@JakeRose Sure. We can make our best effort to go through this without calculus, but if you want rigor, at some point you will need to properly learn calculus somehow. (Maybe some online class) The whole reason calculus exists is very closely related to the conceptual difficulty you are having with this problem.
Anyway, moving on, we have an object moving along some arbitrary curve, subject only to the constraint that the acceleration is perpendicular to the velocity.
@JakeRose Think of it this way. If you push a moving object from behind, so that your force is in the same direction as its existing velocity, then its speed will increase. If you push it from in front, so that your force is in the opposite direction as its existing velocity, then its speed will decrease. OK so far?
Now, imagine that you expose this object to a whole lot of different forces from different directions? Which ones speed it up and which ones slow it down?
There are two cases: the acceleration is zero, or the acceleration is not zero. Let's consider a portion of the motion where the acceleration is nonzero. (Because otherwise, it's just constant velocity motion)
So, that means we should be seeing a lot of vesicles formed from the vinyl cynides in titan's atmosphere, I wonder what they can tell us. Titan is still way too cold for these vescicles to probably do anything though
OK, so... now that we're trying to be somewhat rigorous, we should establish whether the motion proceeds in a straight line or not. Given that the acceleration is perpendicular to the velocity and is nonzero, the motion will not be in a straight line - that is, the path has a finite radius of curvature at every point - but @Jake do you want to go through a demonstration of that, or do you want to just accept it and move on?
@JakeRose No. You should understand "finite radius of curvature" as simply "not a straight line". (A circle has a constant and finite radius of curvature)
Anyway I'll have to think about how to show this without calculus
So, we established that we are looking at the path (or the portion of the path) of a moving object where the acceleration is perpendicular to the velocity at all times, and the acceleration is nonzero, and we accept that this means the path will be curved at all points we are looking at.
The next thing to show is that the acceleration along this path is not constant for any nonzero span of time - that is, it is impossible to find any times $t_i$ and $t_f > t_i$ for which the moving object's acceleration stays constant between $t_i$ and $t_f$. Again, this would be straightforward with calculus but I've got to think about how to justify it without going too deep into the math
(Unless you are also willing to accept this point)
Sure, I mean we can use some algebra, but I'm trying to avoid calculus - the math of infinitesimals
I guess you could say this: we know that the acceleration is perpendicular to the velocity, and we know that the path is not a straight line, meaning that the velocity at the end is not parallel to the velocity at the beginning. If you restrict yourself to two dimensions of space, that also means the acceleration at the end cannot be parallel to the acceleration at the beginning (because acceleration is perpendicular to velocity)
So as for any given time t the velocity is in a different direction and so as a consequence the acceleration is in a different direction
I also should say my calculus abilities arent terrible either so it can be used (although I do thank you for avoiding it as to get purely conceptual understanding)
@JakeRose Yeah, or to be slightly more precise, it's always possible to find two times $t_i$ and $t_f > t_i$ such the velocities at those times ($\vec{v}_i$ and $\vec{v}_f$ respectively) are not parallel, and in turn the accelerations at the times ($\vec{a}_i$ and $\vec{a}_f$ respectively) are not parallel
@JakeRose Well... calculus offers a very neat resolution to this whole issue. I mean, with a solid understanding of calculus it's practically trivial to show that acceleration perpendicular to velocity doesn't change the magnitude of the velocity.
@JakeRose Sure, I mean I can show it now: $\frac{\mathrm{d}}{\mathrm{d}t} v^2 = 2\vec{v}\cdot\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = 2\vec{v}\cdot\vec{a}$ and $\vec{v}\perp\vec{a}\implies\vec{v}\cdot\vec{a} = 0$
As far as the points $t_i$ and $t_f$, they don't have to be very close together, but they could be.
@JakeRose no problem - that last part of the calculus proof is just saying that velocity being perpendicular to ($\perp$) acceleration implies ($\implies$) that the dot product of velocity and acceleration ($\vec{v}\cdot\vec{a}$) is zero.
Anyway, continuing on with the less-mathy argument, I think I may have misstated something earlier.
@JakeRose Yeah, or to be slightly more precise, it's always possible to find two times $t_i$ and $t_f > t_i$ such the velocities at those times ($\vec{v}_i$ and $\vec{v}_f$ respectively) are not parallel, and in turn the accelerations at the times ($\vec{a}_i$ and $\vec{a}_f$ respectively) are not parallel
What I said there is true, but I think the fact we actually need is a little stronger: we need to know that, given any time $t_i$ and the object's velocity at that time $\vec{v}_i$, there is a length of time $\epsilon > 0$, such that the object's velocity does not equal $\vec{v}_i$ at any point in time between $t_i$ and $t_i + \epsilon$
which actually goes right back to what we're trying to prove...
So as for any given time t the velocity is in a different direction and so as a consequence the acceleration is in a different direction
For any time $t$, the velocity for a short time after $t$ is in a different direction than the velocity at $t$. In other words, the object doesn't keep the same direction of velocity for any finite amount of time.
Now, remember, you can only use the Pythagorean theorem in the way you were using it when the acceleration is constant over the time $t$.
And we've shown (I think) that the acceleration cannot be constant over a period of time unless the velocity maintains the same direction over that period time. Right?
@JakeRose Do you accept that any change in a particular directional component of the velocity must arise from acceleration in that direction? That's kind of at the core of it
That's something we haven't proven here, of course
Oh. Well then that makes it easier. If acceleration is perpendicular to velocity, there is no component of acceleration in the direction of the velocity, so the component of the velocity in the direction of the velocity can't change.
That component is the magnitude.
In other words, if the magnitude of the velocity were to change, there would have to be some "forward" or "backward" acceleration at some point. But because the acceleration is always perpendicular to the velocity, there isn't.
@JakeRose You're still using Pythagorean theorem logic there
which doesn't work in this case, remember
Fundamentally, you've got to keep in mind that the process we're talking about here, where acceleration is continuously changing the direction of the velocity, cannot be described by plain old Pythagorean theorem-style vector addition.
It can be approximated using the Pythagorean theorem, but the approximation is not exact. And in this case, the difference between the approximation and the real thing is precisely the difference between having a longer resultant and not.
@JohnRennie For some reason Windows misnamed my main user and now my files are not named Ryan, but something else. Is that something I can't fix without a complete reinstall?
If you open Explorer then on the C: drive there's a folder called Users, and in that are the individual user folders including the incorrect one. In the individual user folders there are a few important ones that you may want to copy.
@0celóñe7 The problem is that it uses folder names based on the first name that you gave the user. So if you change ryan to ryan-ms and create a new ryan, then ryan-ms ends up associated with a folder called ryan, and ryan ends up associated with a folder called ryan2.
The default configuration of my current laptop was that you could start Internet Explorer by placing three fingers on the touch pad. So I'd be staring at the screen, wondering what to do next, when Internet Explorer would suddenly open, and I'd realise that my entire hand was resting on the touch pad. Could that be what you were experiencing?
My problem was that when I bought the computer, the people in the shop installed Windows for me, and created an admin user called Dawood. I had not asked them to do that. What I actually wanted was an ordinary user called Dawood, and a separate admin user. I never managed to do it.
@rob In my years of being on this site, when an otherwise acceptable answer has been posted to a homework like question, users have commented to suggest removal, suggest turning it into more of a hint to help the OP, or downvoted. I have never seen until recently a moderator singlehandedly delete an answer like that. To make matters worse, I was not even politely requested to delete it - it was deleted outright.
@rob In the future, unless something is obviously offensive and requires immediate removal to prevent further offence (which would never be the case, but saying it just to cover all my bases), I expect you to ping me and attempt to resolve the conundrum amicably, not proceed as you have.
@JamalS can't check what happened in your case right now, but we regularly delete full answers to homework questions - I've certainly done so many times - it's not unusual
It's very clumsily worded. I guess it means that given our experience of seeing Flash decline can we use this experience to predict what other technologies are in terminal decline.
So that means, like almost everything in my life, I have until 2020 to get every single thing I have came across in my life to backup into the time capsule
@DawoodibnKareem I've not heard it from any native speaker but I find it fairly regularly from foreign speakers from a variety of linguistic backgrounds.
Mostly spoken, but sometimes written down as well.
I can understand it in a foreigner whose first language is a Romance language, because of the pervasiveness of subjunctive forms in those languages. In my part of the world, these are a rarity. I've only ever heard it said by Americans.
I'm sure that there are many linguists who have studied the distribution of the use of different tenses in different varieties of English.
The "would and should" section on this page seems to support my thesis.
This has been bugging me for some time; I tried to look for previous questions here but my language tools may not be sharp enough to phrase my query correctly so please forgive me if this has already been posted here.
Speakers of English as a foreign language tend to incorrectly use the word "wo...
No - the part that I find fascinating is that you've heard it from people whose first language is neither American English nor a romance language. I'm interested to know why they thought it was right.
As for what goes on on English.SE, those people over there are all a bit weird.
Come to think of it, I'm not sure who the English.SE moderators are at the moment - I think Tom Christiansen and Kit Fox, but I'm not sure whether there are others.
This is a "historical question" more than it is a research question, but was the classical reduction to order-finding in Shor's algorithm for factorization initially discovered by Peter Shor, or was it previously known? Is there a paper that describes the reduction that pre-dates Shor, or is it ...
@JohnRennie the Smithsonian piece is OK if only because they continually make it explicit that they're only parroting content from the New Scientist piece
but then the New Scientist goes and says
> They measured the entire ejection of electrons from a helium atom from start to finish with zeptosecond precision ($10^{-21}$ seconds), marking the smallest time slot ever measured.
