extra time derivatives

Discussion of theories involving time as a dimension, time travel, relativity, branes, and so on, usually applying to the "real" universe which we live in.

extra time derivatives

Postby PWrong » Sat Jul 23, 2005 3:24 am

I've had an idea about generalising the concept of time. Usually we try to introduce extra time dimensions, but I think my idea is somewhat easier to visualise. Essentially I'm proposing a universe with a different number of time derivatives. It might not result in anything interesting, but I'll explain what I mean anyway.

I first had the idea when I noticed that you need 6 numbers to describe an object, 3 for position, 3 for velocity. In N dimensions, you need 2*N. This stems from Newtons laws.
"An object at constant velocity will continue to travel at constant velocity, unless an outside force is acting on it."

So imagine a new universe where Newtons laws are changed, so the first law becomes:
"An object at constant acceleration will continue to travel at constant acceleration, unless an outside force is acting on it."
The formula F = ma would become F = mj, where j is "jerk" or the derivative of acceleration.

In this universe, you would need 3*3 = 9 numbers to describe an object.
More generally, in N dimensions, and T time derivatives, you need N*T numbers.

We could change many other laws of physics using this idea. For instance, the acceleration of light, c is constant for all observers.
The wave equation, x'' + x = 0 would become something like x''' + x = 0.
The solutions to this would still be of the form e<sup>rt</sup> where r would often be complex. So you could still have periodic solutions.

Anyway, it's just an idea, I hope someone finds it interesting.
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Postby wendy » Sat Jul 23, 2005 6:11 am

An object at constant acceleration requires a constant force to it. This force has to be applied elsewhere. One notes that to substain this force, one would have to apply increasing power: since E=M.v²/2 = E(at)²/2. The power would have to come at the rate of Ea²t or something. That makes for eventually infinite power,

The earth in its orbit has a changing accelearation, which requires no substaining force.

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Postby jinydu » Wed Jul 27, 2005 2:13 am

wendy wrote:An object at constant acceleration requires a constant force to it.


That's if you assume Newton's Laws. However, I think that PWrong was thinking about what it would be like to change Newton's Laws.

One important point though, PWrong. In it's most general form, Newton's Second Law states that:

F = dp/dt
where p is the momentum of the particle

In fact, F = ma, is just a special case of that above equation; the case where mass is constant (to see this, apply the Product Rule).

If you want to introduce another time derivative, the general Newton's Second Law would become:

F = d^2p/dt^2

Using the definition p = mv and applying the Product Rule:

F = d/dt (dm/dt * v + m * dv/dt)

F = d/dt (dm/dt * v) + d/dt (m * a)

Now, we apply the Product Rule again to each of the two terms:

F = (d^2m/dt^2 * v + dm/dt * dv/dt) + (dm/dt * a + m * da/dt)

F = (d^2m/dt^2 * v) + (dm/dt * a) + (dm/dt * a) + mj
(again, j means jerk, the third time derivative of displacement)

F = (d^2m/dt^2 * v) + 2(dm/dt * a) + mj

In the special case where the mass is constant, this does indeed reduce down to F = mj. But in the more general case where m is some arbitrary function of time, this modified Newton's Second Law becomes more complicated.

Obviously though, if acceleration were constant when the net force is zero, objects could easily reach arbitrarily high speeds.
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Postby PWrong » Tue Aug 02, 2005 12:12 pm

Good point, I forgot about that. I've looked it some more, and there are actually several ways to define force.

1. F = d^2 (mv) / dt ^2
2. F = d (ma) /dt

These both give F = mj when m is constant.

If we extend to n time derivatives, we get even more options.

F = d^k (m* d^j x / dt^j) / dt^k
where j and k are positive integers and j + k = n

This gives m * d^n x / dt^n no matter what value we choose for k.

Another thing we could do is use mass more than once, or even have different types of mass. For instance in n=3:

Code: Select all

      d        d         d
F =  --- (m_2 --- (m_1  --- (m_0 * x) ) ) )
      dt       dt        dt


generally, in n-derivative time, force is the derivative of (mn * the force used in (n-1)-derivative time)

In our universe, everything happens to have an m_0 of exactly 1, but I don't see why that has to be the case. If all the masses are constant, this boils down to (m_0 * m_1 * m_2 *... *m_n) * d^n x / dt^n

Obviously though, if acceleration were constant when the net force is zero, objects could easily reach arbitrarily high speeds.


True enough, but in our universe we can technically get arbitrarily far away from a starting position, if given enough time, and enough space. With 2 time derivatives, x would always be a quadratic function of time.

If we really wanted to restrict arbitrarily high speeds, we could possibly "wrap" velocity the way we wrap a space around a torus.
If you go past the edge of a torus-shaped universe, you come back on the other side at the same speed. This doesn't break any laws, because that's just the way space is wrapped.

Similarly, if you go above a certain speed, you simply stop instantly, but keep the same acceleration. Hopefully this wouldn't cause any contradiction. You could think of it as a bit like an ordinary circular speedometer. If you keep accelerating, the needle will go all the way round, come back to 0 and keep rising. Everything in the car would experience the same thing, because no "force" is acting, so you wouldn't hit the windscreen.
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Postby jinydu » Tue Aug 02, 2005 1:37 pm

PWrong wrote:If we really wanted to restrict arbitrarily high speeds, we could possibly "wrap" velocity the way we wrap a space around a torus.
If you go past the edge of a torus-shaped universe, you come back on the other side at the same speed. This doesn't break any laws, because that's just the way space is wrapped.

Similarly, if you go above a certain speed, you simply stop instantly, but keep the same acceleration. Hopefully this wouldn't cause any contradiction. You could think of it as a bit like an ordinary circular speedometer. If you keep accelerating, the needle will go all the way round, come back to 0 and keep rising. Everything in the car would experience the same thing, because no "force" is acting, so you wouldn't hit the windscreen.


I'm not so sure about that. It sounds like a scheme like that could conceivably cause problems for the Principle of Relativity (that the laws of physics hold in all inertial frames of reference, and that if one reference frame is moving at a constant velocity with respect to an inertial reference frame, then the former is also inertial).

Speaking of which, in a "F = mj" universe, the Principle of Relativity may have to be modified. If acceleration is constant, then jerk is zero, and hence net force would be zero.

Now, suppose we are in an inertial reference frame (remember that an inertial reference frame is defined to be a reference frame where Newton's Laws hold). Imagine that we're observing an object (call it X), which is moving at a constant acceleration, relative to us. Since X has constant acceleration, we should not be able to detect any net external force acting on X. Clearly, X is made up of many particles. Thus, there are two possibilities:

1) The net external (that is, external to X) force on all of the particles is 0.

2) The net external force is nonzero for at least some of the particles, but these forces cancel each other out when the forces on all the particles are summed together. (Hmm, this reminds me of rotation).

I think I'll only deal with the first case, since it is simpler.

...

Never mind, I give up (for now). What I was trying to establish was that X would also be an inertial frame of reference. This sounds reasonable, since there seems to be a symmetry between X and the postulated inertial reference frame; but I can't come up with a rigorous argument.

Then again, I was taught "F = ma" and "Any reference frame in constant velocity with respect to an inertial reference frame is also inertial" as two seperate laws. Although I have long suspected that it could be possible to derive the second statement from the first, I have never been able to do so anyway.

In any case, if my guess is correct and there is a way to prove the statement about inertial frames from F = ma, then it sounds reasonable that when F = mj, we would have the statement: "Any reference frame in constant acceleration with respect to an inertial reference frame is also inertial". It then follows that X has no way of "detecting it's acceleration", the same way we have know way of "detecting our velocity" when we're standing on a train with a constant velocity (relative to the ground).

Of course, there is a possibility that one or more of your "modified Newton's Laws" universes could be inconsistent. That is, starting from "F = [etc, etc]", it may be possible to derive a result that is inconsistent with some other cherished law of physics, which could be an incentive to abandon that particular version.

I guess I could do some of the classic mechanics problems using F = mj, and see what I get. I'll post again here later.

EDIT: This is offtopic, but I was wondering if it could be possible to enable Latex on this board. It would make equations so much easier to read...
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Postby jinydu » Wed Aug 03, 2005 6:20 am

As you surely remember, the first major topic in Physics class is motion under a constant force. In the typical Newtonian universe, F = ma leads, not surprisingly to a position function that is quadratic in time.

It's not difficult to modify the derivation for a F = mj universe. As before that F (and m) is constant:

F = mj

j = F/m

da/dt = F/m

Integrating both sides sides from 0 to t gives:

a - a0 = Ft/m, where a0 is the acceleration at time 0

a = Ft/m + a0

dv/dt = Ft/m + a0

Again, integrate from 0 to t

v - v0 = F(t^2)/2m + (a0)t

v = F(t^2)/2m + (a0)t + v0

dx/dt = F(t^2)/2m + (a0)t + v0

Integrate one more time from 0 to t

x - x0 = F(t^3)/6m + (a0)(t^2)/2 + (v0)t

x = F(t^3)/6m + (a0)(t^2)/2 + (v0)t + x0

Substituting F/m = j and rearranging into a more "classic" form gives:

x = x0 + v0t + (1/2)a0t^2 + (1/6)jt^3

A few points:

1) The equation of motion looks a lot like that of a particle experiencing a force that changes linearly with time in our F = ma universe.

2) It looks just like a Taylor expansion. This should be easy to generalize to a universe with n time derivatives.

3) There are three "undetermined constants" that are determined by initial conditions. This is perhaps not so surprising, since we started with a third-order differential equation. However, it shows that initial position and initial velocity are not sufficient to completely determine the motion of a particle; initial acceleration is also needed. Thus, Laplace's famous claim:

"If I knew the position and velocity of every particle in the Universe now, I could predict the entire future of the Universe"

would have to be modified to:

"If I knew the position, velocity and acceleration of every particle in the Universe now, I could predict the entire future of the Universe"

I think I'm also going to try looking at what happens to work-energy in a F = mj universe. I suspect that will be more challenging.
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Postby PWrong » Wed Aug 03, 2005 12:28 pm

It looks just like a Taylor expansion. This should be easy to generalize to a universe with n time derivatives.


That's interesting. I noticed the n-degree polynomial, but I hadn't made the connection with Taylor series. It almost suggests that our universe is an approximation to an ideal universe with infinitely many time derivatives...

Anyway, I think making some precise definitions would be a good idea.
An NT universe is one in which
Force in an NT universe (i.e. n time derivatives) is defined recursively, by
F<sub>0</sub> = x (the displacement vector)
F<sub>n</sub>=d/dt ( m<sub>n</sub> * F <sub>n-1</sub>)

where each m<sub>n</sub> can be constant, or a function of other variables.

Let's also define momentum as
p<sub>n</sub> = m<sub>n</sub> * F <sub>n-1</sub>
so F<sub>n</sub> is simply dp<sub>n</sub>/dt

We live in a 2T universe, where m<sub>1</sub> = 1 and m<sub>2</sub> is the mass we are used to.

Speaking of which, in a "F = mj" universe, the Principle of Relativity may have to be modified. If acceleration is constant, then jerk is zero, and hence net force would be zero.


I'm sure that relativity would have to be modified along with everything else. I won't formally learn relativity for a few more weeks, although I've had a look in my textbook. As far as I can tell, it seems a lot easier mathematically than conceptually, since all you do is multiply everything by gamma = 1/sqrt(1- v^2/c^2).

This apparently works for time, displacement, momentum, force and total energy.

It should be possible to derive the formulas relativity in an 3T universe using just Einsteins first postulate, and the 3T version of the second postulate:

"The acceleration of light is the same in all inertial frames of reference,
and this is independent of the motion of the source."

I think I'm also going to try looking at what happens to work-energy in a F = mj universe. I suspect that will be more challenging.

I think you're right there. I've never really understood or liked Conservation of Energy. Apparently when it was discovered, only engineers and chemists cared about it. It comes from Noether's Theorem, which I don't think I'll understand for at least a couple of years.

A likely definition for Work would be the integral of force with respect to distance. But that has units of [W<sub>3</sub>] = kg m^2 s^-3. I can't see any nice formula for energy using that.
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Postby jinydu » Wed Aug 03, 2005 12:50 pm

PWrong wrote:I'm sure that relativity would have to be modified along with everything else. I won't formally learn relativity for a few more weeks, although I've had a look in my textbook. It seems a lot easier mathematically than conceptually, since all you do is multiply everything by
gamma = 1/sqrt(1- v^2/c^2).
This apparently works for time, displacement, momentum, force and total energy.

It should be possible to derive the formulas relativity in an 3T universe using just Einsteins first postulate, and the 3T version of the second postulate:

"The acceleration of light is the same in all inertial frames of reference,
and this is independent of the motion of the source."


Sorry, maybe I wasn't clear. I was talking about the "Principle of Relativity" (something that was known by Galileo in the 1600s), not Einstein's "Theory of Relativity". The Principle of Relativity states that:

"Any reference frame in constant velocity with respect to an inertial reference frame is also inertial."

We can worry about Einstein's Relativity Theory after we deal with Newtonian Theory:wink:

PWrong wrote:A likely definition for Work would be the integral of force with respect to distance. But that has units of [W3] = kg m^2 s^-3. I can't see any nice formula for energy using that.

Energy conservation comes from url=http://en.wikipedia.org/wiki/Noether%27s_theorem]Noether's Theorem[/url], which I don't think I'll understand for at least a couple of years.[/url]


I've seen how to derive K = (1/2)mv^2 from Newton's Second Law and the definition of work. Suppose the starting time is t = a and the ending time is t = b (Please excuse the lack of proper symbols):

W = Integral (F dot dr)

= Integral (mr''(t) dot dr/dt dt)

= m * Integral (r''(t) dot r'(t) dt)

Now comes the trick:

d/dt ((r'(t))^2) = 2r'(t)r''(t)

W = 1/2 * m * (r'(t))^2| (upper bound: t = b, lower bound t = a)

W = 1/2 * m * (r'(b))^2 - 1/2 * m * (r'(a))^2

From there, it's only natural to define

K = 1/2 * m * (r'(t))^2 = (1/2)mv^2

The problem is, I don't know how to do such a trick for a 3T case.

Let's assume that the definition of Work is still the same:

W = Integral (F dot dr)

W = Integral (mr'''(t) dot dr/dt dt)

W = m * Integral (r'''(t) dot r'(t) dt)

Hmm, maybe integration by parts will do the trick!

Edit: Come to think of it though, I don't recall hearing a version of integration by parts, with the dot product replacing ordinary multiplication, for vector-valued functions.
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Postby PWrong » Thu Aug 04, 2005 9:29 am

Sorry, maybe I wasn't clear. I was talking about the "Principle of Relativity" (something that was known by Galileo in the 1600s), not Einstein's "Theory of Relativity".


Oh, ok. In that case, we should just change it to
"Any reference frame in constant acceleration with respect to an inertial reference frame is also inertial." We don't need to redefine an inertial reference frame.

Edit: Come to think of it though, I don't recall hearing a version of integration by parts, with the dot product replacing ordinary multiplication, for vector-valued functions.


Well, the product rule for differentiation works for dot products, so it's not hard to prove that integration by parts works.

So you'd end up with
W = m( r'.r'' - Int( r''.r'' dt)),
where . is the dot product.
This doesn't look any better though.

What about W = Int(F dv) ?

That would give
W = Int( m r''' r'' dt)
K = 1/2 m a^2
This looks exactly like what we want, except there's only one kind of mass.

To introduce the other, let
W = Int(F<sub>3</sub> d(F<sub>2</sub>) )
W = Int(m<sub>3</sub>j d(m<sub>2</sub> v) )
K = 1/2 m<sub>3</sub> m<sub>2</sub> a^2

In general, we can say
W<sub>n</sub> = Int( F<sub>n</sub> . dF<sub>n-2</sub>)

Which gives
E<sub>n</sub> = 1/2 m r<sup>(n-1)</sup>(t)
where m is the product of all the m<sub>i</sub>

The only problem is I don't know how to prove that energy is conserved. For that matter, I can't prove that momentum is always conserved either. :(
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Postby jinydu » Thu Aug 04, 2005 11:36 am

If you use the most general form of Newton's Second Law (in our Universe), Conservation of Momentum is trivial to prove:

F = dp/dt

If, F = 0, dp/dt = 0

Therefore, p is constant

However, if you modify Newton's Second Law to:

F = d^2p/dt^2

it's not hard to see that even when the net external force is zero, momentum may not be conserved (in fact, it could be any linear function of time). This is not really that surprising, since we said earlier that objects under no external force move with constant acceleration.

One possibility for rescuing the Law of Conservation of Momentum could be to redefine momentum:

Define p = ma

Now, Newton's Second Law doesn't need to be changed at all. If we assume that Newton's Second Law remains as the familiar

F = dp/dt

Using the Product Rule, we get:

F = (dm/dt)*a + mj

... and hence recover F = mj in the case where mass is constant.

Energy, on the other hand, could be more difficult to deal with. The key equation that we would probably want to preserve is:

Work Done by Nonconservative Forces = Change in Energy

(or in my professor's memorable words: "Energy is conserved, except when it's not").

Another very important consideration is that our Universe, it is possible to assign a qauntity called potential to any point in a conservative vector field. However, the derivation of this fact, shown in my textbook, ultimately relies on the "dr" present in the definition of work. Thus, I'm pretty sure that if you changed the definition of work to:

Integral of (F dot dv)

... in the best case scenario, the potential energy of an object in a conservative vector field would depend on it's velocity instead of it's position. In the worse case, the whole theory of conservative vector fields and potential energy could come crumbling down.

That's why I prefer sticking with "Integral of (F dot dr)". I'm pretty sure that conservative vector fields should still have a property called potential that depends on position. In another five months, I will have finished a course on vector calculus, and I should have the skills to prove it.
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Postby PWrong » Tue Aug 09, 2005 12:08 pm

That's why I prefer sticking with "Integral of (F dot dr)". I'm pretty sure that conservative vector fields should still have a property called potential that depends on position. In another five months, I will have finished a course on vector calculus, and I should have the skills to prove it.


Ok, fair enough. I've come across conservative forces and potential energy in physics, but I haven't studied them in maths. I think you're about a year ahead of me in maths & physics.

Anyway, we can probably survive without conservation of energy for the time being. What I'd like to do is find out it if there are any forces that will cause an object to rotate. The force should ideally be a function of distance.

It's going to be a bit wierd because in circular motion, jerk should generally be in the same direction as velocity, for the same reason that the acceleration is parallel to the radius vector. So the force will be in the wrong direction.

Then again, since we've changed the principle of relativity, maybe a force can now be a function of velocity. :?

I'm actually getting the feeling that this universe won't turn out much different from ours. All we're doing is replacing nearly every quantity in each formula with it's derivative with respect to time.

If you took a graph of the position of a particle in our 2T universe, and integrated it wrt, it would probably look like the graph of a particle in the 3T universe. Similarly, if you plotted the velocity of a particle in 3T, with certain forces acting on it, it would look like the trajectory of an object in 2T.

I'm not sure if this would be true for situations like collisions and orbits, but it certainly works in the simple case of a constant force.
But if it does, then maybe our universe is actually just the derivative (or integral) of the real thing. :shock:
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Postby jinydu » Mon Aug 29, 2005 4:20 am

PWrong wrote:W = m * Integral (r'''(t) dot r'(t) dt)

Hmm, maybe integration by parts will do the trick!



Unfortunately, it doesn't...

http://www.sosmath.com/CBB/viewtopic.php?t=17644

But in high school, I learned a different derivativation of KE = 1/2 mv^2. It doesn't involve calculus but is only valid in the special case where the force is constant. Still, it might give a hint towards how to do that integral:

Suppose a mass m is moving with a constant velocity v in a particular (constant) direction. We apply a constant force F in the opposite direction to v.

Clearly, by Newton's Second Law, m will experience a constant acceleration, -F/m, in the opposite direction to it's velocity.

There is a well-known kinematics formula that applies to objects with a constant acceleration:

v^2 = u^2 + 2ax
(where v is the final velocity, u is the initial velocity and x is the displacement)

We want the object to come to rest, so v = 0. Also, u = v (not to be confused with the v I just introduced).

0 = v^2 + 2ax

0 = v^2 + 2(-F/m)x

v^2 = 2(F/m)x

x = 1/2 m(v^2)/F

Now, since

Work done on object = Force * Displacement

We have

W = 1/2 m(v^2)/F * F

W = 1/2 m(v^2)

Furthermore, if we assume the Conservation of (Mechanical) Energy (which is only valid when the vector force field is conservative, but this condition is generally ignored in high school):

W = (Kinetic) Energy at start

So

Kinetic Energy = 1/2 mv^2
(I dropped the boldface so that the formula is the way it usually looks)

I could be able to use a similar method (with F = mj instead of F = ma) to arrive at a solution; then work backwards to show that the solution is correct (again, using that integral).

Of course, the trick here is the formula v^2 = u^2 + 2ax. It is not obvious how this can be generalized to an F = mj universe...

I'll have to take a look at the derivation of v^2 = u^2 + 2ax (I do remember that it comes from kinematics alone).

EDIT: I've got it!

Using the definition of velocity and acceleration; integrate acceleration once to get:

v = u + at

Squaring both sides gives:

v^2 = u^2 + 2aut + (a^2)(t^2)

Now, integrate acceleration twice to get:

x = ut + 1/2 at^2

Multiplying both sides by 2a:

2ax = 2aut + (a^2)(t^2)

Substituting this into the first (squared) equation gives:

v^2 = u^2 + 2ax

Note: Things look a bit different from in my original derivation because in this post, x represents displacement, whereas in my original derivation, x represents position.
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Postby jinydu » Mon Aug 29, 2005 4:46 am

Now, I'll try to tackle the F = mj case (I'm making this up as I type). To keep things more in accordance with usual math notation, I'm reverting back to x = position rather than x = displacement.

We start off with j = da/dt. Integrating once (we use the usual assumption that j is constant) gives:

a = a0 + jt

Integrating a second time gives:

v = v0 + a0t + 1/2 jt^2

Integrating a third time gives:

x = x0 + v0t + 1/2 a0t^2 + 1/6 jt^3

I make a note of those three bolded equations, because I'll probably need them later.

Now, suppose an object of mass m is moving with a constant acceleration a (remember that in an F = mj universe, this is what happens when net force is zero). We apply a force F in the opposite direction to a until m comes to rest, after undergoing a displacement d.

Clearly, j = -F/m. Now, we set up our final conditions. Presumably, we would want v = 0 and a = 0 when the object comes to rest (the assumption here is that in order for an object to have no kinetic energy, both it's velocity and it's acceleration must be zero). Already, this is qualitatively different from "F = ma" case, where we only had one condition: v = 0.

Plugging these conditions into the bolded equations gives:

0 = a0 + jt

0 = v0 + a0t + 1/2 jt^2

d = v0t + 1/2 a0t^2 + 1/6 jt^3

Essentially, the problem amounts to eliminating j and t from the third equation.

From the first bolded equation:

t = -a0/j

Plugging that into the third equation:

d = -a0v0/j + 1/2 a0(a0^2/j^2) - 1/6 j (a0^3/j^3)

d = -a0v0/j + 1/2 (a0^3/j^2) - 1/6 (a0^3/j^2)

d = -a0v0/j + 1/3 (a0^3/j^2)

Plugging t = -a0/j into the second equation:

v0 - a0 (a0/j) + 1/2 j(a0^2/j^2) = 0

v0 - (a0^2/j) + 1/2 (a0^2/j) = 0

v0 = 1/2 (a0^2/j)

j = 1/2 (a0^2/v0)

Plugging that into the time independent equation for d:

d = -a0v0 (2v0/a0^2) + 1/3 a0^3 (4v0^2/a0^4)

d = -2 (v0^2/a0) + 4/3 (v0^2/a0)

d = -2/3 v0^2/a0

Using W = F * d = -mjd

W = -m (1/2 a0^2/v0) (-2/3 v0^2/a0)

W = 1/3 mv0a0

Since the work done on the object is equal to it's initial kinetic energy:

KE = 1/3 mv0a0

To write it in a form that looks closer to KE = 1/2 mv^2

KE = 1/3 mva
Last edited by jinydu on Tue Aug 30, 2005 6:39 am, edited 2 times in total.
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Postby jinydu » Mon Aug 29, 2005 6:16 am

Hmm... that result doesn't seem to quite fit in with the more general derivation (involving the integral)...

W = m * Integral (j dot v dt)

= m (va - Integral (a^2 dt))

= mva - m * Integral (a^2 dt)

The factor of 6 doesn't seem to appear here, and there's an extra term proportional to the integral of the square of acceleration...

EDIT: I asked Mathematica, a program on my computer, to attempt to solve those simultaneous equations from the previous post. For some reason, it gave me answer slightly different from the one I derived by hand:

KE = 1/3 mva

So according to the computer, I'm off by a factor of two... But that still doesn't do anything to resolve the questions...

EDIT 2: It took only a few minutes to find my mistake. Originally, I said that (1/2) - (1/6) = 5/12, which is not true. Instead, it should be (1/2) - (1/6) = 1/3. An elementary school error...

Oh well, I fixed my previous post. Now, I get an answer that agrees with Mathematica. All major intermediate steps agree with Mathematica as well.
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Postby jinydu » Tue Aug 30, 2005 6:38 am

Come to think of it, there's something very wrong with a formula like:

KE = 1/3 mva

We would expect that a particle that is not under the effect of a net external force should have a constant kinetic energy.

If the net external force is zero, then jerk must be zero.

If jerk is zero, then acceleration is constant.

If acceleration is constant, velocity (in general) changes.

But here, there is a problem. Since velocity changes, KE = 1/3 mva must also change. Thus, the kinetic energy of a particle can change, even when there is no net external force. This flies in the face of Conservation of Energy.

Since velocity can change without a net external force, it would seem that the correct equation for kinetic energy (if there is a correct one) should be indepedent of velocity...

Hmm, something is not right with my calculations...
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Postby PWrong » Tue Aug 30, 2005 2:36 pm

Presumably, we would want v = 0 and a = 0 when the object comes to rest (the assumption here is that in order for an object to have no kinetic energy, both it's velocity and it's acceleration must be zero).


Your assumption sounds valid, but it's unlikely for any object to have zero velocity and zero acceleration at the same time. a = dv/dt, so when a=0, v is at a maximum or minimum.

More importantly, the assumption doesn't agree with K = mva.
If K = 1/3 mva = 0,
then v=0 OR a=0,
but not neccessarily both.

It might not matter if kinetic energy isn't conserved, as long as the total energy is. We have kinetic and potential energy in our universe, but maybe in F=mj we have other kinds. Of course, I have no idea what they might be. :(
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Postby jinydu » Tue Aug 30, 2005 3:38 pm

PWrong wrote:Your assumption sounds valid, but it's unlikely for any object to have zero velocity and zero acceleration at the same time. a = dv/dt, so when a=0, v is at a maximum or minimum.

More importantly, the assumption doesn't agree with K = mva.
If K = 1/3 mva = 0,
then v=0 OR a=0,
but not neccessarily both.


Hmm, good point there. I hadn't thought of that. So that's why I got an answer that doesn't seem consistent with my assumptions...

It seems then that we have to either drop the assumption that v = 0, or drop the assumption that a = 0. Either way, we've lost on equation. We then would have 2 equations and 3 variables (j, t and d), which doesn't look promising...

I'll have to look at it more later; but it's late at night here, so it will have to wait for tomorrow.
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Postby jinydu » Wed Aug 31, 2005 10:22 am

Hmm, I look at my derivation again, and I see another thing I don't understand. At one point, I say that:

d = -2/3 v0^2/a0

This equation seems to be saying that (since velocity squared is clearly positive), displacement is in the opposite direction to acceleration. I don't see why this should be the case...

The derivation also says that:

j = 1/2 (a0^2/v0)

which implies that jerk should be in the same direction as velocity. Again, I don't see why it has to be like that...

Moreover, both of these constraints appear to be time-invariant. Thus, for instance, if jerk is in the opposite direction to initial velocity, it would appear that the particle will never simultaneously have zero velocity and zero acceleration, no matter how far in the past or future we look. Furthermore, t, d and j (and hence F) are determined by initial conditions. This contrasts with our 'F = ma' universe, where we are free to choose between large force, small stopping distance and small stopping time, or small force, large stopping distance and large stopping time. It would be as if, when we wanted to drain a car's kinetic energy, we had to apply exactly the right force; or else, the car's kinetic energy would never be zero, no matter how long we waited. This does seem to contradict the KE = 1/3 mva formula; if KE = 1/3 mva were true, any jerk could lead to zero kinetic energy, since all we have to do is wait until the acceleration is zero (which of course, always happens eventually for any value of jerk).

In my earlier post, I essentially claimed that in order for a particle to have no kinetic energy, it is necessary that it's velocity and acceleration both be zero. But what happens if I change the word necessary to sufficient? Do one or more of my equations disappear?

In any case, I know that kinetic energy should, in general, not be conserved. But it seems natural to me that the kinetic energy of a particle under no net external force should be conserved...

EDIT: I've thought of this some more, and maybe the answer to my difficulties is that the situation that I used for my derivation wasn't really as general as I thought it was. Basically, part of the derivation involved answering the question: For an arbitrary set of initial conditions and arbitrary (constant) jerk, after what distance does the object have zero velocity and zero acceleration? Implicit in this question, however, is the assumption that such a condition (zero velocity and zero acceleration) exists. On closer inspection, this assumption seems to lead to further subtleties...

I tried to redo the derivation, but this time, I started by eliminating j.

From the third equation:

j = -a0/t

Then, plugging that into the second equation:

v0 + a0t + 1/2 (-a0/t) t^2 = 0

v0 + a0t - 1/2 a0t = 0

v0 + 1/2 a0t = 0

t = -2 (v0/a0)

That equation seems to indicate that if v0 and a0 have the same direction, the instant when the particle had 0 velocity and 0 acceleration had to have been in the past! Furthermore, there seem to be a big problem when the initial acceleration is zero and the initial velocity is nonzero...

Side Note: Though deriving a formula for kinetic energy is problematic in 3T, the corresponding integral (Integral of x''''(t) * x'(t) dt) for a 4T universe is quite doable via integration by parts. If I am not mistaken, in a 4T universe:

KE = m * (vj - 1/2 a^2)

In general, integration by parts appears to work well whenever the number of time derivatives is even; Mathematica easily comes up with formulas for 6T, 8T and 10T; and I think I know how to derive a formula for nT in general, whenever n is even. But when the number of time derivatives or odd, the trick doesn't work, and finding a formula for kinetic energy (if it is possible at all) will require much more ingenuity.

Even - Straightforward
Odd - Not straightforward

reminds me of a certain problem in Mathematics... :wink:
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Postby jinydu » Wed Sep 07, 2005 3:05 am

Well, if I drop the assumption that velocity is zero at the end, I get the system of equations:

a<sub>0</sub> + jt = 0

v<sub>0</sub>t + 1/2 a<sub>0</sub>t<sup>2</sup> + 1/6 jt<sup>3</sup> = d

Together with KE = W = -mjd, that leads to:

KE = m(v<sub>0</sub>a<sub>0</sub> - 1/3 a<sub>0</sub><sup>3</sup>/j)

But that can't be right; we expect the kinetic energy to be independent of the force that it applied to it (and hence the jerk that it experiences).

In fact, I think we can look at an analogous situation for any NT universe:

Let D(n)x be the nth time derivative of position.

Then F = mD(n)x

Assume that initially, the particle has velocity v<sub>0</sub>, acceleration a<sub>0</sub>, jerk j<sub>0</sub> ... and (n-1)th time derivative D(n-1)<sub>0</sub>

Suppose that in the final state, D(n-1)x = 0.

Then the kinematic equations are:

D(n-1)<sub>0</sub> + D(n)xt = 0

v<sub>0</sub>t + 1/2 a<sub>0</sub>t<sup>2</sup> + 1/6 j<sub>0</sub>t<sup>3</sup> + ... + 1/(n-1)! D(n-1)<sub>0</sub>t<sup>n-1</sup> + 1/n! D(n)x t<sup>n</sup> = d

Solving the first equation for t gives:

t = - D(n-1)<sub>0</sub>/D(n)x

Plugging that into the second equation gives:

d = v<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x) + 1/2 a<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x)<sup>2</sup> + 1/6 j<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x)<sup>3</sup> + ... + 1/(n-1)! D(n-1)<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x)<sup>n-1</sup> + 1/n! D(n)x (- D(n-1)<sub>0</sub>/D(n)x)<sup>n</sup>

Now, apply KE = W = -mD(n)xd:

KE = -mD(n)x * (v<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x) + 1/2 a<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x)<sup>2</sup> + 1/6 j<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x)<sup>3</sup> + ... + 1/(n-1)! D(n-1)<sub>0</sub>(- D(n-1)<sub>0</sub>/D(n)x)<sup>n-1</sup> + 1/n! D(n)x (- D(n-1)<sub>0</sub>/D(n)x)<sup>n</sup>)

KE = m(v<sub>0</sub>(D(n-1)<sub>0</sub>) - 1/2 a<sub>0</sub>(D(n-1)<sub>0</sub><sup>2</sup>/D(n)x) + 1/6 j<sub>0</sub>(D(n-1)<sub>0</sub><sup>3</sup>/D(n)x<sup>2</sup>) - ... + (-1)<sup>n</sup>/(n-1)! D(n-1)<sub>0</sub>(D(n-1)<sub>0</sub><sup>n-1</sup>/D(n)x<sup>n-2</sup>) + (-1)<sup>n+1</sup>/n! D(n)x (D(n-1)<sub>0</sub><sup>n</sup>/D(n)x<sup>n-1</sup>))

KE = m(v<sub>0</sub>(D(n-1)<sub>0</sub>) - 1/2 a<sub>0</sub>(D(n-1)<sub>0</sub><sup>2</sup>/D(n)x) + 1/6 j<sub>0</sub>(D(n-1)<sub>0</sub><sup>3</sup>/D(n)x<sup>2</sup>) - ... + (-1)<sup>n</sup>/(n-1)! (D(n-1)<sub>0</sub><sup>n</sup>/D(n)x<sup>n-2</sup>) + (-1)<sup>n+1</sup>/n! (D(n-1)<sub>0</sub><sup>n</sup>/D(n)x<sup>n-2</sup>))

KE = m(v<sub>0</sub>(D(n-1)<sub>0</sub>) - 1/2 a<sub>0</sub>(D(n-1)<sub>0</sub><sup>2</sup>/D(n)x) + 1/6 j<sub>0</sub>(D(n-1)<sub>0</sub><sup>3</sup>/D(n)x<sup>2</sup>) - ... + (-1)<sup>n-1</sup>/(n-2)! D(n-2)<sub>0</sub> (D(n-1)<sub>0</sub><sup>n-2</sup>/D(n)x<sup>n-3</sup>) + (-1)<sup>n</sup>(n-1)/n! (D(n-1)<sub>0</sub><sup>n</sup>/D(n)x<sup>n-2</sup>))

[Note: In the last line, I combined the (n-1)<sup>th</sup> term and the n<sup>th</sup> term; and showed the (n-2)<sup>th</sup> term, although I didn't show it in the previous lines. The first (n-2) terms follow much the same pattern, but the new last term is someone different. Furthermore, the equation turns out to be more even problematic for a 1T universe, see below.]

As can be seen, if we want kinetic energy to be independent of D(n)x (and hence force), the only possibility is that n = 2, our F = ma universe!

But this calculation raises more questions than it answers:

1) The claim that only F = ma leads to kinetic energy that is independent of force seems to contradict my previous post, which showed kinetic energy formulas for even n. In fact, the kinetic energy formula it gives for 4T is:

KE = m(vj - 1/2 a(j<sup>2</sup>/D(4)x) + 1/8(j<sup>4</sup>/D(4)x<sup>2</sup>))

which contradicts the formula that I can derive from integration by parts:

KE = m(vj - 1/2 a<sup>2</sup>)

2) When I apply this formula to a 3T, F = mj universe, I get:

KE = m(va - 1/3 a<sup>3</sup>/j)

When differentiated with respect to time, I get:

m(vj + 1/3 (a<sup>3</sup>D(4)x)/(j<sup>2</sup>))

instead of the expected

mvj

(although, encouragingly, only by a term that is zero when jerk is constant).

3) Something even stranger is going on in the 1T case.

Clearly, there is only one term for kinetic energy:

KE = mxv

Kinetic energy that depends on position? That sounds very weird to me... It means that a moving particle gains (or loses) kinetic energy over time, without any external force!
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Postby PWrong » Fri Oct 14, 2005 3:27 pm

Hey, sorry for the late reply, I got caught up with the Schrodinger equation thread. I also couldn't think of anything to post until now.

I've worked out the whole point of energy is. I never understood the motivation behind it until now. It's essentially a technique for solving differential equations. In 2T, since you almost always have forces that depend on position, you often get this differential equation.
m x'' = F(x)

To solve this, you multiply by x' and integrate
m x' x'' = x' F(x)
1/2 m x'^2 = Int F(x) dx

This is a 1st order equation, and it can easily be solved (although not explicitly). Now, writing the equation in this form immediately shows us the formulas for K and U.

Let K = 1/2 mv^2 and U = - Int F(x) dx.
U will have a constant of integration, which we'll call E, so we have
K + U = E

Now, in 3T it's much more difficult. Let m=1 to make things simple.
We want to solve x''' = F(x), for an arbitrary function F(x).
Failing that, we might be able to at least reduce it to a 2nd order equation.
Rather than look for something arbitrary like K(x',x'') + Int F(x) dx = E,
we'll assume a more general equation.

The most general 2nd order equation that doesn't involve t is
f(x'', x', x) = 0, which I'll rewrite as
f(a, v, x) = 0

We want to find the function f, for a given F. By the way, the units of f are the 3T equivalent of Joules.

Differentiating with respect to t with the chain rule (d represents a partial derivative here):

df/dx * dx/dt + df/dv * dv/dt + df/da * da/dt= 0
df/dx * v+ df/dv * a + df/da * j = 0
but x''' = F(x), so
df/dx * v + df/dv * a + df/da * F(x)= 0

v f<sub>x</sub> + a f<sub>v</sub> +F(x) f<sub>a</sub> =0

So we have a 1st order PDE with 3 variables, which I have no idea how to solve for f. :( Any ideas?
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Postby jinydu » Sun Oct 16, 2005 5:15 pm

Hmm, I don't see how

Int (x' F(x)) dt = Int F(x) dx

(I assume you meant to integrate with respect to time).
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Postby PWrong » Tue Oct 18, 2005 1:37 pm

It's just a substitution. :wink:
Int (x' F(x) dt) = Int (F(x)*dx/dt dt) = Int (F(x) dx)

Or, using the chain rule:
d/dt U(x) = U'(x)*dx/dt = - F(x) x'
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Postby jinydu » Wed Oct 19, 2005 7:09 am

It will probably be at least a few more days before I can look at it in detail; I have lots of homework at the moment.
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Postby jinydu » Fri Mar 17, 2006 9:12 am

PWrong, you mentioned earlier in this thread the possibility of a universe whose evolution is governed by an infinite order differential equation in time. Recently though, I have to come to see why this probably not possible. The argument is surprisingly simple.

Suppose that the universe really were governed by an infinite order differential equation in time. Then we know from mathematics that the general solution to the differential equation would contain infinitely many arbitrary constants whose values would be fixed by the value of the function and all its derivatives at an arbitrary initial time. Therefore, predicting the entire future and past evolution of the universe would require measuring the value of the function and all its derivates at some particular instant in time. But if we were able to make such a measurement, we could simply apply a Taylor expansion on the function and predict the evolution of the universe forever into the future and past; this could be done without any knowledge of physical laws (other than the assumption that the function is well-behaved enough so that the Taylor series converges). Therefore, physics would be redundant.

To put it more concretely, suppose we have a particle moving through space. It is in fact possible to predict the particle's position at any time without any knowledge of physics at all; just measure the position, velocity, acceleration, jerk, time derivative of jerk, etc. at some initial time, and use a Taylor expansion. So what use is physics? Well, for constant mass situations, Newton's Second Law states that

Force = Mass * Acceleration

Thus, if know the force on the particle at a particular instant, and its mass, we can calculate its acceleration. Furthermore, let's assume that force is a function of position and velocity only. Now, suppose that at some initial time t0, we know the particle's position x(t0) and velocity (dx/dt)(t0). Then, since force is only a function of position and velocity, we can calculate the acceleration at time t0. Next, we rely on an important feature of the Taylor expansion: If the value of a function and its derivative at some instant, we can calculate the value of the function an infinitessimal time later. Applying this property, since we know the particle's position and velocity at time t0, we can calculate the particle's position at time t0 + dt. Similarly, since we the particle's velocity and acceleration at time t0, we can calculate its velocity at time t0 + dt. We are now in the same situation we were at the start; we know the particle's position and velocity at some particular instant. Thus, we can repeat the same procedure, to calculate the particle's position and velocity at time t0 + 2dt. In principle, we can repeat the procedure infinitely many times to obtain the particle's position and velocity at any arbitrary time.

Here then is the power of physics: Whereas mathematics on its own (more specifically, the Taylor expansion) require you to the particle's position and all its derivatives at some instant, physics (more specifically, Newton's Second Law) reduces the information you need to only the particle's position and the first derivative of position at some instant. The reason this is possible is because Newton's Second Law is second order in time, and hence requires only two initial conditions. Clearly then, if nature were governed by an infinite order differential equation, there would be no way to predict the particle's motion except to measure the position and all the time derivatives of position at some instant. But by the time you've collected this much information, you could simply apply a Taylor expansion. This would make physics obselete.

Admittedly, this argument isn't totally watertight. In particular, it wouldn't work for special relativity, where it impossible, even in principle, to measure the entire state of the universe at some fixed instant in time, due to the finite speed of light. Still, it is possible to measure any event whose spacetime position (ct, x, y, z) satisfies
(ct)^2 - x^2 - y^2 - z^2 = 0, so it may be possible to make a similar argument.
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Postby PWrong » Mon Mar 20, 2006 3:07 pm

PWrong, you mentioned earlier in this thread the possibility of a universe whose evolution is governed by an infinite order differential equation in time. Recently though, I have to come to see why this probably not possible. The argument is surprisingly simple.

That's a shame. I think I understand your argument. For the taylor series to converge, the force would have to vanish.

While we were talking about this before, I was looking at solutions to 3rd order ODE's, and I realised that basic physics would be much more complicated than in 2T (try solving it in mathematica, with a periodic external force). Maybe the 3T universe is where electrical engineers go when they die, instead of hell :twisted:.

I just managed to show that the solutions to 3rd order equations always grow or decay. They never oscillate nicely (unless one of your arbitrary constants is zero). You need a cubic equation with real coefficients, so that all the solutions are purely imaginary, and you can show that this is impossible.

I thought of another weird universe, by the way. What if Pythagoras's theorem was |r|^3 = |x|^3 + |y|^3 :?:
But apparently this kind of space has already been invented. http://en.wikipedia.org/wiki/Lp_space
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Postby jinydu » Tue Mar 21, 2006 5:50 am

PWrong wrote:That's a shame. I think I understand your argument. For the taylor series to converge, the force would have to vanish.


Actually, my argument was that in a universe governed by an infinite order differential equation, in order to predict the future motion of a particle, you would need to know its position and all the time derivatives of position at some instant; but once you succeded in doing this, you wouldn't need the differential equation, because you could just use a Taylor expansion. Thus, dynamics would be obselete.
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Postby PWrong » Thu Mar 30, 2006 11:59 am

Thus, dynamics would be obselete.

I don't think the universe cares whether physicists understand it. There could be lots of universes where dynamics is obsolete. But I agree that a universe with infinitely many derivatives would be very boring.

I wonder if it's possible to have some mixture of universes. A bit like Kaluza Klein theory is a mix of 3D and 10D space. For instance, if mass is constant, we might have F = m a + t m j, where t is a very small time, like the planck time. That way the units fit.

Maybe we should see whether these universes violate any symmetry conditions. I don't know how to do this, but I'm sure it's important.
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Postby jinydu » Thu Mar 30, 2006 8:53 pm

I'll change your t to a t0 to avoid confusion with the usual meaning of t (time as a free variable).

We can start by checking conservation of momentum. In the absence of external forces, your equation would become:

m d^2x/dt^2 + t0 m d^3x/dt^3 = 0

d^2x/dt^2 + t0 d^3x/dt^3 = 0

According to Mathematica, the solution to this differential equation is:

x = x0 + v0t + a0 t0 (t - t0 + e^(-t/t0) t0)

There are several things in that equation that worry me. They are easier to see when we look at velocity. Differentiating both sides:

v = v0 + a0t0 (1 - e^(-t/t0))

First of all, time shows up in the equation. Thus, velocity is not constant, and momentum is not conserved.

Even worse, look at what happens when t ----> -infinity (the distant past). In that case, the exponential term becomes very large (unless a0 = 0) and causes v to approach positive or negative infinity (depending on the sign of a0) very rapidly. On the other hand, when t ----> +infinity (the distant future), the exponential term disappears and the particle approaches a terminal velocity of v0 + a0t0. In a way, this is good news, because we can define momentum as p = m(v + a t0), and momentum will be conserved (since terminal velocity can't depend on when you do the measurement). But the bad news is that the behavior of the particle at
t = -infinity is very different from its behavior at t = +infinity. This would appear to contradict the principle that nature should make no distinction between the past and the future. That is, if you threw a ball up in the air, made a videotape of its trajectory and showed it to me, and I claimed that the Earth's mass was negative and you had been playing the tape backwards, you would have no way of proving me wrong.
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Postby PWrong » Sat Apr 01, 2006 8:18 am

This would appear to contradict the principle that nature should make no distinction between the past and the future.

Ok, so that's no good.

I've found that there is a way to reduce the third order equation m x''' = F(x) to a second order equation. http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf

Instead of using w(x) = x'^2, we'll substitute K(x) = 1/2 m v^2, the usual expression for kinetic energy. Then we get:

dK/dx dx/dt = dK/dt = m x'(t) x''(t)
dK/dx = m x''(t)
x'(t) d<sup>2</sup>K/dx<sup>2</sup> = m x'''(t) = F(x)
d<sup>2</sup>K/dx<sup>2</sup> = sqrt(m/2) K<sup>-1/2</sup> F(x)
Now we have a really hard 2nd order equation to solve for kinetic energy. :(
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