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The $i \epsilon$ part is introduced in order to define in what direction you go around the pole at $k^2 = m^2$. The choice of $+ i \epsilon$ corresponds to going above the pole at $k_0 = -\sqrt{\mathbf k^2 + m^2}$ and below the pole at $k_0 = +\sqrt{\mathbf k^2 + m^2}$ in the complex plane (where $\mathbf k$ is the trhee-vector part of $k_\mu$).

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Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough $$E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}$$ For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...

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I thought gravity is uniform acceleration, not increasing acceleration.. Position: $y(t) = \frac{1}{2} g t^2$ Velocity: $y'(t) = gt$; Acceleration: $y''(t) = g$;

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Differentiate $\dfrac{dx}{dt}= \alpha \sqrt{x}$ with respect to time again to get: $$\dfrac{d^2x}{dt^2} = \dfrac{\alpha}{2 \sqrt{x}} \dfrac{dx}{dt} = \dfrac{\alpha}{2 \sqrt{x}} \alpha \sqrt{x} = \dfrac{\alpha ^2}{2}$$ Then: $$\dfrac{dx}{dt} = \dfrac{\alpha ^2 t}{2}$$ and: $$x = \dfrac{(\alpha t)^2}{4}$$ for the given initial conditions.

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Well integration is the basic method. But some observation can help too: $$v=\alpha\sqrt x$$ $$\text{Squaring both sides:}$$ $$v^2=\alpha^2 x$$ We know $v^2\propto x$ gives constant acceleration. $\text{Remember} :v^2=u^2+2as.$ So, comparing it with this equation we get $$v^2=0+2\frac{\alpha^2} 2 x$$ So, acceleration =$\alpha^2/2$ and velocity ...

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$\frac{dx}{dt}=\alpha \sqrt{x}$ $\frac {dx} {\sqrt {x}} =\alpha dt$ Integrate to get $x(t)$ as $t=0 \implies x=0$ . Put this in the integral limits . , then differentiate $x(t)$ w.r.t. $t$(time) to get $v(t)$ and $a(t)$ .

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The result you've got would be better known as this: $$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$ In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn. The ...

2

If $h$ is the height about the earth then $$\ddot{h} = -\frac{G M}{(R+h)^2}$$ $$\ddot{h} = \frac{{\rm d} \dot{h}}{{\rm d}t}= \frac{{\rm d} \dot{h}}{{\rm d}h} \frac{{\rm d} h}{{\rm d}t} = \frac{{\rm d} \dot{h}}{{\rm d}h} \dot{h}$$ $$\int \ddot{h}\; {\rm d} h = \int \dot{h}\; {\rm d} \dot{h} = \frac{1}{2} \dot{h}^2 + K$$ $$\int -\frac{G M}{(R+h)^2}\; ... 4 So you were on the right track with integrating over r and over t. Here's how you could do it: The acceleration at any radius, r (if we assume Earth is a point mass) is:$$a=-{GM\over r^2}$$The minus sign is because the acceleration is anti-radial. Then you can do the following:$$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta vthus$$... 0 A little trick required here. Perform a substitution first: $$mv \gamma(v) = u ,$$ hence your integral becomes $$\int \frac{du}{dt}dx ,$$ but notice that $$\frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt}dx ,$$ but \frac{dx}{dt} = v - the definition of velocity! So the integral ... 1 I simply misfactorised the quadratic - I knew it was a stupid mistake. I'm amazed that I didn't see it, but even more amazed that nobody else did! Here is the correct solution.$$\phi_{01}(t,x,y,z) = \frac{1}{2\pi i}\oint d\xi \frac{\xi}{(x^{01'})^2(\xi -\xi_1)^2(\xi - \xi_2)^2}The residue at \xi_1 is \begin{align*} r_1 &= ... 3 Your first mistake is E_x = F_{01} = \phi_{01} You have apparently confused spinor and vector indices here. The identity $E_x=F_{01}$ holds if $0,1$ are interpreted as the vector indices with four possible values corresponding to $0123=txyz$. But then you can't write that it's equal to $\phi_{01}$ because the latter has the spinor indices with two ...

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