Johnny5

So what I want to know is when I drop the ball' date=' does the earth come up to meet it?

[/quote']

The truthful and correct answer is yes, when what happens is viewed from the center of mass frame.

 

Takes only a second to explain.

 

In order to analyze the relative motion of things, one must first choose a frame to view the motion in.

 

I presume you mean simply dropping a ball, and watching it fall to earth.

 

The earth is very massive, compared to a tiny ball. There is a lot more matter in the earth, than a ball. A whole lot more.

 

In order to really grasp the explanation, it will help you to think about an infinitely massive object. No such object exists, yet for some reason this will help you understand.

 

Somewhere in the universe is the center of mass of the universe.

 

Call this location in the universe W.

 

So Omega denotes the center of the universe.

 

Now, suppose that there was an infinitely massive object, shaped in a sphere, and that it currently is at rest relative to location Omega.

 

By Galileo/Newton's first law of motion, the center of mass of this object will remain at rest in this frame.

 

Or to use the term 'distance' to express this, you can say that the distance from Omega to the center of mass of this infinitely massive object is currently constant in time.

 

Now, suppose that you want to accelerate the center of mass of this object, relative to point in space Omega.

 

By Newton's second law of motion, the force F you must apply to give it acceleration a in this frame is related to the mass M of the object as follows:

 

F = M a

 

Since the mass has been stipulated to be infinite, you must apply an infinite force to get that object to move, relative to point Omega.

 

No infinite force can be applied to any object, hence the object will remain at a fixed location in space (relative to Omega) forever.

 

But of course there are no infinitely massive objects, but that's what the math says.

 

Now, think about what you want, which is dropping a ball in the gravitational field of the earth.

 

The earth is far closer to being an infinitely massive object than the ball is.

 

A small force such as you jumping up and down, does almost nothing to "change the position of earth" relative to Omega. (where I am pretending earth is currently at rest relative to the center of the universe)

 

Now, as you hold the ball at a steady height h above the surface of the earth, there are two forces.

 

The force of the earth pulling the ball down, and the gravitational force of the ball pulling the earth up.

 

Now, in between the center of mass of the earth, and the center of mass of the ball, is the location of the center of mass of the two body system.

 

If the two objects had equal massives, the location of this place would be half the distance between them. But of course, the earth is far far far more massive than a little ball.

 

So you are about to release the ball, and you are going to view the motion in the center of mass frame. That is a reference frame at which the center of mass of the two body system doesn't move.

 

Here is what happens.

 

Call the location of the center of the system mass the origin of the reference frame.

 

Once the ball is released, the ball accelerates towards the origin.

 

Also

 

Once the ball is released, the earth accelerates towards the origin.

 

The magnitudes of these accelerations are not equal.

 

But, in this frame, the answer to your question is yes, the earth does "come up to meet the ball."

 

By Newtons third law, the force of the ball on the earth, is equal to the force of the earth upon the ball, but in the opposite direction therefore:

 

F12=-F21

 

Let M denote the mass of the earth, let A denote the acceleration of the earth, let m denote the mass of the ball, and let a denote the acceleration of the ball therefore, the following scalar relation must be satisfied:

 

MA=ma

 

Suppose the ball weighs one kilogram, and accelerates at 10 meters per second squared. Therefore:

 

MA=10

 

The mass of the earth is:

 

6 x 10^24 kilograms

 

Hence:

 

6 x 10^24 A = 10

 

hence

 

A =1.666... x 10^-24 meters per second squared.

 

Now, suppose that initially, you release the ball from a height of say around seven feet above the surface of the earth.

 

In fact, lets make it exactly two meters.

 

So you raise your arm, until the center of mass of the ball is 2 meters above the surface of the earth, and then you let it go.

 

You can use the kinematic equations for constant acceleration to figure out how long it will take the ball to hit the ground.

 

The computation is simple. Here is the formula for the general case of constant acceleration a:

 

D = v0 t + 1/2 a t^2

 

In the formula above, D represents distance traveled, v0 denotes the initial speed, t denotes the amount of time of travel, and a denotes the acceleration.

 

Now in this case the acceleration is ten meters per second squared.

 

The initial speed of the ball, before it is released, in the center of mass reference frame is zero.

 

So here is the formula for D:

 

D = 5 t^2

 

The distance the ball falls is 2, hence:

 

2=5 t^2

 

hence

 

2/5 = t^2

hence

 

4/10 = t^2

 

hence

 

.4 = t^2

 

Now take the square root of both sides of the equation above, and we have:

 

.63245553... = t

 

 

Since we used SI units in the formula, t came out in seconds.

 

So the time of freefall is a little more than half a second.

 

Now, recall the acceleration of the earth in the center of mass reference frame:

 

A = 1.666... x 10^-24 meters per second squared.

 

It's initial speed in the CM frame is zero. Hence the distance it moves whilst falling towards the apple is given by:

 

D = 1/2 a t^2

 

D = 1/2 ( 1.666... x 10^-24 )t^2

 

D = 8.3 x 10^-25 t^2

 

And we know the time till collision is .63 seconds hence:

 

D = 8.3 x 10^-25 (.6324555...)^2

 

Thus:

 

D = 3.3333... x 10^-25 meters.

 

This is a fantastically small distance, in comparison to one meter.

 

The diameter of a hydrogen atom is around

 

1 x 10^-10 meter, which is called an Angstrom.

 

1 x 10^-25 is (1000000000000000) times smaller than that.

 

So thats the distance the earth moves, in the time it takes the ball to fall to earth, when the relative motions of the centers of mass of the earth and ball are viewed in the center of mass frame.

 

Regards

(Of course' date=' some maths purist will come along and point out that a vector is a rank 1 tensor and a scalar is a rank 0 tensor, which is true, but normally in physics (and we are in the physics section) the word tensor is reseverd for rank 2 and above.)[/quote']

 

Ok. You know that actually helps.

 

Regards

circular motion is circular motion.

 

two forces.

 

in this case lift comes into play as well.

 

I wonder how a stoneage people stumbled across this "quirk" in the first place.

 

here is a perfect diagram of the flight.

 

http://www.rangs.co.uk/howaboomworks.htm

Nice picture of the path, not exactly a circle, not an ellipse, but a closed orbit nonetheless.

I have always wondered how a boomerang manages to reach a certain distance and then start back towards its thrower...

 

I never really thought about the mechanics of it much' date=' mostly because im not to great with physics, so does anyone here have an explanation on how boomerangs "work"?

 

-Does it return, because you throw it fast and its spinning fast but as it slows down, it still keeps spinning, causing it to...uhh never mind. [/quote']

 

 

I've never thought about this myself, but as with any question of this nature, one needs to use the laws of physics to answer it.

 

I could try to google for the answer, but then again i could try to figure it out on my own.

 

Lets see.

 

You throw a boomerang in some direction, with some initial velocity v.

 

Now the boomerang itself is spinning very fast, and this matters.

 

What happens, is quite the result of an interaction with the boomerang, and the surrounding air.

 

Now, a boomerang is shaped like a V kind of.

 

So its center of mass does not lie within the object itself.

 

The boomerang approximately travels in a circle, and this is what you want to understand. Why does it do that, instead of just keep going away, in a straight line.

 

Now, the velocity v is a vector, and it is from the origin of some reference frame to the center of mass of the boomerang.

 

The boomerang itself is spinning, and the axis of revolution passes through the center of mass. So you can think of a reference frame moving with the boomerang, with the center of mass permanently at rest at the origin of this frame.

 

So if you treat the motion of the origin of that frame as the place whose position in the thrower's frame as being the point in question, then that place moves roughly in a circle.

 

So let m denote the mass of the boomerang.

 

There is a centripetal acceleration of

 

mv^2/R

 

Where R is the radius of the boomerang's orbit, in the thrower's frame.

 

Now, if there were no air at all, then there would be no centripetal acceleration.

 

 

So we can answer the question the following way...

 

The air exerts a force F on the boomerang, and this force arises because the boomerang is spinning in the rest frame of the air.

 

The boomerang also exerts a force F, on the surrounding air.

 

Using classical physics

 

F = mv^2/R

 

To go any further into the problem, would require the use of aerodynamics. And there are quite a few assumptions in there about air resistance.

 

Such as force is proportional to v.

 

Certain assumptions which hold in one situation, wont in another.

 

You could work air pressure into the solution of the problem too.

 

Now a frisbee doesn't come back to you.

 

So I think the answer lies in the V shape of the boomerang, and the fact that it's spinning, and some formula with air pressure involved in it.

 

At least thats a start anyways.

 

Now I will google to see what others say.

 

Here is a cool site: Physics of a boomerang

 

Here is Bernoulli's answer

 

Here is something i found:

 

Definition: A statement of the conservation of energy in a form useful for solving problems involving fluids. For a non-viscous, incompressible fluid in steady flow, the sum of pressure, potential and kinetic energies per unit volume is constant at any point.

p + qv2/2 + qgy = constant

 

where p is the pressure, q is the density, v the velocity and y the height in a gravitational field of strength g, all measured at the same point. This quantity is then constant throughout the fluid.

 

Bernoulli's principle is used for, but not essential to, lift by airplanes.source

 

To really say you have answered the question, your approach has to predict circular motion. In fact, really it should predict the actual motion.

 

You have to model everything, decide what are the important variables, and ultimately predict a roughly circular path. It's not an easy problem.

 

Regards

 

PS: I can't resist trying to answer it.

 

Here is another link to something which says that the boomerang precesses.

 

I am trying to find a mathematical derivation which predicts circular motion.

 

Here is an exact quote from the source above:

 

When a boomerang is tossed in the correct manner, the wings rotate through the air and react to the aerodynamic and gyroscopic forces. These forces cause the boomerang to circle around and lay down as it returns, until it descends in a horizontal hover. During the flight of the boomerang, the following principles come into play: Bernoulli's relation, gyroscopic stability, gyroscopic precession, and Newton's laws of motion. We shall examine how these forces cause a boomerang to return to the thrower.

 

 

Here is another site which discusses lift.

 

Here is someone who is getting to the heart of it

 

A boomerang does funny things because it is in fact a gyroscope. Aerodynamic forces generate a twisting moment which cause the "gyroscope" to precess and to move on a circular path. (Taken from the link above)

 

All the math needed, for a first approximation to understanding the motion of a boomerang, and thus the answer to your question, is at the site there.

 

Read from "What is a couple?" to equation 8.

 

Start out with this:

 

t = I a

 

The LHS is the greek letter tau, and stands for "torque" or "turning force."

 

On the RHS the letter I stands for moment of inertia, and the letter alpha stands for angular acceleration.

 

I am trying to find a site which gives a clear presentation.

I went and got my textbook "Elementary Classical Analysis"

by Jerrold E. Marsden and Micheal J. Hoffman

I'm not sure if that's the same Hoffman who wrote my linear algebra text.

At any rate he lists the field axioms as:

commutativity of addition/multiplication

associativity of addition/multiplication

additive/multiplicative identities

additive/multiplicative inverses

distributive axiom

non-triviality axiom

He also lists six order axioms

I don't really have a personal version per say, I'm just quoting from memory. If you'd like, I can dig up my old Abstract Algebra book and post those field axioms here. Alternatively, I can just tell you the page number if you already have access to his book; Algebra[/url'] by Michael Artin.

No i don't have access to that book unfortunately.

Hmm.

Well I have a list, why don't you criticise mine.

Let me see...

Let a,b,c denote arbitrary Real numbers.

Closure under addition and multiplication:

a+b is a real number

a*b is a real number

Commutativity of addition and multiplication:

a+b=b+a

a*b=b*a

Associativity of addition and multiplication:

a+(b+c)=(a+b)+c

a*(b*c)=(a*b)*c

Multiplicative and additive identity elements:

There is at least one real number called zero, denoted by 0, such that:

0+a=a

There is at least one real number called one, denoted by 1, such that:

1*a=a

Multiplicative and additive inverses:

Given any real number a, there is at least one real number -a, called the addive inverse of a, also called "negative a" such that:

a+(-a)=0

Given any real number a, if not(a=0) then there is at least one real number 1/a, called the multiplicative inverse of a, also called the reciprocal of a, such that:

a*(1/a)=1

Distributive axiom:

a*(b+c)=a*b+a*c

Axiom of addition:

If a=b then a+c=b+c

Axiom of multiplication:

If a=b then a*c=b*c

Non-triviality axiom:

not(0=1)

Order axioms

For any real number x: not (x<x)

For any real numbers, x,y,z: if x<y and y<z then x<z

For any real number x, there is at least one real number y, such that x<y

For any real number x, there is at least one real number y, such that y<x

And I use some definitions.

Is the list above complete, is it minimal? Is there a preferrable set of field axioms? I'm not clueless as to the answers to my own questions, but I am interested in seeing what others call "the field axioms".

I believe associativity and commutativity are all that's left.

Why don't you just send me all of your field axioms. Yours personally.

That will save us both a great deal of time.

Thank you Dapthar

Yup. My mistake. I've fixed the original post.

Ok, any others you left out?

Originally Posted by Johnny5

You are turning truth into something relative if you do that.

The axioms are not 'absolute truths' as it were' date=' for they cannot be proven to be true, you just have to 'believe' that they are. I suppose it injects a bit of faith into Mathematics.

[/quote']

 

Let me state my position quickly.

 

Mathematicians use binary logic with those axioms. That means they are treating the axioms as statements, assigning them truth value, and then manipulating the axioms, and statements which follow, using logic.

 

Now, if i were to prove that the axioms are meaningless, then all that would be in vain.

 

Now you aren't going to catch me arguing that the axioms are meaningless.

 

I could argue that numbers don't exist, and so on but it's pointless to do that.

 

Instead I permit them to have their intended meaning.

 

So they are statements, and they do have truth value.

By calling them axioms, we are saying that they are true.

 

And so they must be internally consistent.

 

And with so many minds using them, you figure they are.

 

Yet just the other day, I read an article on some Professor Escultera I think was his name, who reached the conclusion that the field axioms lead to a contradiction.

 

Now, there are still groups, fields, and rings, for me personally to contend with, but that can wait till later.

 

Right now I'd simply like to know if

 

not(0=1) is a field axiom.

 

I think such basic questions should be answered, before one moves onto the great mathematically advanced material.

 

Regards

I think that commutators[/url'] can be solved with the methods of algebraic topology. It's just that general relativity uses differential geometry instead. The second choice is better.

Ok thanks.

Originally Posted by Johnny5

Well what is a field supposed to be in the very first place?

A set that is closed under addition and multiplication' date=' that has multiplicative and additive inverses, and multiplicative and additive identity elements.

 

[/quote']

 

 

You left out the distributive axiom or no?

http://mathworld.wolfram.com/FieldAxioms.html

And one more thing...

The above list of axioms at wolfram is not minimal.

There is nothing logically wrong with giving a list which isn't minimal, its just that once the observation is made, the list of axioms can be reduced. From a memory standpoint it is desirable to have the axioms minimized, so from a human standpoint such a thing is desirable, yet there is nothing inconsistent which will happen by listing too many axioms. here is what I mean though:

They give the following statements both as axioms:

Distributivity (addition)

a(b+c) = ab+ac

Distributivity (multiplication)

(a+b)c=ac+bc

And of course they give commutativity of multiplication:

ab=ba

Now, what they do not do is use the universal quantifier to say what they really mean.

And there is a way around that though, which I will use below:

Let a,b,c denote arbitrary real numbers.

Axiom:a*b=b*a

Axiom: a*(b+c)=a*b+a*c

Axiom: (a+b)*c = a*c+b*c

However, the third axiom above is a theorem of a system with the first two axioms.

Theorem: For any real numbers x,y,z: (x+y)*z=x*z+y*z

Let a,b,c denote arbitrary real numbers.

By the closure axiom, b+c is also a real number.

Therefore: a*(b+c)=(b+c)*a by commutativity of multiplication using real number a, and real number (b+c).

Now using the distributivity axiom it follows that the following statement is true in the axiomatic system under discussion:

a*(b+c)=a*b+a*c

Therefore, by the properties of equality the following is true:

(b+c)*a = a*b+a*c

And the numbers a,b,c were arbitrary real numbers, hence the statement above is true for any real numbers a,b,c. That is:

"x in R, "y in R, "z in R: (x+y)*z = z*x+z*y

Which was to be proven. QED

Also note that I used the closure axiom too.

Field axioms for real numbers

Here you see the non-triviality axiom included.

Because it doesn't have to hold in a field. In a field with one element, the additive identity equals the multiplicative identity.

Well what is a field supposed to be in the very first place?

You are turning truth into something relative if you do that.

not(0=1) is some kind of absolute fact.

To say that the additive identity equals the multiplicative identity, changes the meaning of what 0,1 denote.

You cannot de-stabilize what is true, and has a truth value which cannot vary in time for any reason.

Yes, I've made my point.

Regards

Currently, I am trying to make my way through Euclid's book seven, which is on number theory.

I am having trouble with his very first theorem, can anyone here help?

I think the reason I am having trouble with it, is that the English translation isn't as good as it could be.

Here is a link to proposition 1 of book seven:

Euclid's elements, Book 7, Proposition 1

The translation that you see at the website, is not the same as is in Heath. I think Dr. Joyce tried to translate it more clearly on his own.

Anyway, I was reading through a graduate text on linear algebra, and around the time they begin to discuss rings, there is a discussion on "relative primeness" and fields.

So I want to be able to grasp all of this ultimately, and so I think there is a link of the modern information, to Euclid here.

So if anyone can help me decipher this first theorem, that will help me get going on the rest.

Thanks

PS: I was doing this yesterday, and a few things were confusing to me. First, I do not currently understand what is meant by "one number measures another number" does euclid mean "divide evenly"

I think one good example is all it will take.

Let the given numbers be 49,50.

He starts off with:

The less of two unequal numbers AB and CD being continually subtracted from the greater, let the number which is left never measure the one before it until a unit is left.

Actually, I am in the process of answering my own question here.

This is the Euclidean algorithm that Dave went over, a few weeks ago.

I still remember the algorithm.

For some reason Joyce is calling it "antenaresis," don't know where he came up with that.

Joyce clearly says that in modern terminology we say 'divides' not 'measures'.

And the notation is

a|b

for a divides b

Joyce goes on to say:

This proposition assumes that 1 is the result of an antenaresis process. Antenaresis, also called the Euclidean algorithm, is a kind of reciprocal subtraction. Beginning with two numbers, the smaller, whichever it is, is repeatedly subtracted from the larger.

So start with 7,50.

50-7=43

43-7=36

36-7=29

29-7=22

22-7=15

15-7=8

8-7=1

I am reading this now at Wolfram, on the commutator

Wolfram on commutator

They mention tensors if you look down a little.

I was studying GR about a month ago, and tensors came up.

Is the commutator algebra here, related to tensors?

Specifially I am wondering about equation 9 at the wolfram site.

I split this from the Fermat thread because I don't want it going off-topic - this could easily be in a thread of its own

 

Axioms are assumptions that we make - after all' date=' we have to base our initial ideas on some things, and then we can build the rest up from that. In fact, the set of real numbers is an example of something called a field; you can find the axioms at:

 

http://mathworld.wolfram.com/FieldAxioms.html[/quote']

 

Dave is that list complete?

 

For instance I don't see this one:

 

Non-Triviality

 

not(0=1)

Wow! Stop the presses! I guess all those string theorists can stop working now' date=' because Johnny5 has the answer![/quote']

 

When you know the answer just shout it out, like in 3rd grade.

OK' date=' lets restate this...

 

Say a man holds a gun to your head and says the next swan you pull better be white or else. In one hand he has a bag that he has pulled 99 white swans in a row(blindfolded) and has one left. In the other bag he has one swan.

 

Which bag are you going to select?[/quote']

 

I would look him right into the eyes, and ask him which one contains the white swan, and figure out what to do next by watching his eye movements.

This is a good topic for a thread.

However' date=' what I showed can be used to find the sums - as each of the Pascal's triangle parts is a polynomial.

So, for example, to find the sum of squares:

Pascal(x,2) = x^2/2-x/2

Pascal(x+1,3) = x(x+1)(x-1)/6 = x^2/2-x/2+(x-1)^2/2-(x-1)/2+...

x(x-1)(x+1)/3 = x^2+(x-1)^2+... - x - (x-1) - ...

x(x-1)(x+1)/3+x(x+1)/2 = x^2+(x-1)^2+...

x(x+1)(2x+1)/6 = x^2+(x-1)^2+...

-Uncool-[/quote']

 

I am trying to understand what you did there.

 

Just to be sure I understood you,

 

You are saying that if I understand the mathematics of Pascal's triangle, then I could find a formula for the sum of the first N cubes too?

 

1+(2)^3+(3)^3+... N^3 = formula(N)

 

?

 

Quite a few years back, after reading something on the difference calculus by Hamming, I began to do exactly what the original poster asked.

 

I sat there figuring out formulas for the sum of squares, sum of first N cubes, etcetera. And I was able to use what I'd just learned to find formulas, and there was a pattern.

 

I think I went up to 11, or 12 and then stopped. You know...

 

1^11+(2)^11+(3)^11+4^(11)+... (N)^11= formula N

 

The thing about what I was doing though, was that in order to do 12, you needed 11, and so on.

Wait' date=' I'm confused. I thought the definition of a normal operator is one that commutes with its [i']Hermitian[/i] adjoint.

 

Rev Prez

Revprez says the above in post 2, and in post 7

Tom says

No' date=' it's the Hermitian conjugate. You can't take the complex conjugate of an operator.[/quote']

 

So my question is, does Hermitian adjoint mean the same thing as Hermitian conjugate? And is this the thing with the dagger in it? Is that + sign above the A supposed to represent the dagger symbol?

 

I mean the + thing here:

 

You have [A,A⁺] being the notation for the commutator of A (A is an operator) and its Hermitian adjoint/Hermitian conjugate. And the commutator is general defined as: [c,d]=cd-dc

 

So in this case we have:

 

[A,A⁺] = AA⁺- A⁺A

 

And it will commute with its Hermitian adjoint provided

 

[A,A⁺] = AA⁺- A⁺A=0

Raul,

There are three spatial dimensions, no more no less.

Regards

Another electrodynamics question.

You have an infinitely long cylinder made of linear ferromagnetic material of relative permeability K_m, and you place it into an initially uniform B field. Let the cylinder axis coincide with the z axis, and let the B field point in the x direction, or y direction. CURL H = 0

What is the B field inside and outside the cylinder?

I want to use cylindrical coordinates, define a scalar magnetic potential inside and outside the cylinder, and get the B field this way if possible, but I am not sure if it can be done this way.

If there is a better way I am open to that.

Thank you

 

Now' date=' imagine what Wiles must have felt at this stage. He had spent the past 7 years in practical solitary confinement proving a theorem that he had always envisioned solving and that had remained unsolved for 350 years. I think he's perfectly entitled to burst into tears at this point - in fact, I wouldn't have blamed him if he'd just walked out.

[/quote']

 

That is actually quite touching.

 

And one more thing...

 

I praise him just for the effort alone.

 

Regards