# Tag Info

8

When quoting results, there are a few good rules to follow: Avoid rounding errors in intermediate calculations. Write your error to 1 significant figure if your data set is smaller than $10^2$, 2 if it's smaller than $10^4$ etc. Write your estimate and its error with the same number of decimal places. Rules 1. and 3. are simple to understand. Rule 2. ...

8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

5

In both of your solutions, you attempted to use Newton's 3rd law: $$\vec{F}_{1\rightarrow2}=-\vec{F}_{2\rightarrow1}.\tag{Newton's 3rd law}$$ You did this correctly in your first method ("Newton's law method") but incorrectly in your second method ("Kinetic energy method"). In your first method, you explicitly set the magnitude of the forces equal to each ...

4

Take a spin $1/2$ particle with its spin pointing along $\hat{n}$ defined by $$\hat{n}=(\sin{\phi}\cos{\theta},\sin{\phi}\cos{\theta},\cos{\phi})$$ We are measuring the spin along $\hat{n}$ and the operator corresponding to this observable is $\vec{S}\cdot\hat{n}$. $$\vec{S}\cdot\hat{n}=\frac{\hbar}{2}\begin{pmatrix} \cos{\phi} & ... 4 Edit: Note: I have posted another proof of this in another question, here. Those who prefer coordinates may find it slightly more palatable. I gather from your comments that you can do this if you have \mathcal{L}_X\text{Ric} = 0. Thus I will outline a somewhat more general result, assuming a certain identity connecting Killing vectors and Riemann ... 4 This is not a complete answer, but fills in some of the missing pieces Trimok asked about in the comments to Stan's answer: Note that I did not verify that Stan's proof actually works. Re 1)$$ \begin{align} \nabla_Z \nabla_Y X &= Z^\lambda(Y^\mu X^\nu_{;\mu})_{;\lambda} \partial_\nu \\&= Z^\lambda(Y^\mu_{;\lambda} X^\nu_{;\mu} + Y^\mu ...

4

Cool question! Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors. If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined ...

4

So I assume you actually need to prove Poincare invariance of $\int d\tau A_\mu\dot{x}^\mu$ for a particle trajectory, rather than invariance of $A_\mu\dot{x}^\mu$, but the former expression is equal to $\int_a^b dx^\mu A_\mu$, where $a$ and $b$ are the initial and the final points of the trajectory, and Poincare invariance is indeed almost obvious for this ...

4

This is just the potential of a standard harmonic oscillator. The presence of the linear term is just due to the fact that you're using a coordinate system where the minimum of the potential is not at $x=0$. You can re-write it as $$V(x)=\frac{1}{2}mx'^2+c$$ where $x'=x+\frac{\lambda}{m}$ is the coordinate centered at the minimum of the potential and ...

3

Note that if you lower an index of the Kronecker delta, it becomes the metric: $\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$ And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$. Then, the metric terms cancel and you neglect cuadratic terms. That should be enough to solve it.

3

Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with $n$ degrees of freedom and Lagrangian of the form: $$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$ where $T$ is quadratic in ...

3

This question can't be answered because important information is not given. We have no idea how much influence, if any, the wind has on the boat's velocity. In the limiting case, if the boat's wind resistance is 0, then the wind has no effect. The question also asks what the boat would do in still water, but we were never told how still or not the water ...

3

I'd simply like to add to Innisfree's very sound answer. Three points: Don't Round Off in In-Between Steos Innisfree's point 1 is almost "obsolete": I haven't used (and have barely seen) a calculator since the mid eighties - I think the last time was in an undergrad exam: any calculation nowadays is going to be put into something like Mathematica, Maple ...

3

Your question is asking you to calculate the de Broglie wavelength of the electron, which can be obtained from its momentum $p$ via de Broglie's relation $$\lambda=\frac h p$$ where $h$ is Planck's constant. Regarding units: You should work consistently with your units. You can work with the kinetic energy in Joules, $h$ in J s and $p$ in kg m/s. But ...

3

If $L^2$ is $6\hslash^2$ this means that the quantum number $l$ is $2$, hence the quantum number $m$ goes from $-2$ to $+2$, and the possible values of $L_x$ are $-2\hslash, -\hslash, 0, \hslash, 2\hslash$ Conventionally $L_z$ is used instead than $L_x$, but what applies to $L_z$ must apply also to $L_x$ and $L_y$: there are no preferential directions in ...

3

Applying the law of cosines to the triangle $\triangle S_1S_2P$ will yield $$r_2^2=r_1^2+a^2-2ar_1\cos(\angle S_2S_1B).$$ The angle that appears is complementary to $\theta_m$, so you can either use $\angle S_2S_1B=\frac\pi2-\theta_m$ and trigonometric identities, or simply see that $$\cos(\angle S_2S_1B)=\frac{S_1B}{S_2B}=\sin(\theta).$$

3

The question is asking what is the force exerted by the water on either of the two faces of the plate. The net force will be zero as force on either side sides cancel, so your intuition made sense. The force on a side comes from water pressure across the triangular surface. The pressure at any point on the triangle depends on the depth of that point. ...

3

You should realize that the first equation you write gives you the value of $a$. In these kinds of problems, you are always given some force, and you are expected to apply Newton's laws to the problem. Therefore $F_{\mu} = -mg\mu \underbrace{=}_{\text{$2^{nd}$Law}} ma \quad \quad \to \quad \quad a = -g\mu$ Now you know the acceleration, you can find how ...

3

The answer to that is because the moment of inertia is not the same for the solid cylinder than for the hollow one. As you write the formula for the moment of inertia, it depends on the distribution of the mass. The further away the mass is from the rotation axis, the more contributes to the moment of inertia (as in distance squared $r^2$). So, since the ...

3

I'm going to explain roughly what the Born Rule, following Stan Liou's comment. One of the Postulates of Quantum Mechanics relates a mathematical quantity, the wave function (or state $\psi$ of a Hilbert space, $\mathcal{H}$) to a measurable entity, the probability of a given event to happen. The idea goes like this: if you want to measure a quantity ...

2

All forces act in pairs, so let me start by matching them up: Force on $M_1 = F = - M_1$ on Some force providing device Surface on $M_1 = F_1 = -M_1$ on Surface $M_2$ on $M_1 = F_2 = - M_1$ on $M_2$ The values for the forces horizontal components are found using... $F =$ Given (1 Newton) $$F_{sf} \le \mu_{sf} \cdot F_n$$ $$F_1 \le \mu_1(M_1+M_2)g$$ ...

2

I really don’t like that diagram. No, REALLY. I think it conveys a bad intuition that may confuse you. I don’t like what it does with the connexion coefficients (Christoffel symbols). Here’s why. In a general manifold, tangent spaces of course are not comparable at different points. This is in contrast with the situation in Euclidean space, which can be ...

2

Say that we are handed a 1D quantum mechanical system, which satisfies the canonical commutation relation $$\tag{1} [\hat{Q},\hat{P}]~=~ i\hbar~{\bf 1},$$ and handed some choice of eigenstates $|q\rangle$ and $|p\rangle$ for every value of $q,p\in\mathbb{R}$. The eigenstates satisfy $$\tag{2} \hat{Q} \mid q \rangle ~=~q\mid q \rangle, \qquad ... 2 We start from the Lagrangian in spherical coordinates$$L=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2}+r^{2}\sin^{2}\theta\dot{\phi}^{2})+mgr\cos\theta$$The length of the string is d and the system is a constrained one with |\vec{r}|=d. Now, the constraint that is associated with a multiplier \lambda is given by c(r)=r-d=0. At this point we have ... 2 @PPG: Well (−1,−1) is not a solution of your equation, but (1,−1) is... – Adam What he said was: Finding the eigenvectors You did right... but:$$(\hbar/2)\alpha+(\hbar/2)\beta=0\implies\alpha+\beta=0\implies\beta=-\alpha $$Then, the corresponding eigenvector is:  c\left[\begin{array}{c} 1 \\ -1 \end{array}\right]. 2 You are asking about retroreflectors. Some background: In general (implying 3D) and with the assumption that no functional refractive elements are allowed, a hollow corner cube reflector would be the tool of choice. Nevertheless equally useful glass corner cubes exist. Both types are colloquially known as triple-prisms. So that's where your first answer "3 ... 2 If you want to combine the two equation the easiest way is probably taking a "pseudo-derivative" of the equation of status: 2kV\Delta V = n R\Delta T And then substitute \Delta T. I don't find the resulting expression for C_x very illuminating, however the initial hypothesis is pretty weird. 2 Since the Lorentz transformation is valid for any x\in M_{4}, it can be rewritten as \Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get ... 2 There's a more straightforward calculation. In order to travel eastward, the plane's velocity must have a southward component of 60kph to cancel the wind from the south. Since the plane's speed is 100kph, we have the eastward component (in kph) is just:$$v_E = \sqrt{100^2 - 60^2} = 80$$Thus, you are correct; the time required to travel 189km eastward ... 2 First of all, you're overloading your time variable. Let the delay of the launch of the 2nd projectile be t_d while the time variable is t. As you've already correctly written, the equation for the displacement of the 1st projectile is (for t \ge 0):$$s_1 = ut + \dfrac{at^2}{2}$$Now, for the 2nd projection, we have (for t' \ge 0):$$s_2 = ut' ...

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