Hot answers tagged conventions
8
The kilogram is defined by a prototype (the "International Prototype Kilogram", IPK) -- basically, a kilogram is by definition the mass of a metal cylinder sitting in a vault in Paris. People have made a bunch of other metal blocks with almost exactly the same mass (as near as they could get), called "sister copies". To measure a mass extremely accurately in ...
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First, recall what a partition is. A partition of a set $X$ is a way to write $X$ as a disjoint union of subsets: $X=\coprod_i X_i$, $X_i\cap X_j=\emptyset$ for $i\neq j$. When the elements of the set $X$ are considered undistinguishable, what matters are the cardinals of the set only, and we have a partition of an integer number, $n=n_1+\ldots+n_k$. For ...
4
Your second equation, $P(\nu,T) = \frac{2 h {\nu}^3}{c^2}$ $\frac{1}{\exp\bigl(\frac{h \nu}{kT}\bigr) - 1}$ is what is commonly referred to as Planck's law for radiation, although a more standard symbol used is $B_\nu(T)$. This is the energy radiated per time, per area, per frequency interval, per steradian. It is a formula for the 'specific intensity' of a ...
4
There are two separate issues here. (1) Why does it make sense to consider a dipole moment as a vector? (2) Given that it's a vector, why does it make sense to say that it points in this particular direction, rather than the opposite direction.
Intuitively, it makes sense to define a dipole as a vector because when we put it in a field, it aligns itself ...
3
I think you just forgot that the $\int_A^B F\,dl$ is not a scalar expression. Rather it should be written in a form $\int_A^B \vec{F}\cdot d\vec{l}$. Then it comes to the sign of the scalar product:
$$\vec{F}\cdot d\vec{l}=F\,dl\,\cos\theta$$
where the angle $\theta$ is taken between the vector $\vec{F}$ and the direction of the tangent to the integration ...
3
First off, please don't use units with $c\ne 1$ in GR. It makes everything horribly messy.
What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like $\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that $u^\mu=(1,0,0,0)$, not $(-1,0,0,0)$. This is ...
3
Hmm, what if they thought about it this way:
Let's put t for time (latin: tempus) and since time and period are connected (I would think of the period as the smallest time a cyclic phenomena requires to complete one full cycle) let's just denote period with T as t is already taken.
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First, if you're going to keep proper track of covariant and contravariant components, you should lower the index on $B$ and make sure the dummy indices are always of opposite types: $B_k = \varepsilon_{kij} \phi^{ij}$. The reason we can be sloppy in Euclidean space is because of how trivial the metric can be. We can always consider our equations in the ...
3
I) Consider an arbitrary coordinate transformation
$$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$
Let
$$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$
denote the corresponding Jacobian.
Traditionally in physics,
a scalar $\sigma$ transforms as
$$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma, $$
a pseudo-scalar ...
2
Just to put some numbers in: as the wikipedia article states, the stability of the national kilogram standards is of roughly tens of micrograms over a century. This means that the relative stability is of the order of $10^{-11}\textrm{ yr}^{-1}$. This is pretty damn good (if short of the $10^{17}$ stabilities of the standard atomic clocks!) and to replace ...
2
Indeed, a number of websites suggest that the T for period is for time.
Wikipedia: The period, usually denoted by T, is the length of time taken by one cycle, and is the reciprocal of the frequency f:
See also:
Period refers to the time it takes something to happen.
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In the $({+}{-}{-}{-})$ signature, positive values of $V^\mu V_\mu$ are referred to as "timelike" and negative ones correspond to "spacelike" 4-vectors. The velocities are timelike (positive) vectors because the world lines of massive objects must have a more timelike angle – that's because massless objects can't exceed the speed of light.
On the contrary, ...
2
Yes, the leap second are added at the same "instant" in the whole world, whatever is the part of the day on a given place. This is documented by a rather typical screenshot of some "clock" in 2008 according to Chicago's U.S. Standard Central Time:
The confusing second was added right before 6 p.m. – in as big a city as Chicago and similarly in the whole ...
2
This thread on physicsforums elaborates a bit on the difference between Levi-Civita symbols and tensors. Based on that, I conclude...
1) Your index notation formula for the magnetic field should use the Levi-Civita tensor, then. The "symbol" is a convenient thing, but this expression must be written with tensors.
2) Carroll likely made a mistake and ...
2
To formalize the comments as an answer:
The difference between requiring
$$(\alpha u,v)=\alpha(u,v)\quad\text{ (mathematician's definition)}$$
and
$$\langle u, \alpha v\rangle=\alpha\langle u,v\rangle\qquad\quad\,\,\text{ (physicist's definition)}$$
is purely one of convention, and the two definitions are equivalent as $(u,v)=\langle v,u\rangle$. There's no ...
1
"So what if it is an attraction force? How this should influence our calculations
Because you have written the work--energy relation in an incomplete shorthand. The correct version, $$W = \int \vec{F} \cdot d\vec{x} \quad,$$ depends on the relationship between the direction of the force and the direction of the path.
This relationship is the source of ...
1
First of all, we don't usually talk about the direction of propagation of a plane wave in QFT. Plane waves are said to exist at all spacetime coordinates with a certain internal momentum, k. And, in reference to some of the comments, in QFT, we don't normally operate with wavefunctions. We promote wavefunctions to operators and act on states. But in this ...
1
Continuous Fourier analysis, which contains both the Fourier transform and the Fourier series, and which is used in e.g. signal processing, naturally picks the average value of the left and right limits, cf. the Dini-Dirichlet criterion. For the Heaviside step function, this means that
$$\tag{1} H(0)~=~\frac{1}{2} \left(\lim_{x\to 0^-} H(x)+ \lim_{x\to 0^+} ...
1
For analytical calculations, you just can't use $H(0)$ because it is not defined there. Also, you don't need it.
For numerical calculations, situations where a more continuous function might be of advantage; in physics, people often use the Fermi-Dirac distribution (which converges to a Heaviside function for $T=0$; you could call it a Heaviside sequence, ...
1
The inner product used in quantum mechanics is sesquilinear, as opposed to just linear. A good reference to read up on this is Hassani: Mathematical Physics.
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In the US, 1 billion = $10^9$.
The difference is between the Long and short scales. The US uses the short scale, where a billion is $10^9$. In the long scale, a billion is $10^{12}$.
In the short scale, every term after a million (billion, trillion, etc.) is 1,000 times bigger than the previous one. So, million = $10^6$, billion = $10^9$, trillion = ...
1
This is easy to check. The total power frequency over the whole frequency range should match the Stephan-Boltzman law $I=\sigma T^4$ with $$\sigma=\frac{2\pi^5 k^4}{15c^2h^3}$$.
Assuming that you are asking for spectral density of radiation power in unit interval of linear frequency, we can express $I = \int P(\nu,T) d\nu$.
Since $$\int \frac{\nu^3 \, d ...
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