Infinite acceleration?!

[wadi]

Originally posted by pv=nrt
Eternal, I disagree about the atoms at zero kelvin. At that themperature the atoms would be moving so slow they would appear to be still, while actually still moving. I think I remember seeing something about researchers getting down into the 5 K range, but that may be my mind playing tricks on me.

There are two really spiffy ways to do this, one involving step-by-step entropic cooling, and the other one involving some sort of crazy laser. At that tempature you're dealing more with vibrational and rotational energy than translation of the atom.

Originally posted by Telc@t
In the case you have stated, the object travels towards the wall at velocity x will have to come to a stop first before it starts to travel in the opposite direction

No it doesn't. The problem I stated is a common shortcut taken when dealing with collisions between two bodies one >> greater than the other. It assumes that at no time does the body in question (the smaller) have 0 velocity, infact the problem I posed specifically states that. Is this a realistic classical mechanical problem? No. Did I say it was? No. I may suck at physics, but it's nice to know that my reading comprehension's still somewhere in the stratosphere compared to yours. In truth I didn't want to get in to this at all, as my field is more chemistry and less physics, but if I have to....

Think of the simple particle-in-a-box problem dealt with in quantum mechanics . In this problem a particle has a defined energy represented by mass and velocity. As the particle collides with the walls of it's box it must change directions, but as previously stated this is a closed system and all energy is retained by the particle. In this case, you assume the particle shifts from (since this a one dimensional problem) velocity=something to velocity= -something over time 0. I'm wary of going further in depth about this, simply because I hated the fuck out of PChem and wiped most of it out of my brain.

Oh, and somebody mentioned dividing by zero in 7th grade gives you undefined. Too bad when you do it in 10th grade (calc) you get something different. Hooray for ---> 0.

[Epinephrine]

Someone needs to brush up on their physics

-Epi

[Fluffz]

Re: Re: Re: Infinite Acceleration

Originally posted by Sleepy Weasel
Actually velocity is change in distance/change in time, dx/dt. Acceleration is change in velocity over time, the second derivative. And yes your time is very very close to 0 at the initial movement, but the distance is also very very close to 0, so there is nothing remotely close to infinite acceleration about it. If there is movement, time has passed. If time is 0, there has been no movement yet and therefore no velocity yet. Just because we don't really perceive stuff that happens much quicker than say .1 seconds, you can't say something that happens in .00001 seconds happened in 0 seconds.

isnt he actualy right? from a mathematical point of view u use the dx/dt stuff to describe a funktion by splitting it into tiny parts. those parts have no expansion at the time axis but do have a value at the speed axis. this means at the point where the speed changes from 0 to something no time passes at all (since 0 isnt a point on the graph but just a point in the coordinate system no time has passed) -> Infinite Acceleration

some1 correct me please its been some time since i worked with numbers higher than 0 and 1

[EternalEntropy]

Re: Re: Re: Re: Infinite Acceleration

Originally posted by Fluffz
isnt he actualy right? from a mathematical point of view u use the dx/dt stuff to describe a funktion by splitting it into tiny parts. those parts have no expansion at the time axis but do have a value at the speed axis. this means at the point where the speed changes from 0 to something no time passes at all (since 0 isnt a point on the graph but just a point in the coordinate system no time has passed) -> Infinite Acceleration

some1 correct me please its been some time since i worked with numbers higher than 0 and 1

0 is a point on the graph, just like you can have 0 jelly beans in your hands. Again, GRAPH THE FUKING ACCELERATION AND YOU WILL SEE THAT FOR EVERY CHANGE IN SPEED THERE IS A CORRESPONDING TIME INTERVAL, NO MATTER HOW SMALL. THIS IS THE SAME FROM 0 TO WHATEVER. NO INFINITE ACCELERATION.
Good.

[Fluffz]

no sir mr caps writer guy i doubt that since this function does not allow the time to be 0 without being an undefined devision by zero. because of this there cant be a point 0 on the graph since in no time u cant have gained any speed can u?

but the graph has to start somewhere and in this impossible tiny intervall u get an impossible high acceleration (speed / time²).

furthermore no matter how little the speed is the time² will always be smaller.

result: drastical raise of acceleration, and if u make the interval small enough... acceleration = -infinite/-infinite² = 1/-inf = inf

btw u cant draw that

[Fit of Rage]

Originally posted by Fluffz
result: drastical raise of acceleration, and if u make the interval small enough... acceleration = -infinite/-infinite² = 1/-inf = inf

btw u cant draw that

Did that make sense to anyone else? Or is it just the liquor.

[Verthanthi]

Originally posted by poop juice
"No matter how small the time interval, or how slowly an object moves during that interval, it is still in motion and its position is constantly changing" - Peter Lynds

Is that supposed to convince me? There is no proof, certainly no mathmatical proof. You'll have to do better than that. I still think you're trying to argue that there is a point of reference for every object where it is in motion, but that's so trivial that it has NO bearing on any argument and cannot be used in this argument anyway.

Fluffz: Acceleration isn't velocity / time². For starters, it would be (v/t) or (x/t²) but it's neither of those. it's actually (dv/dt) or (dx/dt²). For those who haven't taken any calc, that means change in velocity over change in time. So while the change in time is in fact TEENY TINY, the change in velocity is also TEENY TINY.

[Fluffz]

Originally posted by Verthanthi
Fluffz: Acceleration isn't velocity / time². For starters, it would be (v/t) or (x/t²) but it's neither of those. it's actually (dv/dt) or (dx/dt²). For those who haven't taken any calc, that means change in velocity over change in time. So while the change in time is in fact TEENY TINY, the change in velocity is also TEENY TINY.

no, change in velocity over change in time would be shown by a tiny triangle in front of x and t. the d in front of it comes from the differential calculus. since u cant work with infinite small numbers in reality physicans dont care at all.

nevertheless it was late so i mixed up a few things. the limit value is not -infinite but 0 and yes the formular is x/t².


u still have to see that this function is not defined at time = 0.
so the first defined point is very very close to 0.

now divide 1/10^10 (x) trough (1/10^10)² (t) than try it with ^1000 (closer to 0). the number you get is very high but it is the acceleration close to the first defined point at which it would be infinite high. simple


but on the other hand i never studied maths or something so it could be wrong because basicaly i aint sure if t² is growing faster than x (which it should because of the ²). edit: i think the t² would sooner or later grow faster than any no matter how high (or in this case small) x

Anyway a mathematical proove with limits theory and l'Hospital is far beyond my knowledge

[Verthanthi]

Delta (the little triangle thingy) is used for general rate of change. d is used for instantaneous rate of change. Trust me, it is NOT outside the scope of physics. We used it all the time, and I was only in high school physics.

Furthermore, you've offered me no definitive proof that the there is no t=0. Here is my argument that there IS a t=0.

Let us say the object begins moving right after t=0. You say, then, that the graph of the velocity does not exist until RIGHT after t=0. Then, I ask you, where was the object before it started moving? Well, the object was just sitting there like a penguin on a television. Well, then it had a velocity, it was just 0. So we put a little plot marker ON t=0 with a value of 0. Then I repeat the question for a second before it started moving. Oh hey, again it's just sitting there. So a t= -1 we put another plot point with a value of 0. So there is a graph before it starts moving, it's just flat. And while that means we can't find a function to model this perfectly, the graph still exists.


Anyway, I'm right about it being a = dx/dt² for instantaneous acceleration. Either take calculus or trust me, because I'm right.

[pv=nrt]

Shade is so wrong there is no such thing as negative time. Owned!!

[Fluffz]

Originally posted by Verthanthi
Furthermore, you've offered me no definitive proof that the there is no t=0. Here is my argument that there IS a t=0.

i never sayed that there is no point t=0 or x=0 or v=0 but there is no acceleration at the point t=0.

the first point of acceleration you can describe with dx/dt² is short after t=0. this point has no time interval and there has no time passed before it BUT you can devide trough it without having a division by zero.
furthermore it is smaller than anything else in the formular (thanks to the ²) and since its under the division line everything else grows. (try 1 / 0.0000000001)

u can not imagine that its just a mathematical construct caused by the differential calculus. the limit value for dx/dt² t->0 is infinite. Proove me wrong by a mathematical formular.

dont hunt me down i didnt start this but i think im correct here :/

[Epinephrine]

What you're all forgetting is bullet time. Keanu will show you the way.

Seriously though here's some simple thought experiments that can help you picture why there's no infinite acceleration.

1) Elastic collision - i.e. ball bouncing off a wall

As the ball reaches the wall it does not instantaneously change direction. No sir. The ball actually slows down against the wall as the side of the ball hitting the wall gets minorly deformed (so ball goes from O to |) more or less). As this deformation is happening, you can picture the other side of the ball (the undeformed side) quickly decelerating to zero. When the ball rebounds, the deformation reverses itself and the ball accelerates in the opposite direction. There was no infinite acceleration. If you graph the acceleration of the elastic collision over time you'd get something of a parabola with limits on either side of 0.

2) Acceleration from Zero - The original question.

If you started from rest, and then went to say 1 m/s [EAST], then that means you accelerated [EAST]. Acceleration is roughtly Velocity/Time^2, and since the velocity change was just 1, the acceleration is also small as changing direction takes some time to do.

3) Inelastic collision - For easier picturing think of something "instantly" changing direction.

So you have an undeformable ball (not even sure if such a thing could exist, so pretend it does) moving at 5 m/s [East]. Suddenly it comes to a completely halt and goes 5 m/s [West]. Note that the change in velocity is 10 m/s [West].

Now the argument which proponents of infinite accelration tell us is that this 10 m/s change was done instanteously therefore 10/ZERO seconds^2 = infinite acceleration (or the rest to movement was also done instantenously). The problem with this is that it doesn't happen in zero time. There is no such thing as zero time (just bullet time, ask Keanu). If it did happen in zero time, the theory of relatively would be broken, as infinite acceleration requires infinite energy and would thus accelerate the ball instanteously faster than the speed of light.

Now why can't it be in zero time? This is where my explaination gets a bit unspecific as I'm going by what I read a few years ago. The smallest unit of time it the fermi, which can be characterised at approximately 1.0 * 10^-42 seconds. This is approximately the amount of time it takes for a quantum to change states (something like that). So for the matter to physically change from one direction to another and for that to actually register, it actually takes a bit of time, as that object doesn't physically exist in a different spot until 1.0*10^-42 happens! I'm sure there's also some kind of proof which you can invoke using duality of matter (the duality of matter and energy, and matter as a wave).

Now really those are just thought experiments. Common sense will tell you that things will rarely accelerate that fast, and take considerable amounts of time to do so. You can also calculate the amount of energy necessary to accelerate that fast using F=ma and then converting that to energy and you will realize that at certain accelerations it would be physically impossible for it to happen as there is not enough energy in the entire universe to make that happen.

This is all based on my first year physics class which I took 2 years ago. I guess someone with more knowledge can answer better but I hope my more layman answer helped.

-Epi

[ÆNIMA]

i'm too lazy to read anyone else's post


motion is relative, it's a law of physics that whatever is in motion must stay in motion, unless acted upon an outside force (usually friction) whatever sits still must stay still, unless acted upon by an outside force (like gravity)

why quantum mechanics needs to explain this...i have no idea, it makes sense in itself

[Verthanthi]

Originally posted by Fluffz
i never sayed that there is no point t=0 or x=0 or v=0 but there is no acceleration at the point t=0.

the first point of acceleration you can describe with dx/dt² is short after t=0. this point has no time interval and there has no time passed before it BUT you can devide trough it without having a division by zero.
furthermore it is smaller than anything else in the formular (thanks to the ²) and since its under the division line everything else grows. (try 1 / 0.000 0000001)

Fluffz, if an object moves 1 meter in 1x(10^-5) s as you suggested, that would be an average velocity of 100000 m/s. 100000 m/s in less than a millisecond, is it any wonder you're receiving massive acceleration? It requires massive acceleration.

Alright, let us say this models the function.
x(t) = 0 where t <= 0
x(t) = t^3 + t^2 + t where t > 0

that would mean that for t > 0
v(t) = 3t^2 + 2t + 1
a(t) = 6t + 2

Because the two don't match, when taking the limit you'd just use the right hand limit. Why? Because that's what we care about, the limit RIGHT AFTER t=0 right when it starts moving.

lim+ a(t) as t->0 = 2

One last attempt to convince you, although it isn't an actual proof. Suppose it starts moving at t=-1, then the problem would occur after one second of movement. What if it started moving at t=2, then the problem would occur TWO seconds after movement began. I know your proof is wrong, and I'll admit I can't pin down what it is that you're fucking up, but it's there.

[Verthanthi]

Also, a fun little "proof" that 1=2

let a = b
a² = ab
a² + a² = a² + ab
2a² = a² + ab
2a² - 2ab = a² - ab
2(a² - ab) = a² - ab
2 =1

Easy peasy to disprove, if you know what you're looking for. It's similar to what it is that you're trying to do.