Relativity

[Tom Mattson]

How did you arrive at the coefficients of the three unit vectors?

 

I did it with simple right triangle trigonometry. Theta is the angle that V makes with the z-axis' date=' and phi is the angle that the projection of [b']V[/b] in the xy-plane makes with the x-axis. The derivation is in any calculus book.

 

PS: I can see your argument leads to... a vector is equivalent to the zero vector if and only if its magnitude is zero.

 

That's right.

[J.C.MacSwell]

All right I have a question.

 

Take a body at rest' date=' and not spinning. Its center of inertia is at rest. So we say that the object isn't accelerating in the frame.

 

[/quote']

 

This is only true if there is no net force on it.

[Johnny5]

Using right hand coordinate system:

 

Right hand system

 

Angle from z axis to vector V is theta (which varies from 0 to pi/2)

Then, the angle between i^ and the projection of v onto the xy plane, is phi, which can vary from 0 to 2 pi.

 

So let the origin of the frame be at the center of mass of the body.

 

Any point in space could either be referenced by (x,y,z) or (R,phi,theta)

 

Yes I remember all this.

 

Then simply derive the relations to convert from rectangular coordinates to spherical coordinates.

 

Offhand I don't remember how, but I could re-derive them for myself now.

 

So we have a vector V with one endpoint at the origin, and the other at some point (x,y,z) in space, some distance R away.

 

The direction of this vector is going to be FROM the origin TO the point (x,y,z)

 

[math] \vec V = r \hat r [/math]

 

Ok this is the vector using spherical coordinates, where r is the length of the vector, and r^ is direction.

 

This is the same vector, using rectangular coordinates:

 

[math] \vec V = x\hat i + y \hat j + z \hat k[/math]

 

That projection has length [math] r sin \theta [/math]

 

So now I just have to work out vector V in rectangular coordinates, and then equate to r r^, divide by r, and I should get that formula.

 

The projection is the hypotenuse of a right triangle in the xy plane, with one side of length x, and the other side of length y. Therefore:

 

[math] x^2 + y ^2 = r^2 sin^2 \theta [/math]

 

and also, there is a second right triangle, with hypotenuse r, one side having the length of the projection, and the other side having length z. Therefore, in this frame, it is also true that:

 

[math] (r sin \theta)^2 + z^2 = r^2 [/math]

 

Well yes obviously since [math] x^2+y^2+z^2 = r^2 [/math]

 

Oh I see. Phi is the angle such that:

 

[math] sin \phi = \frac{y}{r sin \theta} [/math]

 

[math] cos \phi = \frac{x}{ r sin \theta} [/math]

 

from which we can readily see the following:

 

[math] \vec V = r sin \theta cos \phi \hat i + r sin \theta sin \phi \hat j + z \hat j [/math]

 

And now, all we have to realize is that

 

[math] cos \theta = \frac{z}{r} [/math]

 

to finally discover that:

 

[math] \vec V = r sin \theta cos \phi \hat i + r sin \theta sin \phi \hat j + r cos \theta \hat j

 

[/math]

 

We can now equate this to rr^, and then divide both sides by r, to finally obtain:

 

[math] \hat r = sin \theta cos \phi \hat i + sin \theta sin \phi \hat j + cos \theta \hat j

 

[/math]

[Johnny5]

Ok, so you can solve for direction, it doesn't matter that the coefficients of i^ j^ k^ were asymmetrical, since the analysis would be the same, if you just renamed the three axis, permuted x,y,z.

 

So back to the problem of vector division.

 

Start with

 

[math] \mathbf{F} = m \mathbf{a} [/math]

 

We are starting off with the case where the center of mass of an object is at rest in some reference frame S. The reference frame is 'good' in the sense that we know the laws of physics in it. It's an inertial reference frame, so therefore our formulas are true in the frame. Whether they be frame dependent formulas, or frame independent formulas is another issue entirely, but all that matters right now is that F=ma is true in S.

 

We can view an acceleration vector, as being a magnitude times a direction. So let us write Newton's second law differently:

 

[math] \mathbf{F} = m a \hat a [/math]

 

Where a is the magnitude of the acceleration vector, and a^ is the direction of the acceleration vector.

 

In the case where a body is at rest, the acceleration vector must be zero. So either a=0 or a^=0.

 

a^ is a unit vector, so its magnitude is 1, and the previous post was meant to show that no vector can have a direction of zero. Hence, it cannot be the case that a^=0, therefore a=0. But presuming that vector division is defined, we can now divide both sides by a^, to obtain:

 

[math] \frac{\mathbf{F}}{\hat a} = m a[/math]

[Tom Mattson]

a^ is a unit vector' date=' so its magnitude is 1, and the previous post was meant to show that no vector can have a direction of zero. Hence, it cannot be the case that a^=0, therefore a=0. But presuming that vector division is defined, we can now divide both sides by a^, to obtain:

 

[math'] \frac{\mathbf{F}}{\hat a} = m a[/math]

 

Now the only question is, how do you define vector division?

[Johnny5]

Now let's try this with vectors' date=' as we attempt to solve for x. We'll have to assume that a "multiplicative inverse" [b']a[/b]^-1 exists for a. I put the word multiplicative in quotes, because it is not clear what kind of multiplication we are dealing with. For now, I'll denote it by the pound sign (#).

 

ax=b

a^-1#ax=a^-1#b

 

Now we have 2 ways to multiply vectors: The dot product and the cross product. Let's see if either one could possibly stand for #.

 

Dot product:

a^-1.ax=a^-1.b

 

Now we hit our first roadblock. What exactly is a^-1 with respect to the dot product? A multiplicative inverse is supposed to have 3 properties:

 

1. It has to return an identity element when multiplied by a.

2. It has to be in the same set as a and the identity.

3. It has to be unique.

 

There is no vector a^-1 that satisfies all these properties.

 

To prove this, assume that an identity element e exists such that a^-1a=e. You could try to define a^-1 as a vector which when put in a dot product with a, yields the scalar "1". But there are two problems with this. First, there are an infinite number of such vectors, which violates #3. And second, vectors aren't closed under the dot product, which violates #2.

 

And of course, if we look at the other case, namely that in which an identity element e satisfying a^-1a=e does not exist, then we have contradicted condition 1 above.

 

Thus, we cannot define vector division from the dot product.

 

 

No it isn't clear what kind of multiplication we are dealing with. Let me deal with a specific problem. Initially, there is an object at rest in some frame S. The center of mass is not translating in S. Now either the object is spinning or not, let us say not.

 

And these are the initial conditions.

 

Let it be the case that an object at rest in this frame will remain at rest in this frame, until an external agent acts upon it.

 

Now we bring into the setup a single contact force F, we push the object say a billiard ball, with a finger.

 

Let the applied external force F, point through the center of mass of the object.

 

1. If the object is on our kitchen floor, the object will begin to translate AND rotate.

 

(It will rotate because of friction between the object and the floor)

 

2. If the object is in the vacuum, then the object will only translate; where we have assumed that the friction between the object and the vacuum is literally zero.

 

Let us carry out the analysis in the vacuum.

 

So we don't have to worry about rotation in this case, UNLESS the applied external contact force does not point right at the objects center of mass.

 

So let us focus on the case where the contact force does not point right at the center of inertia.

 

For simplicity, we are analyzing a spherical solid ball. Let the object have uniform density, so that the center of inertia lies at the center of the sphere.

 

Pick an arbitrary point on the surface of the sphere, where the point of contact is.

 

The applied force is a vector, which has some line of action. We are considering the case where the line of action does not have the exact same direction as the vector from the point of contact to the center of inertia.

 

Let [math] \vec R [/math] denote a vector from the center of inertia, to the point of contact.

 

The line of action of the applied force is not the same line that contains [math] \vec R [/math], so that we have two lines here, with a common point (the point of contact); so they lie in one plane.

 

Draw the tangent to the sphere through the point of contact.

Draw the normal line (line perpendicular to the tangent line, also containing the point of contact).

 

The contact force has to come from exterior to the ball, so there are only 180 degrees spanning the direction the applied force can come from.

 

In the case where the line of action of the contact force is the normal, there will be no rotation after contact is made, only translation. We are considering the general case. So introduce the angle of incidence.

 

Define the angle of incidence [math] \theta [/math] to be the angle between the normal and the line of action of the applied force.

 

So in our problem here, theta is an angle greater than zero (because we are not pushing through the center of inertia), but less than 90 (because the push is coming from outside the sphere).

 

Now, break up the applied force vector F into two components, one which is perpendicular to the normal, and another which points right at the center of inertia (so that it is parallel to the normal).

 

Denote the applied external force as:

 

[math] \vec F [/math]

 

 

[math] \vec F = \vec F_1 + \vec F_2 [/math]

 

Let F1 be portion of F which acts to translate the center of inertia of the object in the CMU frame of reference. Therefore, the magnitude of F1 is:

 

[math] F cos \theta [/math]

 

Where F is the magnitude of the applied force.

 

F2 is the portion of F which acts to rotate this object in the CMU frame.

 

The magnitude of F2 is given by:

 

[math] F sin \theta [/math]

 

So we have:

 

[math] \vec F = F cos \theta \hat a + F sin \theta \hat b

 

[/math]

 

Where a^ is a unit vector which points in the direction from the point of contact to the center of inertia, and b^ is a unit vector which points from the tail of F to the tail of vector F1.

 

Let us focus on just the portion of F, which goes into translating the center of inertia in the CMU frame. This is vector F1. Neglecting F2 for now we have:

 

[math] \vec F_1 = -(F cos \theta) \frac{\vec R}{|\vec R|} [/math]

 

Now, assume that Newton's second law of motion is true, when applied to F1. Therefore, the inertia of the ball, and the resulting acceleration vector of the ball (in the CMU frame), are related by:

 

[math] \vec F_1 = m \vec a [/math]

 

We know the direction of a, it is going to be antiparallel to the direction of [math] \vec R [/math], and even after R vector begins to rotate in the CMU frame, the acceleration vector will still point in the direction from the point of contact (at the moment of contact) to the center of of inertia (at the moment of contact), in the CMU frame.

 

On the other hand, the vector R will now be spinning in the CMU frame. It's tail itravels with the center of mass of the ball, and it's head is always the point of contact, even after contact is made. This vector will be spinning in the CMU frame after contact was made, but before contact was made this vector was static in the CMU frame.

 

It is the final motion of R vector in the CMU frame which we are interested in.

 

The tail of R is always located at the center of inertial mass of the sphere, and we can find its acceleration in the CMU frame by using:

 

[math] \vec F_1 = m \vec a [/math]

 

As for F2, that is the portion of the applied force F, which went into causing the sphere to start spinning in the CMU frame.

 

So the point of contact is now orbiting the center of inertia of the sphere in the CMU frame.

 

And since its orbiting, its direction is continuously changing in time. After the contact force is off, the orbital speed will be constant in time in the CMU frame.

 

However, the direction will be changing, so that the point of contact will no longer be at rest, nor will it be traveling in a straight line at a constant speed in an inertial reference frame.

 

I think (someone correct me if I am wrong)... the point of contact is going to be tracing out a wave shape, in the CMU frame. The frequency and wavelength of it are related to the acceleration given to the object by the applied external force.

 

Now, consider things in the reference frame attached to the object, whose origin is permantly the center of mass of the object. Let the X axis of this frame be the NORMAL line, which was defined way back when we discussed the point of application of the applied force. Let the direction of increasing x coordinates in this frame, be the same direction as the acceleration of the center of mass. Keep in mind that the applied force only lasted momentarily, so that the center of inertia is no longer accelerating, it's traveling in a straight line (the normal line) at a constant speed.

 

[math] v_f - v_i = \vec a t [/math]

 

This formula is true in the CMU frame, and in that frame the objects center of inertia was initially at rest, hence vi=0 so that in the CMU frame the following statement is true:

 

[math] v_f = \vec a t [/math]

 

Where vf is the speed of the center of mass of the object in the CMU frame of reference, and a was the acceleration (in the CMU frame) given to the center of inertia of the object by the applied external contact force. That force only lasted a few moments, afterwards, the speed of the center of mass of the object is now a non-zero constant in the CMU frame.

 

This final speed vf, will now be denoted by V. Now, let us frame switch; transfer into a frame of reference in which the speed of the center of inertia is zero, instead of V.

 

If it is the case that this frame is also an inertial frame of reference, then we can expect Newton's laws to be true in this frame too.

 

Let us call this new frame, frame S`.

 

So the origin of frame S` is moving through the CMU frame with speed V, but the center of inertial mass of the sphere is permanently at rest in S`

 

Let the X axis of S` be coincident to the Normal line, but let the point (0,0,0) be the center of mass of the sphere. Thus, the frame is rigidly attached to the sphere.

 

Since the X axis of S` is coincident to the normal line, the tip of R vector is now tracing out a circle in S`. So this is centripetal acceleration of the point of contact (which travels with the sphere) in the reference frame S`. Centripetal acceleration is given by:

 

[math] \vec a_c = \frac{ v^2_t}{|\vec R |} [/math]

 

Where [math] v_t [/math] is the tangential speed or orbital speed of the contact point in S`, and is not to be confused with V, which is the speed of the center of inertia of the sphere in the CMU frame (reference frame in which the center of mass of the universe is permanently at rest, and Newton's first law is true).

 

The details of the tangential speed vt are related to the portion of the applied force which went into causing the object to begin spinning in the CMU frame in the first place.

 

In other words, they are directly related to F2.

 

Now either S` is an inertial reference frame or not. Right now, the point of contact (which is at the tip of R vector) is tracing out a circle in S`. Acceleration is the time rate change of velocity. Velocity is the product of speed times direction of travel. So acceleration is zero if and only if, neither the speed, nor the direction of travel is changing.

 

Suppose there is a particle located at the tip of R vector, it has some mass m1. The tangential speed vt of this particle is constant in time, but the direction of travel of this particle is constantly changing (the particle, located at the point of contact, is orbiting the center of mass of the sphere).

 

 

 

Ok I think I've worked on this enough to get an idea. It seems to me, that by treating "acceleration" in Newton's second law as "acceleration of the center of inertia" is keeping things too mathematically simplistic.

 

Certainly, the concept of "acceleration of a point" is easier to understand than "acceleration of a rigid straight line," but I seem to be interested in the motion of [math] \vec R [/math] more than just the motion of the center of inertia.

 

So, if I had mathematics to explain the entire motion of R vector, that mathematics would include the motion of the center of inertia as well, but additionally that mathematics would also be telling me how the whole vector is moving through space, not just the tail of R vector (tail of R vector = center of mass of sphere). In other words, I want to know how the "whole vector R" moves through space, after the contact force is applied to the solid sphere.

 

What I was originally hoping, was that somewhere in the solution of this problem, would be a motivation for a way to define vector division.

 

So I am questioning the interpretation of acceleration in Newton's second law. Normally, when we speak of acceleration, we mean acceleration of a point mass, OR acceleration of the center of inertia.

 

In other words, we have Newton's second law, which is this:

 

[math] \mathbf{F} = m \mathbf{a} [/math]

 

But that law only holds if m is independent of time. But more generally, Newtons second law is:

 

[math] \mathbf{F} = \frac{d\mathbf{P}}{dt}[/math]

 

So that if m is a time dependent quantity we have instead:

 

 

[math] \mathbf{F} = m \mathbf{a} + \mathbf{v} \frac{dm}{dt}[/math]

 

If I now modify the problem slightly, instead of a finger pushing the sphere, there is a tiny man standing on the sphere (and he is standing on the point of contact), his knees are bent, and he is getting ready to jump off the sphere with all his might, then the "mass of the sphere changes" because he succeeds in jumping off.

 

So depending on how the contact force arises, we could need the more general form of Newton's second law.

[Tom Mattson]

No it isn't clear what kind of multiplication we are dealing with.

 

Yes' date=' it is clear. In the quoted section, it's the dot product.

 

For vectors [b']A[/b] and B in R³:

 

[math]

\mathbf {A} = A_x \mathbf {i} +A_y \mathbf {j} + A_z \mathbf {k}

[/math]

 

[math]

\mathbf {B} = B_x \mathbf {i} +B_y \mathbf {j} + B_z \mathbf {k}

[/math]

 

the dot product is:

 

[math]

\mathbf {A} \cdot \mathbf {B} = A_x B_x + A_y B_y + A_z B_z

[/math]

 

Also, why are we talking about moving bodies? We're talking mathematics here.

[Johnny5]

You misunderstood. I was agreeing with you, that we don't start off knowing what kind of vector multiplication we are trying to find the inverse for.

 

As to why I am analyzing bodies that can move, it's because

 

1. I need a problem to help me think.

2. Hopefully a problem in motion can motivate the answer.

3. I think an answer lies in this specific problem, or tiny variations of it.

[Johnny5]

Tom, I have an important question.

 

In trying to answer the question about how to define vector division, in the case of force... have you been using F=ma or F=dP/dt?

 

Also, in the equation F=ma, how is inertial mass measured? What operational definition do you use?

[Johnny5]

You are coming at the problem from a purely abstract direction, that's why I worked on that problem all day yesterday. Intuitively, the answer will emerge in the solution of a physics problem, which is why I did that. Today I will look at it more mathematically. I am trying to understand what you were saying.

 

Now let's try this with vectors' date=' as we attempt to solve for x.

[/quote']

 

Ok you are starting off with a simple vector equation...

 

Ax=B

 

Here is how I am interpreting this, so correct me if I am wrong. In the equation above, x is a scalar quantity, and A is an ordinary vector (it has magnitude and direction, it has three components). (I am not thinking of A as a matrix yet).

 

So the LHS is the product of a scalar and a vector, which is just something that would stretch the vector right? In other words, if x=1 the length of A isn't changed by the mathematical operation of multiplication by x, but if x=2, then A x is a vector (so product of a vector and a scalar is a vector), but the new vector is now twice as long as the original, and B is just the name of the 'new' vector.

 

What real physical process could distort a vector?

 

Setting the above question aside, equations are in general, just tautologies.

 

LHS=RHS doesn't give you a whole lot of information, it boils down to A=A every time.

 

But in the case where we multiply a vector by a scalar, there is a 'stretching' involved. Stretching takes time, and the simpleminded equation above makes absolutely no reference to time. I don't know if that matters or not, but the point is, that A could be some vector at some moment in time, then something happens to it to stretch it, or shrink it, then the final vector is B. My point is you didn't say clearly that x is a scalar, but that's what you meant, because vector A times vector B = vector C is wrong. I think its a tensor. There would be nine compents right?

 

Oh wait, multiplication by x could leave the length unaltered, but reverse the direction of x, in the case where x=-1.

 

See you didn't really give me enough information about the intial equation Ax=B.

 

What field is x to be taken from? I just am assuming that x is an element of the real number system. So in the case where x=-1, we have the direction of the vector being changed, in the case where x=1 there is no effect on the vector, and in the case where x is anything else, we have the magnitude changing, and possibly the direction as well.

 

Let A,B be elements of R³.

Let x be an element of R¹.

 

Now, consider the following equation:

 

[math] \mathbf{A} x = \mathbf{B} [/math]

 

Vector A is an ordinary vector in three dimensional space. The components of A, will depend upon what reference frame the vector is viewed from.

 

Let has have set up a three dimensional rectangular coordinate system somewhere in space, so that we can define the vector A, in terms of its components in our rectangular coordinate system. For this specific reference frame, the following statement is true:

 

[math] \vec A = A_x \hat i + A_y \hat j + A_z \hat k [/math]

 

The length of this vector is found from the Pythagorean theorem, and in general for any vector A, we can write this:

 

[math] \vec A = |\vec A | \hat A [/math]

 

(if the notation above for direction of A isn't standard let me know)

 

Using pythagorean theorem (which is true in all frames, where time is a non-issue so that no one needs to worry about whether or not SR is correct) since Pythagoras was right we have:

 

 

[math] |\vec A | = \sqrt{A_x^2 + A_y^2+A_z^2} [/math]

 

(Note the above equation only makes sense after one has chosen a rectangular coordinate system. In general, we could actually measure the length of a physical object, but it isn't clear how one would actually measure the length of "acceleration vector")

 

In other words, a ruler measures real physical distance in space, but it doesn't measure the length of an acceleration vector. I guess that's what an accelerometer measures.

 

Alright, a vector is a mathematical object right from the start. So perhaps you are right, this is mainly a mathematical problem, but I still need physical examples.

 

Alright, distance between points in space is a purely scalar quantity, but there is no physical 'object' which is a vector.

 

But, 'direction' isn't as abstract as one might think.

 

It makes perfect sense to say that there is a place, in the universe where the center of mass is, and my center of inertia isn't there, its somewhere else. And there is a distance between my center of inertia, and the center of inertia of the universe, and that distance is a purely scalar quantity.

 

But, if I start looking for the center of mass of the universe, I want to look in the right 'direction.' Your example yesterday, showed me that I want a direction vector to have unit length.

 

Now I myself, want to focus on F=m a. So let it be the case that inertial mass times acceleration is commutative. Thus, we have this:

 

[math] \mathbf{A}m = \mathbf{F} [/math]

 

This equation is basically identical to the one you originally asked me to solve, only x has been replaced by m, and B has been replaced by F. In this context, that's like saying... "solve for the inertia."

 

In the case of a specific acceleration vector, the specific acceleration vector will be equivalent to the product of its magnitude, and direction. Now, the magnitude of an acceleration vector isn't going to be a real physical distance in space (you can't measure the magnitude of an acceleration vector using a ruler), but on the other hand, the direction of an acceleration vector is going to have a real physical direction.

 

So you need to first choose a unit of acceleration, then you should be able to write an acceleration vector in terms of its magnitude and direction.

 

Let us presume this has been done, for some acceleration vector A. Thus, we have the following true statement:

 

[math] \mathbf{A} \equiv |\mathbf{A}| \hat A [/math]

 

Now we can rewrite the original equation as:

 

[math] |\mathbf{A}| \hat A m = \mathbf{F} [/math]

 

Inertial mass I presume is a scalar, and magnitude of acceleration is a scalar. And provided that the multiplication here is commutative, we can rewrite the previous equation as:

 

[math] \hat A |\mathbf{A}| m = \mathbf{F} [/math]

 

So, as for commutativity, I have presumed that the magnitude of an acceleration vector commutes with its direction. I see no reason to think otherwise. As for whether or not inertial mass commutes with acceleration i don't know, because as of yet there's no operational definition for it.

 

But you say just to deal with one vector A, times one scalar x. So let it be the case that X=m times (magnitude of acceleration vector) So we now have the following equation:

 

[math] \hat A X = \mathbf{F} [/math]

 

Just to make this look more like what you orginally gave me, let F=B. Thus, we have:

 

[math] \mathbf{\hat A} X = \mathbf{B} [/math]

 

I have a question as to whether or not inertial mass, times magnitude of acceleration really is a scalar quantity (it certainly isn't a scalar if inertial mass is a vector, and I have read that in a few places), but setting this concern aside for now, I do fully understand that "direction of acceleration vector" is a real direction in physical space, and that is A^ in the equation above.

 

Now I would like to point out, that you have made me more concerned about the meaning of "magnitude of acceleration vector." How do you measure the magnitude of an acceleration vector exactly? More questions, all arising from the fact that I want to solve a problem in physics, not a problem in mathematics.

 

Also, I previously spoke of the length of a vector, and used the Pythagorean theorem to discuss it. But, the magnitude of an acceleration vector fundamentally isn't a distance in space, so it doesn't apply here.

 

On the other hand, you could do something clever.

 

Seeings as we are assuming that 'direction' commutes, we could think of Newton's second law as follows:

 

[math] \mathbf{F} = \mathbf{M}a [/math]

 

Where a is what we were calling, "magnitude of acceleration vector," but instead of thinking of inertial mass as a scalar, we view it as the vector quantity.

 

The main problem with doing this though, is that intuitively, inertial mass is a scalar. One could think of it as a measure of how many particles are in a body. The earth has more inertial mass than the moon precisely because there is more matter in the earth than in the moon. Simply push things, to discover which is more massive. So in this sense, inertial mass is a property of objects themselves, and has no direction worth speaking of. Intuitively, it doesn't matter which side you push the moon from, it doesnt matter which direction you try to push the center of inertia of the moon, all directions you push from yield the same value for inertial mass, at least intuitively. This is because I assume the vacuum does not offer any resistance to acceleration.

 

Alright let me set all of the physics aside, and just run through your argument. I will focus on the equation you originally gave, but I see no motivation for it yet.

 

Let A,B denote vectors, let x denote a real number.

 

[math] Ax = B [/math]

 

In the case where A is a direction vector, as long as it's not the zero vector, there is a multiplicative inverse which we can denote by A^-1, the problem you are alluding to, is that we don't know what kind of multiplication we are reversing. Let us presume its "dot product reversal" so we have:

 

[math] A^{-1} \bullet A x = A^{-1} \bullet B [/math]

 

From here we get to:

 

[math] e x = A^{-1} \bullet B [/math]

 

we are looking for a vector C, such that:

 

[math] C \bullet A = |C||A| cos (C,A) = 1 [/math]

 

In the case where A is a direction vector, |A|=1, so that we have:

 

[math] C \bullet A = |C|cos (C,A) = 1 [/math]

 

Cos (0) = 1

Cos (90)=0

Cos (180)=-1

Cos (270)=0

Cos (360)=1

 

Let the tail of C be identical to the tail of A, but let the direction of C be antiparallel to A, and let C have the exact same length as A, namely 1. Thus, |C|=1.

 

In this case we would have:

 

[math] C \bullet A = 1 cos (180) = -1 [/math]

 

The vector C, as just defined, is not the dot multiplicative inverse of A, because we got -1 instead of 1.

 

 

But if the vector C had the same direction as the vector A, and the same length, then things would have worked out. In this case, we would have:

 

[math] C \bullet A = 1 cos (0) = 1 [/math]

 

 

And in the general case, where the length of A isn't 1, then the length of C has to be the inverse of the length of A. This will make sense in spherical coordinates. Let A be given by the following:

 

[math] \vec A = R \hat r [/math]

 

Now, let C be given by the following:

 

[math] \vec C = \frac{1}{R} \hat r [/math]

 

Then if we dot C with A, we get the following:

 

[math] C \bullet A = \frac{1}{R} R cos(0) = 1 [/math]

 

You listed three properties that a multiplicative inverse of A has to have. One of them was that it has to return the identity element, when multiplied by A.

 

Any such vector C above, will do this. All that has to happen for C dot A to yield 1, is for C to have the inverse magnitude of A, and be parallel.

 

At first glance, there seem to be an infinite number of possible candidates for vectors which can be the multiplicative inverse of A. This violates criterion 3 which you gave, which is that the multiplicative inverse of A be unique.

 

We can reduce the number of vectors C which are possible candidates for the multiplicative inverse of A, if we insist that the multiplicative inverse of A must have the same line of action of A. Therefore, vectors which are non-collinear with A, but parallel, no longer are acceptable as possible candidates for being the multiplicative inverse of A, they have been excluded.

 

But the line of action is infinitely long, so there are still an infinite number of possible candidates for the multiplicative inverse of A.

 

If finally, I stipulate that the dot multiplicative inverse of A must have the same tail of A, this forces uniqueness upon C i do believe.

 

So not only would the dot multiplicative inverse of A have the same direction as A, it would also have the same line of action as A, and it would also have the same tail as A.

 

So in the case where A was four units long, the length of C would be 1/4 a unit long, and vice versa. And only in the case where |A|=1, would the magnitude of the dot multiplicative inverse of A have the same length, direction, and line of action. In this case the dot multiplicative inverse of A would be A, and could be no other vector in real physical space.

 

You also add a third criterion, which is that

 

2. It has to be in the same set as a and the identity.

 

Now, in what I have just done, the identity element is a scalar quantity, it is an element of the real number system. But C is a vector. So the problem now is that the identity element wasn't in the same set as A.

 

If you remove constraint 2, I showed you how to define a unique dot multiplicative inverse, but if you insist on property 2, then the multiplication we are trying to reverse, cannot be the dot product (as you said).

 

So my question now is, why must we have criterion 2 above?

 

I will ask this in a new post.

 

 

 

 

 

 

 

 

 

 

 

 

We'll have to assume that a "multiplicative inverse" a^-1 exists for a. I put the word multiplicative in quotes' date=' because it is not clear what kind of multiplication we are dealing with. For now, I'll denote it by the pound sign (#).

 

[b']a[/b]x=b

a^-1#ax=a^-1#b

 

Now we have 2 ways to multiply vectors: The dot product and the cross product. Let's see if either one could possibly stand for #.

 

Dot product:

a^-1.ax=a^-1.b

 

Now we hit our first roadblock. What exactly is a^-1 with respect to the dot product? A multiplicative inverse is supposed to have 3 properties:

 

1. It has to return an identity element when multiplied by a.

2. It has to be in the same set as a and the identity.

3. It has to be unique.

 

There is no vector a^-1 that satisfies all these properties.

 

To prove this, assume that an identity element e exists such that a^-1a=e. You could try to define a^-1 as a vector which when put in a dot product with a, yields the scalar "1". But there are two problems with this. First, there are an infinite number of such vectors, which violates #3. And second, vectors aren't closed under the dot product, which violates #2.

 

And of course, if we look at the other case, namely that in which an identity element e satisfying a^-1a=e does not exist, then we have contradicted condition 1 above.

 

Thus, we cannot define vector division from the dot product.

 

[Tom Mattson]

In trying to answer the question about how to define vector division' date=' in the case of force... have you been using F=ma or F=dP/dt?

[/quote']

 

Neither. I've been using ax=b. Why complicate matters by throwing in a derivative?

 

Also, physical interpretations are completely tangent to the issue. The issue on the table is how to define a mathematical operation. Only after it is well-defined can it be used in a physical theory.

 

Also, in the equation F=ma, how is inertial mass measured? What operational definition do you use?

 

I'm not bothering with it for the purposes of this thread. It's a real number, which is all that matters.

[Tom Mattson]

Here is how I am interpreting this' date=' so correct me if I am wrong. In the equation above, x is a scalar quantity, and A is an ordinary vector (it has magnitude and direction, it has three components). (I am not thinking of A as a matrix yet).

[/quote']

 

Yes, a and b are members of R³, and x is a member of R. If it wasn't obvious when I first posted on it, it should be by now.

 

What real physical process could distort a vector?

 

It's not as though the vector is being physically distorted. All we're doing here is multiplying a vector by a scalar, and obtaining a new vector. It's not that complicated.

 

Setting the above question aside, equations are in general, just tautologies.

 

LHS=RHS doesn't give you a whole lot of information, it boils down to A=A every time.

 

If we're talking about vector division, an equation of the type I posted is the only way to do it. Division is the inverse operation of some multiplicative operation. One is defined in terms of the other, so that's the place to start looking.

 

But in the case where we multiply a vector by a scalar, there is a 'stretching' involved. Stretching takes time, and the simpleminded equation above makes absolutely no reference to time.

 

There is no need to mention time. We're doing a math problem here. Also, as I said, we aren't taking some physical object and stretching it at all. If you really insist on physical examples, look at F=ma again. The factor m doesn't "stretch" the vector a to produce the vector F. What does "stretching acceleration" even mean anyway?

It's not like "acceleration" is a rubber band whose ends we can pull apart. No, we just take the quantities and multiply them together to get F.

 

Oh wait, multiplication by x could leave the length unaltered, but reverse the direction of x, in the case where x=-1.

 

See you didn't really give me enough information about the intial equation Ax=B.

 

As I said, it should have been obvious that x is a real variable. If not, then why didn't you ask?

 

Also, the fact that it could be positive, negative, or zero is immaterial. Your distinction between stretching, contracting, and reversing direction has no bearing whatsoever on this problem. We're trying to find the analog of division on the reals, with vectors.

 

Now, consider the following equation:

 

[math] \mathbf{A} x = \mathbf{B} [/math]

 

Vector A is an ordinary vector in three dimensional space. The components of A, will depend upon what reference frame the vector is viewed from.

 

Yes, vector components are always origin-dependent.

[Johnny5]

I have a question. You gave three properties that a multiplicative inverse of a vector A has to have...

 

 

Now we hit our first roadblock. What exactly is a^-1 with respect to the dot product? A multiplicative inverse is supposed to have 3 properties:

 

1. It has to return an identity element when multiplied by a.

2. It has to be in the same set as a and the identity.

3. It has to be unique.

 

There is no vector a^-1 that satisfies all these properties.

 

 

Why is property 2 there?

 

Suppose that you remove that constraint for "multiplicative inverse"

 

Let vector A have its tail at (a' date='b,c) and its head at (x,y,z)

 

Now, define the dot multiplicative inverse of vector A, to be a vector with its tail at (a,b,c), and its head (x`,y`,z`) to be such that:

 

[math'] \mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \hat A [/math]

 

For any vector A (except the zero vector), there is one and only one vector A^-1 so that your property three is satisfied, and property 1 is also satisfied.

 

What do you say to this?

 

 

PS: To be honest with you, I am finding it a bit hard to stop thinking of the magnitude of a vector as having a length in real space.

[Johnny5]

There is no vector a^-1 that satisfies all these properties.

 

To prove this' date=' assume that an identity element e exists such that [b']a[/b]^-1a=e. You could try to define a^-1 as a vector which when put in a dot product with a, yields the scalar "1". But there are two problems with this. First, there are an infinite number of such vectors, which violates #3. And second, vectors aren't closed under the dot product, which violates #2.

 

And of course, if we look at the other case, namely that in which an identity element e satisfying a^-1a=e does not exist, then we have contradicted condition 1 above.

 

Ok I read a little further and I see very clearly where you wrote this

You could try to define a^-1 as a vector which when put in a dot product with a' date=' yields the scalar "1". But there are two problems with this. [/quote']

 

Let me think about this now.

 

Ok you say

First' date=' there are an infinite number of such vectors, which violates #3. [/quote']

 

I showed you a way to crash the number of possible candidates for a dot multiplicative inverse to one, so there is a way to satisfy criterion 1, and 3, but only if you sacrifice part of criterion 2. Namely, that the identity element must be a vector.

 

 

 

...And second' date=' vectors aren't closed under the dot product, which violates #2.

[/quote']

 

 

You know something... vectors aren't supposed to be closed under the dot product, the dot product of two vectors is necessarily a scalar quantity.

 

So I am not sure what you meant with that part.

[Johnny5]

I think it's worth talking about direction a bit.

 

Suppose I am somewhere in space. I can rotate my head, and look in a multiplicity of directions.

 

So given any location in space, we can define a theoretically infinite number of directions, simply by picking another location in space. So, correct me if I am wrong, but to define a direction in space, requires that two points be given to us.

 

Now, it seems to me that the distance between the two points is immaterial.

 

Once we have those two points, they uniquely define TWO different directions. Call one point X, and the other point Y.

 

There is the direction pointing from X to Y, and there is the direction pointing from Y to X.

 

Direction doesn't have length. It has nothing to do with length whatsoever.

 

Now, since two arbitrary points don't uniquely define one direction, we need more information. We need to specify which point is the tail, and which point is the head of a "unit vector."

 

But really, as I said, a direction doesn't have a length. And direction vectors are stipulated to have a length of 1 unit. I'm not sure about what's going on right now, I've never questioned this type of thing before.

 

But my point is, that we need to be given two points in space, one of which is tail, the other head, in order to define direction.

 

My point is only that since we are given two points, I don't see why we cannot use one of them, in order to define the inverse of the multiplication we are talking about. We don't know what that multiplication is in the first place, as you said. But, if we allow it to resemble the dot product, and additionally stipulate that the multiplicative inverse must have the same tail as A, then like I previously explained, given any single vector A (other than the zero vector), there is one and only dot product multiplicative inverse.

 

I shouldn't call it dot product multiplicative inverse at all, since we can dot vectors which don't have the same tail, but I don't know what else to call it.

[Tom Mattson]

I have a question. You gave three properties that a multiplicative inverse of a vector A has to have...

 

Why is property 2 there?

 

It has to be there in order for the set of vectors to be considered a group under multiplication. This is necessary if you want to carry all the properties of multiplication/division on the reals over to vectors' date=' as you seem determined to do.

 

Suppose that you remove that constraint for "multiplicative inverse"

 

In that case, you c

Let vector A have its tail at (a,b,c) and its head at (x,y,z)

 

Now, define the dot multiplicative inverse of vector A, to be a vector with its tail at (a,b,c), and its head (x`,y`,z`) to be such that:

 

[math] \mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \hat A [/math]

 

For any vector A (except the zero vector), there is one and only one vector A^-1 so that your property three is satisfied, and property 1 is also satisfied.

 

What do you say to this?

 

I say that it is nothing more than division on the reals. Going back to Newton's second law, you have:

 

[math]

\mathbf {F} = m \mathbf {a}

[/math]

[math]

m=\frac {\mathbf {F}}{\mathbf {a}}

[/math]

 

which you define to be:

 

[math]

m= \mathbf {F} \cdot \mathbf {a} ^{-1}

[/math]

[math]

m= \mathbf {F} \cdot \frac { \hat { \mathbf {a} } } {| \mathbf {a} |}

[/math]

[math]

m= \frac { \mathbf {F} \cdot \hat {\mathbf {a}}} {| \mathbf {a} |}

[/math]

 

which is just the division of one real number by another.

[Tom Mattson]

You know something... vectors aren't supposed to be closed under the dot product' date=' the dot product of two vectors is necessarily a scalar quantity.

 

So I am not sure what you meant with that part.[/quote']

 

I know that vectors aren't closed under the dot product, and indeed that was the point. In order to have group properties of vectors under multiplication, we have to have closure. That is one reason you can't use the dot product to define vector division.

[Tom Mattson]

Direction doesn't have length. It has nothing to do with length whatsoever.

 

Then you have to change your definition of a vector from:

 

[math]

\mathbf {V} = (magnitude) \cdot (direction)

[/math]

 

to something else.

 

Honestly Johnny5, I don't see the point in carrying this exercise any further, so I'm done with it.

 

Best of luck with it,

[Johnny5]

Honestly Johnny5' date=' I don't see the point in carrying this exercise any further, so I'm done with it.

 

Best of luck with it,[/quote']

 

Ok, but something tells me that it's important. I would have liked to see some of your own ideas, which you discarded for some reason or other. But thank you.

[pljames]

Tom , Thanks you have made my day. pljames