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The kinetic energy of a particle whose motion is described by $\textbf{r}(t) = \left(x(t),\,y(t),\, z(t)\right)$ is, at the point $(x,y,z)\in\mathbb{R}^3$, defined as $$ T(x,y,z) = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) = \frac{1}{2}m\,\textbf{v}\cdot\textbf{v} $$ and yes, the square of a vector means (by abuse of notation) its scalar ...


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The velocity is always, by definition, the derivative of the position $\textbf{r}(t)$ with respect to the time and likewise for the acceleration $$ \textbf{v}(t)=\frac{d}{dt}\textbf{r}(t),\qquad \textbf{a}(t)=\frac{d}{dt}\textbf{v}(t)=\frac{d^2}{dt^2}\textbf{r}(t). $$ The position is, in turn, always a function of the time, although often not explicitly ...


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Acceleration is defined as the derivative of velocity with respect to $t$: $$a=\frac{dv}{dt}$$ It is the instantaneous change of velocity. Just like velocity is defined as the instantaneous change of position $r$: $$v=\frac{dr}{dt}$$ If you agree that: $$a=-\frac{GM}{r^2}$$ then it is a simple thing to exchange $a$ with its definition $dv/dt$.


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Because by definition acceleration is change in velocity over time. The fact that acceleration is a function of radius changes nothing. Basically what the equation says is that acceleration as a function of radius is equal to some formula. We then know from the definition of acceleration that it is change in velocity over time. If this helps clear it up: ...


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Here taking $|0\rangle$ and $|1\rangle$ as orthonormal basis for 2 dimensional hilbert space. Now $|00\rangle ,|01\rangle,|10\rangle,|11\rangle$ are orthogonal to each other ( take the inner product of any two it will be zero, eg. $\langle 00|01 \rangle= \langle0|0\rangle \langle 0|1\rangle =0 $ ). Thus any vector of a 4 dimensional dimensional hilbert space ...


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$\newcommand{\ket}[1]{\left| #1 \right>}$Note that you can write a tensor product as a matrix in the following way: $$A\otimes B = \begin{pmatrix} A_{11}B & \ldots & A_{1m}B\\ \vdots & \ddots & \vdots\\ A_{m1}B & \ldots & A_{mm} B \end{pmatrix}$$ where $A$ is a $m\times m$ matrix and $B$ is a $n\times n$ matrix. Notice that ...


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Your second option is correct. It is a simultaneous eigenket of all three commuting operators. And that is how it should be interpreted. To be explicit, the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$ Is a triple of equations $\hat{x}\mid \mathbf x'\rangle = x'\mid \mathbf x'\rangle$ $\hat{y}\mid \mathbf ...


2

Short answer His notation is not ambiguous because the expression $$V^{'\mu} \equiv \Lambda^\mu_\nu V^\nu$$ can only mean sum along the $\nu$ component. Since $\Lambda$ is a representation of the Lorentz group, it is a linear operator, hence it can only act on a vector by the usual way that matrices act on vectors. Hence the above is unambiguous. Longer ...


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You're getting tripped up by summation notation. Whenever you have a repeated index, this means that that index is to be summed from 1 to 3: $$ \delta_{ij} \delta_{ik} \equiv \sum_{i=1}^3 \delta_{ij} \delta_{ik}. $$ You're right that there are two terms in this sum where $i \neq j$, and so the contribution to the sum from these terms is zero. But the ...


1

I think you're tripping on repeated indices/Einstein notation here. If an index is repeated, you're supposed to sum over it. So $\delta_{ij} \delta_{ik}$ seems like it would be zero, except that one term in the sum will have $i = j$ and another will have $i = k$. If they're the same, you get a $1 \cdot 1 = 1$ term, if they're different you get $1 \cdot 0 + 0 ...



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