# Tag Info

5

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): $$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$ $$= \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} ... 3 I am not sure what is the path C you are integrating over? In your definition you evaluate U(C) which in the present case of force is independent on the explicit path you choose but still depends on initial and final point, i.e. U(p_1,p_2). In your final result it seems you are actually 'walking' three times the path p_1=(-\infty,y,z) to ... 3 Yes, you are right. Indeed, since z is a rotational symmetry axis, it defines an eigenspace of the inertial operator I. Since I is a symmetric linear operator, it admits an orthonormal basis of eigenvectors, one such vector is {\bf e}_z. This unit vector can be completed into a basis of eigenvectors just adding some pair of unit orthogonal ... 3 since P is a constant and can be taken outside of the integral There is no reason whatsoever why p should be a constant, unless specified so; in particular, in your exercise the task is to find a solution for isothermal transformations. For gases and fluids p is a function of the volume and other variables as well, therefore the equation becomes$$ ... 3 Yes, there are an infinite number of solutions, though your teacher will want you to choose the most obvious one. When the force does work on the mass, that work can be converted into two forms: the potential energy of the object the kinetic energy of the object If you apply a force of$800g$then once the object has been raised the 2.4m it will still ... 3 Here is the procedure:$KE = 0.5mv^2\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$So the question becomes,how do we find the derivative of$v^2$with respect to time? One can easily see that$\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$(Notice how the$dv$cancels top and bottom) Therefore,$\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ...

2

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.

2

A Lagrangian can easily be written down for a relativistic particle in a curved spacetime (i.e., under the influence of gravity.) Specifically, the "action" is the proper time between two events along a particle's world-line, and the particle's trajectory will extremize the proper time between these events: $$S = \tau = \int \sqrt{ - g_{\mu \nu} dx^\mu ... 2 No that is not what you must prove. It is not true that if u^a is hypersurface orthogonal then \nabla_a u_b = \nabla_{(a}u_{b)}. In fact this is only true if u^a is geodesic. If u^a is hypersurface orthogonal then, by definition, u_{[c}\nabla_b u_{a]} = 0. Writing this out we have$$ 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

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

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 ...

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

What data do you have for linear motion? Your equations are correct if you have the acceleration as a function of time and the orientation is constant. The angular accelerometer can give you the angles as a function of time with integration. Unfortunately, drift can be a problem. The received wisdom is to use an accelerometer (linear or angle), integrate ...

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 ... 1 Equation (2.4.6): T(z)X^\mu(0)\sim \frac{1}{z}\partial X^\mu(0) means that the RHS is the most singular term of the LHS. T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4} So \begin{align*} T(z)X^{\mu}(0) & =-\frac{1}{\alpha'}:\partial X^{\nu}(z)\partial X_{\nu}(z):X^{\mu}(0)\\ & =-\frac{2:\partial ... 1 CuriousOne writes in a comment that single atom is visible "if it is illuminated properly," which is correct. It's possible to construct a trap for a single atom with transparent windows, and to illuminate that atom so that it fluoresces. Note that this was impossible thirty years ago and is nontrivial today. What your textbook author almost certainly has ... 1 Your last equation is a quadratic in t. The a is simply \sin(\theta).g. You can then solve it with the usual formula for a quadratic equation. There are two solutions to a quadratic, and that's because if you go into negative time you'd be pulled down by gravity any get to the new X position. This solution, of course, wouldn't apply to your ... 1 Is this the correct way to find the derivative of kinetic energy? K=\frac{1}{2}m v^2 \\ $$So:$$ \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right) $$If the mass does not change over the time, then$$\frac{dm}{dt}=0$$And finally$$ \frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right) $$So simplifying:$$ ...

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