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First of all: you and the people in your course are most certainly not the only students encountering this problem. In my 2nd and 3rd university year I had it myself. The reason is that physicists use abusive notation. And they do it a lot. Mathematicians have less of a problem with these things. In this answer I will try to use intuitive terms and be ...


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Perhaps this will elucidate your first point: $\frac{dy}{dx}=\tan \alpha$ is in fact the gradient of $y=f(x)$ at $(x,y)$. Of course this also means: $dy= \tan \alpha dx= \frac{dy}{dx}dx$ Divide both sides by $dt$: $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ $v_y=\frac{dy}{dx}v_x$ Also: $v^2=v_x^2+v_y^2$


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We know that the Levi-Civita connection satisfies $\nabla_a g_{bc} = 0$ and the product rule. The definition of the inverse metric $g^{ab}$ is $g^{ab}g_{bc} = \delta^a_c$. Therefore, we have: $$\begin{align} 0 &= \nabla_a \delta^b_c \\ &= \nabla_a (g^{bd}g_{dc}) \\ &= (\nabla_a g^{bd}) g_{dc} + g^{bd} \nabla_a g_{dc} \\ &= (\nabla_a g^{bd}) ...


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Whenever you see something like $\ln(X)$ or $\exp(X)$ where $X$ is a dimensionful quantity, whoever wrote that expression was too lazy to carefully state that you are meant to non-dimensionalize $X$ by introducing a scale $X_0$ for it, and replacing $X$ by $X/X_0$. Now $X/X_0$ is dimensionless, and you can apply $\ln$ or $\exp$ or whatever other function to ...


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The equilibrium thermodynamic state of a single phase substance of constant composition is determined by specifying two intensive properties. Note that, from the equation of state, V and T uniquely determine P (where V is the specific volume).


1

In general you're right; if we know $V(0,0, z)$ we can only get its $z$-derivative. But the author is implicitly assuming the electric field is directed along the $z$ axis, because of the rotational symmetry of the problem. Therefore, the $z$ component of the gradient is all we care about.


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For any function of multiple variables, say, $f(x, y, z, \ldots)$ you can always write the following: $df = \big( \frac{\partial f}{\partial x} \big)_{y, z, \ldots} dx + \big( \frac{\partial f}{\partial y} \big)_{x, z, \ldots} dy + \big( \frac{\partial f}{\partial z} \big)_{x, y, \ldots} dz + \ldots$ In fact, the constancy of the other variables is ...


2

It implies that total energy is conserved along the paths specified by the trajectories, except for whatever explicit changes are written in the Hamiltonian. It is not sufficient to get the rest of the physics because there may be many energy-conserving paths. For example, for the free-particle Hamiltonian, the energy is the kinetic energy and the predicted ...


2

$$\frac{d\mathcal{H}}{dt}= \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}+\frac{\partial\mathcal{H}}{\partial t}$$ And physically this is related to how the Hamiltonian experienced, $\mathcal{H}(t)=\mathcal{H}(q_i(t),p_i(t),t),$ changes as you move around in phase space, compared to how the Hamiltonian ...


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For $\mathcal{H}(q,p,t)$ to have in any meaningful sense a partial derivative $\partial_t$ that is different from the total time derivative, you have to consider it applied to a curve $(q(t),p(t))$ in phase space. The equation $$ \frac{\mathrm{d}\mathcal{H}}{\mathrm{d}t} = \partial_t \mathcal{H}\tag{1}$$ then says that $\mathcal{H}$ is "constant" along the ...



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