Proving Limit of t to 0 as a/v Goes to Infinity/1

[yuiop]

Hi,

I am trying to prove that t goes to zero in the limit that a goes to infinite in this equation:

$$1/a \, \, arcsinh(aT)$$

or alternatively that t goes to zero in the limit that v goes to 1 in this equivalent equation:

$$t = T \sqrt{(1-v^2)}/v \, \, arcsinh(v/\sqrt{(1 - v^2)})$$

Just for the record, T is physically related to a and v by the following equation:

$$T = \frac{v}{a\sqrt{1-v^2}}$$

Now I have tried applying L'Hopital's rule to try and obtain a form that is not indeterminate, but it gets into an alternating pattern that makes it clear that that no matter how many repetitions of the rule are applied it is not going to to resolve to a determinate limit in this form.

Now if I take the hyperbolic sine of both sides I appear to be able to find a limit for sinh(t) which is equal to 0. Is it valid to then claim that the limit of t itself is zero when a=infinite or v=1?

In other words, if it can be shown that:

$$\lim_{v\to 1} sinh\left ( t \right ) = 0$$

does it automatically follow that:

$$\lim_{v\to 1} t = arsinh( 0) = 0$$

?

If not, any other ideas of how to find the limit of t in this example?

 

[Mute]

If you start with

$$t = \frac{\mbox{arsinh}(aT)}{a}$$

(with T constant) and take the limit a -> infinity, you should not get an alternating pattern when doing L'Hopital's rule. Do you know/are you using the correct derivative for arsinh?

$$\frac{d}{dx}\mbox{arsinh}(x) = \frac{1}{\sqrt{1+x^2}}$$

(P.S. - some trivia: the correct term for inverse hyperbolic functions is arsinh, arcosh, etc. The 'ar' stands for area, whereas for regular trig functions arcsin, arccos, etc, the 'arc' stands for... well, arc. You wrote arsinh the last time, so maybe you already know this and the first two arcsinh's were habit from writing arcsin. :) )

 

[yuiop]

If you start with

$$t = \frac{\mbox{arsinh}(aT)}{a}$$

(with T constant) and take the limit a -> infinity, you should not get an alternating pattern when doing L'Hopital's rule. Do you know/are you using the correct derivative for arsinh?

$$\frac{d}{dx}\mbox{arsinh}(x) = \frac{1}{\sqrt{1+x^2}}$$

(P.S. - some trivia: the correct term for inverse hyperbolic functions is arsinh, arcosh, etc. The 'ar' stands for area, whereas for regular trig functions arcsin, arccos, etc, the 'arc' stands for... well, arc.)

I considered using the method you suggest, but the trouble is with acceleration (a) going to infinite the acceleration time T goes to zero, so x= a*T = [itex]\infty*0[/tex] so I do not know whether to take the limit of x to infinity or zero. ?

 

[yuiop]

I considered using the method you suggest, but the trouble is with acceleration (a) going to infinite the acceleration time T goes to zero, so x= a*T = [itex]\infty*0[/tex] so I do not know whether to take the limit of x to infinity or zero. ?

Actually, I have just noticed that I defined T as:

$$T = \frac{v}{a\sqrt{1-v^2}}$$

so when a goes to infinite and assuming v goes to 1 (where 1 is the speed of light) then the limit of T is:

$$T = \frac{1}{\infty \sqrt{0}}$$

so this implies x is:

$$x = a*T = \frac{\infty*0}{\infty \sqrt{0}}$$

which does not help much. hmmm... where does that leave us?

(P.S. - some trivia: the correct term for inverse hyperbolic functions is arsinh, arcosh, etc. The 'ar' stands for area, whereas for regular trig functions arcsin, arccos, etc, the 'arc' stands for... well, arc. You wrote arsinh the last time, so maybe you already know this and the first two arcsinh's were habit from writing arcsin. :) )

Yep, I meant arsinh throughout. My bad


<EDIT> I wil have to think about your suggestion of using T=constant. It certainly makes things simpler and just might work. If the terminal velocity (v) at the end of the acceleration period in coordinate time (T) is unknown (or unassumed), then I guess T can be a constant. Infinite acceleration is physically impossible, so I guess it might be possible that v is not limited to the speed of light in this impossible situation. Is there any other way to find limits in these sort of situations, besides L'Hopitals rule?

 

[Mute]

I assumed T was meant to be constant as you gave the two forms of t and an expression which relates T to a and v. One of T, a and v has to be constant, otherwise the limit is not well defined. If a -> 0 and v -> 1 independently, you can't evaluate the limit. If you plug in for T, you have

$$t = \frac{1}{a}\mbox{arsinh}\left(\frac{v}{\sqrt{1-v^2}}\right)$$

If you take v-> 1, a->infinity, you have no way of determining the limit unless you know how a depends on v as it tends to the limit. Does it go as a=1/(v-1)^2? a=exp(1/(v-1)^2)? How a depends on v is very important to what the limit turns out to be. Hence, I assumed that the expression relating T, a and v was defined for T constant, since you were interested in the a and v limits.

This is similar to problems in statistical mechanics, for example, where one takes the number of particles to infinity, as well as the volume they are contained in, but done in such a way that the total density remains constant: $\rho = N/V$ is fixed as N and V grow large, implying $N = \rho V$.

 

[yuiop]

I assumed T was meant to be constant as you gave the two forms of t and an expression which relates T to a and v. One of T, a and v has to be constant, otherwise the limit is not well defined. If a -> 0 and v -> 1 independently, you can't evaluate the limit. If you plug in for T, you have

$$t = \frac{1}{a}\mbox{arsinh}\left(\frac{v}{\sqrt{1-v^2}}\right)$$

If you take v-> 1, a->infinity, you have no way of determining the limit unless you know how a depends on v as it tends to the limit. Does it go as a=1/(v-1)^2? a=exp(1/(v-1)^2)? How a depends on v is very important to what the limit turns out to be. Hence, I assumed that the expression relating T, a and v was defined for T constant, since you were interested in the a and v limits.

The particular physical example we are trying to analyse in a physics thread, is t when a particle or rocket accelerates almost instantaneously, implying infinite acceleration a(a) and infinitesimal coordinate turn around time (T). Initially we were assuming infinite acceleration meant the terminal speed had to be the speed of light, but upon reflection, going from 0 to +0.4c in "almost no time at all" implies "almost infinite" acceleration, without v being necessarily equal to c=1, where c is the speed of light.
Plugging in the numbers for progressively smaller (T) and larger (a) definitely shows a trend towards proper time (t) going to zero, but proving it in the limit is proving frustratingly slippery. So in this case, you can take the terminal velocity (v) after the acceleration phase to be a constant, but that leaves T and a dependent on each other. For finite termianl v, what is the limit of t as a goes to infinite and T goes to zero? Does any of that information help any?

$$t = \frac{1}{a}\mbox{arsinh}\left(\frac{v}{\sqrt{1-v^2}}\right)$$

If you take v-> 1, a->infinity, you have no way of determining the limit unless you know how a depends on v as it tends to the limit. Does it go as a=1/(v-1)^2? a=exp(1/(v-1)^2)? How a depends on v is very important to what the limit turns out to be. Hence, I assumed that the expression relating T, a and v was defined for T constant, since you were interested in the a and v limits.

Now that I have realized v is finite and constant and v<1, a is defined by:

$$a = \frac{v}{T\sqrt{1-v^2}}$$

and we are looking for the limit of t as T goes to zero for constant v<1.

 

[yuiop]

I am trying to prove that ...
... t goes to zero in the limit that v goes to 1 in this equivalent equation:

$$t = T \sqrt{(1-v^2)}/v \, \, arsinh(v/\sqrt{(1 - v^2)})$$

Taking into consideration my musings in the last post, it would probably be better to pose the question as what is the limit of t as T goes to zero for constant v<1 in the above equation.

<EDIT>Actually, I think I have answered my own question now. The limit of t in the above equation is zero when T goes to zero, in the given conditions, without requiring L'Hopital's rule. Would you agree?

 

[starthaus]

Hi,

I am trying to prove that t goes to zero in the limit that a goes to infinite in this equation:

$$1/a \, \, arcsinh(aT)$$

It is very easy, this is an indeterminate case of $$\infty / \infty$$ so u will need to apply l'Hospital rule one time.

 

[yuiop]

It is very easy, this is an indeterminate case of $$\infty / \infty$$ so u will need to apply l'Hospital rule one time.

and what answer do you get?

See my argument in post #3 as to why it is not this simple.

Also, are you saying that:

$$arsinh(\infty*0) = \infty$$

Can you justify that?

 

[starthaus]

and what answer do you get?

you need to learn the l'Hospital rule and you'll find out.

Also, are you saying that:

$$arsinh(\infty*0) = \infty$$

Nope, what gives you this idea?

 

[yuiop]

you need to learn the l'Hospital rule and you'll find out.

Considering I introduced L'Hospital's rule in post 1 of this thread and considering you were obviously unaware of the rule in the other thread when you said t is indeterminate in the case when the acceleration is infinite, your answer is pathetic.

Also, are you saying that:

$$arsinh(\infty*0) = \infty$$

Can you justify that?

Nope, what gives you this idea?

We were discussing the proper time of a particle that reverses direction in infinitesimal coordinate time T.

This implies that in the limit that (T) goes to zero,

$$1/a*arsin(aT) = \frac{arsinh(\infty*0)}{\infty}$$

Since you are claiming that:

$$1/a*arsin(aT) = \frac{arsinh(\infty*0)}{\infty} =\frac{\infty}{\infty}$$

this suggests that you have concluded:

$$arsinh(\infty*0) = \infty$$

Then again it is hard to know what you have concluded, because you are being deliberately vague and unhelpful as usual.

 

[Mute]

Taking into consideration my musings in the last post, it would probably be better to pose the question as what is the limit of t as T goes to zero for constant v<1 in the above equation.

<EDIT>Actually, I think I have answered my own question now. The limit of t in the above equation is zero when T goes to zero, in the given conditions, without requiring L'Hopital's rule. Would you agree?

If you regard v as fixed while taking t to zero, then by

$$t = T \frac{\sqrt{(1-v^2)}}{v}\mbox{arsinh}\left(\frac{v}{\sqrt{(1 - v^2)}}\right)$$

yes, t goes to zero as T goes to zero, as the arsinh is just a constant.

 

[starthaus]

Considering I introduced L'Hospital's rule in post 1 of this thread and considering you were obviously unaware of the rule in the other thread when you said t is indeterminate in the case when the acceleration is infinite, your answer is pathetic.

You are getting personal again.

This implies that in the limit that (T) goes to zero,

Not for physical cases, only for your unphysically contrived case.

$$1/a*arsin(aT) = \frac{arsinh(\infty*0)}{\infty}$$

L'Hospital rule applies for one variable only, I thought that you claimed you knew calculus.

Since you are claiming that:

$$1/a*arsin(aT) = \frac{arsinh(\infty*0)}{\infty} =\frac{\infty}{\infty}$$

Once again, I am not claiming the nonsense you write above. What I am claiming is that the case
$$\frac{arcsinh(aT)}{a}=\frac{\infty}{\infty}$$ for $$T$$ finite and non-zero and $$a-> \infty$$ can be solved through a simple application of l'Hospital rule in one simple iteration. If you knew l'Hospital, you would have long solved it.

this suggests that you have concluded:

$$arsinh(\infty*0) = \infty$$

No, this is your own personal misunderstanding.

Then again it is hard to know what you have concluded, because you are being deliberately vague and unhelpful as usual.

Well, in order to know what I have concluded, you need to learn basic calculus, until you do so, we can't have a meaningful discussion. You need to learn how to calculate the limit:

$$\frac{arcsinh(aT)}{a}$$ for $$T$$ finite and non-zero and $$a-> \infty$$

The above is a simple exercise in high school calculus. The answer is 0. Now, all you have to do is to learn calculus.

 

[starthaus]

the trouble is with acceleration (a) going to infinite the acceleration time T goes to zero,

Err, no. T is fixed , doesn't "go to zero" (unless you cling to your unphysical scenario contrived in the other thread). This is one of the reasons why you are getting meaningless answers.

 

[yuiop]

Well, in order to know what I have concluded, you need to learn basic calculus, until you do so, we can't have a meaningful discussion. You need to learn how to calculate the limit:

$$\frac{arcsinh(aT)}{a}$$ for $$T$$ finite and non-zero and $$a-> \infty$$

The above is a simple exercise in high school calculus. The answer is 0. Now, all you have to do is to learn calculus.

So you finally agree that JesseM's claim that proper time goes to zero as the acceleration goes to infinite.

Below is a reproduction of your post #160 in other thread https://www.physicsforums.com/showthread.php?t=422350&page=10

(I have changed all occurrences of coordinate time $T_a$ to T to be consistent with this thread.)
It is clear that you did not know L'Hopital's rule at the time you wrote that post (Presumably you found out about it when you Googled the term having read my post in this thread) or you were being dishonest in your claim that jesseM's claim that the proper time goes to zero when the acceleration is infinite, when you secretly knew all along that he was correct.

This is actually an example of you making a pretty clear physics error--do you not understand that as the acceleration approaches infinity the time needed to change from one cruising velocity to another approaches zero (instantaneous acceleration), which means the proper time elapsed in the acceleration phase approaches zero too?

Not so fast.

$$v=\frac{aT}{\sqrt{1+(aT/c)^2}}$$

so, $$a->oo$$ implies $$v->c$$, meaning $$\gamma->oo$$

so you are rushing to conclusions. You have a lot more work to do in order to find the answer.

you get $$\frac{2c}{a} \, \, arcsinh(\frac{v}{c*\sqrt{ 1 - v^2/c^2}})$$. Clearly, this does approach 0 in the limit as a approaches infinity, since a appears only in the denominator of the expression outside.

Actually this statement is pretty amusing, speaking of blunders. $$T$$ is the duration the rocket accelerates (see the wiki page), there is no reason whatsoever why $$T$$ should correlate to $$a$$.

I see your error, it is a repaet of the error that I flagged above:

T = 2v/(a*sqrt[1 - v²/c²])

What happens when $$a->oo$$? You are forgetting that , according to your starting point (see the formula for v), $$v->c$$ so, the limit is undetermined (not that this even the correct way of calculating $$T$$) . Your math has deserted you today.

It is clear from the bolded text, that JesseM is talking about an instantaneous change of velocity in an infinitesimal time period, so we are talking about the limit of T going to zero in this context.

Your claim that $$a->oo$$ implies $$v->c$$ is false.

Using:

$$v=\frac{aT}{\sqrt{1+(aT/c)^2}}$$

for finite non zero T.

$$\lim_{a \to \infty} (v) = \frac{aT}{\sqrt{1+(aT/c)^2}} = \frac{\infty}{\infty}$$

Applying L'Hopital's rule:

$$\lim_{a \to \infty} (v) = {\sqrt{1+(aT/c)^2}} = \infty$$

and not c as you claim.

-----------------------------------------

 

[starthaus]

So you finally agree that JesseM's claim that proper time goes to zero as the acceleration goes to infinite.

I never disagreed, I even pointed out to him the correct proof based on the application of the l'Hospital rule. See this post.

It is clear that you did not know L'Hopital's rule at the time you wrote that post

LOL.

 

[starthaus]

Your claim that $$a->oo$$ implies $$v->c$$ is false.

Using:

$$v=\frac{at}{\sqrt{1+(aT/c)^2}}$$

You mean
$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

right? The above was just one of your "typos", correct?

for finite non zero T.

$$\lim_{a \to \infty} (v) = \frac{aT}{\sqrt{1+(aT/c)^2}} = \frac{\infty}{\infty}$$

Applying L'Hopital's rule:

$$\lim_{a \to \infty} (v) = {\sqrt{1+(aT/c)^2}} = \infty$$

Err, you don't know how to apply the l'Hospital rule, you got the wrong result.

and not c as you claim.

-----------------------------------------

Your math is unravelling very quickly, the above limit

1. does not require the application of the l'Hospital rule, it can be calculated directly by dividing both the denominator and the numerator by $$a$$
2. The limit is indeed c, and not $$\infty$$
3. Your incorrect attempt at applying the l'Hospital rule shows that you don't know basic calculus. If you insisted in applying it (it isn't really needed , see point 1) and if you applied it correctly, you would have recovered the limit to be...c!

 

[yuiop]

This implies that in the limit that (T) goes to zero,

Not for physical cases, only for your unphysically contrived case.

We were considering an extreme situation of acceleration in a very short period but they are not completely physically contrived. Extreme accelerations of particles are routinely achieved in particle accelerators and probably in nature too in Supernova explosions and in collisions in the upper atmosphere from high energy particles.

Also, remember when we talk about "instantaneous acceleration" we do not actually literally mean infinite acceleration in exactly zero seconds. By "approaching infinite" we mean $1/\delta t$ and not exactly 1/0 and by "approaching zero" we mean an infinitesimal time period $\delta t$ where is the time period is sufficiently small that it can be considered negligible, but we don't mean "exactly zero". This is fundamental to calculus and the theory of limits.

$$1/a*arsin(aT) = \frac{arsinh(\infty*0)}{\infty}$$

L'Hospital rule applies for one variable only, I thought that you claimed you knew calculus.

I am aware of that. If you look again you will see that I had not applied L'Hopital's rule in the above statement.

Since you are claiming that:

$$1/a*arsin(aT) = \frac{arsinh(\infty*0)}{\infty} =\frac{\infty}{\infty}$$

Once again, I am not claiming the nonsense you write above. What I am claiming is that the case
$$\frac{arcsinh(aT)}{a}=\frac{\infty}{\infty}$$ for $$T$$ finite and non-zero and $$a-> \infty$$ can be solved through a simple application of l'Hospital rule in one simple iteration. If you knew l'Hospital, you would have long solved it.

I was already aware of L'Hopital's rule and that is why I mentioned it in the first post of this thread.Mule has already solved this for us for the case where T is non zero and finite independent constant. What I am disputing with you whether T is an independent constant. I have already shown in my last post, that if T is considered an independent non zero and finite constant then when the acceleration is infinite, the terminal velocity is infinite and not c as you have assumed.

If a particle accelerates from 0 to 0.8c with an acceleration of 10⁶*299,792,458 g = 9.8*10⁶ lightseconds/second², then T is a function of acceleration and the terminal velocity and not an independent constant. In this example usings units of c= 1 ls/s,

$$T = \frac{v}{a\sqrt{1-v^2/c^2}} = \frac{0.8}{9.8*10^6*0.6} \, \approx 1.036*10^{(-7)} \, \, seconds$$

 

[starthaus]

I was already aware of L'Hopital's rule and that is why I mentioned it in the first post

Yet you have no clue how to apply it. Actually, you don't even know how to calculate basic limits that don't even require the application of the rule. If you want to get involved in physics you need to learn math. The way you operate right now precludes you from learning physics. This is a friendly advice: learn math.

 

[yuiop]

1. does not require the application of the l'Hospital rule, it can be calculated directly by dividing both the denominator and the numerator by $$a$$
2. The limit is indeed c, and not $$\infty$$
3. Your incorrect attempt at applying the l'Hospital rule shows that you don't know basic calculus. If you insisted in applying it (it isn't really needed , see point 1) and if you applied it correctly, you would have recovered the limit to be...c!

I would ask you to demonstrate the above claim, but I know it is not in your nature to be helpful or clarify what you mean when asked.

 

[starthaus]

I would ask you to demonstrate the above claim, but I know it is not in your nature to be helpful or clarify what you mean when asked.

You would need to take a calculus class first. All the claims in my previous post are easily understood by someone who has basic understanding of limits. Since you demonstrated that you don't, it would be difficult, if not impossible, to explain that to you.

 

[yuiop]

You mean
$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

right? The above was just one of your "typos", correct?

I meant:

$$v=\frac{aT}{\sqrt{1+(aT/c)^2}}$$

to be consistent with the definition of T as coordinate time interval that the acceleration takes place over in the non accelerating frame. In the other thread the same quantity is variously represented by t or $T_a$. Yep, I missed one, when quoting from the other thread.

 

[starthaus]

I meant:

$$v=\frac{aT}{\sqrt{1+(aT/c)^2}}$$

to be consistent with the definition of T as coordinate time interval that the acceleration takes place over in the non accelerating frame. In the other thread the same quantity is variously represented by t or $T_a$. Yep, I missed one, when quoting from the other thread.

Same difference, if you knew how to calculate limits, you would get the correct result (hint: c).

Try this:

$$v=\frac{aT}{\sqrt{1+(aT/c)^2}}=c \frac{1}{\sqrt{1+(c/aT)^2}}$$

 

[yuiop]

1. does not require the application of the l'Hospital rule, it can be calculated directly by dividing both the denominator and the numerator by $$a$$
2. The limit is indeed c, and not $$\infty$$
3. Your incorrect attempt at applying the l'Hospital rule shows that you don't know basic calculus. If you insisted in applying it (it isn't really needed , see point 1) and if you applied it correctly, you would have recovered the limit to be...c!

That works if T is an independent constant, but it is not because it depends on a.

 

[starthaus]

That works if T is an independent constant, but it is not because it depends on a.

Not in the general scenario. This is one of the many reasons why you aren't able to calculate any meaningful limit and you keep running around in circles.