it's a very "physics" argument because when you try to formalize it it doesn't make a whole lot of sense (how would the dot product of a vector with anything that's not a vector even be defined?)
the idea is simply that you know that $v\cdot w$ is a scalar when $v$ and $w$ are both vectors. Now for some (deeply unclear) reason you are in doubt that $\nabla \phi$ is a vector, and you reassure yourself: "I know that $\mathrm{d}\vec r$ is a vector and that the result of this dot product is a scalar, so $\nabla\phi$ must be a vector, too, because it it transformed in any way but that of a vector, then the whole expression couldn't transform as a scalar anymore"
hm I am confused. we have as a starting point that $R(v_1 \cdot v_2) = R(c) = cR$. and we know that under a rotation $v_1 \mapsto Rv_1$? and we want to show that under a rotation $v_2 \mapsto Rv_2$?
@ACuriousMind :P well i agree that one needs to define the dot product (which already has a definition) to begin to start. but then I don't understand what you say in this message.
effectively it's just this: If you know that $v_i w^i$ is invariant under rotation, and you know that $v_i\mapsto R_i^j v_j$, then the only option is that $w^i \mapsto {R^{-1}}^i_j w^j$, i.e. $w^i$ transforms like a co-vector, so $w_i$ transforms like a vector
if you try to make this argument in abstract and not index language it will make no sense, at least not in a way I can see
like from what I'm reading, it seems like $\vec{\nabla}\cdot\vec{A}$ is invariant under rotation because like in the gradient, $\vec{\nabla \phi} \cdot \vec{dr}$ we have $\frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy +... = d\phi$
wait that's not the same thing
$\vec{\nabla}\cdot \vec{A}$ is not a vector woops.
Isnt the divergence fundamentallt rotationally invariant. As a quantity it takes information from every direction and adds it up and ao should be invariant under rotations
@Obliv it means the same thing as always: You jut rotate the vector $\vec r$.
which is another reason the "dot product argument" is a bit weird, because this directly induces how $\mathrm{d}x$ and $\frac{\partial}{\partial x}$ transform via the chain rules
@EmilioPisanty not entertaining - too few pictures for my image-addled Instagram mind
Hi all, does anyone have any good book recommendations on machine learning theory? I'm a physics PhD student and I'd like to increase my knowledge from what I've learned in an introductory course on neural networks during my Master's. If there are code examples in the book, the language isn't very important. I'm proficient in C++ and Matlab, but other languages such as Python should be fairly easy to pick up.
@WaveInPlace The Schwarzschild r coord is defined in a way that minimises the problem due to changing spacetime curvature.
So it's defined in terms of the circumference of a circle, or the area of a sphere, as measured by a local observer in freefall.
But there's an additional subtlety. The Schwarzschild coords are defined in terms of an observer in flat spacetime, so the Schwarzschild coords become equivalent to Minkowski coords. There's a passage about that on Wikipedia that I quote here: physics.stackexchange.com/a/552874/123208
> A Schwarzschild observer is a far observer or a bookkeeper. He does not directly make measurements of events that occur in different places. Instead, he is far away from the black hole and the events. Observers local to the events are enlisted to make measurements and send the results to him.
> The bookkeeper gathers and combines the reports from various places. The numbers in the reports are translated into data in Schwarzschild coordinates, which provide a systematic means of evaluating and describing the events globally. Thus, the physicist can compare and interpret the data intelligently.
Practically speaking, you can't just dangle a measuring tape into a black hole to measure the distance to its centre. ;) And even if you try measuring distances by bouncing photons off stuff near a BH & timing how long it takes, it's tricky, because c is only locally constant. And photon trajectories can get complicated. physics.stackexchange.com/a/680961/123208
> Painlevé wrote to Einstein to introduce his solution and invited Einstein to Paris for a debate. In Einstein's reply letter (December 7), he apologized for not being in a position to visit soon and explained why he was not pleased with Painlevé's arguments, emphasising that the coordinates themselves have no meaning.
Still, Painlevé's "raindrop" coords are nice, and in some ways are more natural than Schwarzschild coords.
Someone who speaks French should edit Painlevé's Wiki page en.wikipedia.org/wiki/Paul_Painlev%C3%A9 Some of the titles of his works have been given odd translations. I wouldn't translate "mémoire" as "memory". I guess in English we'd be more likely to use "note".
I was actually tweaking my Schwarzschild photon trajectory plotting code a few hours ago. Here's the plot for a trajectory with deflection = 60°. The impact parameter is r_s × 3.62001264240871876
Eg, on my diagram above, the observer is at some large distance along the +X axis. The photon comes in along the red curve, bends around the BH, and heads along the blue curve. So the observer sees it at a distance of b above the X axis.
Oops. I meant to say that $u$ in $b^2=u^2-u^3$ is the $u$ correspond to the point where the trajectory is closer to the BH, the purple point where the trajectory switches from red to blue.
When $b$ is large, $b\approx r+1/2$. So for normal stars, they're very close to each other.
It's easy to calculate b from r. But it's a bit painful calculating r from b when they're small. I (mostly) use Newton's method, but you have to give it very good initial approximations for b<2.6, or it goes crazy.
Calculating the angle phi from u involves an elliptic integral of the first kind. I don't know much elliptic integral theory, but I know enough to do that stuff. ;)
I sometimes do it by solving the cubic. But that's also annoying because it requires complex numbers, so you get tiny imaginary bits that have to be removed.
You have to add two cube roots. In theory, the imaginary parts cancel. But with finite precision arithmetic you get tiny imaginary components left over.
I remembered wrongly. The trigonometric case is when you have 3 real roots. The case with complex numbers, you can convert it into hyperbolic functions, and that should not involve imaginary components at all.
It is beyond time for meow to go to bed, but when I do have some time leftover, I might help you formulate a direct solution.
Here's a plot related to the deflection by the Sun during the May 1919 eclipse that Eddington photographed on the African island of Principe. According to JPL, the distance from Principe to the Sun at maximum eclipse was 151646958.17 km.
@user85795 I'm using dark mode on my Android phone. Light mode is rather painful. I have to use the blue light filter when reading PDFs. It's still unpleasant, but it helps, a little.
The numbers above the vertical gridlines are impact parameters, in arcminutes. The angular radius of the Sun was ~15.77 arcmin
> The three expeditions travelled to the Crimea in the Russian Empire to observe the eclipse of 21 August. However, the First World War started in July of that year, and Germany declared war on Russia on 1 August. The German astronomers were either forced to return home or were taken prisoner by the Russians. Although the US and Argentine astronomers were not detained, clouds prevented clear observations being made during the eclipse.
I have looked at the history of all eclipse experiment and it's a real clown show
It is a very hard experiment to do since you have to 1) take a before and after photo the previous and following night, such that your telescope does not move At ALL 2) you have to do the whole thing during the eclipse in like 5 minutes
Also basically any small mechanical or thermal perturbation will change the photo
Best spots for observation are also not typically at an observatory
@Slereah You really need shots ~6 months before &/or after, when the stars aren't anywhere near the Sun. The Eddington exhibition decided it was too much hassle to do those shots in Principe, so they approximated using English photos.
