Ant on an expanding balloon

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  • #1
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This isnt homework but nobody is going to believe that so here it is:

Homework Statement


at t=1 (or any nonzero value of t)
an ant starts moving across a balloon at velocity
v=c=1 (c is the speed of light)​
the radius of the balloon as a function of time is given by
r(t)=nct (n is a constant much larger than 1)​

how far does the ant move as a function of time?

Homework Equations





The Attempt at a Solution


the angular velocity of the ant is
ω(t) = v/r(t) = c/nct = 1/nt​
integrating to get angular displacement I get
θ(t)=log(t)/n​
multiplying by radius to get distance I get
d(t) = r(t)*θ(t) = nct*log(t)/n = ct*log(t)​

The trouble is that this cant be right.
if n>>1 then the ant can never get all the way around the balloon
therefore the angular displacement θ must approach a limit as time goes to infinity.
But log(t) increases without limit.

I've gone over and over it
it seems too simple to be wrong but it must be.
 
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Answers and Replies

  • #2
DaveC426913
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Can the balloon physically expand as fast as nct? Or is that an assumption?
 
  • #3
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that is assumed
 
  • #4
DaveC426913
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that is assumed

I am not sure you can assume this. If the balloon's expansion is limited then your paradox might never occur.
 
  • #5
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the point is that the balloon is much expanding faster than the ant can walk
so the ant cant walk all the way around it.

there is no paradox. My math must be wrong somehow

the trouble still remains even if the radius of the balloon were as low as r(t) = ct/2π
the circumference would then be c(t) = 2πct/2π = ct
 
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  • #6
cepheid
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the point is that the balloon is much expanding faster than the ant can walk.

there is no paradox. My math must be wrong somehow

Part of the problem seems to be the initial singularity (the fact that the balloon has 0 radius) at the beginning of the expansion. Your angular velocity blows up at t = 0 if you assume that the ant has always been moving at c "around" the surface of the balloon.


[tex] \theta(t) = \frac{1}{n}\int_0^t \frac{d\tau}{\tau} [/tex]

[tex] = \frac{1}{n} [\log(t) - \log(0)] [/tex].

The logarithm of 0 is not defined. This integral is not defined.

I'm curious what happens if you stipulate that the ant has to start at rest and start moving at c only after a finite time. OR you could stipulate that the balloon has a finite radius at t = 0.
 
  • #7
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that was stipulated but I assumed that the equation would be the same.
 
  • #8
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I'm more concerned by the fact that log(infinity)=infinity

for large values of n it should by finite
 
  • #9
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log does have the nice property of equaling zero at t=1
so I originally tried to make that the point where the ant starts moving.

actually I dont see any trouble with doing that,
you just have to adjust the distance scale to match it so that c=1
 
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  • #10
cepheid
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that was stipulated but I assumed that the equation would be the same.

Sorry, my bad. So if we just consider things from t = 1 onward, then it is indeed true that θ(t) = log(t)/n.

So it seems like the answer is that the ant will make it all the way around in a finite time equal to t = exp(2πn). So the time required to make a full circle increases exponentially with n, but it is still finite.

Why are you so convinced that this answer cannot be right?
 
  • #11
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because some points on teh balloon will be receding from the point of origin of the ant faster than c (much faster)

consider the rate at which the circumference is increasing.

c(t) = 2πr(t)=2πnct

a point on the opposite side of the balloon will be receding at half that speed

space is opening up between the ant and that opposite point much faster than the ant can move forward.
therefore it should be receding from the ant.

now the total distance the ant travels will increase without limit
but the angle θ(t) will not.
 
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  • #12
cepheid
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Your expression for the distance travelled s(t) is wrong. It's NOT true that s(t) = θ(t)r(t). Instead, we must write:

ds = r(t) dθ = r(t) (dθ/dt) dt = r(t)ω(t) dt = c dt

s(t) = ∫ds = ∫cdt = ct

So, that's progress. We found one mistake.
 
  • #13
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thats just the velocity of teh ant.

you also have to consider taht the space already traveled by the ant is expanding.
 
  • #14
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according to someone on another forum



ds/dt = r dθ/dt + θ dr/dt.

the first part is the velocity of the ant
the second part is the expansion of space.

but the second part just seems to reduce to what I already have above.

I cant believe that such a simple looking problem has not only baffled me but even you guys arent sure about it.
 
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  • #15
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θ(t) is teh total angular distance traveled.

it seems to me that if you know the radius then it should be trivial to figure out teh total distance
 
  • #16
Dick
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Your expression for the distance travelled s(t) is wrong. It's NOT true that s(t) = θ(t)r(t). Instead, we must write:

ds = r(t) dθ = r(t) (dθ/dt) dt = r(t)ω(t) dt = c dt

s(t) = ∫ds = ∫cdt = ct

So, that's progress. We found one mistake.

That 'correction' is dead wrong. The original analysis is correct, even it SEEMS paradoxical. s(t) is supposed to be the displacement from the original point at time t. Remember the space behind the ant is growing as well as the space in front of the ant. The ant will circle the balloon. Very slowly if n is large, but it will.
 
  • #17
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so even though a point opposite the origin of the ant is indeed receding from teh ant
it will recede slower and slower till it begins to move toward the ant?
 
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  • #18
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how can that be?

the rate at which a point opposite the origin of the ant is receding depends only on the amount of space between it and the ant.
The amount of space is increasing.
Therefore the rate at which it is receding should increase not decrease.
 
  • #19
Dick
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I am not sure you can assume this. If the balloon's expansion is limited then your paradox might never occur.

The 'c' is a red herring. This isn't a relativity problem. Even it it were there are probably regions of our universe that are receding from us faster than light. We just haven't seen them yet. And may not ever, if the universe is really in accelerated expansion.
 
  • #20
Dick
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how can that be?

the rate at which it is receding depends only on the amount of space between it and the ant which is increasing.
Therefore the rate at which it is receding should increase not decrease.

Slow down. You worked it out correctly. You already pointed out the flaw in reasoning like that in post 13. You have to consider the expansion of the space behind the ant as contributing to his progress.
 
  • #21
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no wait. the total amount of space being added to the balloon remains the same with time.

but since the balloon is getting larger the amount of space being added per space is getting less.

That solves it.

thank you
:-)
 
  • #22
Dick
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balloon.

Part of the problem seems to be the initial singularity (the fact that the balloon has 0 radius) at the beginning of the expansion. Your angular velocity blows up at t = 0 if you assume that the ant has always been moving at c "around" the surface of the
[tex] \theta(t) = \frac{1}{n}\int_0^t \frac{d\tau}{\tau} [/tex]

[tex] = \frac{1}{n} [\log(t) - \log(0)] [/tex].

The logarithm of 0 is not defined. This integral is not defined.

I'm curious what happens if you stipulate that the ant has to start at rest and start moving at c only after a finite time. OR you could stipulate that the balloon has a finite radius at t = 0.

The problem said you start at positive time.
 
  • #23
cepheid
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The problem said you start at positive time.

Yeah, we sorted that out already.
 
  • #24
Dick
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no wait. the total amount of space being added to the balloon remains the same with time.

but since the balloon is getting larger the amount of space being added per space is getting less.

That solves it.

thank you
:-)

I really don't understand a lot you are saying, but if you are happier, that's good.
 
  • #25
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The 'c' is a red herring. This isn't a relativity problem. Even it it were there are probably regions of our universe that are receding from us faster than light. We just haven't seen them yet. And may not ever, if the universe is really in accelerated expansion.

http://en.wikipedia.org/wiki/Doppler_effect

If the source moving away from the observer is emitting waves through a medium with an actual frequency f0, then an observer stationary relative to the medium detects waves with a frequency f given by

0f66e50f3ffb9ae08eb06b31a31e2886.png


if v=c then f = f0/2

this would correspond to a redshift of 1

http://en.wikipedia.org/wiki/Recessional_velocity

[URL]http://upload.wikimedia.org/wikipedia/en/math/2/0/5/205268021556951f829a4bbb3c2986d2.png[/URL]

http://en.wikipedia.org/wiki/Hubble_constant

H0 = 70.8 ± 1.6 (km/s)/Mpc if space is assumed to be flat, or 70.8 ± 4.0 (km/s)/Mpc otherwise

a megaparsec is about 3,000,000 light years

300,000 / 70.8 = 4237

4237 * 3,000,000 light years = 12,700,000,000 light years
 
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  • #26
Dick
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Yeah, we sorted that out already.

Sorry, there were a lot of posts in this thread when I came in. Guess I missed those. Sorry again.
 
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  • #28
cepheid
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I don't know what the hell the Doppler effect has to do with anything here. In any case, Dick was right that your derivation of θ(t) was totally correct. The only hiccup was when we started to consider s(t), s being the arc length (distance travelled around the balloon). It's true that:

ds = rdθ + θdr

the first term representing the increase in distance due to the motion around the circle, and the second term representing the increase in distance due to the expansion alone. There is a term that is second order in the infinitesimals (dθdr) and we ignore it as being negligibly small.

So this means that the rate of increase in distance is given by:

[tex] \frac{ds}{dt} = r(t)\frac{d\theta}{dt} + \theta(t)\frac{dr}{dt} [/tex]

The first term on the right hand side (the rate of motion through space) is just 'c' (this was given). For second term, we know that θ(t) = ln(t)/n (we worked this out). We also know that dr/dt = nc. So the expression becomes:

[tex] \frac{ds}{dt} = c + \frac{nc}{n}\ln(t) [/tex]

[tex] = c[1 + \ln(t)] [/tex]

Now, apparently (according to Wolfram Alpha):

[tex] \int_1^t c[1+\ln(\tau)]\,d\tau = ct \ln(t) [/tex]

which means that you had it right all along. That's embarrasing, but at least I redeemed myself. And the answer to your "paradox" was just that ds/dt ≠ c, but was rather greater than it for any t > 1, which means the ant can catch up to points that are receding from it at > c.

EDIT: and I just realized I was pretty stupid to say that s(t) would not equal θ(t)r(t). But I redeemed myself. Or, at least, that's what I'll keep telling myself.
 
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  • #29
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Forgive me if this is dumb, but I'm interested in this problem.

If the ant is moving at v(t)=c, that means x(t)=ct

You gave in the problem that the expansion of the balloon is r(t)=nct.

We're going to assume that the path traced by the ant is completely about the equator of the balloon. (It wasn't specified in the problem, but if the path were anything other than this, then this problem would be a lot more complicated.)

The balloon has a circumference of 2πr at its equator. Let's view this in the linear case. In other words, let's transform the polar path to rectangular coordinates with the domain [0,2πr). That means that the ant is always 2πr(t)-x(t) from reaching the end of the domain (or equivalently, completely orbiting the balloon). We'll call this distance from the "goal" d(t).

d(t) = 2πr(t)-x(t) = 2πnct-ct = (2πn-1)ct.

Now we can find the critical case in which the balloon expands just as quickly as the ant travels. If n=1/(2π), then we have just that. For other values, it's better to examine the ant's position based on the angle, and we can just make the transformation back onto polar coordinates by letting x(t) = r*θ(t). This is because the definition of x(t) is its position from its starting point.

θ(t) = x(t)/r(t) = ct/(nct) = 1/n

This disagrees with the answer you guys found and agree with, but it makes sense, since in the case found in the original post, there is no n at which the ant can ever travel exactly as fast as the expansion of the balloon, but that's not true.

Also, in the case originally found, the position is time dependent. What we should have is that θ(3)/θ(2) = θ(2)/θ(1), since the ant is moving linearly and the circumference is increasing linearly. Obviously, the above relation does not hold for logs.

Are you sure nothing has gone wrong here?

Edit: I'm sure now that I'm not understanding the problem correctly. The question I should ask is: HOW is the balloon expanding? Does the expansion contribute to the ant's progress along balloon, because if it doesn't, I'm pretty sure my derivation holds. However, if it does, then the ant's velocity isn't always a constant, so something must be going wrong here.
 
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  • #30
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the ant is carried along on the surface of the balloon.

thats why its easier to use angular velocity.
 
  • #31
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Let θ=0 be the angle at exactly the top of the balloon. If the balloon is expanding uniformly, then none of the points on the balloon can be moving.

Let the ant be not crawling and be at the angle θ=0. Then it is still at angle 0 at any two points in time, and its velocity is zero because it's still 0 distance away from 0.

Now let the ant be not crawling and be at the angle θ=π/4. Then it is still at angle π/4 at any two points in time. However, it's now a linear distance away from θ=0 than it was previously. Therefore, it has a nonzero linear velocity, which violates the fact that we chose its linear velocity to be v=0 for this case, since the ant is not crawling.

Notice that both situations are actually identical, so the ant has both 0 velocity AND it has linear velocity. This is a serious contradiction.

Here's another case. Suppose the ant starts at position 0. It moves to a position 1 unit away from its original rest point and stops moving. However, when it stops moving, its linear displacement is still increasing due to the expansion of the balloon. This violates the fact that we can choose the ant's linear velocity.

The real problem here is that the question isn't well-defined. That's why you found a paradoxical answer. You need an absolute reference frame, and by using angular coordinates exclusively, you've effectively chosen that absolute reference frame to be at the center of the sphere. (This is like choosing the ant to be its own reference frame when its moving.) This is problematic, because then the ant is not in control over its absolute velocity, which we wanted to be true in the problem. (Otherwise, why even bother giving the ant a velocity?)

Forgive me if this sounds a bit rude, but I'm not sure why you asked people for help if you already knew the answer that you seek. I see this problem as a case of modifying the question to correspond to an answer that you want. I guess if you were interested in making test questions you would want to know this, but math is about elucidating concepts, not making them more ambiguous.

Oh and wth does the Doppler effect have anything to do with this?
 
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  • #32
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θ=0 is by definition the point at which the ant begins to crawl
 
  • #33
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teh velocity of the ant is its motion relative to the balloon immediately below it.
 
  • #34
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there is no paradox.

the problem has been completely solved.

the original forumulas were right.

I was simply wrong is believing that it had to be wrong.
 
  • #35
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I know there is no paradox. There is, however, ambiguity, but whatever. I'm going to forget that this problem exists.
 

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