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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 ...


0

Force is just a product of mass and acceleration and to calculate the force required to move the earth is quite easy. Though incredibly difficult to produce. IF we take the mass of earth as 5.9736 x 10^24 kg. And the acceleration 1 meter per second per second. And that is where you are wrong when you said 1 meter per second which is the unit for speed or ...


0

Your question makes sense if you replace "force" by "impulse". And that impulse is simply the mass of the Earth times $1 {\rm m\,s^{-1}}$. Moreover, you need to make the qualification: how much impulse is needed to change the Earth's motion state so that it is moving at $1 {\rm m\,s^{-1}}$ relative to its motion state now. According to the estimates of this ...


0

Let the node between R2 and R4 be a, and node between R3 and R1 be b. Then By voltage Divider Rule: $ Va = (R2)/(R2+R4) * E $ $ Vb = (R3)/(R3+R4) * E $ And then subtract them. You will get your answer.


0

First be aware that, so far, you have been dealing with projectile velocity as a 2-dimensional phenomenon. This is going to have to change. In order to conform to standard notation, you'll need to refer to altitude (vertical motion) as the z-coordinate, while horizontal will be handled by x and y coordinates. For ease of calculation, you can assume that you ...


0

I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. [...] $ \gamma^3~m_0~\ddot{x} = -k x $ Since your question is specificly about harmonic motion, we might (instead) insist on this solution $$x[~t~] := x_{\text{max}}~\text{Sin}[~\omega~(t - t_0)~],$$ ...


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) ...


0

Allowing the bar to pivot about point G, you will see that R1 (if it was the only force) would rotate the bar clockwise and R2 (by itself) would rotate the bar counterclockwise. The direction of rotation is determine by the direction of the force (up or down) and where it acts in relation to the pivot (right or left).


0

I would do a hand analysis at first. You have two torques on the arrow. One is the drag torque, which is maximum when the arrow is transverse to the orbital velocity, as in your initial condition. The second is gravity gradient torque, which will be maximum when the arrow is horizontal. Compute each of these for your arrow. If one is much greater than ...


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.


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 - ...


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 ...


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

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 ...


0

For the first question, you do not need to interchange $\nabla_{\alpha}$ and $\nabla^{\alpha}$. We simply have $$ \nabla_{\alpha}\varphi\nabla^{\alpha}\nabla_{\beta}\varphi \\ = \eta^{\alpha\gamma}\nabla_{\alpha}\varphi\nabla_{\gamma}\nabla_{\beta}\varphi \\ = \nabla^{\gamma}\varphi\nabla_{\gamma}\nabla_{\beta}\varphi \\ = ...


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 ...


0

We can call the first object P, and the second object Q. Then we can say that the location of the particles can be notated as $ <P_x, P_y>, <Q_x, Q_y> $ Therefore, the distance, $d$, between A and B is $d=\sqrt{(A_x-B_x)^2+(A_y-B_y)^2}$. So, because $x(t)=v_x*t+x(0)$, and $v=<U_1*\frac{B_x-A_x}{d},U_1*\frac{B_y-A_y}{d}>$ Combining all ...


0

The short answer is, yes. The slightly longer, but slightly more accurate answer is yes, but not noticeably. As you say, increasing humidity by evaporating existing water will cause the mass of the vapor to move to a greater distance from the earth's center, and this will increase the moment of inertia, reducing the rotation rate. Consider that the ...


0

@RandyWelt Note that, first, theta cannot be pi/2 for a this specific problem. There exist no $theta, J, M, r$ that solve the time dilation equation for $theta = \pi/2$. Even in my calculations, I was careful to take for example $\pi/2 < \theta \leq \pi$ or $0 \leq \theta < \pi/2$.Also, as an aside, for a Kerr metric, it is singular at points ...


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 ...


0

I think there is some confusion on your part here. a = 0.998 represents the "speed" at which the black hole is rotating, which is taken to be by Kip to be 99.8% the speed of light, i.e., it is a = 0.998*c. The "a" in the calculations for the Kerr metric, is very different. It is a length scale. This can be seen for example by considering a spherical shell ...


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 ...


0

$M=10^8M_\odot,$ so $r_s=$ $2GM/c^2=$ $2.95\times 10^{11}m.$ And $\alpha = J/Mc,$ has units of distance, and we can set it to $\alpha =0.998 r_s/2,$ which is very close to the critical value of $r_s/2.$ Now you can put $d\tau=$ 1 hour, $dt=$ 7 years, and $ \theta=90° $ into $$\left(\frac{d\tau}{d\ t}\right)^2=1-{\frac{2GMr ...


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 $$ ...


0

You are making confusion between operators and their representations on the position basis. Your hypotheses 1) and 2) are wrong (more ill-interpreted): as you can easily notice the left hand sides contain an element in the Hilbert space whereas the right hand sides contain numbers (functions evaluated in a point $x$, therefore a number). To answer your ...


0

There are a couple of ways of knowing the wavelength of laser pointer. 1) Using Snell'law (law of refraction): A light passing the border between two media whose refractive indexes vary. The incident light PO of wavelegth, $\lambda_{1}$ travelling in a media of refractive index, $n_{1}$ is refracted in to another media of refractive index, $n_{2}$, with a ...


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 ...


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} ...


0

I think your error is in assuming that $E_{n+1} - E_{n}$ is proportional to $n$. At least, I assume you assumed it; it's the only way I can see that you could go from the statement $$ E_{n+1} - E_n \propto E_n^{-1/2} $$ to the statement $$ E_n \propto n^{-2}. $$ Really, what the first proportionality above implies is that $$ \frac{dE}{dn} \propto ...


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

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 ...


0

Your entire analysis is entirely correct and complete (and in my opinion the best way to go about it). There's no more information that need to be derived about this issue. Other analyses are just a different representations of the same facts.


0

With this transformation $\rho = kr$ you can't possibly change the form of the equation, because it's a scale transformation that does nothing special to the derivatives, so I think that you have computed the derivatives wrong.


0

Hint: Calculate the vertical velocity as a function of time ($v_y=-g\cdot t$). Two vectors are perpendicular when their dot product is zero, so $$v_{x1}\cdot v_{x2} + v_{y1}\cdot v_{y2}=0$$ Now $v_{y1} = v_{y2} = gt$ since both are free falling. So you find that $gt = \sqrt{12}$ when they are perpendicular. From this, you calculate the time; and from the ...


0

It you are working with second quantization (quantum field theory), then what you wrote under "1." and "2." is incorrect. Your physical degrees of freedom are not coordinates $x$, but rather field variables $\phi$. So, $$ \left| 0 \right> \sim \exp \left\{ -\frac{1}{2} \int d^3 x \phi(x)^2 \right\} $$ This object should be called wave functional ...


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: $$ ...


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.


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 ...


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 ...


0

Two hints: Constant of integration! Redo your calculations, and remember that when you integrate, you introduce a constant $C$ whose value is to be determined by the boundary conditions that you've stated. $v=\frac{dx}{dt}$


0

There 3 rather simple formulas which allow to compute the parameters that are asked for. $L'=\mu/(2\pi) \cdot \ln(b/a)$ for the inductivity per meter L' $C'=2\pi\epsilon/(\ln(b/a))$ for the capacity per meter C' $Z= Z_0/2\pi \sqrt{\frac{\mu'}{\epsilon'}}\ln(b/a)$. b is the outer diameter and a is the inner diameter of the coaxial line $Z_0 = ...


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

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 ...


0

Using distributions is a trick that people use whenever dealing with systems that do not have the required smoothness and integrability conditions and is in general only a mathematical technique to nevertheless solve those problems. Maxwell equations, to start with, require both sides to be differentiable (at least a few times) and integrable and when you ...


0

Start from $d^4p = dp_{\mu}\,dp^{\mu}$ which is manifestly Lorentz invariant. In order to obtain the actual measure to integrate against you have to pair this up with the mass-shell condition for the particle to have positive energy and lie on the mass-shell hyperboloid $$ dp_{\mu}\,dp^{\mu}\,\theta(k^0)\,\delta(k^2-m^2). $$ Use now $k^2={k^0}^2 - k_ik^i$ ...


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 ...


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 ...


0

The "blacker" things are the less they reflect. That doesn't necessarily mean they are less visible - just less reflective. Matt coatings are probably your best bet. There is "the blackest surface there is" - nanotubes that trap the incident light and don't let it out again. Essentially they make the photons interact with the surface multiple times - if you ...


0

I would really like to express the distance from the point of mass to the axis via sin/cos: $$4 * \int_{0}^{\pi \over 2} r * \cos(\phi) d\phi \tag{1*}$$ You shouldn't express the distance of a point mass as an integral. That doesn't make sense. And why do you have the factor of 4. That doesn't make sense either. Based on your description, the ...



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