limv(t) = a(t)·t
t→0
to get v(0)=0 which we all agree to. What we really want is v(0+) Here is the acceleration from the definition of dv/dt
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
v(t+dt) = v(t) + ∫a dt v(0+) = v(0) + ∫a dt v(0+) = ∫a dt Physics Monkey10-18-05, 10:29 PMIn the interval from t to t + dt, the velocity changes by dv = a(t*) dt. The mean valua theorem tells us that for some t* in the interval, this equation is exact no matter what size dt is. Of course we have in mind that dt is small (compared, say, to other timescales in the system). If we neglect terms of order dt² and higher, then the it doesn't matter where we evaluate the acceleration in the interval (t,t+dt). In other words, a(t+dt) dt = a(t) dt = a(t+.5dt) dt, etc and the error in each equal sign is of order dt². Thus as long as the trajectory is differentiable, it doesn't matter where you evaluate the acceleration when computing the change in velocity in the limit i.e. when you do an integral. This is a fundamental result of the theory of integration, but it is nontrivial as can be seen in stochastic differential equations were the point of evaluation in the interval does matter (leading to two inequivalent prescriptions which are the Stratonovich and Ito calculi). Aer10-18-05, 10:46 PMPhysics Monkey, are the following equivalent?
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
a(t) = limv(t+Δt)-v(t-Δt)
Δt→0 Δt
Aer10-18-05, 10:49 PMAnyway, my point is, that v(0+)=v(t+Δt) as Δt→0. and from the definition of a(0), we see that v(0+) cannot be zero, or else a(0) must be zero. Pete10-18-05, 10:50 PMv(t+dt) = v(t) + ∫a dt v(0+) = v(0) + ∫a dt v(0+) = ∫a dt You need some bounds on that integral. If you do it right, I believe you'll find that v(0+) = 0 Aer10-18-05, 10:50 PMWhich equals zero. Definately not! Pete10-18-05, 10:51 PMWhat does it equal, then? Pete10-18-05, 10:52 PMAnyway, my point is, that v(0+)=v(t+Δt) as Δt→0. and from the definition of a(0), we see that v(0+) cannot be zero, or else a(0) must be zero. Correction: v(0+)=v(0+Δt) as Δt→0 Your conclusion is a crock. Look at the definition of a limit, and try again. What's the limit of aΔt as Δt→0? This is basic stuff. Aer10-18-05, 10:53 PMv(0+)=v(0+Δt) as Δt→0 Good job, yes, t=0 was a given here. What's the limit of aΔt as Δt→0? v(0+) ≠ aΔt as Δt→0. v(0+) = ∫a dt What a crock. Look at the definition of a limit, and try again. I did and I explained to you that your limit was finding v(0), not v(0+). What does it equal, then? Have we defined any values? I don't believe we have other than to say that a(0)≠0. Do you agree that:
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
Pete10-18-05, 10:57 PMI did and I explained to you that your limit was finding v(0), not v(0+). Yes, you explain a lot of things. Poorly and incorrectly. Congratulations on discovering that v(0) = v(0+). Do you agree that:
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
Yep Aer10-18-05, 11:02 PMYes, you explain a lot of things. Poorly and incorrectly. Congratulations on discovering that v(0) = v(0+). Wrong, my explanation is correct and yours is wrong. Do you agree that:
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
Yep Which can be written in the following notation:
a(0) = limv(0+)-v(0)
Δt→0 Δt
So given your conclusion that v(0+)=v(0), a(0)=0. Are you saying now, that B has not accelerated according to A? Physics Monkey10-18-05, 11:07 PMAer, The two expressions are not the same. The first of your expressions is the definition of the derivative, and the second is actually twice the derivative. To see this you can add and subtract v(t) and group terms: v(t+dt)-v(t) gives one copy and v(t) - v(t-dt) gives another. Aer10-18-05, 11:12 PMAer, The two expressions are not the same. The first of your expressions is the definition of the derivative, and the second is actually twice the derivative. To see this you can add and subtract v(t) and group terms: v(t+dt)-v(t) gives one copy and v(t) - v(t-dt) gives another. OK, I didn't think they were the same but wasn't sure since they are limit cases. What about the following (they may seem equivalent but I don't think they are, or rather, t in a(t) is slightly differently defined for each):
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
a(t) = limv(t)-v(t-Δt)
Δt→0 Δt
Physics Monkey10-18-05, 11:13 PMIf one defines v(0+) as the limit dt -> 0 of v(dt) then v(0+) = v(0), this is essentially the definition of continuity. Of course, v(dt) is not zero for any finite dt (provided a is monotone), but in the limit it is zero if v(0) is zero. According to the mean value theorem, and neglecting higher order terms in dt, all the following are true v(dt) = a(0)dt = a(dt)dt = a(.5dt)dt, etc. Pete10-18-05, 11:13 PMWrong, my explanation is correct and yours is wrong. "Yes it is!" "No it isn't!" "Yes it is!" "No it isn't!" Which can be written in the following notation:
a(0) = limv(0+)-v(0)
Δt→0 Δt
I really don't think so. You're trying to take the limit on the top half of the equation before taking the limit on the bottom half. This form is closer to the truth:
a(0) = v(0+)-v(0)
lim Δt
Δt→0
It's still not completely correct (it is no longer possible to evaluate the function). I have no idea why you'd want to use that 0+ notation - it just doesn't seem useful. Physics Monkey10-18-05, 11:16 PMActually, those two are equivalent. In general anything of the form v(t+ c dt) - v(t- (1-c) dt) gives the derivative dv/dt at t (c in [0,1]). Pete10-18-05, 11:18 PMAer, You still haven't figured out the limit of aΔt as Δt→0. Pete10-18-05, 11:20 PMThe idea of 0+ is not my idea. You really think that is some idea that I cooked up? No, I certainly did not make up the notation and it is commonly used. Can you give me a reference to somewhere it is used? I'd be interested in seeing it in a useful context. Physics Monkey10-18-05, 11:26 PMPete, You will find the notation 0+ and 0- in the very famous book "A Course of Modern Analysis" by Whittaker and Watson. It was conventionally used to denote one sided limits and otherwise employed in the analysis of discontinuous functions. More recent treatments tend to use the notation ε and - ε for 0+ and 0-. In either case, the notation conventionally means that one should take the limit at the end of the day i.e. f(0+) = f(ε ) is short for lim x -> 0 (from above) f(x). Pete10-18-05, 11:27 PMv(0+) ≠ aΔt as Δt→0. v(0+) = ∫a dt This is informative. In this statement, you're implying that v(0+) depends not only on a(0), but on a changing value of a. But that disagrees with what you said earlier: The velocity at t=0+ is dependent on the instantaneous acceleration at t=0. It looks like you don't know what you think. Aer10-18-05, 11:28 PM"Yes it is!" "No it isn't!" "Yes it is!" "No it isn't!" I've not said one of the above phrases. Are you quoting yourself perhaps? I really don't think so. You're trying to take the limit on the top half of the equation before taking the limit on the bottom half. This form is closer to the truth:
a(0) = v(0+)-v(0)
lim Δt
Δt→0
It's still not completely correct (it is no longer possible to evaluate the function). I have no idea why you'd want to use that 0+ notation - it just doesn't seem useful. That is a weak argument. You take the limit of the top and bottom separately in this situation anyway. You are too boggled down with the 0+ notation. It is just shorthand when you are talking about limits (And is useful when the concept is not limits itself which unfortunately this thread has fallen too). Aer10-18-05, 11:32 PMv(0+) ≠ aΔt as Δt→0. v(0+) = ∫a dt This is informative. In this statement, you're implying that v(0+) depends not only on a(0), but on a changing value of a. But that disagrees with what you said earlier: The velocity at t=0+ is dependent on the instantaneous acceleration at t=0. It looks like you don't know what you think. Are you seriously suggesting that a, because I wrote it on the right side of the integral means it cannot be a constant? If you don't think it is a constant, then what is a, as a function of t? In this case, for v(0+), I think a is constant from the definition of a. Pete10-18-05, 11:34 PMPete, You will find the notation 0+ and 0- in the very famous book "A Course of Modern Analysis" by Whittaker and Watson. It was conventionally used to denote one sided limits and otherwise employed in the analysis of discontinuous functions. More recent treatments tend to use the notation ε and - ε for 0+ and 0-. In either case, the notation conventionally means that one should take the limit at the end of the day i.e. f(0+) = f(ε) is short for lim x -> 0 (from above) f(x). Thanks PhysicsMonkey, that makes sense. So it's immediately obvious that if v is continuous, v(0+) = v(0), right? Aer,
a(t) = limv(t+Δt)-v(t)
Δt→0 Δt
... can be written in the following notation:
a(0) = limv(0+)-v(0)
Δt→0 Δt
Not correct, since you're mixing notations and taking a limit twice. Try this:
a(0) = v(0+)-v(0)
0+
Aer10-18-05, 11:36 PMActually, those two are equivalent. In general anything of the form v(t+ c dt) - v(t- (1-c) dt) gives the derivative dv/dt at t (c in [0,1]). Yes, but what about when v(t=0)=0, v(0-)=0, but v(0+)=c? One definition will give a(0)=0 and the other a(0)=d where c and d are constants. Physics Monkey10-18-05, 11:37 PMSo it's immediately obvious that if v is continuous, v(0+) = v(0), right? Right. So long as v(0+) is defined as I have said, any contnuous function satisfies v(0-)=v(0+)=v(0) as a matter of definition. Pete10-18-05, 11:38 PMAre you seriously suggesting that a, because I wrote it on the right side of the integral means it cannot be a constant? If you don't think it is a constant, then what is a, as a function of t? In this case, for v(0+), I think a is constant from the definition of a. If a is a constant, then your first statement is false: v(0+) ≠ aΔt as Δt→0 and the second is unnecessary. Have you figured out the limit of aΔt as Δt→0 yet? Aer10-18-05, 11:39 PM
a(0) = limv(0+)-v(0)
Δt→0 Δt
Not correct, since you're mixing notations and taking a limit twice. Try this:
a(0) = v(0+)-v(0)
0+
I never intended to expand v(0+) back into a limit, but only to say that it is a value and plug that value in. In that respect, I am not taking the limit twice. Physics Monkey10-18-05, 11:40 PMAer, if v(0+) does not equal v(0) then the velocity is discontinuous and the derivative is not defined. Alternatively, if you're a physicist who likes to play it fast and loose, one can define the derivative to be a delta function. Either way, the derivative is highly singular and then it does matter where you evaluate your limits as you have pointed out. Pete10-18-05, 11:42 PMI never intended to expand v(0+) back into a limit, but only to say that it is a value and plug that value in. In that respect, I am not taking the limit twice. Then you're using the notation incorrectly. The notation 0+ directly implies taking the limit. Pete10-18-05, 11:43 PMI think this discussion is done. It's been nice chatting with you, Aer. You too, Physics Monkey. Aer10-18-05, 11:54 PMAer, if v(0+) does not equal v(0) then the velocity is discontinuous and the derivative is not defined. Alternatively, if you're a physicist who likes to play it fast and loose, one can define the derivative to be a delta function. Either way, the derivative is highly singular and then it does matter where you evaluate your limits as you have pointed out. Well, we do have a discontinuous function. dv/dt is 0 for t<0 and dv/dt is a constant for t≥0. Is that not a discontinuous function? If we take the limit from separate sides of t=0, we get different results. How can we get this if as you just claimed, v(0+)=v(0)=v(0-). Pete10-19-05, 07:58 AMIs that not a discontinuous function? It's a continuous function with a discontinuous derivative. If we take the limit from separate sides of t=0, we get different results. You get different results for dv/dt (which is discontinuous), but not for v (which is continuous). Aer10-19-05, 02:17 PMIt's a continuous function with a discontinuous derivative. dv/dt is the function I was talking about, the function that is defined from the limit I gave which is merely the definition of a derivative. dv/dt is a discontinuous function as I said the first time. It is not "a continous function with a discontinuous derivative" although, I am sure you mean, v is a continous function with a discontinous derivative.. You get different results for dv/dt (which is discontinuous), but not for v (which is continuous). Precisely what I said, dv/dt is discontinous. Physics Monkey10-19-05, 05:49 PMYes, dv/dt is discontinuous in your idealization, but v is still continuous and so v(0) = v(0+) = v(0-). Post ReplyCreate New Thread