Jim's maths and physics problems

[Twiglet]

If anyone is interested.
On the way maths and science interacts.


This is the general form of the schrodinger equation for a particle in a potential well.

schrod.jpg

The term on the l.h.s. contains the symbol i.

This stand for the square root of -1.

This number does not exist.

Can you explain why it is there?

Five or six senteces.

I've posted my answer to myself.[/quote]

It springs from the fact the wavefunction is described as a plane wave propagating through space, the expression for which takes the form psi=A exp i(kx-wt)
The particular form you quoted lends itself to producing the solution easily (i.e. find A etc)
More generally the identity from elementary complex variable theory Ae ix= A(Cos x + iSinx) allows for easy distinction between real and imaginary parts of the equations, as it's much easier to deal with exponential functions than Sin/Cos identities.

[colubridae]

For what it's worth here's mine

Maths is an invention of human thought. Maths’ sets and ‘rules’ map very well onto the real world and with real world measurements can be used to make predictions.

Most of the real world follows the maths of real numbers. Eg squareing a negative quantity results in a positive value.

Some of the physical quantities occurring in QM in the real world do not follow these rules. They follow them with one exception, when squaring negative values the result is negative and converts to a real number.

The values in the first term of the schrodinger equation follow these rules. Why no one knows.


Why maths maps onto the operation of the real world no-one knows either.

Your answer is much better than mine.

[Twiglet]

For what it's worth here's mine

Maths is an invention of human thought. Maths’ sets and ‘rules’ map very well onto the real world and with real world measurements can be used to make predictions.

Most of the real world follows the maths of real numbers. Eg squareing a negative quantity results in a positive value.

Some of the physical quantities occurring in QM in the real world do not follow these rules. They follow them with one exception, when squaring negative values the result is negative and converts to a real number.

The values in the first term of the schrodinger equation follow these rules. Why no one knows.


Why maths maps onto the operation of the real world no-one knows either.

Your answer is much better than mine.

Yours is fine col. The probability of finding a particle in a particular state is psi.psi*, which is a real number which must always be between 0 and 1. The wavefunction has imaginary components, but the probabilities produced are always real.

[colubridae]

For what it's worth here's mine

Maths is an invention of human thought. Maths’ sets and ‘rules’ map very well onto the real world and with real world measurements can be used to make predictions.

Most of the real world follows the maths of real numbers. Eg squareing a negative quantity results in a positive value.

Some of the physical quantities occurring in QM in the real world do not follow these rules. They follow them with one exception, when squaring negative values the result is negative and converts to a real number.

The values in the first term of the schrodinger equation follow these rules. Why no one knows.


Why maths maps onto the operation of the real world no-one knows either.

Your answer is much better than mine.

Yours is fine col. The probability of finding a particle in a particular state is psi.psi*, which is a real number which must always be between 0 and 1. The wavefunction has imaginary components, but the probabilities produced are always real.

Thanks.

My QM textbook was French and Taylor - started to lose it round about 'probability currents'

psi.psi* normalised to 0-1 is the probability of finding the particle there.

I have racked my brain over many years trying to give semantic meaning to "The square root of an unnormalised probability"

[Twiglet]

My QM textbook was French and Taylor - started to lose it round about 'probability currents'

psi.psi* normalised to 0-1 is the probability of finding the particle there.

I have racked my brain over many years trying to give semantic meaning to "The square root of an unnormalised probability"

Oddly enough my first year undergrad book was French & Taylor too. Second year was Rae and final year Gasiorowicz (sp).

All thats being said is the particle has a number of possible states it can occupy (each with a different real number). When you add all possibilities together, they must come to 1 (normalisation). You square the wavefunction to obtain the probability because the equation is formulated in real and imaginary terms to make solving it easier, and squaring always obtains the real term which you wanted.

[colubridae]

I never went beyond the french and taylor. So you can guage mye level from that.

Nice explanation. Thanks.

[Pappa]

I used to be able to do this stuff. Sadly it has been so long since I needed to I have forgotten the lot and replaced it with other random junk.

B3TA has a lot to answer for.

[lpetrich]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

[JimC]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I'll look forward to an exposition on this one, it is somewhat beyond this Year 11 physics teacher, though I hope that when I see the process explained, I will understand (and enjoy) the steps...

[JimC]

No one has had a go at this one yet:

Two 20 kg blocks are held, side by side, at the top of a long inclined plane (60 degrees to the horizontal), ready to slide down when released.

Block 1 will have a frictional force opposing motion when it slides of 100 N
Block 2 will have a frictional force opposing motion when it slides of 50 N

Block 1 is released, then Block 2 is released 5 seconds later. Calculate:

a) the time after the release of block 1 when block 2 catches up with it
b) the distance down the slope where this happens
c) the difference in velocity between the 2 blocks at that point

Take g = 9.8 m/s/s (none of this wussy approximation to 10! )

If no one has in the next day or so, I will give my solutions...

Hint - the inclined plane has to be very, very long...

[Twiglet]

I've just done the solution to this, but I am going to hang back on posting it, because I'd like to see if farsight can solve it. I'm happy to PM the solution to you Jim, within 24 hours and suggest anyone else does the same.

This problem is a great test, because unlike most of the quantum ones like solving hydrogen atom orbits, the solution can't be wiki'd easily lets see if farsight can put his money where his mouth is.

NB my solution is now PM'd to Jim.

[JimC]

I've just done the solution to this, but I am going to hang back on posting it, because I'd like to see if farsight can solve it. I'm happy to PM the solution to you Jim, within 24 hours and suggest anyone else does the same.

This problem is a great test, because unlike most of the quantum ones like solving hydrogen atom orbits, the solution can't be wiki'd easily lets see if farsight can put his money where his mouth is.

NB my solution is now PM'd to Jim.

Got it mate, a little different to mine...

Will check it tomorrow, to eliminate the faint chance that I made a mistake...

[lpetrich]

I've put all the problems and their solutions into a Mathematica notebook, so I can easily get results. If using Excel is cheating, then using Mathematica is even worse cheating. But it's easy to see the equations that one's using in it.

In the first problem, the velocity factorizes as (4 + t) (-7 + 3 t)

In the copper-and-water one, the reason that the copper seems so wimpy is its smaller volume and its smaller heat capacity.

In the octane-combustion one, its' 1 kg of octane and 3.50 kg of oxygen making 3.08 kg of carbon dioxide and 1.42 kg of water, the opposite of what JimC had posted earlier.

I have a solution to the inclined-plane-and-blocks one.

[JimC]

I've put all the problems and their solutions into a Mathematica notebook, so I can easily get results. If using Excel is cheating, then using Mathematica is even worse cheating. But it's easy to see the equations that one's using in it.

In the first problem, the velocity factorizes as (4 + t) (-7 + 3 t)

In the copper-and-water one, the reason that the copper seems so wimpy is its smaller volume and its smaller heat capacity.

In the octane-combustion one, its' 1 kg of octane and 3.50 kg of oxygen making 3.08 kg of carbon dioxide and 1.42 kg of water, the opposite of what JimC had posted earlier.

I have a solution to the inclined-plane-and-blocks one.

Just checked my previous answer for the octane one:

for every 1 kg of octane, 3.50 kg of O2 is consumed, and 1.42 kg of CO2 and 3.08 kg of H2O are produced

I wrote the damn things in the wrong order...

And please give you solutions to the plane problem, Twiglet can give his, and I'll give mine (which I must re-check). There's even a chance they may be the same!

[lpetrich]

I first calculate the accelerations of the two objects:

a = a(gravity) - a(friction)
a(gravity) = a(vertical)*sin(elevation angle) = 8.48705 m/s2
a(friction) = force(friction)/mass = {100 newtons / 20 kg, 50 newtons, 20 kg} = {5, 2.5} m/s2

a = {3.48705, 5.98705} m/s2

Each velocity is v = a*(t-t0) and each position is x = (1/2)*a*(t-t0)2, where x = v = 0 at t = t0.

Setting the two x's equal gives an equation for t with two solutions: 2.8358 s and 21.1124 s. The first one is for before the second object is released, so the appropriate time is the second one. At that time, both distances are 777.147 m.

When they meet, the two objects' velocities are 73.62 and 96.4657 m/s, and their difference is 22.8458 m/s.