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The equation of motion of the particle is $$m \ddot{x}(t) = F(t)$$ where $x(t)$ is the position and $F(t)$ is the force. In the situation you describe, ("suddenly I hit the particle"), the force as a function of time can be written as $F(t) \propto \delta (t)$, with $\delta$ the Dirac distribution. Integrating once, you obtain that $$\dot{x}(t) \propto ... 0 What's wrong with this is the phrase 'suddenly I hit the particle'. What you are assuming by this is that you have some hard-edged and rigid object which you crash into the particle, which is also hard-edged. But you don't have either of those things: what you have is something which is both not rigid and whose surface is actually a little bit of EM field ... 0 This diagram shows part of a surface cut up into small elements (shaded pink) of area dA. The normal to the area is the vector \vec {dA} and is of magnitude dA. What is required is the projection of the magnetic field \vec B onto the vector \vec{dA} which is B_\bot = B \cos \phi in the diagram. To do this the dot product is taken \vec B ... 0$$ \int \mathbf{B} \cdot \,d\mathbf{A} $$just means$$ \int B\,dA\,\cos\theta. $$d\mathbf{A} points in the direction perpendicular to the surface, so if this is in the xy plane, it'll be in \mathbf{\hat{z}}. I.e. your usual integration over the area ("integrand") weighted by a factor of \cos\theta. \theta is the angle with the z axis. Keeping ... 0 The integration variable is the surface vector: flux is an oriented thing. It is the "normal" integration of the scalar product of the field and the surface vector:$$\Phi=\iint_S \vec{B}\cdot\vec{\mathrm dS}$$Or, if \vec{n} is the normal to the surface at the point of integration:$$\Phi=\iint_S \vec{B}\cdot\vec{n}\ \mathrm dS$$0 Your derivation is almost correct, and as I'll show below, one does sometimes make use of it. However, it relies on an unwritten assumption that makes it applicable only in a limited range of situations. That assumption is that the heat capacity c doesn't depend on the temperature. It is this assumption that allowed you to take the factor of mc outside ... 0 Your equation is correct only if:$$\mathrm dQ = mc\,\mathrm dT$$which is not generally true, indeed, common sense tells you that a change in temperature leads to conclusion that an object being heated up. But we do not encounter gases much in our life, which could be regarded as a general case. In reality your assumption is generally false, a good example ... -1 Here is a link that might be useful http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/temper2.html#c2 0 This is just complementary to Alex's answer. For the second integral the book provides an analysis in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral$$ \frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}} $$At the limit r\rightarrow \infty, the effect ... 0 A differential like dx in an equation like$$dz = dx + dy + dx^2 + dx\,dy is shorthand for a quantity that will eventually go to zero in a limit. If a differential is divided by another differential, then it has a chance of resulting in a finite, non-zero quantity. Let's assume $x$, $y$, and $z$ can be parameterized by a variable $t$. If we divide the ...