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The point is there's several $W$'s. $$\rho_i=\sum_j W(\mathbf{p}_i-\mathbf{p}_j,h)=W(\mathbf{p}_i-\mathbf{p}_1,h)+W(\mathbf{p}_i-\mathbf{p}_2,h)+\ldots$$ dropping the mass as they do in the paper. The thing is they're treating the particles as indistinguishable e.g. they all have equal mass, so the form of the functions/constraints applied to them are the ...
This theorem can be used to prove Archimede's Principle in a region with a non-uniform gravitational field. The weight of the displaced fluid is $$\vec W=\int_\Omega \rho \vec g(\vec r)~\mathrm d\Omega.$$ Let us consider a body fully immersed. Then the buoyancy force is given by $$\vec B=-\oint_\Gamma p(\vec r)~\mathrm d\vec \Gamma =-\int_\Omega\vec\nabla p~... 2 If you very slowly increase the volume of an isolated system, then the internal energy will drop. The ratio between the drop in energy and the volume increase is, by definition, the pressure. The system can be in one of many possible energy levels, these energy levels decrease if we increase the volume (the larger the volume of a system the more closely ... 1 Why your computed average acceleration is wrong? the average acceleration is defined as: \overline a=(v_2-v_1)/(t_2-t_1) where the v's are instantaneous speeds. If you start with zero initial speed you can simplify it to \overline a= v /t  v is still the instantaneous speed at t. For a constant acceleration v=at so in this case you ... 1 The glitch in your logic is that you supposed acceleration to be distance/time squared while using your formula:$$ \Delta x = v_0t+\frac{1}{2}at^2 $$The above formula gives$$ a= \frac{2\Delta x}{t^2}$$supposing that vo is equal to zero. This makes the acceleration equal to the average acceleration in your case since the car is moving in one direction ... 1 It's just like a directional derivative in standard Cartesian coordinates. In \mathbb{R}^3, if I have a function f : \mathbb{R}^3 \rightarrow \mathbb{R}, the gradient \nabla gives me a map \nabla f : \mathbb{R}^3 \rightarrow \mathbb{R}^3. For any vector \vec{v} \in \mathbb{R}^3, I can define a map \vec{v} \cdot \nabla f : \mathbb{R}^3 \rightarrow ... 2 Since your question is how to gain an intuitive understanding of what the two equations \nabla \cdot \nabla \mathbf{F} = 0 and \nabla \times \nabla f=\mathbf{0} mean, I will discuss the equations in a simple setting. The simpler, more intuitive setting I am thinking of is a discrete setting: the honeycomb lattice, shown below. \hskip 2 in Let me ... 3 Let us begin with$$\mathrm{div} \left[\mathrm{curl} \mathbf{A}(\mathbf{x})\right]=0 Consider the electrostatic field, it has sources to emerge from (positive charges) and sinks to go into (negative charges). Such a field has a non-zero divergence. When we say that the divergence of $\mathrm{curl} \mathbf{A}(\mathbf{x})$ is equal to zero, this means ...