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1

The two body problem can be made equivalent to the one body problem. Say you have a mass $m_1$ at $\vec{r}_1$ and a mass $m_2$ at $\vec{r}_2$ interacting gravitationally. Now focus your attention on the center of mass $\vec{r}$ and the difference vector $\vec{r}_{12}=\vec{r}_2-\vec{r}_1$. Then the kinetic energy of the system can be written as ...


3

First, they do not transform in an actual "representation" in the sense of a linear representation of the group of coordinate transformation since their behaviour under a coordinate transformations $x\mapsto y(x)$ is given as $$ {\Gamma^\alpha}_{\beta\gamma} \overset{y(x)}\mapsto \frac{\partial x^\mu}{\partial y^\beta}\frac{\partial x^\nu}{\partial ...


4

The Christoffel symbols do not transform under any representation. The reason for this is that they do not transform linearly, which puts them out of the game altogether. The transformation law is $$ \tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde ...


1

The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ...


1

If there is no fulcrum, if there is no fixed pivot, a body will always translate. Such point does not exist. If the body is free, it can only translate but can never only rotate: if the impulse is at the Center of Mass linear velocity will be 100% , the minimum percentage of translational velocity is 25%, and it is reached at one tip of the rod. No matter ...


0

If the observer is not in free-fall, the metric-tensor $g_{\mu,\nu}(s)$ at the observer's position, expressed in local coordinates around the observer, will not be $\eta_{\mu,\nu}$. Your first assumption about the path $(\gamma)$ is wrong. I guess what you are aiming at is the notion of the space of coordinates around a point, which is indeed a flat space ...


3

Let $M$ be your spacetime, a smooth manifold equipped with (pseudo) Riemannian metric (for example $\mathbb{R}^{(1,3)}$ for special relativity). The set of reference frames is the frame bundle over $M$, usually denoted $FM$. Explicitly a frame at point $p$ in $M$ can be viewed as an ordered orthonormal basis (with respect to the the inner product defined ...


0

you can either evalute the integral numerically or search for a good coordinate transformation to evaluate it "by hand". This transformation is the one to spherical coordinates, though. Also you could try this: \begin{align} Q & = 2 \pi A \int_{-\infty}^{\infty}\int_0^{\infty} r \cdot H(R^2-r^2-z^2)~dr dz \\ & = 2 \pi A ...


2

I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ...


4

You were pretty close already. There is a handy table on Wikipedia for a variety of coordinate systems. But for the polar system: $$ \vec{\nabla} \cdot \vec{U} = \frac{\partial U_r}{\partial r} + \frac{1}{r} \frac{\partial U_\theta}{\partial \theta} $$ and you can look up the curl in the same table. These can be derived from the Cartesian definitions by ...


1

The columns of a 3×3 rotation matrix contain the coordinates of the local xyz axes (expressed in world coordinates). With Euler angles (321) you apply the elementary rotations $R_Z$, $R_Y$, $R_X$ in sequence to form the local → world rotation matrix. That is $$ E = R_Z(\varphi) R_Y(\psi) R_X(\theta) $$ This is interpreted as each rotation occuring about ...


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Let's say you do the calculation $C_{final}=R_xR_yR_zC_{o}$, where the $C$s are coordinates. The first rotation is about the $z$ axis of $C_{o}$ and will produce a new coordinate system in which the new $z$ is the same as the old, but the $x$ and $y$ axes are different. The next rotation will be about the new $y$ axis and will produce a newer, new $x$ and a ...



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