# Ant on an expanding balloon

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?

## 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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DaveC426913
Gold Member
Can the balloon physically expand as fast as nct? Or is that an assumption?

that is assumed

DaveC426913
Gold Member
that is assumed
I am not sure you can assume this. If the balloon's expansion is limited then your paradox might never occur.

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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cepheid
Staff Emeritus
Gold Member
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.

$$\theta(t) = \frac{1}{n}\int_0^t \frac{d\tau}{\tau}$$

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

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.

that was stipulated but I assumed that the equation would be the same.

I'm more concerned by the fact that log(infinity)=infinity

for large values of n it should by finite

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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cepheid
Staff Emeritus
Gold Member
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?

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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cepheid
Staff Emeritus
Gold Member
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.

thats just the velocity of teh ant.

you also have to consider taht the space already traveled by the ant is expanding.

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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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

Dick
Homework Helper
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.

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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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.

Dick
Homework Helper
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.

Dick
Homework Helper
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.

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
:-)

Dick
Homework Helper
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
$$\theta(t) = \frac{1}{n}\int_0^t \frac{d\tau}{\tau}$$

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

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.

cepheid
Staff Emeritus
Gold Member
The problem said you start at positive time.
Yeah, we sorted that out already.

Dick
Homework Helper
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.

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 if v=c then f = f0/2

this would correspond to a redshift of 1

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

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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