joshuaronis

I guess that'll come with time. Anyways, thanks a lot David! I'll probably be back with more questions when I get into formal epsilon-delta proofs!

Actually, wait a sec...no, the more I think about it, that does seem impossible. If the derivatives in the $x$ and $y$ directions are zero, and the slope is not $0$ in any combination of those directions...then it MUST have a sharp bend...I think...bleh, I just wish I could visualize it a bit better.

...and then downwards as we rotate around, although its still differentiable.

Got you! That's pretty cool though, that there IS a way to prove that functions are differentiable...I mean, at least to me who hasn't had much exposure to it, it seems a little magical that we could somehow show that it doesn't stick out at angles in ANY direction. Also, although I understand it a lot better now, it still seems like there would be a way for the partial derivatives to (as an example) be zero in the $x$ and $y$ directions, but the surface curves a little bit upwards and....

Look at my update to the question, and let me know what you think!

@DavidK thank you!!! Yes you're right. And I think you also predicted correctly that I have no formal introduction to multi-variable calculus yet, just references here and there from a physics book. However, I now understand your answer a LOT better after watching some lectures on YouTube!

Hey David. Yep, that's what I meant...but, lets consider just convex functions then...look at my update to the question! How can I tell that lines on the plane in some other direction (besides the $x$ and $y$ directions) won't cross the surface?

... since the plane agrees with the slope of the surface in those two directions, how do I know that lines on the plane won't cross the surface if we draw them in another direction?

Okay, let me try again. As someone pointed out in the comments, that the tangent line didn't cross the function only applied if the function is convex (or concave) like $x^2$. Lets try making a tangent line to $f(x)=x^2$. If the tangent line agrees with the slope in the one direction it can agree with the slope (the $x$ direction), I know that the tangent line won't cross the parabola. However, for the graph of $f(x,y)=x^2$, although I know lines on the plane won't cross the function when moving in the $x$ direction or in the $y$ direction...

So tangent lines drawn on the plane in those two directions will be tangent to the surface. But, how can I intuitively see that tangent lines drawn on the plane in ANY direction will be tangent to the function at that point? Also, in my previous comment, I meant "take the function $f(x) = x^2$" the first time. Thanks again.

Hmm...but I can't see it even for very simple functions. In your answer, you say that we don't have to do a delta-epsilon proof in single-variable analysis. But, for example, take the function $f(x,y)=x^2$. If a tangent line agrees with the slope $2x$ at a certain point...well, there's only one slope to agree with, so I KNOW that the line will be tangent. But, now take the surface $f(x,y)=x^2$. A tangent plane at a certain point must have a slope of $2x$ in the $x$ direction and $0$ in the $y$ direction.

Thanks David! Alright, if I understand your answer, the definition of a function being differentiable at a certain point is that it has a tangent plane. Once we identify a plane that agrees with two slopes at that point, that plane MUST be the tangent plane at that point, since the two slopes uniquely identify the plane. Okay, I accept that. However, at the end you say that there are other ways to show differentiability without having to check the slope...is there a way to connect those "ways" with the tangent plane "definition" intuitively? Thanks!

@knzhou is really good too...he's been helping me, but I've been bothering him too much with high school physics questions recently. K, I'll check back here later. Thanks anyways!

I'd post it on the main site...but, there's too many there already, and I've already posted like 3 questions in the span of a day that are all very similar

Hey, anyone have time to help me with a relativity thought experiment?

I hope I'm explaining myself well there - at first, I thought the answer would just be $\frac{L_PV}{C^2}$ seconds ahead...but it can't be, since the Alice that sees THAT Bob (the one that sees her pass the left clock at $t=0$ on the left clock) must be some distance behind the left clock...

Question 1: Thte Bob which Alice sees as she passes the left clock is $\frac{L_PV}{C^2}$ seconds ahead from the Bob that sees Alice as she passes the left clock. Additionally, the Alice which Bob sees as she passes the left clock is......how far ahead from the Alice which sees THAT Bob?

Hmm...sorry, the above isn't very coherent, and I haven't posed any questions well...I'm still organizing my thoughts, but just typing this out is helping me a lot. I'll try thinking a bit more and seeing if I can pose concrete questions.

So I understand all that...pretty much, I understand everything from Alice's point of view, but I'm confused about Bob's point of view...when Alice first passes the left clock, Alice sees $\frac{L_PV}{C^2}$ seconds into the Bob that sees her future. However, from the point of view of the Bob that sees her, he is seeing $\frac{L_PV}{C^2}$ into Alice's future...oh, wait, that makes sense! The Alice that sees the Bob that sees her passing the left clock hasn't yet passed the left clock!

Sure enough, $\frac{L_P}{V\gamma^2} + $\frac{L_PV}{C^2} = $\frac{L_P}{V}$$ They agree on the time on the right clock once Alice gets there.

Well, anyways, lets continue. When Alice reaches Bob, he will have measured a time of $\frac{L_P}{V}$ to have passed. Alice will have measured a time of $\frac{L_P}{V\gamma}$ to have passed for her, while she will have seen Bob age by $\frac{L_P}{V\gamma^2}$, which is less than Bob felt himself age by, but is possible because the Bob she saw at when she first passed the left clock was $\frac{L_PV}{C^2}$ seconds into the future of the Bob that was looking at her as she passed the left clock.

There goes my first question...what does the Bob which Alice is seeing as she passes that left clock see? Presumably, he sees her some distance ahead from the left clock...right? How far ahead?

At the instant that Alice passes Jack's left clock, she sees Bob's right clock $\frac{L_PV}{C^2}$ seconds ahead from his left clock. What this means is that for Alice, at the instant she passes the left clock, the Bob she is looking at is $\frac{L_PV}{C^2}$ seconds older than the Bob that is seeing her pass that left clock actually is.

At a time $t=0$, Alice passes by his left clock, and she's travelling towards him at a really, really fast velocity $V$.

Say that Bob has a "left clock" and a "right clock". They are a distance of $L_P$ away from one another, according to him, and are synchronized with one another, according to him. Bob is stationed at his right clock.

.....okay, scratch all that...back to the basic Alice and Bob scenario, because I confused myself again... :(

Hmm...actually, nevermind. I was thinking maybe there was some way for Jill to see Jack react to her sending the message, and then not send it to see what would happen...but now I can't come up with it.

Lets say that $\frac{L_PV}{C^2}$ time intervals into the future, a bee stands on his shoulder. Bob is allergic to bees. So, Alice sends a light pulse to Bob that indicates to him to brush the bee off, so that the bee doesn't sting him.

Therefor, the Bob she sees at a distance of $\frac{L_p}{\gamma}$ away (due to length contraction) is a Bob $\frac{L_pV}{c^2}$ time units (whatever they are) in the future from the Bob that's currently there, seeing her a distance of $L_p$ away from him.

Lets say Bob is $L_p$ away from Alice, as measured by him, and she's travelling towards him at a really, really fast velocity $V$.

@knzhou alright, that wasn't so long, but I slept on it, and now this whole Alice being in Bob's future thing is really blowing my mind.

Will do! Thanks again, I'll probably be bothering you again with questions soon, as we're starting relativity in class soon! Adios and (idk where u are, but its night here) goodnight!

...Alice measures passes for her. $\frac{L_p}{v\gamma^2}$ - The time which Alice measures passes at the location $L_P$ away from Bob. $\frac{L_pV}{c^2}$ - the amount by which Alice's now is in Bob's future (or he is in her past, same thing) at a distance of $L_P$ (according to Bob) in the direction in which Alice is travelling as she passes him, and correspondingly the amount by which Alice is in Bob's past (or he is in her future, same thing) as she gets to that distance away from him, at his position. Am I missing anything, or is that all? Thanks again @knzhou !!!!!!!!!

@knzhou awesome! Thank you for helping me thus far! I want to just ask these last questions to make sure I'm getting everything correct...*(I'd start a chat, since I know we're not supposed to have such long comments, but I don't know how, and its also just not a normal question...I just want to see if I have things right):* Alice flies past Bob, as described in the situation in the comments above, at a velocity of $V$, and travels a distance (as measured by Bob) of $L_p$ away from him. $\frac{L_p}{v}$ - The time which Bob measures passes. $\frac{L_p}{v\gamma}$ - The time which...

...must've passed for Bob, while at the same time it was possible for Bob to conclude that if $\frac{L_p}{v}$ seconds happened for him, $\frac{L_p}{v\gamma}$ seconds passed for her? How can it be that Bob thinks more time passed for him than for her, and Alice thinks more time passed for her than for him? AND NOW I FINALLY UNDERSTOOD! When Alice was first passing past Bob, events happening for her NOW a distance of $\frac{L_P}{\gamma}$ away (from her point of view) and a distance of $L_P$ away (from Bob's point of view) were still in Bob's future! Bob still had to wait $\frac{L_Pv}{c^2}$...

...for those events to happen! Since she was travelling towards a location where, in her NOW was still in Bob's future, its entirely possible that she concludes more time passes for her than passes for Bob, while Bob concludes that more time passed for him, because as she moved away from him, she was literally moving into locations that were in Bob's future but her now! Its like when Alice flies past Bob, she's not only moving away from him in time, she's also moving towards his future! What the heck?!! Relativity is so cool!

@knzhou holy shish kebabs I just had a breakthrough in understanding. Okay, so Alice is shooting past Bob in her space-ship at a velocity of $v$. She's going to travel a distance of $L_p$ according to Bob, and a distance of $\frac{L_p}{\gamma}$ according to her. The time it will take for her to travel that distance is $\frac{L_p}{v}$ according to Bob, and $\frac{L_p}{v\gamma}$ according to her. Alright, so far so good...however, what was confusing me was how it was possible that from her point of view, if $\frac{L_p}{v\gamma}$ seconds passed according to her, $\frac{L_p}{v\gamma^2}$ seconds...

@knzhou In the tree's frame,the area enclosed by the parallelogram made by two parallel lines slanted lines, each signifying events at the same place for the car, and two other parallel slanted lines, each signifying events at the same time for the car, stays the same when we make those lines vertical and horizontal respectively (when we switch into the car's space time diagram).

@knzhou Thanks, I will! I gave up with the illustrated guide though :( ....I got too confused...I'm going through Morin's now, I really like his lively style. I didn't really like SpaceTIme Physics, because his approach was wayyy too convoluted (almost like he needs props to explain things...I don't know why the author brings in so many external references. Do you have any other recommendations? Thanks!

@knzhou (and I'm putting this in a separate comment because this is the heart of my question) I understand that when we switch into the car's diagram, lines at the same time for the car must become horizontal, and lines at the same place must become vertical. But why is it that when we switch into the car's diagram, we need to do so in such a way that the areas stay the same? Thanks!

@knzhou so I took your advice started reading an illustrated guide to relativity...and I'm pretty confused. I'm on page 96...(in case you have the book)...where we have a space time diagram from the point of view of a tree, and a car moving past the tree. I understand that slanted lines parallel to the car's world-line in the tree diagram are in the same place from the car's point of view, since they correspond to points that are all the same distance away from the car at any point in time. Additionally, I understand that slanted lines, slanted in such a way that two photons shot from the...

...origin when the car passed the tree are the same distance away from the car at any point in time, correspond to points that are all happening at the same time for the car. So, events that are moving to the right as time passes for the tree, are staying in the same location for the car, and events that are happening later in time for the tree that are in the direction the car is moving, are already happening for the car.

@knzhou thank you for the advice. I'll let you know how it goes!

Okay I rewrote it:

Fixed it!

Alright, will do!

And also, does anyone know the answer? Thanks!

why?

And it got downvoted...