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# [bi|sur|in]jective coordinates

Discussion in 'Physics & Math' started by RJBeery, Sep 29, 2010.

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1. ### AlphaNumericFully ionizedRegistered Senior Member

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6,699
What a cop out attitude.

The greatest teacher in the world couldn't explain the Dirac equation properly to someone who hasn't done calculus and linear algebra, unless they also explain calculus and linear algebra. There's a reason the education system in every country and university in the world teaches certain topics in a particular order, the later courses require understanding of the earlier ones.

If you or RJ don't make any attempt to learn anything relevant its not then the fault of the people willing to try to explain some things to you, its yours. Forums are not the place to get a full understanding of something, they are to help fill in the gaps where you get stuck. You aren't willing to put in any effort at all and then you blame the people who are willing to spend some of their own time helping you?!

Yet another example of your completely skewed view of how science is done.

3. ### Guest254Valued Senior Member

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1,056
Well that's not very nice. What did I tell you about biting the hand that feeds you?

Aside from all that, do you understand the mathematics people have explained to you? Did you understand the link on coordinate charts? If the answer to both these questions are "yes", then you should have no problem proving that every manifold can be mapped bijectively to Euclidean space. Finally, then, you can see how to deal with a very specific example - say: the pseudo-Riemannian manifold corresponding to the Schwarzschild solution to the Einstein equations.

5. ### przyksquishyValued Senior Member

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3,186
Well hopefully you've more or less understood the point then: merely asking for a bijection is not a significant restriction, nor do most bijections make for useful coordinate mappings.

This "clique" paranoia is getting a little old.

Well not just anything, eg. there's no bijection between the real and natural numbers, but it's not a restriction when it comes to manifolds. In general if you've got a patchwork of charts covering a manifold, you'll always be able to shrink, translate, or deform their coordinate images in such a way that they don't overlap, and that'll get you a bijection defined on the entire manifold. That much isn't difficult. If a mathematician says that it is impossible to define a coordinate chart that covers an entire manifold, it's because they're asking for considerably more than just a bijection.

Kruskal coordinates accomplish this for spherically symmetric black holes. You can represent every point in, on, and out of the event horizon on a single Kruskal chart, including an outside observer arbitrarily far away ("at infinity" isn't a point in space-time). For the eternal Schwarzschild black hole you can even fit a bonus white hole and parallel universe all in the same chart.

7. ### RJBeeryNatural PhilosopherValued Senior Member

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I don't know if you followed the black hole thread, przyk, but this was all covered in depth. I really appreciate your contribution though because your sincerity to engage is obvious.

To say "at infinity isn't a point in space-time", I would agree completely. The 'infinite observer' stipulation in the BH OP was there so that no gravitational effects were acting upon him. I could've given Observer A an initial velocity precisely = escape velocity of the BH and, presuming quantum gravity exists, he would come to rest WRT the BH at a finite distance. I wonder what quantum gravity would do to Kruskal coordinates?

Anyway...observers A and B are hovering over a large BH. Observer B claims that they could free-fall beyond the EH in finite time as calculated from their current location, and using Kruskal coordinates produces a finite answer of T1. Observer A claims it simply isn't true, because he prefers Schwarzschild coordinates, and that from their current location nothing crosses the EH in finite time. To settle the argument, Observer A space-walks out of the ship with a very powerful jetpack and free-falls towards the EH until such time that B's clock says T1...THEN he turns on his jetpack and flies back to the ship to give B an "I told you so". Pure logic here tells me something is amiss.

So what's the problem? Maybe you can explain but is it possible that Kruskal coordinates are bastardizing the term "time coordinate"?

8. ### Green DestinyBannedBanned

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1,211
Is that the best you can do?

Overall, if you can't explain science atleast to a certain degree, that's not the fault of the student.

9. ### Green DestinyBannedBanned

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1,211
Let me rephrase that one. Cannot teach it so it can be understood. That's more to the point.

10. ### RJBeeryNatural PhilosopherValued Senior Member

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"You do not really understand something unless you can explain it to your grandmother." -- Albert Einstein

Anyway, GD, I'd really like to hear someone's response to the following so I respectfully ask that you issue a cease-fire with the other posters

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Class!

12. ### przyksquishyValued Senior Member

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Er, why? What does quantum gravity have to do with this?

No idea. I don't know the first thing about quantum gravity. But presumably not much except near the singularity where quantum gravity effects dominate.

I think the problem is that you're assuming a well-defined notion of simultaneity when you say what amounts to "A turns his jetpack on when B's clock reads T1". In flat spacetime, A could send B a signal when he turns around, and B, knowing A's distance and the speed of light, could work out that A really did turn his jetpack on when B's clock read T1. This kind of reasoning isn't necessarily meaningful in curved spacetime, where there are no globally defined inertial coordinate systems. The Schwarzschild and Kruskal charts differ in their notion of "coordinate simultaneity" (ie. when distant events get the same value of the time-like coordinate) and I don't know any basis for preferring one over the other.

If we let A free-fall into the blackhole, there are only two coordinate system independent ways I can think of in which B could say that A enters the black hole in finite or infinite time. The first sense is that B continues to receive light signals from A forever, although they're increasingly redshifted and attenuate over time at an exponential rate. For the second sense, you ask "Could B, at any time, send A a signal telling him to stop, or fly in after him to stop him?". It turns out that the answer to this is "no" - if A left the ship and allowed himself to free-fall into the black hole, there would be a finite time on B's clock after which he would be powerless to do anything to rescue A.

I'd say if anything the Kruskal coordinates are better behaved in this respect than the Schwarzschild coordinates. Outside the event horizon, the Schwarzschild t coordinate is a time-like coordinate and r is a space-like coordinate. Inside the event horizon their roles are reversed: t is actually a space-like coordinate while r is time-like. They twist right on the event horizon. Kruskal coordinates don't have this problem: the "u" coordinate is always time-like and "v" is always space-like.

Last edited: Oct 4, 2010
13. ### AlphaNumericFully ionizedRegistered Senior Member

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6,699
Clearly Guest has explained this stuff 'to a certain degree'. Clearly people have explained calculus and basic physics to you 'to a certain degree'. The fact you still don't understand it is not our faults, as even with the best explanation in the world there's only so much you can explain using short forum posts. This is why universities have things like libraries and lectures, as many things require a great deal of time and effort to even get close to fully understanding.

Cranks always trot that one out as an excuse for their own lack of understanding, its not because the crank is thick and too lazy to put any effort in, its because someone hasn't spoon fed them properly yet

I can explain the superficial concepts of string theory in a short post but I couldn't convey the specifics someone would need to know to actually do some string theory with anything less than several hundred, or more likely thousands, of hours of detailed explanation. Einstein in that quote clearly wasn't referring to the details of any physics or mathematics, after all he said something along the lines of "Whatever your problems with mathematics, mine are worse", in reference to the complexity of the work he did.

The concept of coordinates is easy to explain, including to grandmothers. The specifics of how you go about defining charts on non-trivial topological spaces is not.

Clearly you understands the concepts of coordinates, that's something simple enough to spoon feed someone as most people are familiar with latitude/longitude or grid references on a map. Topological spaces and the defining of additional structure to make a manifold are sufficiently outside the realm of common experience that you have to put in some work yourself, you have to be willing to make an effort and you can't go blaming other people for your own short comings.

Clearly some things require a lot of time to learn and understand. Doctors spend years learning before they are allowed near patience. Soldiers must be trained before going into combat. Clearly some things take time to learn else we'd not have universities. If that quote of Einstein is supposed to be applied to everything does that mean all lecturers at all universities don't understand what they teach? After all a typical lecture course will involve dozens of hours of lectures, as well as supervisions, revision periods, group discussion, homework.

If someone is only interested in the superficial details then that quote of Einstein is fine since it can be viewed as defining 'superficial details' as those understandable to anyone. If someone is interested in the details, like bijections in coordinates or energy levels relating to the Dirac equation then it is on them to put in some time and effort. If you think that quote applies to all things then you're naive and not a little ignorant of the real world. Learning anything worthwhile is more than being spoon fed facts or small snippets of information, its about putting time and effort in. That's why things like degrees or any kind of academic qualification are valued, they demonstrate someone is willing to put in time and effort and are capable of learning things.

You two should try it some time, actually sitting down and reading a book or a resource on the thing you don't understand, that's how other people do it. Rather than asking someone else immediately finding the information yourself can be rewarding. It also means that as you search for the answer you learn things you didn't think about before, broadening your horizons a little each time.

Or you could just piss your time down the drain by making shit up, posting it in the hopes of deceiving a few people who know even less than you and then throwing a hissy fit when someone who bothered to open a book on the subject points out you're wrong on just about everything you say......

Last edited: Oct 4, 2010
14. ### rpennerFully WiredValued Senior Member

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I don't believe this is a true quote. I think it is a paraphrase of a Russian proverb and is not actually attributed to a saying or writing of Albert Einstein (either the physicist or the actor who now uses Albert Brooks).

15. ### RJBeeryNatural PhilosopherValued Senior Member

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I believe "differing notions of simultaneity" is irrelevant here. The only requirement is that A and B can both agree what B's clock will read, strictly using Kruskal coordinates, when A "crosses the event horizon". If such a calculation is not possible, or if such calculation indicates that B's clock will read "infinity", then the claim that "Kruskal coordinates allow for crossing the event horizon in finite time" is a meaningless statement.
OK, how would you calculate that time? You say this rather emphatically, like it is a well-accepted conclusion, but I would be surprised if this was the case.
Actually the U and V Kruskal components both rely on space-like and time-like parameters, R and T, on both sides of the EH which may make it appear to behave itself but is also why I made the "bastardization" comment.
Just curious, but are the "Alpha Rules" named after you? Because I've already asked for specific examples to justify making asinine comments like this and I'm about to invoke an "Alpha Rule challenge" that you either defend this statement or stop making it.

Last edited: Oct 4, 2010
16. ### przyksquishyValued Senior Member

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3,186
"When" means "at the same time as", which requires a notion of simultaneity. You can work out at what value of the u coordinate A crosses the event horizon, and you can ask what time B's clock will read when B's worldline reaches that same value of the u coordinate, but that's all you'd be doing. There is no priviledged sense in which you can claim two events whose u coordinates coincide are actually happening "at the same time".

You don't need to calculate anything. Kruskal coordinates have the nice feature that light rays travel along straight diagonal lines, just like on a Minkowski diagram in SR. This makes the Kruskal chart particularly well suited to studying causal relationships: you can just draw the past and future light cones of an event like you would in SR. If you consider the situation on the Kruskal chart, you'll see that as B advances along his world line, the point where A crosses the event horizon soon drops out of B's future light cone. Beyond that, B has no hope of pleading with or rescuing A.

No they don't. As I said, the Schwarzschild r and t coordinates are only space-like and time-like (respectively) outside the event horizon. Inside it, they're still called r and t, but in GR it's the space-time metric, and not their names, that determines whether they're really space- or time-like coordinates.

Also it's wrong to say the u and v coordinates "depend" in any way on the Schwarzschild coordinates. Of course the Kruskal and Schwarzschild charts overlap and you can write each chart's coordinates in terms of the other's in their common domain, but you don't have to. You could elect to show directly that the black hole metric in Kruskal coordinates is a valid solution to the Einstein field equation, and never care about Schwarzschild coordinates. The fact that the black hole is usually first presented in Schwarzschild coordinates is just a pedagogical choice that reflects history, not necessity.

Your comment only makes sense if you're assuming that the r and t coordinates are in some sense the "true" space and time coordinates. They aren't. There are no globally defined inertial coordinate systems except in flat space-time.

17. ### RJBeeryNatural PhilosopherValued Senior Member

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4,189
Hmm maybe a picture would help. Would you happen to have one? I'm not saying this isn't the case, but a visual aid helps me internalize a new concept. I'm not convinced it matters but I'll explain why in a moment**...
OK wait a minute. Look at this statement:
"Using Kruskal coordinates, an outside observer will claim that a free-falling object will cross the event horizon of a black hole in finite time."
- Do you agree with this statement?
- Does "finite time" apply to the single temporal dimension, traditionally measured by clocks and denoted t, or does it refer to the Kruskal time coordinate V?
- If "finite time" refers to V, what does it mean for the time coordinate to contain what we traditionally call the space coordinate of r as a parameter?? I'm sorry, but this, plus your quote above, sounds like a claim that "space and time coordinates" are arbitrary labels and, if that's true, I could claim that my Przyk Coordinates contain no singularity at the EH of a BH because the Przyk time coordinate "kumquat" is a constant equal to 7. (Forgive the snarkiness, but maybe you see my point. Or maybe I'm drunk again.

)

** As for why I think this is all academic, it's because my original thought experiment is valid for any arbitrary value of t < infinity. Theoretically, a powerful enough jetpack could rescue a free-falling observer from an event horizon such that a distant observer will have aged by any arbitrary amount when they met up again.

18. ### przyksquishyValued Senior Member

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There are general Kruskal diagrams of the black hole available online (see eg. the relevant Wikipedia article). I don't know of a diagram that specifically illustrates the scenario I have in mind. I could construct one, but I'm not sure there would be much point. If you're already familiar with the Kruskal chart it's an easy exercise: you draw a world line for A that falls past the event horizon, draw a world line for B that stays outside the event horizon, look at the point X where A crosses the event horizon, and consider whether X is always in B's future light cone. If you're not already familiar with the Kruskal chart, the diagram won't be very convincing because I'll just be handing down its general features to you and you won't know where they're coming from.

It's a bit vague as stated. Which time? Kruskal time? Proper time? Is the outside observer making his claim based on an observation or measurement of some kind (he never sees the object cross the event horizon unless he himself crosses the event horizon), or based on a theoretical understanding of black holes and Kruskal coordinates (he calculates the Kruskal time at which he predicts the object will cross the horizon, and finds that it is finite)?

As the person asking the question, that's up to you to decide. Short answers: the free-falling object crosses the event horizon for a finite value of the Kruskal v coordinate, no value of the Schwarzschild t coordinate (it blows up), and a finite proper time as measured by a clock attached to the object. An outside observer detecting signals emitted by the free-falling object never detects a last signal.

They are. The only coordinates that you could really claim have "physical significance" are the coordinates of inertial reference frames, which coincide with what clocks and rulers would actually measure. In curved space it's impossible to construct a globally defined inertial coordinate system. You can do it locally in the vicinity of any point, but your coordinates lose their physical significance with distance. That's why it isn't a problem if eg. Schwarzschild coordinates blow up on the event horizon. They don't represent anything physical there anyway.

There are quantities such as proper times and the scalar curvature which are independent of how you define your coordinates. These are the ones you have to worry about if you find them taking nonsensical values. For example, the scalar curvature is infinite at the black hole singularity, and meddling with the coordinates can't eliminate this.

So...?

19. ### RJBeeryNatural PhilosopherValued Senior Member

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I'm trying to pin you down into agreeing that the Schwarzschild infinite time calculation to cross the EH (from the outsider's perspective) is not some mathematical anomaly whose meaning is up for interpretation. Can we agree that quantum gravity would allow for such an indisputable calculation?

Also (and I have to make it clear that I'm making this statement based on the conviction of my understanding rather than my direct experience) I continue to have doubts that Kruskal coordinates will ever produce a outsider's ("clock") time after which an infalling body could not be rescued from an event horizon. This is why I asked for a picture because I believe you are mistaken on this point.

20. ### przyksquishyValued Senior Member

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Why should I agree with this?

Indisputable calculation of what? The infalling object does not cross the event horizon in a finite value of the Schwarzschild t coordinate. Nobody is disputing that. The outside observer will never receive any signals sent by the infalling object as it crosses the event horizon. If you're literally talking about the light that reaches his eyes, the outside observer only sees the infalling object cross the event horizon after "infinite" time has accumulated on his watch (ie. he never sees it happen, as long as he himself steers clear of the event horizon). Nobody is disputing that either. I really have no idea what you think quantum gravity is supposed to change or why you're bringing it up.

Well if you really insist:

This is the relevant part of a Kruskal chart of an eternal black hole. The singularity (S) is represented by the gold hyperbola at the top, and the event horizon (EH) by the dashed diagonal red line. The blue lines depict the world lines of two observers A and B (labelled). They start off together then separate. A follows a trajectory into the black hole (not necessarily free-fall). B stays at roughly constant Schwarzschild radius from the black hole. A crosses the event horizon at the point I've marked x. I've also marked a point y on B's world line. The future light cone of y is delimited by the dotted grey diagonal lines. Clearly, x isn't in the future light cone of y. The components of the black hole metric in Kruskal components are finite (except at the singularity), so finite distances along world lines as depicted on the Kruskal diagram correspond to finite accumulated proper times. B will reach point y in a finite time according to his watch.

EDIT: I think I might be able to guess what's troubling you, actually: B never receives any confirmation that A fell past the event horizon (unless B himself crosses the event horizon). B could indeed always find out later that A turned his jetpack on at the last minute, or that someone else rescued A. What I'm saying is that, assuming A stayed on course into the black hole, there would be a finite time on B's clock after which A's fate would be out of B's hands. If A accidentally left his jetpack behind, B would have a finite amount of time to notice and arrange a rescue mission before A became irretrievable.

Last edited: Oct 6, 2010
21. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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I have a question about Kruskal coordinates that's based on another nagging question I've had for a long time. When I took my one course in graduate GR, we covered the static, symmetric Schwartzschild solution, and then you discuss questions like what happens when a small particle falls into a black hole of this type. Now the way we were working, we were pretending the particle has no effect on the surrounding spacetime curvature, and hence it would take forever to enter the black hole.

From the particle's POV, time passes normally and it passes the event horizon in a finite time period, so to keep track of the motion in this reference frame, after an infinite amount of time has passed outside the black hole, this is where we needed to introduce Kruskal coordinates. Now my question: as I understand it, when you add mass to a black hole, its event horizon grows to eventually swallow that added mass, hence it does not take infinitely long for a particle to cross the horizon as seen from the outside (I believe this is how two black holes are able to merge into one). If this is true, are Kruskal coordinates still needed? Are they still relevant? Are they still a good approximation?

22. ### RJBeeryNatural PhilosopherValued Senior Member

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Przyk, this is great, thanks for taking the time to do this. I have a couple questions:
1) What program did you use to make this? Did you "free hand" it? I just want to know if this is a depiction of your understanding of Kruskal coordinates or a generation from a Mathematica-type modeler.
2) Are you certain that light cones apply to a graph such as this? I'm still trying to digest what you wrote about "time" and "space" labels being just that, arbitrary labels; but I'm having SERIOUS problems accepting that a light cone doesn't have a very specific meaning related to "moving" in a specific ratio of certain spacial dimensions over a temporal one. U and V both contain spacial and temporal parameters. For example, what would the world line of A look like?

23. ### przyksquishyValued Senior Member

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It's based on my understanding of Kruskal coordinates. I used Dia to create it, so it's inexact in some respects. For instance the singularity should be a hyperbola, as should B's world line if B maintains a constant Schwarzschild radius. I just used bezier curves that reproduced their qualitative features. A's world line is pretty much arbitrary - A can do anything he wants as long as he doesn't go faster than light. I don't know off-hand exactly what a free-fall trajectory would look like, except that it would cross the event horizon at a finite point, which is all that's necessary. Overall, the diagram is maybe about as accurate as what I'd draw on paper.

By the way, see the edit to my last post for a possible clarification (and if you're wondering, I'm using a rather flaky internet connection at the moment - I really didn't see you post an hour and a half before my edit).

Well they're arbitrary in the sense that you can always choose to parameterise space-time in any way that pleases you. In general, coordinates alone aren't useful. In order to do physics, you need coordinates and an expression of the space-time metric in those coordinates. The metric keeps track of how stretched or deformed your coordinate system is compared with an inertial coordinate system.

In flat space-time it's usually easiest to just stick with inertial reference frames. SR inertial coordinate systems have a number of familiar properties: the time coordinate coincides with the time clocks at rest measure, the spatial coordinates coincide with the distances rulers measure, light travels in straight lines, and so on. In curved space you can define coordinate systems which retain some of these properties globally, or all of them locally, but not all of them globally.

The light cone has a specific meaning related to the trajectories of light rays. In an SR inertial coordinate system, light rays travel along straight diagonal lines, (eg. x = ct). In a more general coordinate system, you can still predict the trajectories that light rays will follow, but they won't necessarily be linear in the coordinates, and a plot of the light cone actually might not look much like a cone. Kruskal coordinates are defined in such a way that light cones do look like cones: the trajectories of radially directed light rays in Kruskal coordinates are all of the form $u \,=\, \pm v \,+\, b$.

And u and v are themselves respectively spatial and temporal parameters. There is no one true temporal or spatial coordinate. All you can do in general is distinguish between space-like, time-like, and light-like parameters. This in itself is nothing new: it's a feature of special relativity. Even in pre-SR Newtonian mechanics, symmetry under Galilean transformations (x[sup]'[/sup] = x - vt) meant that there wasn't a priviledged spatial plane.

And as I said earlier, the Schwarzschild r and t coordinates aren't respectively space-like and time-like everywhere. They switch roles inside the event horizon.

In Kruskal coordinates? Under specific conditions (eg. free-fall)? I'm not following what you're asking for.


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## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 9 - Sections 9.1-9.6 - Integrated Review - Functions and Properties of Logarithms: 9

Not one-to-one

#### Work Step by Step

We know that a function is one-to-one if each x-value of the function corresponds to only one y-value and each y-value corresponds to only one x-value. The given function is $y=3$, which is a horizontal line, such that y is equal to 3 for all x-values. Therefore, this function is not one-to-one, because the y-value of 3 corresponds to more than one value of x.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

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# Math Help - Inversely proportional

1. ## Inversely proportional

Have I done this correctly?

f is inversely proportional to d

when d=50, f=256
Find value of f when d=80

Figured out a constant of 25/125

so

f=50/1*125/25=250

Is thsi correct?

2. its incorrect
since f is inversely proportional to d we have
f=k/d where k is a constant
now as f=256 when d=50 then
k=fd=50*256
so value of f when d=80
f=(50*256)/80=160

3. f is inversely proportional to d ...

$f = \frac{k}{d}$

... where k is the constant of proportionality.

substitute in the given values to solve for k ...

$256 = \frac{k}{50}$

$k = 50 \cdot 256 = 12800
$

$f = \frac{12800}{d}$

... finish the problem

4. ah I did d/k.

Thanks.


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# Deriving the expression of the Lorentz force using the language of a mathematician

This is a continuation of my previous “complaining” post on the language difference of mathematician vs physicists regarding the same thing.

First keeping in mind that non-conservative only matters in macroscopic level, in atom level every type of force is conservative. Maxwell Faraday’s equation implies that the electric field $\boldsymbol{E}$ is non-conservative when the magnetic field $\boldsymbol{B}$ varies in time, and is not expressible as the gradient of a scalar field, and not subject to the gradient theorem (path independence of a gradient field) since its rotational is not zero.

Here the goal is to derive the equation of motion from Lagrangian of a particle moving in an electromagnetic field with charge $q$, which is described by the Lagrangian

$L=\frac{m\boldsymbol{v}^2}2+\frac qc\boldsymbol{v}\cdot\boldsymbol{A}(\boldsymbol{r},t)-q\phi(\boldsymbol{r},t).$

The electric and magnetic fields $(\boldsymbol{E}(\boldsymbol{r},t)$ and $\boldsymbol{B}(\boldsymbol{r},t)$ correspondingly$)$ are related to “potentials” $\boldsymbol{A}(\boldsymbol{r},t)$ and $\phi(\boldsymbol{r},t)$ introduced above as follows:

$\boldsymbol{E}=-\frac1c\frac{\partial\boldsymbol{A}}{\partial t}-\nabla\phi,\qquad\boldsymbol{B}=\nabla\times\boldsymbol{A}.$

The equation of motion obtained from the Lagrangian basically is the Euler-Lagrange equation:

$\displaystyle\frac{\partial L}{\partial {\boldsymbol{r} }} - \frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\boldsymbol{r}'}}\right) =0, \tag{1}$

whereas the $\dfrac{\mathrm d }{\mathrm dt}$ represents taking the total derivative (time derivative, plus the convective derivative w.r.t. the spacial variable, or so to speak). Yet prime only stand for the time derivative usually.

Plugging $L$ into $(1)$ we have: (the usual postulate is that $\boldsymbol{v}$ is not explicitly a function of $\boldsymbol{r}$), and sing the identity

$\boldsymbol{v}\times(\nabla\times\boldsymbol{A}) =\nabla(\boldsymbol{v}\cdot\boldsymbol{A})-(\boldsymbol{v}\cdot\nabla)\boldsymbol{A},$

we have

\begin{aligned} \frac{\partial L}{\partial {\boldsymbol r}} &= \nabla \left( \frac{q}{c} \boldsymbol{v}\cdot \boldsymbol{A} - q\phi\right) \\ &= q\left(\frac{1}{c} \nabla(\boldsymbol{v}\cdot \boldsymbol{A}) - \nabla \phi \right) \\ &= q\left(\frac{1}{c} \boldsymbol{v}\times( \nabla\times\boldsymbol{A}) + \frac{1}{c} (\boldsymbol{v}\cdot \nabla) \boldsymbol{A} - \nabla \phi \right). \end{aligned}

And

$\frac{\partial L}{\partial {\boldsymbol{r'}}} = \frac{\partial L}{\partial {\boldsymbol{v}}} = m\boldsymbol{v} + \frac{q}{c}\boldsymbol{A}.$

Now taking the total derivative:

$\frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\boldsymbol r'}}\right) = \frac{\partial }{\partial t}\left( \frac{\partial L}{\partial {\boldsymbol r'}}\right) + \left(\frac{\mathrm d \boldsymbol{r} }{\mathrm dt}\cdot\nabla\right)\left( \frac{\partial L}{\partial {\boldsymbol r'}}\right).$

Notice the total derivative of $\boldsymbol{r} = (x(t),y(t),z(t))$ is $\boldsymbol{r}’$, hence:

$\frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\boldsymbol{r'}}}\right) = \frac{\partial }{\partial t}\left(m\boldsymbol{v} + \frac{q}{c}\boldsymbol{A}\right) +(\boldsymbol{v}\cdot \nabla ) \left(m\boldsymbol{v} + \frac{q}{c}\boldsymbol{A}\right) \\ = m\frac{\partial^2 \boldsymbol{r}}{\partial t^2} + \frac{q}{c}\frac{\partial \boldsymbol{A}}{\partial t} + \frac{q}{c}(\boldsymbol{v}\cdot \nabla )\boldsymbol{A},$

where we used that the convective derivative of $\boldsymbol{v} = \boldsymbol{v}(t)$ is zero.

Now put everything back to $(1)$, we have:

$\displaystyle m\frac{\partial^2 \boldsymbol{r}}{\partial t^2} = - \frac{q}{c}\frac{\partial \boldsymbol{A}}{\partial t} +\frac{q}{c} \boldsymbol{v}\times( \nabla\times\boldsymbol{A} )- q\nabla \phi. \tag{2}$

Equation $(2)$ is basically the motion equation of the Lagrangian.

Express the force acting on the particle in terms of electric and magnetic fields only (i.e. the equation of motion should have the form of the Newton’s second law and contain fields $\boldsymbol{E}$ and $\boldsymbol{B}$ but not the ‘potentials’ $\boldsymbol{A}$ and $\phi$).

If the potentials are replaced using the $\boldsymbol{E}$ and $\boldsymbol{B}$ respectively, $(2)$ will become: $m\frac{\partial^2 \boldsymbol{r}}{\partial t^2} = q(\boldsymbol{E} + \boldsymbol{v}\times \boldsymbol{B}).$ This is nothing but the famous Lorentz force formula, a.k.a. the Newton second law for a particle moving through an electromagnetic field.

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1. Math
2. Calculus
3. find the linear approximation of fx x14 at x...

# Question: find the linear approximation of fx x14 at x...

###### Question details

Find the linear approximation of f(x) = x1/4 at x = 625 and use this to get an estimate for 4√610.


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# If the fifth gas giant in the early Solar System was completely ejected, where would it be now?

I've read about the possible 5th gas giant in the Solar System, and about its ejection about ~100 million years after the formation of the Solar System. However, I have not seen anything about its potential whereabouts. Assume this gas giant (let's call it G5) was completely ejected from the solar system. Would G5 still remain in the vicinity (100 ly or less) or would it be very far away from the Sun right now?

• Do you have a rough time for when G5 started on its escape trajectory? If it left 4 billion years ago, it would be over 220 lightyears away by now, using the simple radial parabolic trajectory equation. Of course, that answer is for a simple 2 body trajectory, and totally ignores the galaxy's gravitational field, but I guess it's a reasonable 1st approximation. Feb 1 at 6:32
• @PM2Ring: Your comment is actually an answer, if not the answer, isn't it? Feb 13 at 9:06
• @B--rian I didn't post it as an answer mostly because I couldn't find a time for when G5 was supposedly kicked out of the Solar System. I'm still hoping for the OP to give us that info... Feb 13 at 10:26
• @fasterthanlight Did I guess your source correctly? Feb 13 at 10:41
• @B--rian Generally, New Scientist isn't considered to be a great reference. (Neither is Wikipedia, but at least WP articles have citations to proper references). Fortunately, that NS article has this link to the preprint of the article by David Nesvorny, Young Solar System's Fifth Giant Planet?, Sep 2011. The abstract doesn't say when G5 was ejected, but I've had a quick look at that article, and on p.3 it says "it is often assumed that the instability occurred at the time of the Late Heavy Bombardment (LHB) of the Moon some 3.9 Gy ago". Feb 13 at 12:39

It is for similar reasons that, even though the Sun was probably born in a cluster of $$\sim 10^4$$ stars, none of those siblings have been firmly identified.


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How to calculate the X Y coordinates of an ellipse with only the X and Y radius length?

I have an ellipse where the radius of x-axis = 100 and y-axis = 30. I have 3 objects where I want to evenly distribute it along the ellipse.

I have already done this for a circle where both axis' are the same, ex: radius x-axis=100 and y-axis = 100. I did:

• evenSpace = 360/3.

Object1 (x,y) = (100 * cos(120*180*1) , 100 * sin(120*180*1).

Object2 same thing but instead of '1', it's 2.

Object3 same thing but instead of '1', it's 3.

This worked out fine for a circle, but with an ellipse I can't get it to work because the x-axis radius is much longer than the y-axis. Any help people? Thanks in advance.

-
You will need to use an elliptic integral (of the second kind) for this. – J. M. Jan 5 '12 at 3:21
@J.M. Can you give more detail, how this will work? – draks ... Apr 5 '12 at 10:53
@draks: the arclength of an ellipse requires an elliptic integral of the second kind for representation. I distinctly recall posting something in another question how to obtain the elliptic integral, but I can't find it at the moment. – J. M. Apr 10 '12 at 3:43
It depends on what the OP means by "evenly distributed". The responders are using "equal arc length", but another possibility could be "equal area of sectors from the center". These are the same for a circle, but different for an ellipse. – marty cohen Jul 16 '13 at 5:22

Under the usual parametrization of ellipse $x=a\cos t$, $b=\sin t$, the arclength function comes out in terms of the elliptic integral of second kind as $$\int_0^t \sqrt{a^2\sin^2 s+b^2\cos^2 s}\,ds = b\int_0^t \sqrt{1-(1-a^2/b^2)\sin^2 s}\,ds$$ which in Sage is b*elliptic_e(t,1-(a/b)^2). I divided this ellipse in three arcs of equal length:
a=100


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# The most highly concentrated hydrochloric acid is 12 M. What volume of a 12 M stock solution of...

## Question:

The most highly concentrated hydrochloric acid is 12 M. What volume of a 12 M stock solution of HCl should be diluted in order to produce 6.0 L of 0.50 M HCl?

## Hydrochloric Acid:

Hydrochloric acid (HCl) is a strong acid that is commonly found in laboratories and is also a component of stomach (gastric) acid. In the stomach the acid is used to help kill microorganisms and denature proteins so that enzymes can break them down.

We are asked what volume of 12 M HCl stock solution is needed to prepare 6.0 L of 0.50 M HCl. As this is an example of a dilution problem, we can use the following formula to solve the problem:

{eq}\rm c_1V_1 = c_2V_2 {/eq}

Where:

{eq}\rm c_1 {/eq} is the concentration of stock solution
{eq}\rm V_1 {/eq} is the volume of stock solution
{eq}\rm c_2 {/eq} is the concentration of diluted solution
{eq}\rm V_2 {/eq} is the volume of diluted solution

Rearranging the formula to solve for {eq}\rm V_1 {/eq} and substituting in the values:

{eq}\begin{align} \rm V_1 &= \rm \dfrac{c_2V_2}{c_1}\\ &= \rm \dfrac{(0.50\:M)(6.0\:L)}{12\:M}\\ &= \rm 0.25\:L \end{align} {/eq}

Therefore 0.25 L of the 12 M stock solution is needed.

Calculating Dilution of Solutions

from

Chapter 8 / Lesson 5
69K

Learn what a solution is and how to properly dilute a new solution from a stock solution. Learn the dilution equation that combines molarity, the volume of stock solution and desired solution to determine how much stock solution is needed for the new solution.


text:
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# Quick Physics torque question - Need a check

• Jan 4th 2009, 02:50 PM
Intrusion
Quick Physics torque question - Need a check
So I've gotten an answer, but my percent error was huge, which is why I want to make sure that I did it right. Can someone do this real quick and tell me what they get?

http://i42.tinypic.com/dxk9wl.png
[not drawn to scale]

Thanks!
• Jan 4th 2009, 03:03 PM
skeeter
net torque caused by the 10 gram mass ?

(.010)(9.8)(.48) = .047 N-m

I get x = 2.4 cm to balance the system using the 200 gram mass.
• Jan 4th 2009, 03:07 PM
Intrusion
Thanks, that's what I got. I guess it's meant to have a huge percent error. What could be the cause of a 70% error? (the "actual" value should have been 8m with a 200 kg mass)
• Jan 4th 2009, 03:08 PM
Mush
Quote:

Originally Posted by Intrusion
So I've gotten an answer, but my percent error was huge, which is why I want to make sure that I did it right. Can someone do this real quick and tell me what they get?

http://i42.tinypic.com/dxk9wl.png
[not drawn to scale]

Thanks!

$M_1 = F_1 \times d_1 = 0.48m \times 0.01(9.81) = 0.047Nm$

$\Sigma M = M_1+M_2 = 0.047-(x\times 0.2 \times 9.81) = 0$

Hence:

$x = \frac{0.047}{0.2\times 9.81} = 0.024m$
• Jan 4th 2009, 03:08 PM
RossBrons
Since 200 is 10 x 20

It stands to reason that 48/20 = x

Thus 2.4 cm

EDIT ---------------------------------

• Jan 4th 2009, 03:11 PM
Mush
Quote:

Originally Posted by Intrusion
Thanks, that's what I got. I guess it's meant to have a huge percent error. What could be the cause of a 70% error? (the "actual" value should have been 8m with a 200 kg mass)

But it isn't a 200kg mass, it's a 200g mass. 8m is ridiculous compared to the tiny torque exerted by the 10g mass.
• Jan 4th 2009, 03:13 PM
Intrusion
Quote:

Originally Posted by Mush
But it isn't a 200kg mass, it's a 200g mass. 8m is ridiculous compared to the tiny torque exerted by the 10g mass.

Yeah, I meant a 200g mass, not kg.
• Jan 4th 2009, 04:33 PM
Mush
Quote:

Originally Posted by Intrusion
Yeah, I meant a 200g mass, not kg.

Even at that, 8m would produce far too much torque.

10g mass is placed 48cm away, to balance this a 200g mass would be a lot closer than 48cm... and as you know 8m is far from being closer than 48cm.
• Jan 4th 2009, 05:24 PM
Intrusion
Alright, I'll have to enlighten my teacher then and let him know his "actual" answer is wrong. Thanks!


text:
COMPGROUPS.NET | Post | Groups | Users | Stream | Browse | About | |

### Problems using tiff2pdf libtiff...

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Hello,

I want to convert multipages tiff files to pdf,

I use this command line:
tiff2pdf  input.tif

and I obtain this message:
tiff2pdf : Can't open temporary output file for writing to stdout.

If I specify an output file name, the program runs but it don't create any
file...

I tried to convert my file from ycbcr to rgb, changed compression, but
there's no change.

Have you an idea?

Cedric

 0

See related articles to this posting

C�dric wrote:
> Hello,
>
> I want to convert multipages tiff files to pdf,
>
> I use this command line:
> tiff2pdf  input.tif
>
> and I obtain this message:
> -: Error writing TIFF header
> tiff2pdf : Can't open temporary output file for writing to stdout.
>
> If I specify an output file name, the program runs but it don't create any
> file...
>
> I tried to convert my file from ycbcr to rgb, changed compression, but
> there's no change.
>
> Have you an idea?

Sounds like the disk is full or you do not have writing permission in
that particular directory.

Please double-check and file a bug report in case this is not the problem.

Ralf

 0

"Ralf Koenig" a �crit dans le message de news: ...

> Sounds like the disk is full or you do not have writing permission in
> that particular directory.
>
> Please double-check and file a bug report in case this is not the problem.
>
> Ralf

I checked with the installation instructions,
I added and modified environnement variables,
and I have space on my disk but it changes nothing....

Nevertheless, I feel that the problem is with the set of my windows,
my OS is Windows 2000 Pro.

Please, could you give me the path of the "particular directory"
that could be at the origin of the problem?

Cedric

 0

"C�dric" <cedric@somewhere.com> wrote:

>I use this command line:
>tiff2pdf  input.tif
>
>and I obtain this message:
>tiff2pdf : Can't open temporary output file for writing to stdout.

The code calls GetTempPath and GetTempFileName, standard Windows
routines. Check your TEMP and TMP environment variables, and make sure
they specify the correct and complete path name of a writeable
directory.
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using problem
How can I use the 'using' to fix this problem? If impossible with 'using', what I can do? ---------------------------------------------- class A { public: virtual char *read(int size) = 0; char *read(char *str) { return 0; } }; class B : public A { public: virtual char *read(int size) { return 0; } }; int main() { B b; b.read("throws an error"); return 0; } ---------------------------------------------- THANKS Chameleon wrote: > How can I use the 'using' to fix this problem? > If impossible with 'using', what I can do? > ------------...

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Hello, take a grammar G with alphabet {0,1} such that (the word problem for) the language L_G={w in (0+1)*|w\in L(G)} is very complex, say in PSPACE or some higher complexity class. Now consider an "easy" problem EASY like: "Is an element of L_G in L_G?". Well, this seems to be fairly easy because the answer is "yes" in any case. Hence the problem should be of small complexity. But if one models such a decision problem one has to construct a language L with alphabet A such that the word problem for L reflects exactly the decision problem which one is interested...

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text:
# Luogu P1113 chores (DP, topology sorting)

Keywords: C++ Algorithm Graph Theory

[Title Description]
John's farm has a lot of chores to complete before milking the cows. Each chore takes a certain amount of time to complete it. For example, they have to gather the cows, drive them into the barn, wash their breasts and do some other work. It is necessary to finish all the chores as soon as possible, because there is more time to squeeze out more milk. Of course, some chores can only be carried out when others are completed. For example, only when the cow is driven into the cowshed can it start to wash its breasts, and the cow can't be milked before washing its breasts. We call these work preparations for completing this work. At least one chore does not require preparation. The work that can be completed at the earliest is marked as chores 1 1 1. John has something to do n n A list of n chores, and this list is in a certain order k ( k > 1 ) k(k>1) The preparation of K (k > 1) is only possible in chores 1 1 1 to k − 1 k−1 k − 1.
Write a program from 1 1 1 to n n n read in the job description of each chore. Calculate the shortest time for all chores to be completed. Of course, unrelated chores can work at the same time, and you can assume that John's farm has enough workers to complete any number of tasks at the same time.

[input format]
The first 1 1 Line 1: an integer n n n. Number of chores that must be completed ( 3 ≤ n ≤ 10 , 000 ) (3 \le n \le 10,000) (3≤n≤10,000)；
The first 2 2 2 to ( n + 1 ) (n+1) (n+1) line: shared n n n lines, each line has some use 1 1 1 space separated integer, representing:

• Job serial number( 1 1 1 to n n n. Is ordered in the input file);
• Time required to complete the work l e n ( 1 ≤ l e n ≤ 100 ) len(1 \le len \le 100) len(1≤len≤100)；
• Some preparatory work that must be completed, and the total amount shall not exceed 100 100 100, by a number 0 0 End of 0. Some chores do not need to be prepared, only describe a separate task 0 0 0, no extra spaces will appear in the whole input file.

[output format]
An integer representing the minimum time required to complete all chores.

[input example]

7
1 5 0
2 2 1 0
3 3 2 0
4 6 1 0
5 1 2 4 0
6 8 2 4 0
7 4 3 5 6 0

[output example]

23

[analysis]

① DP analysis
Because tasks can be concurrent, if a task has a precursor, the optimal solution is to start immediately after the end of its latest precursor, and the task k k The precursor node of k must be less than k k k. So when reading in, you can select the largest transfer from its precursor. The final answer can be updated at the same time.

② Topological sorting analysis
We set w [ i ] w[i] w[i] stands for the second i i Time required for i chores, t w [ i ] tw[i] tw[i] indicates the second i i The earliest completion time of i chores, p r e pre pre indicates the precursor node of the chore, then t w [ i ] = m a x { t w [ p r e ] } + w [ i ] tw[i]=max\left\{ tw[pre]\right\}+w[i] tw[i]=max{tw[pre]}+w[i], finally, the maximum of the earliest completion time of each chore is the earliest time to complete all chores r e s = m a x { t w [ i ] } res=max\left\{ tw[i]\right\} res=max{tw[i]}.

[DP code]

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 10010;
int last[N], res;//last[i] is the earliest completion time of chore I
int n;

int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
int t, pre, s = 0;
cin >> i >> t;
while (cin >> pre, pre) s = max(s, last[pre]);//Find the latest completed precursor
last[i] = s + t;//Start chores as soon as the front drive is completed at the latest i
res = max(res, last[i]);//The answer is the maximum of the earliest completion time of all chores
}
cout << res << endl;
return 0;
}

[topology sorting code]

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int N = 10010, M = 1000010;
int e[M], ne[M], h[N], idx;
int w[N], in[N], tw[N], res;
int n;

{
e[idx] = v, ne[idx] = h[u], h[u] = idx++;
}

void topSort()
{
queue<int> Q;
for (int i = 1; i <= n; i++)
if (!in[i]) Q.push(i), tw[i] = w[i];//The earliest completion time of chores without precursor nodes is the time to complete itself
while (Q.size())
{
auto t = Q.front();
Q.pop();
res = max(res, tw[t]);//The earliest completion time of the node in the queue must have been updated
for (int i = h[t]; ~i; i = ne[i])
if (!--in[e[i]]) Q.push(e[i]), tw[e[i]] = max(tw[e[i]], w[e[i]] + tw[t]);
else tw[e[i]] = max(tw[e[i]], w[e[i]] + tw[t]);
}
}

int main()
{
cin >> n;
memset(h, -1, sizeof h);
for (int i = 1; i <= n; i++)
{
cin >> i >> w[i];
int x;
while (cin >> x, x) add(x, i), in[i]++;
}
topSort();
cout << res << endl;
return 0;
}

Posted by cjl on Wed, 17 Nov 2021 17:04:04 -0800


text:
# Question #4c668

There are 20 balls in total. So the probability of a red ball is 12 out of those 20, hence $\frac{12}{20}$ which can be simplified to $\frac{3}{5}$.
The process is the same with the black balls, and the simplified answer there is $\frac{2}{5}$.


text:
## Accurate estimation of cosmic shear error bars

The light that we observe that is emitted by far away galaxies travels towards the Earth not through straight trajectories, but rather through perturbed ("curvy") ones as the photons' paths are bent by the gravitational lensing effect caused by the matter that exists between the Earth and the galaxies. This induces tiny distortions in the shapes of the observed galaxies, which can appear more elliptical than what they truly are. Measurements of these distortions are called cosmic shear data, and are one of the most powerful ways to probe the large-scale matter distribution in the Universe.

Like in any analysis of cosmological data, in cosmic shear too one must carefully estimate the statistical errors of the measured quantities. In many areas of Physics this estimation can be done by repeating an experiment over and over again, and then take the statistical error to be the variance (or spread) between the outcome of the different experiments. In cosmology, however, we cannot afford the same luxury since we have only one Universe at our disposal, i.e., one ''experiment''. One way to go about this problem is to run many $N$-body numerical simulations of structure formation in the Universe, and evaluate the variance across this ensemble of simulations. This works in principle, but it is very computationally demanding. In fact, the number of simulations that is thought to be necessary for the analyses of future cosmic shear data makes this method computationally prohibitive, even with current state-of-the-art supercomputing facilities.

Another way to tackle the problem is to calculate directly the expected variance analytically. A problem that existed here was that the ingredients that enter the calculation were not all very well understood theoretically and could not be calculated accurately. This is the problem that I have tackled in a series of papers [8-11] for the most important cosmic shear statistic called the power spectrum. The statistical errors in cosmic shear analyses are encapsulated in a mathematical quantity called the ''covariance matrix'', which has three main physical contributions:

\rm{Covariance} = \rm{G} + \rm{SSC} + \rm{cNG},

• The Gaussian term (G) is a term that is relatively easy to evaluate, and this has been known for a long time.
•  I have shown in [11] that the super-sample covariance (SSC) term can be evaluated completely and accurately using a formalism called ''Response Approach to Perturbation Theory'' that I have developed in [9].
• The connected non-Gaussian (cNG) term is the hardest to evaluate completely, but I have shown using the response approach in [10] that it is possible to evaluate at least the majority of its contribution.

These works culminated in the analysis presented in [8], where I have carried out a simulated cosmic shear likelihood analysis of the upcoming Euclid and Vera Rubin surveys to quantify how accurately do we need to know the cNG term. The main finding from this work is that dropping the cNG contribution from the analysis has less than a 6$\%$ impact on the error bar of the cosmological parameters; see Fig.1. This meets already very comfortably the target accuracy requirements for these surveys, and it is quite an aggressive approach in that we completely neglected the cNG contribution: accounting for it with existing analytical calculations, even if approximate, will result in an even better performance. Our work in  [8] demonstrated, for the first time, that analytical approaches to the covariance are sufficient and efficient enough for future cosmic shear surveys.

[1] A. Barreira, D. Nelson, A. Pillepich, V. Springel, F. Schmidt, R. Pakmor, L. Hernquist, and M. Vogelsberger, Mon. Not. R. Astron. Soc.488, 2079 (2019), arXiv:1904.02070 [astro-ph.CO].

[2] A. Barreira, G. Cabass, D. Nelson, and F. Schmidt, JCAP, 005 (2020), arXiv:1907.04317 [astro-ph.CO].

[3] A. Barreira, G. Cabass, F. Schmidt, A. Pillepich, and D. Nelson, JCAP, 013 (2020), arXiv:2006.09368 [astro-ph.CO].

[4] A. Halder and A. Barreira, arXiv e-prints, arXiv:2201.05607 (2022), arXiv:2201.05607 [astro-ph.CO].

[5] A. Barreira, G. Cabass, K. D. Lozanov, and F. Schmidt, JCAP, 049 (2020), arXiv:2002.12931 [astro-ph.CO].

[6] A. Barreira, JCAP, 031 (2020), arXiv:2009.06622 [astro-ph.CO].

[7] R. Voivodic and A. Barreira, arXiv e-prints , arXiv:2012.04637 (2020), arXiv:2012.04637 [astro-ph.CO].

[8] A. Barreira, E. Krause, and F. Schmidt, Journal of Cosmology and Astro-Particle Physics, 053 (2018), arXiv:1807.04266 [astroph.
CO].

[9] A. Barreira and F. Schmidt, JCAP, 6, 053 (2017), arXiv:1703.09212.

[10] A. Barreira and F. Schmidt, JCAP, 11, 051 (2017), arXiv:1705.01092.

[11] A. Barreira, E. Krause, and F. Schmidt, JCAP, 6, 015 (2018), arXiv:1711.07467.

[12] T. Baker, A. Barreira, H. Desmond, P. Ferreira, B. Jain, K. Koyama, B. Li, L. Lombriser, A. Nicola, J. Sakstein, and F. Schmidt, arXiv
e-prints (2019), arXiv:1908.03430 [astro-ph.CO].

[13] A. Barreira, B. Li, W. A. Hellwing, C. M. Baugh, and S. Pascoli, JCAP, 027 (2013), arXiv:1306.3219 [astro-ph.CO].

[14] A. Barreira, M. Cautun, B. Li, C. M. Baugh, and S. Pascoli, JCAP, 028 (2015), arXiv:1505.05809 [astro-ph.CO].

[15] A. Barreira, A. G. Sánchez, and F. Schmidt, Phys. Rev. D, 94, 084022 (2016), arXiv:1605.03965 [astro-ph.CO].

[16] J. Renk, M. Zumalacárregui, F. Montanari, and A. Barreira, JCAP, 020 (2017), arXiv:1707.02263 [astro-ph.CO].

[17] A. Barreira, B. Li, A. Sanchez, C. M. Baugh, and S. Pascoli, Phys. Rev. D, 87, 103511 (2013), arXiv:1302.6241 [astro-ph.CO].

[18] A. Barreira, B. Li, C. M. Baugh, and S. Pascoli, JCAP, 059 (2014), arXiv:1406.0485 [astro-ph.CO].