I simply asked a question and AN comes pounding around.
I would assume if he had the answer, he would just give it.
You did not ask a
simple question but rather a
leading question that already assumed physics contrary to special relativity therefore you were trolling pseudo-physics instead of precise and communicable predictive models of observable phenomena. You also misstate the burden of proof. It is your burden to describe and argue for your thesis, while AlphaNumeric has no burden to respond to you no matter how wrong you are. You simply don't matter enough for your wrong ideas to demand a reply. Likewise you don't matter enough to get an education in logical reasoning, math and physics when by rights you should pay for such an education so you will properly value the time taken to explain topics and assign homework to you.
And, you have no idea what I am talking about as usual.
Actually, I know exactly what you are saying. I also know it's nonsense.
Seriously? Special relativity has the postulate that the speed of light is fixed at c and you are complaining that if you fix $$t_0, x_0, y_0, z_0, x, y, \quad \textrm{and} \quad z$$ that there is only one solution for the light moving from $$(x_0, y_0, z_0)$$ to $$(x,y,z)$$? If there was more than one solution that would violate the postulate that the speed of light is fixed at c and would prove Special Relativity inconsistent.
So now you are reduced to complaining that Special Relativity is mathematically self-consistent. You might as well complain that 2 + 3 = 3 + 2. The alternative is nonsense and therefore you advocate nonsense.
If we consider pure Euclidean space, then given any x or y then z can be any real number. It is absolutely independent. Any child knows this.
That's completely the wrong way to think about it. If you consider a "pure" 3-dimensional Euclidean space then you have distinct points. If you have any two distinct points, you can classify all the points in the space as either co-linear with the two points or not-co-linear with the two points. The set of all points which are colinear with the two points is a 1-dimensional sub-space of the full 3-dimensional space. In this 1-dimensional sub-space you can describe all the points in relation how far along the line they are in terms of the first two points. $$(\xi -O ) = k_{\xi}(A -O)$$ describes the point $$\xi$$, which must be on the line, in terms of the orgin O and the directed distance to the first point A and a scalar multiplier, k. This can also be written as $$\xi = O + k_{\xi}(A-O)$$.
If you have two points and a third point which is not co-linear with the first two, you can similarly form a 2-dimension sub-space of all points co-planar with the three points. If you have four points, not all mutually co-planar, you can form a 3-dimensional sub-space which is identical to the original space. Now if you have 4 points, not all mutually co-planar, you can assign one as an origin and the others as foundation for basis vectors. You learn that any point in the whole 3-dimensional space can be written in terms of the points O, A, B and C as: $$\xi = O + k_{\xi}(A-O)+ \ell_{\xi} (B-O)+ m_{\xi}(C-O)$$. This description of the location of $$\xi$$ in terms of three numbers (k,l,m) is the Euclidean base concept behind coordinate systems.
The Cartesian coordinate system has an origin and three basis vectors, $$\hat{x}, \hat{y}, \hat{z}$$ so that any point in the 3-dimensional space can be written in terms of the orgin O as $$\xi = O + x_{\xi} \hat{x} + y_{\xi} \hat{y} + z_{\xi} \hat{z}$$ where the basis vectors are all of the same unit length: $$\hat{x}^2 = \hat{y}^2 = \hat{z}^2 = 1$$ and are all mutually orthogonal so that $$\hat{x} \cdot \hat{y} = \hat{x} \cdot \hat{z} = \hat{y} \cdot \hat{z} = 0$$. This in turn gives the Cartesian coordinate system its greatest strength over the klm arbitrary system -- you can easily figure the distance between points based on just the differences of coordinates.
$$ \left| \xi - \eta \right| = \left| \left( O + x_{\xi} \hat{x} + y_{\xi} \hat{y} + z_{\xi} \hat{z} \right) - \left( O + x_{\eta} \hat{x} + y_{\eta} \hat{y} + z_{\eta} \hat{z} \right) \right| = \left| \left( x_{\xi} - x_{\eta} \right) \hat{x} + \left( y_{\xi} - y_{\eta} \right) \hat{y} + \left( z_{\xi} - z_{\eta} \right) \hat{z} \right|
\quad = \sqrt{ \left| \left( x_{\xi} - x_{\eta} \right) \hat{x} + \left( y_{\xi} - y_{\eta} \right) \hat{y} + \left( z_{\xi} - z_{\eta} \right) \hat{z} \right|^2 } = \sqrt{\left( x_{\xi} - x_{\eta} \right)^2 + \left( y_{\xi} - y_{\eta} \right)^2 + \left( z_{\xi} - z_{\eta} \right)^2 } $$
So in summary, a N-dimensional space (or sub-space) requires N (possibly Cartesian) coordinates to describe each point. And an unbounded N-dimensional space (or sub-space) places no limits on what those coordinates might be. But it is incorrect to claim coordinates are "absolutely independent" if we put constrains on them. If we consider the 1-dimensional subspace which is colinear with points O and A, then in klm coordinates, the l and m coordinates are always zero. If we consider the 1-dimensional subspace which is colinear with points (1,2,3) and (2,4,8) in a Cartesian coordinate system, if we know any one coordinate we know the other two. Any two of the follow equations completely constrain two of the coordinates in terms of the first known.
$$10x - 5y = 0 \\ 10x - 2z = 4 \\ 5y - 2z = 4$$
Alternately one may describe the constraint is that every point on the 1-dimensional subspace can be written in terms of a single invented coordinate k.
$$\xi = O + ( k (2-1) + 1 ) \hat{x} + ( k (4-2) + 2 ) \hat{y} + ( k (8-3) + 3 ) \hat{z} = ( k + 1 ) \hat{x} + ( 2 k + 2 ) \hat{y} + ( 5 k + 3 ) \hat{z}$$.
This is basic geometry that has been part of physics since Newton wrote the first book on the topic.
Now, let us consider space-time where only light flashes and light spheres are present.
You mean light cones. A sphere is defined by its fixed radius, while a cone has a radius which is not fixed.
To be a real geometry, all dimensions must be independent of one another.
Here you confuse "dimension" which is a geometric description of a space with "coordinates of a point" which are just convenient labels determined in relationships to other points.
However, the equation below under SR must hold true in the world of only light flashes.
$$c^2 (t - t_0)^2 - (x - x_0)^2 - (y - y_0)^2 -(z - z_0)^2 = 0$$
So, if I pick x, y, z, then t is determined and is therefore not independent
Correct. The light cone in 4-dimensional space-time is a 3-dimensional subspace. By picking 3 coordinates you have eliminated all the degrees of freedom.
We can write the Cartesian coordinates of every point in the 3-dimensional subspace by using 3 spherical coordinates θΦr thus proving it has 3 degrees of freedom.
$$t = t_0 + \frac{r}{c} \\ x = x_0 + r \cos \phi \cos \theta \\ y = y_0 + r \cos \phi \sin \theta \\ z = z_0 + r \sin \phi$$
Or alternately in xyz:
$$t = t_0 + \frac{1}{c}\sqrt{(x - x_0)^2 + (y - y_0)^2 +(z - z_0)^2} \\ x = x\\ y = y \\ z = z $$
as required by a true geometry.
You don't know geometry or how to support your claims in a discussion.
Now is this true of false?
Part true, part false.
Remember, in a pure Euclidean space, if I pick x, y, then z can be anything.
In an unbounded 3-dimensional space or subspace, you can pick 3 coordinates from their full range which may or may not be the whole set of real numbers. Typically, adding an additional equality as constraint reduces the dimensionally of the allowable points by 1. So a 3 dimensional space with 2 equations constraining coordinates results in a 1 dimensional space (but for some exceptional circumstances). The constraint on the propagation speed results in creating a 3-dimensional subspace of a 4-dimensional space-time.
Anyway, can you and AN answer this with all of your "deep" understanding? I get confused by all this confusion.
Thanks in advance.
I think this understanding of algebraic geometric is a prerequisite for taking calculus instruction in high school or first year university and that calculus is a prerequisite for even Newtonian physics. So it's not so very "deep".
