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This may be the argument: you have $N$ particles, and for each one you can put it on the left side or on the right side. Each of these choices, for each particle, leads to a different microstate. There are $2^N$ possible choices you can make for how to distribute the $N$ particles between the left and right halves of the box (assuming the particles are ...

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The ratio between avaible microstates is given by the following ratio: $$\frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$ where $N$ is the number of ...

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If e.g. we consider a 1D non-relativistic free particle with kinetic energy $$\tag{1} T(q,v,t)~=~\frac{1}{2}mv^2$$ the information that the velocity is a constant$^1$ $$\tag{2} v~\approx~\text{constant},$$ only came afterwards from solving the Lagrange eqs. $$\tag{3} \left(\frac{d}{dt}\frac{\partial T}{\partial v}-\frac{\partial T}{\partial ... 1 Yes. The action functional S[q]=\int \mathcal{L} dt is something that can be applied to ANY [differentiable] path. So \mathcal{L} is to be understood as an object which can really take n any value of q and \dot{q} independently. This makes it so that \mathcal{L} really is a function of two variables, and so that it really does have to be defined ... 5 As pointed out by lemon, two angles are enough to specify a direction in a three dimensional coordinate system, but another is needed to specify a complete coordinate transformation. You can think of a rotation transformation in three dimensions as a mapping between two different coordinate systems. Two angles are needed to specify the relative pointing ... 0 I really recommend 4 books in 3 steps 1)Mathematics for physics, Michael Stone Paul Goldbart 2)Modern Mathematical Physics, Peter Szekeres 3)Geometry for Physics, T. Frankel with An introduction to Manifolds, Loring W. Tu After all, The Road to Reality, Roger Penrose 1 Your question is "Is there an infinite series of higher derivatives of position for this to work?" Answer: No. Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one. From the question: "A change is velocity is acceleration, so the value of the ... 2 The definition suggested by joshphysics and clarified by Qmechanic already exists in the literature under then name of representation operator. This is discussed in, e.g., Sternberg's Group Theory and Physics, as well as the somewhat more elementary text An Introduction to Tensors and Group Theory for Physicists by Jeevanjee. 1 Let \phi:\mathbb{R}^3\times\mathbb{R}\rightarrow\mathbb{R}^3\times\mathbb{R} be a Galilean transformation. Let R\in\mathrm{O}(3), \tau\in\mathbb{R} and \mathbf{v},\mathbf{y}\in\mathbb{R}^3. The goal of this exercise is to show that$$\phi: \begin{pmatrix} \mathbf{x} \\ t \end{pmatrix}\mapsto \begin{pmatrix} R & \mathbf{v} \\ 0 & 1 ...

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you can either evalute the integral numerically or search for a good coordinate transformation to evaluate it "by hand". This transformation is the one to spherical coordinates, though. Also you could try this: \begin{align} Q & = 2 \pi A \int_{-\infty}^{\infty}\int_0^{\infty} r \cdot H(R^2-r^2-z^2)~dr dz \\ & = 2 \pi A ...

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