Jim's maths and physics problems

[lpetrich]

I read that paper, and it focused on how some Kaluza-Klein sort of theory may behave. KK theories divide space-time into 4 "big dimensions" and extra "small" dimensions.

However, I'm concerned here with what happens with n "big" and coequal space dimensions. What happens to orbits of test particles around point sources? What numbers of dimensions allow stable orbits? See if you can find the solution in the nonrelativistic classical limit.

Though if you are masochistic enough to want some extra challenge, you could try the nonrelativistic quantum-mechanical case, using the appropriate version of Schroedinger's equation.

[JimC]

This thread has been hijacked by the tertiary sector!



Not that I really mind...

However, back to upper secondary level science.

Calculate the mass of CO2 that would be found in a pressurised tank with a capacity of 0.7 m3, at a pressure of 20 atmospheres and a temperature of

20 degrees C

Take the relative molecular mass of CO2 as 44

[newolder]

I read that paper, and it focused on how some Kaluza-Klein sort of theory may behave. KK theories divide space-time into 4 "big dimensions" and extra "small" dimensions.

Correct. However, this theory does not constrain the size of the extra dimension like K-K:

As we know, the original version of Kaluza-Klein theory assumes as a postulate that the fifth dimension is compact. However, in the case of the induced-matter theory (IMT) this requirement has been dropped.

However, I'm concerned here with what happens with n "big" and coequal space dimensions.

So were those authors.

What happens to orbits of test particles around point sources?

In 5-d they follow geodesics (postulate 3)

We then add a third postulate concerning the motion of particles and light:
The paths corresponding to the motion of free-falling test particles and light rays are
geodesic lines in the 5D fundamental Ricci-flat space M5.

; observed within 4-d they deviate because of source masses outside the scope of the brane.

What numbers of dimensions allow stable orbits? See if you can find the solution in the nonrelativistic classical limit.

My (intuitive) guess would be all numbers of dimensions up to 3 space + 1 time.

Though if you are masochistic enough to want some extra challenge, you could try the nonrelativistic quantum-mechanical case, using the appropriate version of Schroedinger's equation.

It's a challenge to get back home in good time on the bike! I leave this sort of work to the junior physicists, these days.

What is the tertiary sector, JimC?

P.S. Using the time independent Schrödinger equation gives a family of eigenstate solutions with harmonic Fourier decomposition.
TISE: E Ψ(x) = -ħ2/2m d2Ψ(x)/dx2 + V Ψ(x)
Let the potential vanish and rearrange...
d2 Ψ/dx2 + E 2m/ħ2 Ψ(x) = 0
Let 2mE/ħ2 = -k2 and we have an harmonic oscillator across space.

[JimC]

newolder wrote:

What is the tertiary sector, JimC?

At least in Australian educational circles, there are 3 sectors in education: Primary school (up to 11 or 12), secondary school (12 to 17 or 18) then a whole array of post-secondary educational institutions such as Universities, Institutes of technology and various training colleges, collectively known as the Tertiary sector...

My teaching duties (and thus the problems I set here) are generally from upper secondary maths and science, my bread and butter...

But of course I don't really mind the invasion of the hyper nerds!

[lpetrich]

Here's the rest of the solution of my dimensionality-of-space-time problem:

Trigger Warning!!!1! :

The force law is f(r) = f0*r1-n
where I've put in the minus sign because a stable orbit requires an inward force.

Let's ignore the mass of a test particle, by setting it to 1. The polar-coordinate equations of motion (radius r, position angle a) are:
d2r/dt2 - r*(da/dt)2 + f(r) = 0
r(d2a/dt2) + 2(dr/dt)(da/dt) = 0

The second one yields angular-momentum conservation:
da/dt = h/r2
and the r-a equations are, for u = 1/r,
d2u/da2 + u - 1/(h*u)2f(1/u) = 0

For 1 space dimension, there is no position angle, and the equation of motion is purely "radial":
d2r/dt2 + f0 sgn(r) = 0
giving
r = v0*t - (1/2)*f0*t2
for r > 0, and a reversed-sign expression for r < 0.

All orbits are bound and stable.

For n > 1, the u-a equation becomes
d2u/da2 + u - (f0/h2)u3-n = 0
A circular orbit with u = u0 has
h = sqrt(f0u02-n)
and perturbing it with
u = u0*(1 + e*cos(g*a))
gives
g = sqrt(4 - n)

This means that bound orbits will be stable for 2 or 3 space dimensions.

For 2 space dimensions, one can integrate the radial equation of motion to get
(1/2)((dr/dt)2 + (h/r)2) + f0 ln(r/r0) = E
That equation yields conservation of energy.

I've been able to integrate it in the radial case (h = 0); I could easily do it with Mathematica. The time is
t = sqrt(pi/(2*f0))*r0*erf(sqrt(log(r0/r)))
where erf is the error function and r0 is the maximum radius. I am unable to do the integral in the nonradial case, with h != 0.

All orbits are bound and stable.

For 3 space dimensions, the solutions should be easy to find in some celestial-mechanics text.

All bound orbits are stable, with some orbits being unbound (flybys).

In the flyby case, h = b*vi, where b is the "impact parameter" (offset of aim from center point), and vi is the velocity at infinity.

For more than 3 space dimensions, there are no stable bound orbits. Flybys are only possible for h greater than some value that's some power of the energy. For n = 4, h > sqrt(f0)

One can find parallel results in quantum mechanics, using Schroedinger's equation. The wavefunction for the n = 1 case is an Airy function, and I've been unable to solve the n = 2 case. The n = 3 case is solved in textbooks of quantum mechanics, and for n = 4, there is a rather odd solution. For n > 4, there is a rather severe singularity at the origin, something like exp((r0/r)(n-4)/2).

This result has implications for what sort of Universe could allow us to come into existence. For n >= 4 space dimensions, there won't be stable planetary orbits and stable atoms. So we live in a Universe with the maximum number of space dimensions that can allow us to exist.

Here's a somewhat easier problem that is related to it. Imagine a Universe with more than one time dimension. Will there then be a well-defined direction of time?

Hint: consider the possible space-time trajectories of an object: x = x(T), t = t(T), where T is the object's internal or proper time, and x and t are the space and time dimensions. The object's velocity through space-time satisfies the condition
(sum of time dimensions) (dt/dT)2 - (sum of space dimensions) (dx/dT)2 = 1.

[mistermack]

Quite so, thanks col.

Farsight and mistermack - please take note - this is what an admission of a mistake looks like.

I stand corrected. I got the maths a bit wrong.

Oddly, I don't feel stupider or diminished for admitting it

Objection : I posted my thread and invited correction. It didn't come. In fact, in the end I had to announce that I'd seen at least one flaw in my own argument, that's still an admission in my book. And still nobody else has seen it to date.
You point it out, I'll admit any mistake and say thanks. Why wouldn't I? If you point out a mistake, you've increased my knowledge and understanding.
But don't just give me '' it's wrong somewhere '', any child could say that.
.

[lpetrich]

Calculate the mass of CO2 that would be found in a pressurised tank with a capacity of 0.7 m3, at a pressure of 20 atmospheres and a temperature of

20 degrees C

Take the relative molecular mass of CO2 as 44

I'll use the ideal gas law:

P = (den*R*T)/m
finding
den = (P*m)/(R*T)

Also,
mass = den*volume

Pressure (P) = 20 atm = 2.026*106 pascal
Gas constant (R) = 8.3145 J/mol/K
Temperature (T) = 20 C = 293.15 K
Mol Wt (m) = 44 g = 0.044 kg
Density (den) = 39.3 kg/m3
Volume = 0.7 m3
Mass = 27.5 kg

[JimC]

Using a similar approach (the ideal gas law, which I take as PV = nRT, where R = 8.31), and P as 2.026 x 106, I get:

n = 582.46 moles, which I then multiply by 0.044 to get 25.63 kg

Very close to your answer, but too different for small rounding errors etc...

Someone else needs to do it, then we can apply the "3 computers" rule!

By the way, lpetrich, can I recommend that you add some lines between lines when you use super or subscript - it tends to produce more ledgible posts...

[lpetrich]

I discovered a typo in one of the numbers in my solution expression, and when I corrected it, I found:

Density = 36.6 kg/m3
Mass = 25.6 kg

[JimC]

I discovered a typo in one of the numbers in my solution expression, and when I corrected it, I found:

Density = 36.6 kg/m3
Mass = 25.6 kg

My next problem will not be difficult in abstract, but fiendishly complex... :twisted:

It will involve some circuit elements in parallel, some heating certain materials, others melting ice, and still others being electric motors of a given efficiency raising masses to given heights, with the final answer being to calculate the overall resistance and current...

[lpetrich]

Here's the solution to the time-dimensionality problem:

Trigger Warning!!!1! :

Imagine more than one time dimension, and imagine an object stationary in the space dimensions. For two time dimensions t1 and t2, let
u1 = dt1/dT
u2 = dt2/dT

We find

u12 + u22 = 1

and likewise for more time dimensions. Imagine that some force causes this trajectory:

t1 = (1/w)*cos(w*T)
t2 = (1/w)*sin(w*T)
u1 = - sin(w*T)
u2 = cos(w*T)

That satisfies the above equation, and it also demonstrates that there is no well-defined direction in time -- the particle follows a circular trajectory in the two time variables.

Now turn to one time dimension. Let u be the space-component velocity:
ux = sqrt((sum of space dimensions) (dx/dT)2)

That makes the time-component velocity ut = dt/dT either:
ut = sqrt(1 + ux2)
or
ut= - sqrt(1 + ux2)

So ut <= -1 or ut >= +1, with a forbidden region between -1 and +1. This means that it is not possible for ut to continuously change from a positive to a negative value, thus it is always one time or the other.

That is what gives time its well-defined direction.

[lpetrich]

Now for a simpler one.

Let's say that you want to make table salt from concentrated sodium hydroxide and hydrochloric acid. Let's say that you are starting out with 1 liter of 1 mole/liter (M) solution of each, and that both of them completely dissociate. The pH will to be 7, since it's - log10([H+]), since the solution is neutral ([H+] = [OH-]), and since the concentration equlibrium with water is - log10([H+]) - log10([OH-]) = 14. Both [H+] and [OH-], these ions' concentrations, are in M.

What happens to the pH if you put in 1% too much hydrochloric acid? 1% too much sodium hydroxide? 1% too little of either?

Let us ignore complications like Activity (chemistry) - Wikipedia.

[JimC]

Now for a simpler one.

Let's say that you want to make table salt from concentrated sodium hydroxide and hydrochloric acid. Let's say that you are starting out with 1 liter of 1 mole/liter (M) solution of each, and that both of them completely dissociate. The pH will to be 7, since it's - log10([H+]), since the solution is neutral ([H+] = [OH-]), and since the concentration equlibrium with water is - log10([H+]) - log10([OH-]) = 14. Both [H+] and [OH-], these ions' concentrations, are in M.

What happens to the pH if you put in 1% too much hydrochloric acid? 1% too much sodium hydroxide? 1% too little of either?

Let us ignore complications like Activity (chemistry) - Wikipedia.

That is a good one, certainly in the realms of upper secondary chemistry. I'm going to wait for Monday, where I can check something in my chem textbook, which is at school - I'm a little rusty on equilibrium problems...

I always get my advanced science Year 10s to do at least a couple of acid/base titrations every year - nothing like hands-on chemistry with a need to be precise and accurate...

Still working on my electricity/heat capacity problem, but it will come...

[lpetrich]

The solution:

An excess of 1% of hydrochloric acid means 1.01 moles [H] and 1.00 moles of [OH] in 2.01 liters of water. This yields [H] - [OH] = 0.01 moles. The combined quantity of water is 2 liters. Repeating this for the other scenarios gives

HCl excess: [H] - [OH] = 0.005 M
NaOH excess: [H] - [OH] = - 0.005 M
HCl deficiency: [H] - [OH] = - 0.005 M
NaOH deficiency: [H] - [OH] = 0.005 M

Let us now translate this into pH. For that, we need the equilibrium equation:
[H]*[OH] = Keq2
where Keq is about 10-7 M.

For [H] - [OH] = X, the solutions are:

[H] = (X/2) + sqrt(Keq2 + (X/2)2)
[OH] = - (X/2) + sqrt(Keq2 + (X/2)2)

For |X| >> Keq, one ion or the other dominating, we have these approximations:

X >> Keq:
[H] = X
[OH] = Keq2/X

X << - Keq:
[H] = Keq2/(-X)
[OH] = (-X)

So the pH values become:
HCl excess: 2.3
NaOH excess: 11.7
HCl deficiency: 11.7
NaOH deficiency: 2.3

[lpetrich]

Another one:

How many grams of oxygen is necessary for burning each gram of these fuels to completion? That is, producing CO2 and H2O.

Hydrogen: H2
Methane: CH4
Long-chain hydrocarbons: effectively (CH2)n - treat as CH2
Carbon: C

Atomic weights:
H: 1 - 1.0079
C: 12 - 12.011
O: 16 - 15.999

Gasoline, kerosene, diesel fuel, fuel oil, etc. are all long-chain hydrocarbons.