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4

Okay, the system is in equilibrium so all the forces must be balanced. First consider the weights We have $W$ downwards from the centre of each of the 3 balls (assuming their centres of mass are at their geometric centres). We also have $N_A = N_B = \frac{3}{2}W$ upwards from the points of contact between the plane and balls $A$ and $B$ respectively. ...


3

We consider the integral: $$\sum_{i\lt j}\int_{t_0}^{t_f} \mathbf{F}_{ij}(\mathbf{r}_j(t))\cdot(\mathbf{r}'_j(t) - \mathbf{r}'_i(t))dt$$ For a rigid body, the distance between any two masses is always held constant, a fact that we can express as: $$\vert\mathbf{r}_i(t) - \mathbf{r}_j(t)\vert^2 = \Delta_{ij}$$ or $$(\mathbf{r}_i(t) - ...


3

I don't think the explanation in the book is clear, but you can just ignore it and get the correct factor $S=\frac{1}{8}$ as follows. Start by drawing 5 isolated vertices, two labelled ones with degree one and three unlabelled 4-valent vertices. Then $$ S=\frac{1}{3!}\left(\frac{1}{4!}\right)^3\times C $$ where $C$ is the number of contraction schemes ...


3

There is indeed a way to construct the projection-operators, when you only know the operator itself and its eigenvalues. The derivation can be found in Julian Schwingers "Quantum Mechanics: Symbolism of Atomic Measurement" and leads to $$ P_j = \prod_{i\neq j} \frac{A-a_j}{a_i - a_j}, $$ where the product goes over all distinct eigenvalues $a_i$ of $A$. ...


3

What am I doing wrong? The expansion of $e^x$ is: $$ e^x=1+x+x^2/2+x^3/3!+\ldots $$ From expanding the expression: $$ \left<\phi_x\phi_y\exp{\left(-\frac{\lambda}{4!}\int dz \phi_z^4\right)}\right>\;, $$ the third order term is: $$ \left< \phi_x\phi_y\frac{1}{3!}{\left(\frac{-\lambda}{4!}\right)}^3\int dz \int dw \int dv ...


2

Your result only holds in the $U/t\gg 1$ limit. A brute-force method to obtain the result is by exact diagonalization in the Fock space basis, where all the operators are represented as matrices. Let us introduce the Fock states $|n_{1\uparrow}n_{1\downarrow}n_{2\uparrow}n_{2\downarrow}\rangle$ where $n_{i\sigma}=0,1$ denotes the fermion number on site $i$ ...


2

Contrary to your previous question Problem understanding the symmetry factor in a feynman diagram the roles of the three vertices are not different so you have from the expansion of the exponential a $1/3!$ but it does not get compensated by the choice of role assignement. Here this choice amounts to deciding who connects directly to $x$ and $y$ and that's ...


2

Observe the potential lines for a moment. You will find that for equal change in distance, there is equal change in potential. Means, if I move 0.5 m to the left, the potential increase is 10 V. In other words, we have equidistant equipotential lines which is a graphical way of denoting uniform field. Whenever you see straight equipotential lines, it means ...


2

We choose one of the $4$ z-fields to contract with the single x-field. We then choose one of the remaining $3$ z-fields to contract with one of the $4$ w-fields. The remaining two z-fields just contract with themselves. Now choose one of the remaining $3$ w-fields to contract with the single y-field. (Here is where we have to be careful). There are $2$ ...


2

Assuming $\Delta$ is the Laplacian operator, i.e. $-\Delta=D_x^2$, where $D_x=-i\partial_x$, this goes as follows (but the result is different from the one you give). Choose a suitable dense domain of $L^2$ where $x$ and $-\Delta$ are well defined, e.g. the functions that are $C_0^\infty$. Let $\psi\in C_0^\infty$, then $$e^{itD_x^2}xe^{-itD_x^2}\psi=x\psi ...


2

I looked up leap second in Wikipedia. It is a second added (usually) to clocks to keep them in sync with the atomic clock. Civilian clocks use Coordinated Universal Time (UTC), sometimes erroneously called Greenwich Mean Time (which no longer exists). Atomic clocks use International Atomic Time (TAI). UTC and TAI are in sync. Civilian clocks tick at the ...


2

$g^{\mu\nu}$ is the inverse of the metric $g_{\mu\nu}$; and the expression is for a general metric, not for the Minkowski metric. In 4 dimensions, if $g = \det(g_{\mu\nu})$ $$\det(\sqrt{-g} g^{\mu\nu}) = (\sqrt{-g})^4\det(g^{\mu\nu}) = \frac{g^2}{\det(g_{\mu\nu})} = g$$


2

$$\vec{X}_{cm} = \frac{1}{M} \sum_i m_i \vec{x}_i$$ where $m_i$ is your mass of particle $i$, $\vec{x}_i$ is the position of particle i, and $M$ is your total mass.


2

The first one is basically derived from the velocity-time graph. As the acceleration is constant the area under the graph represents displacement (s). Thus by the formula to get the area under a trapezium. $$\text{Area} =\frac{1}{2} ( \text{sum of the parallel sides}) \times ( \text{vertical height})$$. We have $$S = \frac{1}{2} ( v + u)\cdot t$$.


2

First We'll see the FBD (Free Body Diagram)and we get: Where Fc is normal force acting (Force of contact). From FBD of 5 Kg block (By newton's 2nd law) $$F-F_c = ma$$ (1) From FBD of 10 Kg block $$F_c=Ma$$ Solving above equations we will get: $$F=ma+Ma$$ $$a=F/(m+M)$$ Putting the values you may get your result and your resultant force.


2

As per @lemon mentioned : You can treat the sphere as a single point (this follows from point 1 of the shell theorem). So you just need to integrate along the length of the ring. So I just did like that only. Taking sphere as point mass then point mass will experiences one gravitational force of attraction that is: $$df=cos\theta$$ As vertical ...


1

Since the eigenvalues are not degenerate, the correction to the energy level $E_n$ is just $\langle n | \gamma B | n \rangle $. It's easy to see that the correction $\delta E_n$ is $\gamma \hbar$ for all $n$. The correction is good if $$ \frac{\delta E_n}{E_n}\ll 1 $$ That's $$ \frac{\gamma}{\omega(n+\frac{1}{2})}\ll 1 $$ For all $n$. And this is guaranteed ...


1

You are really asking about the reaction forces felt on a rod when you push on its end. For a rod, you can work with a quantity called the reduced mass (see for example this excellent answer for the derivation). As long as the rod is balanced on its end, the reduced mass tells you exactly how much greater the inertia of the cart appears to be: $$m_r = ...


1

I'm a little new here, so I can't comment yet. However, I do think you're supposed to indicate what you've already tried, so please try to give this problem you're best shot before looking to my answer below and next time, give some indication that you've put some effort into the problem. ...


1

You are making this rather hard for yourself. You correctly solved for the velocity, which is of the form $$v(t) = c_1 e^{-\alpha t} - \frac{g}{a}$$ where $a = mk$ and $c_1$ is found from the initial conditions. Integrating this expression should just give you $$x(t) = -\frac{c_1}{a} e^{-at} - \frac{gt}{a}$$ I think that because you ended up splitting ...


1

Consider the total energy of the particle $$ E=\frac{mv^2}{2}+mgh $$ Then (assuming $k>0$): $$ \dot{E}=mv\dot{v}+mgv=mv[-g-mkv+g]=-m^2kv<0 $$ So when the particle is thrown up and returns to a given height it has less energy than when it was first there. Since the potential energies are the same the speed has fallen. That is it comes down slower than ...


1

Hint: The center of mass of pieces A and B moves in the same path as the intact shell would. (This arises from the conservation of momentum.) Edit: The gravitational force is only acting along the vertical direction. So there is net acceleration only in the vertical direction. Looking at the horizontal one, if we neglect things like air resistance (which ...


1

Rewrite the Shouten identity by pulling down $\nu$: $$ 0 = \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda} + \delta_\mu^\rho \eta_{\chi\nu} \epsilon^{\sigma\tau\lambda\chi} + \cdots$$ Then plug this into the left hand side of the equation you marked with (?). You will get four terms, all having the correct structure of the right hand side (one free index at ...


1

Suppose you have a constant acceleration $a$. Acceleration is the derivative of velocity with respect to time, so we have: $$ \frac{dv}{dt} = a $$ and if we integrate this we get: $$ v = at + C $$ where $C$ is the constant of integration. To find $C$ we note that when $t = 0$ the velocity is equal to $u$ (the initial velocity) so that means $C = u$ and ...


1

So there is no way I can relate this result to Sin(kx) of equation (3.43) with the spatial part $e^{i\vec k \cdot \vec x}$. Where is the mistake on my calculations? Underneath the $d^3k$ integral you can rewrite $$ e^{i\vec k\cdot \vec x}sin(\omega_k t) $$ as $$ \sin(k\cdot x) $$ with the understanding that $k_0=\omega_k$. This is true because you ...


1

As CuriousOne inexplicably said in the comments, but not as a formal answer, you should use this equation: $$\vec{L}=m{\vec{r}}\times{\vec{v}}$$ This is the standard equation for angular momentum in vector form. Once you have your angular momentum vector, you can get the individual components. You can see how to take a cross product here. If you need to ...


1

You have a compound pendulum so you need to consider its moment of inertia, by using the parallel axis theorem on its parts


1

Since you know that the field inside a conductor is zero, you can apply Gauss' Law for flux to say that any spherical surface lying inside the conductor cannot have any flux through it, so that the enclosed charge is zero. Therefore, you need -2.3mC of charge from the conductor on the inner surface to make sure that the enclosed charge is zero. For the ...


1

The two questions are slightly different. Each individual measurement of $L^2$ or $L_z$ will return an eigenvalue. In this case, you have only one possible measurement for $L^2$ (corresponding to $l=1$), but you have two possible measurements for $L_z$; 2/3 of the time you'll get $m=1$, and 1/3 of the time you'll get $m=0$. The expectation value, on the ...


1

Look at a free body diagram. With red and the contact normal forces, with pink the friction forces, and with gray the gravity forces. If in the end any of the friction forces come up to being negative, then flip the orientation.



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