Timeline for Does Euler number $e$ have a role in kinematics?
Current License: CC BY-SA 4.0
15 events
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| Dec 22, 2018 at 22:51 | comment | added | BioPhysicist | @Sierra in that system $\omega$ can't represent an average velocity or period because the units are not the same, plus the displacement and velocity are growing exponentially. In that system $\omega$ just characterizes the exponential growth. I guess you could say in a time $1/\omega$ the displacement and velocity increase by a factor of $e$ though. | |
| Dec 22, 2018 at 22:37 | comment | added | Sierra | So... focusing on $F=kx$ (non-oscillatory motion), where there is no cycle, may the reason to still use $\omega$ be that it is regarded as a sort of average velocity within the relevant time lapse, which would thus act as the period? | |
| Dec 22, 2018 at 17:59 | comment | added | BioPhysicist | @Sierra It seems like your are confusing angular frequency with angular velocity (which makes sense since they are both typically denoted by $\omega$). Motion under the force $F=-kx$ exhibits ocsillations on a straight line with an angular frequency $\omega=\sqrt{k/m}$. When $F=kx$ then $\omega$ still characterizes the motion, but it's not viewed as a frequency. Neither example involves an angular velocity $d\theta/dt$, which it seems like you think $\omega$ represents here. | |
| Dec 22, 2018 at 17:44 | comment | added | Sierra | Once again I said "straight line" when I meant "non-oscillatory". Anyhow, I think that you answered: if the F were not restorative, if the motion were not oscillatory, you would not use \omega, you would use v, right? | |
| Dec 22, 2018 at 11:28 | comment | added | BioPhysicist | @Sierra In knzhou's example if $F=-kx$ then $\omega=\sqrt{k/m}$ would represent an angular frequency and the motion would still be along a straight line. We can call $\omega$ an angular frequency because in that case there would be oscillations. When $F=kx$ the force is no longer restorative, but it's still useful to used the defined variable $\omega=\sqrt{k/m}$. In other words, $\omega$ helps determine a quality of the motion, and then depending on that motion we can call it an angular frequency. | |
| Dec 22, 2018 at 8:25 | comment | added | Sierra | But how do you adjust to different conditions? If we must introduce here another exponent ω, like knzhou did, that looks fine to me, that'd be the way to tweak e to reflect particular conditions (it plays the role of interest rate, right?), but ω usually denotes angular frequency and here we have a velocity in a straight line… | |
| Dec 22, 2018 at 8:23 | comment | added | Sierra | Yes, if the infinite terms of the sequence converge to 0 (like here, since factorials always grow faster than powers), the sequence can converge to a finite value (e^t in this case). The question was as Zeno-like one: whether the infinite terms is a list of things to be done, which would be impossible, or a way of describing what has been done. If you tell me that dwindling acceleration/jerk/snap/crackle… formula describes a person’s displacement when he is stepping more and more on the gas pedal in proportion to r, that is fine, that would answer the question. | |
| Dec 10, 2018 at 17:24 | comment | added | BioPhysicist | @Sierra Yes you would get those values ad infinitium. This isn't an issue because infinite sums can still converge. | |
| Dec 10, 2018 at 17:10 | comment | added | Sierra | My question is that this series, with value e or e tweaked however you may imagine, reflects a situation with infinite derivatives that looked unreal. I then wondered if the case mentioned by knzhou does fit in spite of all, why is it and if another example would be one of those I mention (someone stepping on gas or brakes in proportion to displacement…). Do you get in those case jerk, snap, crackle, pop and so on ad infinitum and of dwindling values? | |
| Dec 10, 2018 at 17:05 | comment | added | Sierra | I am aware that e is just 2.718 + infinite decimals... and you can find a similar number anyhwhere (eg in simple interest), although probably not with exactly with same digits if the underlying situation does not match with what you call its dynamics. I am also aware that its dynamics are a function whose rate of change is itself. In fact, the Newton series that is included in the OP has the characteristic that for each term the derivative is the previous one. This series assumes 1 in x, v and t and gives e but you can tweak e's exponent and adapt to any other situation... | |
| Dec 10, 2018 at 14:09 | comment | added | BioPhysicist | @Sierra What is more interesting, and what I thought you were going for, is looking at instances where $e$ is involved in the underlying dynamics of the problem. When there is something changing whose rate of change is proportional to itself. In other words, as just a number there really is nothing special about $e$. It is just a real number like all other real numbers. It becomes important when you start considering "special" functions like $e^x$ whose rate of change is equal to itself. | |
| Dec 10, 2018 at 14:07 | comment | added | BioPhysicist | @Sierra I mean getting the actual position or velocity to be $e$ is pretty contrived. In my example you can find a time where $x$ or $v$ is equal $e$ (with appropriate units). Even objects moving at a constant velocity will have a time where the distance traveled is equal to $e$. For example, for a particle moving with a constant velocity $v=e$, then after one second $x=e$. For a particle moving at constant velocity $v=e/2$, then after two seconds $x=e$. There are many possibilities where you can find some point in time where a value is $e$. | |
| Dec 10, 2018 at 8:34 | comment | added | Sierra | Thanks, in the end I accepted the other answer because there e appears alone in the solution and signifying displacement. But can you please clarify this? In the last comment I asked knzhou if he could build a similar example with a car speeding up. I could ask from you a similar clarification: what if the drag force were proportional to displacement, what if a car is progressively stepping on the brake in proportion to x? If x starts at 1 and v starts at 1, after t=1, would x be e? | |
| Dec 3, 2018 at 23:52 | vote | accept | Sierra | ||
| Dec 9, 2018 at 8:36 | |||||
| Dec 3, 2018 at 22:54 | history | answered | BioPhysicist | CC BY-SA 4.0 |