Johnny5

I would not start with any of the things you mentioned. I would start from simple kinematics. It would begin with the general expression for the derivative of a vector. If A is some (time-dependent) vector in some reference frame that has angular velocity w then dA/dt is given by:

 

dA/dt=(dA/dt)_rot+wxA.

 

Where:

 

(dA/dt)_rot is the derivative of the vector as observed in the rotating frame.

How do you come up with that formula? I believe I saw something resembling it at a NASA site' date=' I'm gonna check to see if i can find it again. The article was on the coriolis force.

I actually found it, its for centrifugal force, not coriolis force, here it is

NASA: Derivation of centrifugal force

Look down and you will see this:

(d /dt)_i = (d /dt)_r+ w x

That is Tom's expression practically.

How in the world do you come up with it is my question?

Let both operators operate on vector A, and you get:

dA /dt = (dA /dt)_rot+ w x A

So the LHS is the time derivative of vector A, in as seen from some external inertial frame of reference, and there is an equivalence.

So I guess what I am asking for, is to see a proof of the equivalence.

I read the article at NASA before, and didn't know where they came up with the formula above. I'm going to read it again.

If you were going to teach someone about physics involving rotating frames of reference, where would you begin?

In other words where is the best place to begin?

I was thinking about starting off with the definition of angular momentum, and going from there.

L=R X P

From there you could go on to cover torque, and moments of inertia. The problem is I'm not sure the best place to start. The moment of inertia tensor is not for a beginner, so I need someplace else.

Any thoughts?

PS: My goal is to understand gyroscopes in particular, with a view towards quaternion algebra, and how it is used to avoid gimbal lock. Also, I want to tie the work here, to SO(3), which I am discussing with Tom Mattson in another thread.

There is just so much about rotating frames of reference out there, it seems difficult to compile it all, and make sense out of it, but that's what I am trying to do. So any suggestions at all would be most welcome.

Thanks

Every presentation of group theory I have seen defines a group <G' date='*> as a set G with some binary operation * defined on it. And every presentation contains the following 3 group axioms:

 

1. Associativity must hold.

2. There must exist an identity element.

3. There must exist an inverse under * for each element in G.

 

The only possible source of conflict I can imagine is that some references list a fourth axiom:

 

4. G must be closed under *,

 

whereas other references include that as part of the definition of a binary operation on a set, rather than as part of the definition of a group itself.[/quote']

 

Ok, thank you. I didn't know that about closure.

 

Is that true for any arbitrary binary relation on a set?

 

 

Let A denote a set, and let R denote a binary relation on A.

 

if x,y elements of A, then x-R-y is an element of A, for any x,y.

 

?

 

Actually, no that cannot be true. Strike the question above.

 

A mathematical operation like multiplication or addition, is not a binary relation. They are "binary operations" as you say which are different.

 

I think then, that it is best to include closure as a group axiom.

What are the axioms of a group?

I suspect I know them, but i've read conflicting answers from time to time.

I was just thinking: some theoreticions believe that the universe will eventually collapse under its own gravitational pull and result in a Big Crunch? Some of these theoreticions would go even further and postulate that after this Big Crunch, the universe will once again be in the same state it was in before the Big Bang, and so another Big Bang will occur. If this were true, then the universe is simply a perpetual series of Big Bangs and Big Crunches, and it not only makes sense to project this series into the future but into the past as well, which means that the most recent Big Bang was not the absolute beginning of the universe but only the beginning of the current cycle. In effect, it would mean that there isn't necessarily a beginning since this series could be retro-projected eternally. Is there any reasoning to this idea?

Nietzsche pondered it.

Time had a beginning, regardless of this theory, it wouldn't matter what cycle we were in now.

In order to even attempt to reason about time, and draw conclusions with certainty, one needs to start off with statements which are true, and known to be true.

How would anyone know there was a first moment in time?

I would suggest simply build up an axiomatic theory of time, and use that as an axiom.

It is possible to construct an argument in which you deduce that time had a beginning, but it's somewhat contrived.

Also, Stephen Hawking recently tried to prove that time had a beginning using umm, oh yeah... using the concept of imaginary time, but the problem with that boils down to using square root of -1.

No matter how you look at it, the negation of the fact that time had a beginning, leads to a multiplicity of contradictions. But you need a highly sophisticated pre-developed, axiomatic theory of time.

Regards

PS: Here is one idea i had, but eventually decided the argument was too complex to carry out mathematically...

Imagine everything running in reverse, from right now. Visualize the planets reversing their spins, and moving backwards along their orbits, broken things coming back together, go back more and more.

If we take current macroscopic astronomical observations seriously, as we go more and more back into the past, relative to now, things will be getting closer and closer together. Perhaps a computer could be used to project how far back in time all of the material was concentrated into one mass.

Now comes the idea, but as I said it's too complex to prove useful.

Eventually, there would come a condition of the universe, at which the reversal of motion would cause some problem. Like, for example, the temperature of the supermass approaches infinity if there was no first moment in time. Something along these lines.

But you aren't going to deduce the answer this way, for a lot of reasons.

But my point is, time did have a beginning, how you argue that fact is up to you.

now i think i have figured out that the emf that will be induced in the wire when it is spun around in the magnetic field created by the magnets will be given by this equation...

How did you get that formula?

The point is that' date=' as long as there is rotational invariance, the physics is unchanged after rotating the physical system by an angle A, and that this is no different than rotating the axes by an angle -A.

[/quote']

 

I understand that.

 

The formulas which are being used to express the "laws of physics" are the same. Still thinking about it.

 

I'm not quite sure how I would go about showing "rotation invariance."

 

OK i get everything you said...

 

After you have rotated through some angle, the laws of physics are still the same locally. You aren't saying anything more than that.

 

"the physics is unchanged after being rotated through angle A"

 

yes i get that.

 

I'm not sure it's actually true, but i get it.

 

What I certainly get, is that we want to formulate the laws of physics in such a way, as so that they are the same after the rotation.

 

We don't want rotations to change the laws of physics so to speak.

 

But you know what...

 

A spinning frame is a non-inertial frame.

 

The laws of physics do have a different mathematical form there.

 

What say you to that?

 

PS: Actually, I don't want the thread to digress that much, I would like it to remain basically on track, and it is about SO(3), so I'm going to run through your derivation of v`, using complex numbers.

What physics? We're talking math here.

 

Anyway your statement that the physics will be different is not true in general. Any closed system (that is any system that is not influenced by work' date=' forces, or moments from external sources) is rotationally invariant, and hence rotating the system has no effect on the physics of that system.

[/quote']

 

I have a simple way to show you what i mean.

 

Suppose that you are standing on the moon, and you are looking straight ahead of you, and you see the sun off in the distance, but otherwise just moon horizon, and black sky.

 

Now, suppose that you have in your hand, a steel cross, to represent an x axis and a y axis, and suppose you are holding it out in front of you, so that the origin of it covers the sun.

 

You can still see the sun, because its too big to totally block, but you get the idea.

 

Now, think of the axes of this steel cross as vectors. The steel thing is finite, so your "vectors" have a finite length, they also have mass too, but ignore that, and electrical properties, but ignore all that too.

 

Now, the two axes of the steel cross currenly lie in one XY plane.

 

Now, i say, "rotate the steel cross 90 degrees" in the XY plane.

 

What do you do? Do you rotate it counterclockwise or clockwise?

 

Lets suppose you rotated it counterclockwise.

 

Ok so you can remember this.

 

YOu can remember the state of the universe, how it ended up. You were standing on the moon, looking right at the sun, and initially the x axis of an steel cross was parallel to the moon's horizon, and the y axis was perpendicular to the moon's horizon, then you rotated the steel cross 90 degrees counterclockwise and stopped.

 

In what was done here, i will say that "the vector was rotated"

 

Now, suppose that there is literally no gravity on this moon, so that if you release the steel cross, it will not fall to the moons surface, it will hover where it is in relation to your eye, and continue to block out the sun for the most part. Or if it does fall, let it do so imperceptibly slowly.

 

Now, instead of rotating the steel cross, rotate the frame.

 

Rotate the frame by 90 degrees.

 

Is the final state of the universe the same as before? Or something different?

 

When I say rotate the frame, I don't mean rotate the cross.

 

In other words, you are turned.

 

And you are turned in such a way that:

 

The x axis of the cross now appears vertical to you, and the y axis horizontal.

 

And if you want, let the moon have been rotated too, so that your feet are still on it.

 

There is a difference in the final state of things.

 

Because the cross wasn't touched, it's X axis still points to alpha centauri.

 

In the first case, when the steel cross was rotated, the x axis of the cross went from pointing at alpha centauri, to pointing to the constellation ursa minor.

 

This is the physical difference i was thinking of.

 

Regards

The operations are indeed interchangable if you remember to negate the angle.

But there is a difference in the physics. But see i know you know that, as well as I do. Also, i began reading an article on Geometric algebra, that seems related somehow.

Kind regards

Neither classical trigonometry nor quaternions is the way to go. The quickest' date=' most direct route to the rotation matrices is to use complex exponentials. This is the case for 2 reasons:

 

1. Complex numbers in C[sup']1[/sup] can be represented as vectors that obey the same algebra as real vectors in R².

2. Exponentials are easier to work with than trigonometric functions by several orders of magnitude.

 

That said let's develop our formula.

 

Let z=x+iy be a complex number with the usual vector representation. Its polar form is z=r exp(iB), where B is the angle the vector makes with the polar axis. Now rotate the vector clockwise (remember this is the same as rotating the axes counterclockwise) by an angle A. The new complex number is z'=r exp[i(B-A)]=r exp(iB)exp(-iA)=x'+iy'.

 

Now the question is: What are x' and y'? Noting that r exp(B)=x+iy, we have:

 

x'+iy'=(x+iy)[cos(A)-i sin(A)]

x'+iy'=[x cos(A)+y sin(A)]+i[-x sin(A)+y cos(A)]

 

or...

 

x'=x cos(A)+y sin(A)

y'=-x sin(A)+y cos(A)

 

And we're done.

Thanks Tom

PS: Rotating the axes isn't the same as rotating the vector, but I like what you did here. I'm also going to have a look at this quaternion thing, i think thats a goofy name but whatever, ummm in the thread above, with Diana Gruber and someone else having a semi-friendly discussion, eventually Diana quoted some formulas obtained by Gibbs, which i am going to have a look at, but from what I gathered, the main thing about Gibbs' approach was that he avoided using the square root of negative one. The thread was on the whole... very illuminating. Also, there was mention of Gimbal lock.

Regards

Whoops' date=' I sure did change the problem from one post to the next, didn't I? The matrices I cited are in fact for rotating the [b']axes[/b] counterclockwise by an angle A. If the angle is negative it means that you are rotating clockwise. Note that a counterclockwise rotation of the axes will give you the same answer as a clockwise rotation of the vector.

 

Ok. "The matrices you cited are in fact for rotating the axes counterclockwise by an angle A." Hmmmm

I found this, and was wondering about it. Id like to hear some thoughts on this.

Progammer lashes out at quaternions

From what I have gathered, a computer programmer, Diana Gruber, wrote an article about how terrible it is to use quaternion algebra to develop the rotation matrices. What made the whole thing notable, was because she works for some 3D gaming site, or something along those lines.

I read through her article, which was subsequently lambasted in the post at the link above, and in her article, she actually goes through extensive effort, using classical trignometry, to develop rotation formulae.

But, then I read the post at the link above, and the person there mentions SO(3), here:

Diana Gruber has written a very interesting article about Quaternions for the Gamedev site; if you haven't read it yet' date=' you can find it here. It's certainly worth reading as one person's perspective on the issue. Unfortunately, I also find it to be inaccurate (or at least misrepresentative) on several counts. Mrs. Gruber is welcome to her own opinions on whether quaternions are worthwhile for her own programming efforts, but at the same time obviously a lot of people _have_ implemented quaternion-based rotations for their graphics and physics engines, and physicists (not just mathematicians) have been actively using them since Hamilton discovered them (as the most convenient representation of the double cover of SO(3) by S^3, to put it in the most technical terms) so I think there's at least some empirical evidence that they _do_ indeed do something 'that couldn't be done more easily using more traditional mathematics', to use her phrase. But she's also wrong about some of the specifics; here's why I think so.

 

Firstly, Mrs. Gruber compares building a rotation from a quaternion to building a rotation from a rotation angle and an axis; I won't reproduce the formulas but you can get them from her article. Her implication is that building from a rotation angle is easier, but let's count. [/quote']

 

 

I'd like to have a look at the quaternion approach.

 

Anyone know it?

It is my understanding that inertial space has to be flat.

Let me see if i understand you right.

Are you saying that if there is an inertial frame somewhere, a frame in which Newton's law of inertia is true, then using the formulas from GR, the conclusion is that space is flat?

Do I have that right?

Regards

You will not get my answer because you are not solving the same problem. As I said my answer is for what happens when you rotate the frame, not the vector.

Ok, I see that now. In your initial post, you said rotate the vector, and then later you said, rotate the axes.

they are not the same problem.

Moving this to the algebra forum since it has little to do with classical mechanics.

No problem Dave.

As I said my answer is for what happens when you rotate the frame, not the vector. If you rotate the vector instead of the frame then you will obtain an answer that is equivalent to replacing A by -A in my answer.

This is quite subtle Tom.

Ok I'm gonna draw a new diagram, and see if i can get your answer.

Start by looking at a vector v=<x' date='y> in the xy-plane. Then rotate the axes by an angle A and see what the new components are. You should get:

 

[b']v[/b]'=<xcos(A)+ysin(A),-xsin(A)+ycos(A)>

I have a question. Is A necessarily positive?

Orientation Methods

We have vector <x,y> in unprimed frame S.

The same vector is <x`,y`> in frame S`

Frame S` has the same origin as frame S, but is rotated slightly by angle A.

Now, if A=0, then S=S`, but it is not the case that A=0.

I made a drawing, and I have A = 30 degrees, and the vector V is in the first quadrant of S, at about 80 degrees to the x axis of S, and 50 degrees from the x` axis of S`.

As is mentioned in a few other posts' date=' the hypersphere does not constitute an inertial frame, which means you can't blindly apply SR to it.

[/quote']

 

I was aware of this Dr. Swanson, because, Bonzo is traveling in a space ship which isn't subjected to any external forces, and hence should move in a straight line at a constant speed, in Gonzo's frame F, which was stipulated to be inertial. Yet because the universe was assumed hyperspherical, Bonzo did in fact travel a curved path in Gonzo's inertial frame.

 

That pushes the confusion to definition of inertial frame vs definition of hyperspherical universe.

 

I'm not sure of the definition of 'hyperspherical universe' but then again, the universe isn't hyperspherical, in the sense that the enterprise, guided using a gimbal to keep it moving straight, will not return to the sound buoy.

 

Regards

Why would you try and say that (r^2+r+1) is equal to zero? That would invalidate the whole expression ie 8 . 0 = r^3 . 0 which would be 0=0, we get nowhere. Also if a=0 then we get 0=0 again. Since the 2 expressions that cancel are identities i didnt think it would matter what value they have, as they will always cancel to leave r^3=8.

It is an issue with complex numbers. Let me go back and show you why.

8(1+r+r²) = r³ (1+r+r²)

 

The equation can be true if (1+r+r²) =0.

 

In the case where (1+r+r²) is not equal to zero' date=' you can divide by it to obtain:

 

8 = r[sup']3[/sup]=r*r*r

 

Whence it follows that r is equal to the cubed root of 8, which is 2.

 

As for the case where

 

1+r+r² =0

 

You can use the quadratic formula to solve for r, and see what happens.

 

root1 = -1+ (1-4)^1/2

root2 = -1- (1-4)^1/2

 

As you can see, the discriminant is negative, which means the two roots above are complex numbers, not real numbers.

Now, some have a problem with the complex numbers, and others do not, but regardless, it has been highly developed over the past few centuries.

So here is what i did, during the solution of the problem.

First, I started with the information given, as true.

9(a+ar+ar²) = (a+ar+ar²+ar³+ar⁴+ar^5[/sup'])

^{You did as well:}

^{Q: the sum of the first 3 terms in a GP is 9 times less than the sum of the first 6 terms' date=' find the ratio.}

 

^{Here is my solution:}

 

^{since: 9(a+ar+ar^2)= a+ar+ar^2+ar^3+ar^4+ar^5[/quote']}

 

^{Then, i had to recall the definition of a GP, which i already knew. Again, and this is from memory, a geometric progression is a sequence of numbers, where the next term in the sequence is obtained from the previous one, by multiplication of a fixed number r, called the common ratio. Now, you chose 'a' to denote the first term of the sequence, so, so did i.}

 

^{Now, there was no mention about whether or not a=0, or r=0, or both. Just not enough information given. So I just decided to give an answer which covered all possibilities.}

 

^{(0,0,0,0,0,0...) is a sequence I believe, according to the definition of a sequence. I'll check right now to see, but I already do think it is. And if not, I certainly could adjust the definition of sequence, to allow it to be. Hold on, let me grab a link.}

 

^{Ok, I looked and didn't find a definition, but I know there is one in my Real analysis text. But the basic idea behind one can be seen by considering the sequence of odd numbers.}

 

^{Starting with 1, we have:}

 

^{(1,3,5,7,9,11,13,...)}

 

^{And we can put that sequence in 1-1 correspondence with the sequence of natural numbers:}

 

^{(1,2,3,4,5,6,7,...)}

 

^{Let A(n) denote an arbitrary term of the sequence: (1,3,5,7,9,11,13,...)}

 

^{ane let n denote an arbitrary term of the sequence: (1,2,3,4,5,6,7,...)}

 

 

^{The two are related by: A(n) = 2n-1}

 

^{So the basic idea, is that the elements of the sequence are ordered, a first, a second, a third, and so on.}

 

^{Now, in the example above, consider Dn.... "change in n."}

 

^{If you study, or have studied the finite discrete difference calculus, you will have come across the difference operator, defined below:}

 

^{Df(x) = f(x+h) - f(x)}

 

^{In the notation above, h is called the step size. In the sequence (1,2,3,4,5,...) the step size happens to be one, but this does no have to be the case.}

 

^{Also, the difference operator, defined above, is used to define the derivative in the differential calculus.}

 

^{So the point I am currently trying to make, is that all that is important in the notion of "sequence" is that the terms come in an order, you know like how things are ordered in time. Nothing more than this.}

 

^{So, from that, (0,0,0,0,...) is a sequence.}

 

 

^{Hence, if a=0, and not(r=0), then the following sequence is a geometric progression, with common ratio r:}

 

^{(0,0r,0rr,0rrr,0rrrr,...)}

 

^{which is equivalent to}

 

^{(0,0,0,0,0,...)}

 

^{Also, possible is this case:}

 

^{not(a=0) and r=0}

 

^{(a,a0,a00,a000,...)}

 

^{which is equivalent to}

 

^{(0,0,0,0,...)}

 

^{In mathematics, it is always important to know what definitions you are using, so that you can explain yourself, at the very least.}

 

^{So, like i said, i believe (0,0,0,0,0,..) is a sequence, so i don't think you can just ignore the two cases I just showed above, without more information given in the problem.}

 

^{So I just handled all the possible cases, which leads to not just dividing by 1+r+rr, because I did not assume that it isn't zero.}

 

^{In other words when I got to the following statement:}

 

^{8(1+r+r2) = r3 (1+r+r2)}

 

^{The statement above was certainly true, regardless of what r could be equal to.}

 

^{All is fine up to there, and I was certain that all was fine.}

 

^{But, there is a logical reason why you cannot divide by zero.}

 

^{If you logically analyze the field axioms of the algebraic operations, you see that if you are using certain field axioms together, that permitting yourself to divide by zero, allows you to reach the conclusion that 0=1 and not(0=1).}

 

^{And I am freely using the field axioms, so i have to be careful whenever I divide by something with a letter in it.}

 

^{So here is what the logic is then...}

 

^{8(1+r+r2) = r3 (1+r+r2)}

 

^{Statement above is true for certain.}

 

^{If (1+r+r2)=0, then RHS=0 and LHS =0, which makes statement true, as we know it is. Yet if that is the case, then we cannot divide by it, we have to leave things as they are.}

 

^{At this point, I didn't know anything about the roots, i simply used the quadratic formula to have a look at them.}

 

^{The quadratic formula can be obtained by completing the square, you learn how to do that, before you memorize the quadratic formula, because the one is used to deduce the other.}

 

^{It's been years since I've actually derived the quadratic formula by completing the square, but i still remember how.}

 

^{So, using the quadratic formula I found this:}

 

^{You can use the quadratic formula to solve for r' date=' and see what happens.}

 

^{root1 = -1+ (1-4)[sup']1/2}

root2 = -1- (1-4)^1/2

 

As you can see, the discriminant is negative, which means the two roots above are complex numbers, not real numbers.

 

Up till then, i didn't already know that the roots of that polynomial, were necessarily complex. But then at this point i did.

 

I looked back at your question, and didn't see any reason why r couldn't be a complex number.

 

So that finished off the case where (1+r+r²)=0.

 

Then I covered the mutally exclusive case, where the polynomial above is not equal to zero. In this case you can divide through it, and then you can rapidly see that r=2.

 

The two cases above are mutually exclusive, and collectively exhaustive, so i was done.

 

To summarize: Unless there is a reason why (0,0,0,0,0..) cannot be a sequence, or unless there is a reason why r cannot be a complex number, I don't see anything wrong with a comprehensive answer.

 

There was nothing at all wrong with what you did.

 

You just asked to see alternative approaches to answering the question, so i tried to see how I would solve it. I happen to like the geometric series, which is where i encountered arithmetic/geometric progressions in the first place, and this was many years ago.

 

Kind regards

Start by looking at a vector v=<x' date='y> in the xy-plane. Then rotate the axes by an angle A and see what the new components are. You should get:

[/quote']

 

Ok this is all that I've read of your answer so far, and I'm just going to try and do this much without reading the answer yet.

 

Consider an arbitrary position vector in an XY plane.

 

It's tail is located at the origin (0,0), and the tip is at some arbitrary point in the plane, say (x,y).

 

There are many ways we can express this vector.

 

V=V(x,y) = <x,y>=xi+yj=xe₁+ye₂

 

There should be a little hat ^ over the einheitsvektors (sp?) but latex isn't working so boldface will have to do.

 

So we have an arbitrary vector, in an arbitrary plane to start off with.

 

Now, we want to be able to discuss, mathematically, a rotation of this vector in the plane, not a frame rotation, the vector is to rotate.

 

So we are going to have it rotate through an angle A.

 

So consider the circle, of radius |V| (magnitude of vector V), with center at the origin.

 

The distance from any point on the circumference of said circle, can be expressed in terms of the coordinates on the axes of the frame, by using the Pythagorean theorem.

 

Before the vector rotates, the following statement is true in the frame:

 

|V| = (x²+y²)^1/2

 

Now, after the vector has rotated through the angle A, the statement above is false in the frame, but another statement which used to be false is now true, namely:

 

|V| = ((x₂)²+(y₂)²)^1/2

 

Where

(x₂, y₂)

 

is the point in the frame, where the tip of the vector V is now, after being rotated through angle A.

 

I drew a little diagram, to help me out. At this point, I want formulas with A in them.

 

x₂ =

y₂ =

 

Ok I got stuck at this part for a little while, but i think i figured out. What I am going to do, is use the law of cosines. There will be a vector triangle.

 

One side of the vector triangle will be the initial vector, then you add the vector (whose magnitude is the length of a chord), and then you get the final position vector v`.

 

This way, I can write the final vector in terms of A, since by the law of cosines the square of the chord length C satisfies:

 

c^2 = R^2 +R^2 - 2R^2 cos(A)

 

where R=|V|=radius of the circle.

 

If this doesn't work, then I'm gonna need help on this part, but it should work. I don't know if I'll get Tom's answer but we'll see.

Rotate the vector v=<x' date='y,z> about the z-axis by an angle of 90 degrees. You would take the rotation matrix for the z-axis and plug in phi=90:

 

      [0  -1   0]
R[sub]z[/sub](90)=[1   0   0]
      [0   0   1]

 

You would find the rotated vector [b']v[/b]' by multiplying R_z(90) and v, in that order.

Alright, this is a place to start.

Rotation Matrix

R_z ( F )

You start with a three dimensional rectangular coordinate system, and x axis, a y axis, and a z axis, meeting at the origin (0,0,0).

Let V(x,y,z)=<x,y,z>=xi+yj+zk denote an arbitrary position vector in the frame.

Now, we want to "rotate that vector about the z axis, by 90 degrees."

I'm not sure how to visualize this.

My question is how do you come up with the rotation matrix in the first place.

I don't see it just yet.

I understand the last part though.

Once I have R_z ( F ), i let phi =90 to get:

R_z ( 90 )

Which is the matrix you gave.

Then matrix multiplication gives the answer to the question.

R_z ( 90 ) V = answer

Mathematically they are special only in the sense that they are a special case of O(3). One could just as well regard as "special" the subgroup of O(3) whose determinants are -1.

 

 

 

The rotation matrices are members of SO(3). I don't recall if they are the only members' date=' so someone else can answer that one. But a rotation matrix must be orthogonal because orthogonal matrices are [i']norm-preserving[/i]. That is, the length of a vector does not change under an orthogonal transformation. Since rotations do not alter the norm of a vector, orthogonality is a necessary condition for rotation matrices. A further necessary condition is that the determinant has to be +1 instead of -1. If the determinant is -1 it turns out you do not get a continuous transformation, and continuity is als a necessary condtion for rotations. Elements A of O(3) with det(A)=-1 are used for discrete transformations, such as space inversion (parity).

Ok, thank you.

Can you show me a basic example of something, which forces me to use 90% of the ideas here?

PS: Just so you know, the reason I asked the question, is because I am trying to understand gyroscopes, precession, nutation, spin, roll, pitch, yaw, Euler angles, that kind of thing. In the process of trying to understand gyroscopic motion, I came across SO(3), and it seemed connected to most of the things i just listed, as well as something called quaternions, developed by William Rowan Hamilton. So I am trying to learn all these new things, just to understand how a gyroscope works.

If I could understand just this one thing, I would probably understand the whole of classical mechanics, at least that is how it is beginning to look to me. Another thing of possible interest, is that all of this seems related to the Sagnac effect, which Geistkiesel and Tom Swanson and I were discussing in some other thread. I now realize that mechanical gyroscopes can be replaced with optical ones, which are based upon the "Sagnac effect." And Dr. Swanson also mentioned, that the linear sagnac effect was somehow related to the Michelson Morely experiment.

So I can see some kind of synthesis that can happen, if I learn the right things. I expect to use linear algebra to understand the gyroscope, possibly quaternions, possibly Euler angles, and all of that stuff seems centered on SO(3).

You've neglected the case where a hyperspherical universe violates one of the assumptions of SR.

Can you be more precise?

They would both be travelling at the same v wrt the observer, and have the same dilation.

Hmm i need a moment. let me answer the previous thing, and I will come back and think about this.

Johnny5' date='

Help me out here.

Assuming time dilation and your statement are true that after each complete the round trip of the universe, moving in straight line, A and B will see the other's watch slower than his own. As the ships pass each other at the point where earth was located when A and B started their voyage, each releases a piece if paper on which is printed the time of the A and B watches (each assumed slower than the other). What are the times printed on each piece of paper?[/quote']

 

The point of my response, was meant to imply that it cannot be the case that both SR is correct, and the universe is hyperspherical.

 

Here is the logically complete answer for you:

 

1. SR false and not (universe is hyperspherical)

2. SR true and not(universe is hyperspherical)

3. SR false and universe is hyperspherical

 

The possible choices are reduced by one, since both cannot be true.

 

So you can now do a case by case logical analysis.

 

Without knowing the exact meaning of "universe is hyperspherical" its difficult to carry out the logical analysis, but i think it has something to do with your direction isn't changing, yet you can return to your starting point.

 

This kind of space is impossible, but was the subject of a star trek episode.

 

The space ship entered some kind of highly warped region of space, and they couldn't get out, and had no way to know what direction they were moving in, because they couldn't see any stars no matter what direction they looked.

 

They dropped a sound buoy, that pinged the ship, and they began to travel away from the buoy in a straight line at a constant speed.

 

Sound pulses or whatever, were shot from the buoy in all directions, and obviously some hit the hull of the spaceship, and when that happened, there was a 'ping' recorded that everyone could hear.

 

As the ship moved further and further away from the buoy, the loudness of the 'ping's decreased, until they could no longer hear the buoy.

 

Now, they knew for sure that they were traveling in a straight line away from the buoy, and here's how.

 

On the bridge of the enterprise, they had a gyroscope.

 

The spin axis of the gyroscope was aligned with the ship's motion.

 

Had the enterprise strayed in any direction, the gimbals would have reoriented, but left the direction spin axis alone.

 

That's the special feature of a gyroscope, so they had one, and made sure that the gimbals never moved.

 

In other words, if the spin axis, started pointing at different objects on the bridge, they would know they were veering off a straight line path for certain, even though they couldn't see any stars.

 

They could then quickly compensate, so that the spin axis again, pointed at the front of the bridge. So the point is, they could control their speed, and their direction, and keep both constant in time.

 

 

In this episode they did this.

 

And yet, after a few minutes of moving away from the buoy, in a straight line at a constant speed, they began to hear a faint pinging sound, which got louder and louder, until they were back where they started, right next to the buoy.

 

Obviously, our real universe cannot be like this, it is truly impossible.

 

So if that kind of thing can happen in a hyperspherical universe, then the universe for sure isn't hyperspherical.

 

So that would eliminate #3 above.

 

But, if the meaning of hyperspherical universe isn't quite what i said, then some other analysis would be needed.

 

For example, suppose that as one of the twins moved along at a constant speed, his direction slowly changed, so that he really did travel in a circle.

 

And suppose his brother didn't move, i.e. his brother was the sound buoy.

 

So we are imagining the radius of the universe to be enormous, but circumnavigable.

 

So over any short amount of time, the tangential speed v of one of the twins is constant in time, but the direction of motion is slowly varying, and the centripetal acceleration of this particular twin is given by:

 

v^2/R

 

where R is the radius of the universe.

 

I don't think centripetal acceleration affects SR formulas, but I'm not sure.

 

The reason I'm not sure, is simply because i've never thought about it before, not because I don't know how to carry out the analysis.

 

So, in the case where the one twin remained at rest in some frame F, and the other twin entered a rocket, was given a brief impulse, and then just coasted for a very long time...

 

What SR says is undeniable...

 

Lets say that Gonzo remained at rest in inertial reference frame F, and just to help us visualize, let it be the case that the center of mass of Gonzo is also at rest relative to the center of the universe. Just to help you imagine.

 

Then identical twin Bonzo, boards a spacecraft, and the engines are turned on for just a few seconds.

 

Then the fuel runs out, and Bonzo and the spaceship just coast, at a constant speed v.

 

Suppose that the impulse given to Bonzo+spaceship was as follows:

 

Initial speed of Gonzo relative to Bonzo =0, then force is applied for three seconds, then all fuel used up, and no force is ever applied to ship again.

 

Final speed of Gonzo relative to Bonzo = .9999 c

 

Thus, Bonzo is smushed, but presume his wristwatch is still intact, and can circumnavigate the universe. Or if you want Bonzo to live, pretend his ship was superconductive, and that inertial forces are really due to external magnetic fields, and that advanced technology prevented him from being smushed, even under such extreme but brief acceleration. in either case the watch was undamaged.

 

 

Regardless of what you chose, what SR says about time dilation is very very easy to understand, and i will now explain what follows if time dilation formula is true:

 

Let Dt denote an amount of time as measured by Gonzo's watch.

 

Let Dt` denote an amount of time as measured by Bonzo's watch.

 

If there is no such thing as time dilation then

 

Dt = Dt`

 

Which means that if their watches were synchronous at the start of Bonzo's voyage, then upon his circumnavigation of the universe, what he writes on his paper will match what his brother writes as well.

 

At most, any discrepancy would be due to some effect which happened during the acceleration phase of Bonzo's trip, but that only lasted for a few seconds, and would be small.

 

So thats what would be the case if SR is false, if SR is true, then you have the following formula, which Gonzo has to use...

 

Dt = Dt` (1-vv/cc)^-1/2

 

keep in mind, the LHS is what is measured by Gonzo's watch, a 24th century digital timex.

 

Now, the delta t` on the RHS is what Gonzo is supposed to compute to be the amount of time that passes aboard his brother's spaceship. In other words, spaceship time. v denotes the relative speed of Gonzo to Bonzo, or equivalently the speed of the ship in reference frame F.

 

c is the speed 299792458 meters per second.

 

You can now infer that time passes slower aboard the ship, relative to external reference frame F's time.

 

Now, recall that the impulse only lasted for three seconds, but that afterwards, the relative speed of Gonzo to Bonzo was .9999 c.

 

Hence we have this here:

 

Dt = Dt` (1-.99980001)^-1/2

 

Since (.9999)(.9999)=.99980001

 

Now, 1-.99980001 = .00019999

 

so that we have:

 

Dt = Dt` (.00019999)^-1/2

 

So that we have:

 

 

Dt = Dt` (5000.25)^1/2

 

 

since

 

1/.00019999 = 5000.25 and 1/x = x^-1 for any x

 

Now we just take the square root to obtain the following statement, which must be true in the Gonzellian frame F:

 

Dt = Dt` 70.7

 

So, for the sake of using numbers, and not letters, let us suppose that after 20 billion years have passed according to Gonzo's watch, his brother finally is coming into view.

 

What does special relativity theory say that Bonzo's watch will read?

 

Well before trying to answer this, let's figure out the radius R of the universe first. I think i've introduced enough information for that to be computable now.

 

We know the path traveled by Bonzo was a circle, we are stipulating that there were no perturbations to the uniform circular motion of his ship for any reason.

 

Letting R denote the radius of the universe, the distance traveled by Bonzo, in Gonzo's frame is given by the Archimedian formula :

 

C = 2pR

 

We have stipulated that the amount of time of the Bonzo's whole trip, as measured by Gonzo's 24th century digital timex, was 20 billion years.

 

Now lets convert that to seconds.

 

20 billion years = 6.3 x 10¹⁷ seconds

 

Now, Bonzo's tangential speed v, was stipulated to be .9999c throughout the duration of his ship (well all but three seconds of it anyways).

 

So, we do have enough information to compute the radius R of the universe.

 

v=.9999c = distance traveled/time of travel

 

hence:

 

.9999c = 2pR/(6.3 x 10¹⁷ seconds)

 

and c = 299792458 meters per second hence:

 

.9999(299792458 m/s) = 2pR/(6.3 x 10¹⁷ s)

 

This is one equation in one unknown, the unknown is R, hence we can solve for R, as an explicit number, doing so we find that:

 

R = 3.0005 x 10²⁵ meters

 

So, Gonzo has assumed that the time dilation formula is true, hence Gonzo must use the following formula:

 

Dt = Dt` 70.7

 

And the time, as measured by Gonzo's wristwatch was:

 

6.3 x 10¹⁷ seconds

 

Hence, Gonzo reaches the conclusion that:

 

6.3 x 10¹⁷ = Dt` 70.7

 

From which it follows that:

 

8.9 x 10^15 seconds = Dt`

 

Which we can transform to years.

I want to solve a problem, to help me recall some more linear algebra.

I want to solve a matrix system, and I want the answer to be a straight line in three dimensional space.

So intuitively, i know this:

If two planes are parallel they have absolutely no points in common, but

If they are not parallel then they do have points in common, and

the set of all points they have have in common lie on one and only one infinite straight line.

Now, i know that the form of an equation for a plane in three dimensional space is:

Ax+By+Cz=0

So choose two planes from the set of planes, but make sure they are not parallel.

At this point, i want to solve a system of simultaneous linear equations, and write my answer in the form of an arbitrary straight line in 3D space.

This is the part I cannot remember how to do.

Thank you

Question: What is so special about the elements of O(3) whose determinant is +1?

Also, what does SO(3) have to do with rotations?

Regards

Please don't type everything in caps - it comes across as being quite rude

I thought his keyboard was broken.