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I would normaly understand an Integral $\int d^3\mathbf{r}$ as a volume integral over the whole space $ \mathbb{R}_3$, where I would understand the bold r there as $\vec{r}$ . I have also seen $\int d^3 \vec{r}$ meaning the same volume integral over $ \mathbb{R}_3$. Or even $\int d~ \vec{r}$ with the superscript dropped (I do not like this one but I have ...


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The notation $\mathrm d^3r$, often also $\mathrm d^3\mathbf r$, is generally understood to indicate a three-dimensional volume integral, as you correctly surmise. If $\mathbf r=(x,y,z)$ then you could also denote that as $\mathrm dx\,\mathrm dy\,\mathrm dz$, or as $\mathrm dV$ if it is clear what the integration variable is. The notation $\mathrm d^3\...


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If we look at nature from an empirical side, we can deduce the four Maxwell equations from experiments as: $$\nabla\vec{E}=4\pi k \rho$$ $$\nabla \times \vec{E} = - k'' \frac{\partial}{\partial t}\vec{B}$$ $$\nabla\vec{B}=0 $$ $$\nabla \times \vec{B} =4 \pi k' \vec{j} + \frac{k'}{k} \frac{\partial}{\partial t}\vec{E} $$ The choice of the value of those ...


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Potential energy stored in a body is relative. We have to first choose the potential at a finite point or infinity. In the case given above, we take potential energy to be 0 at the centre of the Earth. So according to the relation, $PE = mgh$ where $h$ is the height from the centre of the Earth. Generally, we take height from the surface of the Earth and ...


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I don't have enough reputation to comment but please note that Weinberg uses the (-,+,+,+) metric, which means you need a change of sign in $$e^{iP.x}$$. The field transforms as, with $x'=x+a$, $$\phi'(x')=U^{-1}(a)\phi(x')U(a)=\phi(x)$$. For (+,-,-,-) signature this is implemented by $U(a)=e^{iP.x}$, see Peskin & Schroeder, page 26 for example. In ...


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A rotation is of the form $$\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$ A reflection is of the form $$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & - \cos(2\theta) \end{bmatrix}$$ If we want to find a $2-$dimensional representation of a $3-$dimensional rotation then we can ...


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The ampere is used, among other things, to derive the volt, which is the electric potential difference that a current of one ampere has to flow across in order to deliver a power of one watt. (The watt itself is defined mechanically, from second, meter and kilogram). One volt is around the same order of magnitude as the potential differences commonly ...


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The kilogram is currently defined, as you well note, as exactly the mass of the International Prototype Kilogram, and by definition the mass of the IPK is $1\:\mathrm{kg}$ with zero error. This definition took over from the previous one (the mass of $1\:\mathrm{dm}^3$ of water at $4°\mathrm C$ and sea-level pressure) because the previous one was hard to ...


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This definition isn't obsolete, by definition, because we are still using it. There are, however, efforts underway to redefine the kilogram. One proposed definition is based on the watt balance, basically defining the kilogram in terms of the electrical units volt and ampere. Another proposed definition, called the Avogadro Project, is in terms of the mass ...


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If you are measuring y at some value x, and both quantities have uncertainty, then in principle you should show the uncertainties on both axes. In some circumstances you might omit the x error bars. This would be the case if the y value depends on x such that $$\Delta y \gg |dy/dx| \Delta x,$$ where $dy/dx$ is your best estimate of the gradient of $y(x)$. ...


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If we know-and we are sure about-the relationship between y and x: If $y=f(x)$, you must propagate the uncertainty of $x$ to $y$. That is, since $y$ is analogous to $x$, then the uncertainty in $y$ is analogous to the uncertainty of $x$. So, you use a method that you can find in detail in Taylor's book "An Introduction to Error Analysis: The Study of ...


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As in ptomato's answer optical path length is generally expressed with units of length. However, a related quantity is optical path difference for a system or rays, which measures the degree of optical aberration (not to be confused with stellar aberration). Optical path difference is the RMS deviation of the optical path length of rays through a system's ...


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Optical path length generally has units of length, as its name implies. I haven't seen the dimensionless quantity $OPL/\lambda_0$ referred to as optical path length.


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W.u stands for Weisskopf unit: [ref 1, ref 2]. Despite being called a 'unit', it does not have a universal value; the value of the Weisskopf unit depends on the mass number of the nucleus in question and which transition the nucleus is undergoing ($E\lambda$ or $M\lambda$). The references contain expressions for the value of a Weisskopf unit in terms of $A$ ...


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In nuclear Physics estimates can be made using the shell model of the nucleus of the gamma ray transition rates in excited nucleii and such estimates are named after Victor Weisskopf. A measured rate is compared with the Weisskopf estimate and the ratio is said to be in Weisskopf units (W.u.).


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When you write "energy given out" or "energy gained" you are expressing a choice of sign for the energy transfer that you are specifying, and each of them is different. Notice that in your scenario A loses energy, so it's "energy given out" is positive, but it's "energy gained" would be negative. Employing a sign convention means always stating changes of ...


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You have added the negative sign in front of your integral and then put in the cos(180) as well. Pick one. Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.


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The force exerted on a unit positive charge is $ \dfrac {kq}{r^2} (+\hat r)$ and so an external force $ \dfrac {kq}{r^2} (-\hat r)$ must be applied to move the charge. If the step is $d\vec r$ then the work done by the external force in moving that step is $ \dfrac {kq}{r^2} (-\hat r) \cdot d\vec r = -\dfrac {kq}{r^2} dr$. You have moved from $\vec r$ to $\...


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This definition doesn't depend on the metric signature convention. Note that in definition of $\gamma$ the metric doesn't appear anywhere. It is defined purely in terms of "3-vectors" and "3-scalars" measured by particular observer. So it is impossible for metric to appear here explicitly.



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