# Tag Info

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In relativity we deal with spacetime which is a 4-dimensional structure called a (pseudo-Riemannian) manifold. It includes space and time together. Meaning that time is just another direction in spacetime, perpendicular to the three dimensions of space. In the manifold we have a metric which describes all of the geometric properties of spacetime. It ...

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Yes, since force is a vector quantity you can have a negative force - simply meaning a force acting in the opposite direction to whatever you have defined as positive.

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The Lorentz transformation where we consider motion in the x-direction only, is given by the equations $$\tag 1 x’ = \gamma (x - vt)$$ $$\tag 2 t’ = \gamma (t -\frac{vx}{c^2} )$$ $$y’=y$$ $$z’=z$$ where $$\tag 3 \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ The question is, can these equations $$x' = 1.25x - 0.75t$$ $$t'=1.25t -0.75x$$ $$y'= y$$ $$z' = z$$ ...

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You should not try to think of the $\delta$ as a "usual" function when saying that (2) is undefined at $r \rightarrow 0$, because, in fact, $\delta$ is not a function. It is a distribution. You can view them as kinds of weights to integrate test functions, as in the following formula: $$f(x) = \int\delta(y-x)f(y)dy$$ From this definition, two ...

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In special relativity one traditionally imagines that spacetime is filled with a network of clocks that are Einstein synchronized. The clocks define an inertial reference frame, and the time coordinate of an event is the reading on the nearest clock. When dealing with curved spacetime you usually can't Einstein synchronize clocks, and it's usually not ...

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It is indeed correct that the radial acceleration, $a_r = 0.$ Your mistake is in thinking that $\dot{r}$ is the only contributing factor to $a_r$, whereas from the very formula you have cited for $a_r$, it is not: there is a $-r\dot{\theta}^2$ term as well. In other words, looking at $\dot{r}$ is not sufficient to infer $a_r$. Use the correct formula for the ...

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In an equation there is no inherent distinction between dependent and independent variables. There are to my knowledge only two contexts where the distinction makes sense. Experimental: In an experimental context the independent variable is the one that the experimenter is controlling in the experiment. It is the treatment. For example, if the experimenter ...

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You are using a too implicit formalisation where the physical content is not evident. In classical physics the spacetime is made of the following structures (a) a smooth four-dimensional manifold $M$, the spacetime of classical physics, (b) a surjective non-singular smooth map $T : M \to \mathbb{R}$ called absolute time. $T$ is defined up to additive ...

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No it doesn't. The spatial coordinates here are $x$ and $y$, which appear in the arguments of the sine and cosine functions. It seems strange to me that you are this far into a course in quantum mechanics and have not yet learned that the wavefunction is the spatial probability amplitude. To obtain the spatial probability density, you take the square ...

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In an axisymmetric system, the gravitational force on a particle is going to be radial. It therefore exerts zero torque about the origin, so angular momentum about the origin is conserved: $$L = mr^2\dot{\varphi} = \text{const.}$$ Now differentiate both sides with respect to time to get $$2mr\dot{r}\dot{\varphi} + mr^2\ddot{\varphi} = 0,$$ which upon ...

3

A geodesic is a curve $\gamma$ which extremizes the path length functional $$S[\gamma]= \int d\lambda \ L(\gamma,\dot \gamma) =\int d\lambda \sqrt{g_{\mu\nu}(\gamma)\dot\gamma^\mu\dot\gamma^\nu}$$ The Euler-Lagrange equations are $$\frac{\partial L}{\partial \gamma^\alpha} = \frac{d}{d\lambda}\frac{\partial L}{\partial \dot\gamma^\alpha}$$ $$\implies (\... 3 How does one determine the independent and dependent variables? It's totally relative. In F = k\Delta l all three variables can be considered dependent or independent, depending on your purpose. E.g. in \Delta l=\frac{F}{k}, \Delta l would now be considered the dependent variable. Now suppose you studied a set of different springs, so that:$$k=\frac{...

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The claim that linear algebra is a gauge theory doesn't even make sense to me. It's like saying that calculus is a Lagrangian theory. However, one can frame a general gauge theory as a manifestation of basis independence for each fiber of a vector bundle associated to some principal bundle or another. In that sense, the claim is backward, if anything. I ...

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Mathematically Space-Time is a 4-dimensional Lorentzian manifold $L$. Coordinates of an event, i.e. $(x_{\mu})=(ct, x, y, z)$ only make sense when you define an Atlas $\mathcal{A}={(\mathcal{U}_{\alpha}, \phi_{\alpha})}$ on $L$ where $\mathcal{U}_{\alpha}\subset L$ and $\phi_{\alpha}:L\rightarrow\mathbb{R}^4$ is a homeomorphism (i.e. continus functions ...

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By distributive property: $$\vec{A} \times \vec{B} = \\ = \left( A_x\hat{i}+A_y\hat{j}+A_z\hat{k} \right) \times \left( B_x\hat{i}+B_y\hat{j}+B_z\hat{k} \right) = \\ = A_xB_x \hat{i} \times \hat{i} + A_xB_y \hat{i} \times \hat{j} + A_xB_z \hat{i} \times \hat{k} + \\ + A_yB_x \hat{j} \times \hat{i} + A_yB_y \hat{j} \times \hat{j} + A_yB_z \... 2 No. Linear Algebra (LA) is not a gauge theory. LA is a field of mathematics, whereas a gauge theory is a field in physics. In a large sense one can say that gauge theories contain the concept of {{ local}} gauge transformations. However, in Linear Algebra the transformations are all global, they cover the whole space. 2 I haven't thought about canonical transformations in a while so I thought it would be fun to give some hopefully constructive comments on this question. Note that we can use simple physical reasoning to determine the relationship between the new and old coordinates (p,q)\leftrightarrow(P,Q). We are told that the new Hamiltonian H' is only a function of ... 2 d\tau is an invariant of the metric. dt is the time measured by a clock in a particular frame, as dx is the length measured in this frame. It is like saying that in 2D spatial cartesian coordinates, ds^2 = dx^2 + dy^2. ds here is invariant by rotation. And we can also say that for any orientation of ds, there is a particular frame so that ds^2 = ... 2 You're getting tripped up in two different ways here. (1) When you reason about the sizes of the terms in the K-S metric, you're ignoring the fact that the expression you're using is written using a mixture of variables from the two coordinate systems. This point is easier to understand with a simpler example. Suppose that ds^2=e^{-r}dX^2, where r=\ln X. ... 2 Let M be a smooth manifold and (P,\pi,M,G) a principal G-bundle over M. Let \omega be a connection one-form and let \sigma : U\subset M\to \pi^{-1}(U) be a local section which locally trivializes the bundle by means of h:U\times G\to \pi^{-1}(U):$$h(x,g)=\sigma(x)\cdot g.\tag{1}$$With h we can pullback \omega to understand how it is ... 2 The object will only contract by a factor of$$V'=\frac{V}{\gamma}=V\sqrt{1-\frac{v^2}{c^2}}$$where V' is the measurement* made from our "stationary frame" (from which the object's frame is moving at relative velocity v) and V is the measurement made from the object's rest frame. * While V usually indicated Volume, the same relationship ... 2 The Lorentz contraction is a statement about how a given object, which should be regarded as a region of spacetime (not just space since the object continues to exist over time), will be determined to have different spatial dimensions depending on which inertial reference frame is being adopted in order to specify how spacetime is to be divided into "... 1 This is a case where you'd actually need limits. Consider$$ f(x):= \begin{cases} (x-1)^2,\quad x\neq 0 \\ 1,\quad x=0\\ \end{cases} $$Notice how f(x) is identically equal to g(x)=(x-1)^2. However, if you simply differentiate each piecewise component, you run end up with f'(0)=0, whereas g'(0)=-2. That's a clear contraction. However, if you were to ... 1 I don't think there are any r\to\infty asymptotic timelike observers. There seems to be a limit of that sort in some coordinates, like Eddington-Finkelstein (u,r), but it's an illusion. Conformal spatial infinity is a single point with no time dimension, and its causal past only contains past infinity. The r\to\infty limit of stationary worldlines on ... 1 You are correct in expecting to need two coordinates to specify position in the orbital plane. Distance from the central body supplies one of them. The other is simply the angle \theta. The two numbers give you the position vector in the polar coordinates (r,\theta). In order to convert them to the Cartesian coordinates (x, y) you can use these ... 1 The general delta function in 3D curved space is$$ \frac{\delta^{3}({\bf x}-{\bf x}_c)}{\sqrt{ g }} $$where g is determinant of metric. For spherical coordinates, you should have \sqrt{g}=4\pi r^2, if you have spherical symmetry. Thus you should get$$ \frac{\delta( r- r_c)}{4\pi r^2} $$Edit: Delt function in spherical coordinates. Since \sqrt{g}=r^2\... 1$$\vec{y'}=\left[ \begin {array}{c} \sin \left( \theta \right) \sin \left( \phi \right) \\\cos \left( \phi \right) \\ \cos \left( \theta \right) \sin \left( \phi \right) \end {array} \right] $$\Rightarrow$$\tan(\Phi)=\frac{\sin(\Phi)}{\cos(\Phi)}=\frac{\vec{y'}_x}{\vec{y'}_y}=\frac{\sin(\theta)\,\sin(\phi)}{\cos(\phi)}$$thus:$$\sin(\Phi)=\sin(\...

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Your second law is applied in an inertial reference frame using polar coordinates. Although the unit vectors (radial and tangential) in polar coordinates have constant magnitude (one), they change direction in time. If you used a non-inertial reference frame attached to (and moving with) $m$, in that frame the radial and tangential coordinates of $m$ are ...

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the action function in the HJ formalism is dependent on time through the upper bound. $$S(q_i,t,\alpha_i,t_0)=\int_{t_0}^tL(q_i,\partial_tq_i,t')dt'$$ in cases of Lagrangian that is independent (explicitly) of time then the constant of integration conjugate to $t$ let's call it $\alpha_t$ is simply the Energy. read more here: https://en.wikipedia.org/wiki/...

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A total derivative of a function $f(x_1,x_2,\cdots ,x_n)$ is defined as $$df\equiv \sum_{i=0}^n\left(\frac{\partial f}{\partial x_i}\right)dx_i.$$ In your situation, the total differential of a function $r_i(q_1,\cdots ,q_n,t)$ is given by $$dr_i= \sum_{j=0}^n\left(\frac{\partial r_i}{\partial q_j}\right)dq_j+\frac{\partial r_i}{\partial t}dt,$$ or \boxed{...

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