodesic. If $u^a$ is hypersurface orthogonal then, by definition, $u_{[c}\nabla_b u_{a]} = 0$. Writing this out we have $$ u_c \nabla_b u_a - u_b \nabla_c u_a + u_a \nabla_c u_b -u_c ...


2

The relation you ask about is just a reshuffling of the components. Writing out the indices we have $$ \Theta_1^T C \, \Gamma_{\mu} \Theta_2 = (\Theta_1^T)_a C_{ab} \, (\Gamma_{\mu})_{bc} (\Theta_2)_c = - (\Theta_2)_c (\Gamma_{\mu})_{bc} C_{ab} (\Theta_1^T)_a = - (\Theta_2^T)_c (\Gamma_{\mu}^T)_{cb} (C^T)_{ba} (\Theta_1)_a $$ where the minus sign in the ...


2

Let's do this explicitly for both cases. For these examples, the classical formula for the geodesic curvature $k_g$ suffices. Let $\gamma(t)$ be a curve in a surface $S \subset \mathbb{R}^3$, and let $n(t)$ be the unit normal to $S$ at the point $\gamma(t)$. Then $$ k_g = \frac{\ddot{\gamma}(t) .(n(t) \times \dot{\gamma}(t))}{|\dot{\gamma}(t)|^3} $$ First ...


2

There's an error in the notes you posted. The tortoise coordinate is usually defined via $$ \frac{dr}{dr_*} = 1 - \frac{2m}{r} = f \neq \frac{1}{f}. $$ Note that the correct definition is given in eq. (42) of your link. I suspect this will fix your problem.


2

This is the picture of your problem So point M movement is described by $x_M = a \sin(\omega t)$. This is a system with one degree of freedom and for the coordinate that completely describes this system we will use angle $\theta$. Let us describe $x$ and $y$ position of a pendulum at any moment $$x_m = x_M + r \sin(\theta)$$ and $$y_m = r \cos(\theta)$$. ...


1

The final equation turns out to be: $$ l \ddot{\phi} - a \omega^2 \sin{\omega t} \cos\phi + g\sin\phi = 0$$ Now, as Aleksandar has suggested, all that remains is to put in the condition of small oscillations i.e. small values of $\phi$. Therefore \begin{align} \sin\phi &\approx \phi \\ \cos\phi &\approx 1 \end{align} Your equation reduces to $$ ...


1

You know from the question that the forces are in the same direction as the sides of the equilateral triangle, so they are at 60 degrees to each other. They all have different magnitudes so cannot possibly form an equilateral triangle if placed "end-to-end". To work out the answer, you can calculate the horizontal and vertical components of each of the ...


1

You conclude that constant acceleration is appropriate for one of a few reasons: You read in the question text that you should treat it that way. If (1) does not apply, you read in the question text that some physical situation obtains and you know or suspect that this situation is usually well represented by a constant acceleration. If neither (1) nor (2) ...


1

Whatever force you like. Since force just determines the rate change of velocity, you can use a massive force for a trillionth of a second or a tiny force for a long time period. However, if changed the velocity of the Earth (relative to the Sun) of 1 m/s, you'll would cause an impulse – change in momentum – on the Earth of $5.972×10^{24}$ kilogram-meters ...


1

There is a subtle difference between saying $(2,2)$ and $2\otimes 2$. In the latter case we are thinking of both reps as transforming under the same element of the group $SU(2)$. In the former case we are thinking of $(2,2)$ as transforming under the Lorentz group, which contains two distinct copies of $SU(2)$. Call one copy the $L$ copy and the other the ...


1

You don't need to use the metric of the hemisphere. This is because the pullback of arbitrary forms onto the submanifold is the trivial pullback operator. All you need to do is apply the projection operator. Therefore, the extrinsic curvature tensor is just $K_{ab} = - \gamma_{a}{}^{c}\gamma_{b}{}^{d}\nabla_{c}n_{d}$, where $\gamma_{ab}$ is the metric of ...


1

It can be shown that $\omega_{ab} = 0 \Leftrightarrow \omega^a \equiv \epsilon^{abcd}u_b \nabla_c u_d = 0$. The latter quantity is known as the twist (or vorticity). In a local inertial frame it is easy to see that $\vec{\omega} \sim \vec{\nabla}\times \vec{v}$ where $\vec{v}$ is the 3-velocity field of the flow. This lends to the following interpretation ...


1

The derivative of $x f(x)$ is $x f'(x) + f(x)$, and here you're integrating $k \int_{-\infty}^\infty dx ~ x ~ f'(x)$ for some constant $k$, and some complicated function $f$. When you integrate this by parts, you raise $f' dx$ and lower $x$ to find:$$k \int_{-\infty}^\infty dx ~ x ~ f'(x) = k \left[x ~f(x)\right]_{-\infty}^{~\infty} - k \int_{-\infty}^\infty ...



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