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The general and easy rule to remember the sign convention for convex and concave mirrors and lenses or any optical component is thinking in terms of power. The power of an optical element is calculated by 1/focal length and the unit is dioptre (1/m). It is negative for components which diverge parallel incident rays of light and positive for components ...

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This plano concave lens has the same object and image positions as your silvered plano convex lens. Hence the same focal length. The sign convention comes from the sign convention of the u and f distances.

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There is no $1/2$ because such factors arise whenever there are products of the same field. For example, you probably have seen the interaction term in $\phi^4$ theory written as $\frac{\lambda}{4!}\phi^4$. This is because when taking functional derivatives to get the Feynman rules, these combinatoric factors will arise from the derivative hitting any of the ...

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This is mostly because we're usually more interested in the spatial part of a plane wave than in the temporal part, so that plane waves are most convenient when written as $$e^{i(\mathbf k\cdot\mathbf r-\omega t)}. \tag 1$$ The normalization follows from this choice. In general terms, it's hard to call which factor has more weight. There are plenty of ...

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A real scalar field has one degree of freedom...Here we have two degrees of freedom (two real scalar field) and we treat $\phi$ and $\phi^{*}$ as independent field. When we put the lagrangian in Euler-Lagrange's equation we generate a factor of 2 and 1/2 cancel it,e.g. $m^{2}\partial_{\phi}(\phi^{2})=2m^{2}\phi$..and similar for the derivative term... But ...

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There's a problem in this equation: $$d\vec{l}=dy \cdot -\vec{j}$$ here $dy$ needs a minus sign. It is easier to see this writing the path. The path of integration is parameterized by: $$\vec{l}=\vec{P}+(\vec{N}-\vec{P})\frac{y-y_P}{y_N-y_P} = \vec{P}+\hat{j}(y-7a) \\ y_p \le y \le y_N$$ Therefore $$d\vec{l}=\hat{j}dy$$ and $$\int_P^N E_y \cdot ... 0 The right hand rule is simply a mathematical convention. We use right handed coordinate systems. We could make the left hand rule true if we wanted to, but we would need to adapt our equations to the new convention. 1 Your copy of Verma has already defined gravitational potential energy previously in (11.3) The gravitational potential energy of a two particle system is$$U(r) = -\frac{Gm_1m_2}{r} \tag{11.3}$$where m_1 and m_2 are the masses of the particles, r is the separation between the particles and the potential energy is chosen to be zero when the ... 1 Your teacher is wrong. The gravitational potential V(x) is generally defined as potential energy per unit mass i.e. V(x) \equiv \dfrac{U(x)}{m}. So for the points where U(x) is zero, V(x) is zero and vice-versa by definition. EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a ... 0 You want to use$$ \hat x= i\hbar\frac{\partial}{\partial p} $$in the momentum basis. This means that$$ <p|\hat x|\psi>= i\hbar\frac{\partial}{\partial p} <p|\psi> $$Thus, by hermiticity of \hat x, we evaluate$$ <x|\hat x|p> = (<p|\hat x|x>)^*  =(i\hbar\frac{\partial}{\partial p} <p|x>)^*  ...

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Note that the sign of the expression $$\frac12 \;k \; (x_f^2 - x_i^2)$$ depends on your choice of coordinate system. If you put $x_f=0$, then the result is negative; if you put $x_i=0$ then the result is positive. But the work done should be the same, regardless of the choice of coordinates. So what is happening? Simply, when we say $F = -kx$ we have ...

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Prahar has already given a good answer. Here we will instead focus on the pertinent Lie group (as opposed to the Lie algebra and its generators). The Lie group $SL(2,\mathbb{C})$ is the double cover of the restricted Lorentz group $SO^+(3,1)$, cf. e.g. this Phys.SE post. The fundamental/defining representation $V\cong\mathbb{C}^2$ of the Lie group ...

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There's something wrong with your sign permutations in the Hodge star operator calculation. If $F = B + E \wedge dt$, then, in 2D, $F = B dx \wedge dy + E_x dx \wedge dt + E_y dy \wedge dt$, as you wrote yourself. Now, let us take our initial Hodge star as $\star dx \wedge dy = dt$. This means that $\star dt \wedge dx = dy$ and $\star dy \wedge dt = dx$, ...

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This question arises because of a subtlety involved in the choice of variables and infinitesimals we use in the definite integration to find the work done by the force. If the block is at the coordinate $x$ and is moving towards the origin then the work done by the spring on it during the small interval of time in which the block travels a distance of $dl$ ...

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Notice, we know $$x_f<x_i$$ $$\iff x_f^2<x_i^2\ \ \ \ \ ( \forall \ \ \ x_i, \ x_f\ge0)$$ $$\iff x_f^2-x_i^2<0$$$$\iff k(x_f^2-x_i^2)<0 \ \ \ \ \ ( \forall \ \ \ k>0)$$ Now, the work done by the spring on the block is $$\int_{x_i}^{x_f}kx\ dx=\frac{1}{2}k(x_f^2-x_i^2)<0$$ Thus, the work done by the spring on the block is negative. It ...

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