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2

$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ is wrong. The delta function is spherically symmetric, and thus has no θ dependence. Simply use: $$d^3(q⃗ )=\frac{δ(q)}{2\pi q^2}$$ instead. Use the Jacobian when you switch coordinate systems (from Cartesian to spherical) ($r^2 \sin(\theta)$), and you should get the result.


4

Hints: In mathematics, a distribution is usually only defined wrt. smooth testfunctions. However the function ${\bf q}\mapsto({\bf q}\cdot{\bf p})^2/q^2$ is not continuous at the origin ${\bf q}={\bf 0}$. Nevertheless, we can e.g. try to evaluate the triple integral using the following representation of the 3D Dirac delta distribution $$\tag{1} ...


1

First go to spherical coordinates: $$ A(t) = \frac{1}{2\pi^3}\int_0^{\infty}\text{d} p\int_0^{2\pi}\text{d} \phi \int_0^{\pi}\text{d}\theta\text{ } p^2 \sin\theta e^{-it\sqrt{p^2+m^2}} e^{ip\cos\theta|r-r_0|}, $$ and perform the trivial integral over $\phi$. Subsequently substitute $y=\cos\theta$ such that $\text{d}y=\sin\theta\text{d}\theta$ and integrate ...


2

I suppose your title would rather be ; "... if not, why ?". In momentum space (which I assume is the focus of your question), these do not appear; the integrals are made of Feynman propagators $(p^2+m^2+i\epsilon)$ in the denominator and in the numerator you find various tensor contractions of your loop momentum (or momenta) with the external momenta / ...


1

We start with the integral $$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$ Since we are integrating over $\vec{r}$, we can without loss of generality, arrange for $\vec{r}'$to lie along the $+Z$ axis, so that $\vec{r}'=r'\hat{z}$. Then the angle between $\vec{r}$ and $\vec{r}'$ is the standard angle (in spherical coordinates) ...


0

To solve the integral, I came across the multipole expansion $$\frac{1}{\left|\vec{r}-\vec{r}_{0}\right|}=\sum_{l=0}^{\infty}\frac{r_{<}^{l}}{r_{>}^{l+1}}P_{l}\left(\cos\sphericalangle\left(\vec{r},\vec{r}_{0}\right)\right),\;\; r_{<}=\min\left(r,r_{0}\right),\; r_{>}=\max\left(r,r_{0}\right)$$ With $\vec{r}_0=\vec{r}^{\prime}$ one has ...


1

Let me simplify the problem: assume a container of unit LENGTH along the X axis, assume that we have only 3 molecules instead of 5, and that the molecules move only along the X axis, i.e. they don't have a transversal velocity. Now, what you need to find (in my understanding) is the probability to find the molecules 1 and 2 separated by a distance R. ...


2

The first mistake is directly taking $dq=\lambda dy$. If you look closely, as you go up from the origin to higher y, you find that charge density per unit y increases. Take instead $\lambda$ as charge per unit circumference of the wire. i.e $\lambda={dq \over dl}$ = $\lambda={dq \over rd\theta}$ Now, $$dE= {k\lambda dl \over r^2} \cos{\theta}$$ ...


0

$ \newcommand{\r}{\mathbf{r}} \newcommand{\F}{\mathbf{F}} \newcommand{\g}{\mathbf{g}} \newcommand{\t}{\boldsymbol\tau} $Let me start by defining $\g(\r)$ to be a position dependent force per unit mass. Then the force per unit volume $d\F$ is given by $d\F(\r) = \g(\r) \rho (\r) dV$. The torque per unit volume $d\t$ is given by $d\t = \r \times d\F$. The ...


2

You don't need to introduce photon's fictive mass: dimensional regularization can be used for IR divergences as well as for UV divergences. First, use gamma-matrices identities, $$ \gamma_{\mu}\gamma^{\alpha}\gamma^{\mu} = (2 - d)\gamma^{\alpha} ,\quad [\gamma_{\mu}, \gamma_{\nu}]_{+} = 2g_{\mu \nu}, \quad g_{\mu}^{\mu} = d $$ and do simple that you've ...


2

The job of calculus is to handle quantities that vary over the domain of the problem at hand. Often, and particularly in introductory physics, we care about quantities that vary in time. We cannot put them into our equations as constants. We also often care about the interrelationship of these quantities. So for example now velocity is given by ...


1

How else can you recover displacement from a non-linear velocity without integrating? How else would you calculate acceleration of a non-linear velocity without differentiating? ADDED: If your question is how to solve $\displaystyle v(t) = \frac{dx}{dt} = b\sqrt{x}$, let us know.


0

So in your calculation, you have set $\epsilon$ to be zero. Since the result should be independent of the gauge you choose, why not calculate under $\epsilon=0$, which is equivalent to take the propagator as $-\frac{g^{\mu\nu}}{k^{2}}$? If you still want to calculate with a brute force, I think the problem may arise from the Feynman parameterization, where ...


2

I'll write my comments here as a full answer, as suggested by Floris. I won't use the moment of inertia tensor: it's simpler from pure angular momentum of each point particle. We know that $$\vec{L} = (\vec{r} \times \dot{\vec{r}})\,m .$$ So, for a point particle, $$d\vec{L} = (\vec{r} \times \dot{\vec{r}})\, dm .$$ Noting that $\rho = \frac{dm}{dV}$, ...


5

One may only talk about a discrete sum over $k^\mu$ vectors if all the spacetime directions are compact. In that case, $k^\mu$ is quantized. If the spacetime is a periodic box with periodicities $L_x,L_y,L_z,L_t$, then $V=L_x L_y L_z$ and $T=L_t$. The component $k^\mu$ in such a spacetime is a multiple of $2\pi \hbar / L_\mu$ (I added $\hbar$ to allow any ...



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