# Apparant 2 directions

Discussion in 'Pseudoscience' started by chinglu, Aug 13, 2013.

1. ### rpennerFully WiredValued Senior Member

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Hard details part 1
For any four events, A, B, C, D prove that $\left< B-A , D -C \right>$ is constant under the Lorentz transform. We do this by recognizing the output of the inner product is a number so we can demonstrate that this number is the same after transformation.
That is we seek to prove $\left< \Lambda B- \Lambda A , \Lambda D - \Lambda C \right> - \left< B-A , D -C \right> = 0$. Or in matrix form,remembering that $\eta = { \small \begin{pmatrix} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} }$:
$\left( \Lambda B - \Lambda A \right) ^{\textrm{T}} \eta \left( \Lambda D - \Lambda C \right) - \left( B - A \right) ^{\textrm{T}} \eta \left( D - C \right) = \left( B - A \right) ^{\textrm{T}} \Lambda^{\textrm{T}} \eta \Lambda \left( D - C \right) - \left( B - A \right) ^{\textrm{T}} \eta \left( D - C \right) = \left( B - A \right) ^{\textrm{T}} \left( \Lambda^{\textrm{T}} \eta \Lambda - \eta \right) \left( D - C \right) =^? 0$
Which is true for all A,B,C and D only if $\Lambda^{\textrm{T}} \eta \Lambda = \eta$.

Let $K = \begin{pmatrix} c & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ and let $D = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ then $K^{-1} = \begin{pmatrix} c^{-1} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ and $K^{\textrm{T}} = K$. Then we see $D^2 = 1$ and that $\eta = KDK$.

Let $Q = \begin{pmatrix} 0 & b_1 & b_2 & b_3 \\ -b_1 & 0 & r_{12} & r_{13} \\ -b_2 & -r_{12} & 0 & r_{23} \\ -b_3 & -r_{13} & -r_{23} & 0 \end{pmatrix}$ with all parameters real numbers. Then it follows that $Q^{\textrm{T}} = - Q$.

Let $P = \begin{pmatrix} 0 & b_1 & b_2 & b_3 \\ b_1 & 0 & r_{12} & r_{13} \\ b_2 & -r_{12} & 0 & r_{23} \\ b_3 & -r_{13} & -r_{23} & 0 \end{pmatrix} = Q D$ with all parameters real numbers.
Then it follows that $\left( P^{\textrm{T}} \right)^n = \left( P^n \right)^{\textrm{T}}$.
Then it follows that $\exp P = \sum_{k=0}^\infty \frac{1}{k!} P^k$ is finite and $\left( \exp P \right) ^{\textrm{T}} = \exp \left( P^{\textrm{T}} \right)$.
Then it follows that $\left( \exp - P \right) \left( \exp P \right) = \left( \exp P \right)\left( \exp - P \right) = I$.

If $\Lambda = K^{-1} \left( \exp P \right) K$ then $\Lambda^{\textrm{T}} = K \left( \exp \left( P^{\textrm{T}} \right)\right) K^{-1}$ and thus:
$\Lambda^{\textrm{T}} \eta \Lambda = K \left( \exp \left( P^{\textrm{T}} \right) \right) D \left( \exp P \right) K = K \left( \sum_{k,\ell = 0}^{\infty,\infty} \frac{1}{k! \ell !} \left( P^{\textrm{T}} \right)^k D P^{\ell} \right) K = K \left( \sum_{k,\ell = 0}^{\infty,\infty} \frac{1}{k! \ell !} \left( - D Q \right)^k D \left( Q D \right)^{\ell} \right) K = K D \left( \sum_{k,\ell = 0}^{\infty,\infty} \frac{-1^k}{k! \ell !} \left( Q D \right)^{k + \ell} \right) K = K D \left( \sum_{k,\ell = 0}^{\infty,\infty} \frac{-1^k}{k! \ell !} P^{k + \ell} \right) K = K D K = \eta$
Because for fixed $m = k + \ell > 0$ We have $\sum_{k = 0}^{m} \frac{-1^k}{k! (m - k) !} = 0$

Setting $b_1 = \tanh^{-1} \frac{v}{c}$ and all other components to zero, we recover:
$\Lambda = \begin{pmatrix} c^{-1} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cosh \tanh^{-1} \frac{v}{c} & \sinh \tanh^{-1} \frac{v}{c} & 0 & 0 \\ \sinh \tanh^{-1} \frac{v}{c} & \cosh \tanh^{-1} \frac{v}{c} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
which is our pedestrian Lorentz boost in the X direction.

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3. ### chingluValued Senior Member

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Fair nuff.

Assume only c units for distances and velocity. Assume v=3/5c and y=10, z=0. Also, that means \gamma=5/4.

I will not respond to anyone else in this thread because they do not explain first that the one slw moves 2 different directions from the unprimed point (0,10,0).

First take dx/dt using x=+-sqrt(t^2-100). So, the SLW moves in a positive and negative direction from (0,10,0) in the unprimed frame.

Next, use LT to predict the direction the SLW moves from the view of the primed frame.

First, understand if the SLW expands in the unprimed frame it expands primed because there is only one SLW. Hence, if t increases, t' increases for all clocks in the respective frames.

So, if we say the SLW is at unprimed (0,10,0) and it expands, then we can conclude t' increases.

Now, consider LT for x.

x = (x'+vt')\gamma.

Now, since we are timing light, x'= +-sqrt(t'^2-100).

At unprimed (0,10,0), x'<0, hence use x'= -sqrt(t'^2-100).

Then, using LT t = (vt'-sqrt(t'^2-100))\gamma.

dx/dt' = ( v - t'/sqrt(t'^2-100)).

It is a trivial task to prove if t'>10, then dx/dt'<0. Since the SLW is at (0,10,0), then it is at primed (0,12.5,0), Hence, t'>10.

So, if we are close to unprimed (0,10,0) then LT claims dx/dt'<0.

Hence, LT claims the one and only one SLW moves in a negative x direction for all t'>10.

So, near unprimed (0,10,0), as the SLW expands, the light postulate in the unprimed frame claims the SLW moves in a positive x direction along y=10.

LT on the other hand, claims that same SLW on any expansion only moves in a negative x direction if t'>10. Since t'>12.5 is near unprimed (0,10,0) , then LT claims the SLW only moves in a negative x direction with further expansion, which is a contradiction.

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5. ### chingluValued Senior Member

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Explain specifically how this proves the unprimed frame and primed frame agree on the direction the SLW travels from unprimed (0,y,0) in the positive x direction.

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7. ### rpennerFully WiredValued Senior Member

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The correct way to say this is:
Assume arbitrary units for time and distance.​
The correct way to say this is:
Assume the Lorentz transform relating coordinates for the same space-time events in two different inertial coordinate systems can be written in matrix form as:
${ \Huge \Lambda }= \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} c^{-1} & 0 & 0 \\ -\frac{3}{4} c & \frac{5}{4} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \left(\frac{3}{5}\right)^2}} & -\frac{\frac{3}{5} c}{c^2 \sqrt{1 - \left(\frac{3}{5}\right)^2}} & 0 & 0 \\ -\frac{\frac{3}{5}c}{\sqrt{1 - \left(\frac{3}{5}\right)^2}} & \frac{1}{\sqrt{1 - \left(\frac{3}{5}\right)^2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \cosh \, \tanh^{\tiny -1} \, \frac{3}{5} \quad & -c^{-1} \, \sinh \, \tanh^{\tiny -1} \, \frac{3}{5} \quad & 0 & 0 \\ -c \, \sinh \, \tanh^{\tiny -1} \, \frac{3}{5} \quad & \cosh \, \tanh^{\tiny -1} \, \frac{3}{5} \quad & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t' \\ x' \\ y' \\ z' \end{pmatrix} = { \Huge \Lambda } \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix}$
And let there be a wire parallel to the X axis with fixed coordinates y = 10 and z = 0.
Further, let there be a spherical wave front moving outward from the point (x=0,y=0,z=0) at uniform speed c.​
So we have most of what we need for my geometric demonstration in my post [POST=3100132]A space-time hyperbola, hurrah![/POST].
We have a spherical light pulse, $\mathcal{V}$ originating at O and it encounters the world-sheet of the wire's past present and future $\mathcal{U}$ forming a 1-dimensional space-time hyperbola $\mathcal{H}$ that we can calculated explicitly.
$O = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} , \quad A = \begin{pmatrix} 0 \\ 0 \\ 10 \\ 0 \end{pmatrix} , \quad B = \begin{pmatrix} 0 \\ 1 \\ 10 \\ 0 \end{pmatrix} , \quad C = \begin{pmatrix} 1 \\ 0 \\ 10 \\ 0 \end{pmatrix} \\ \mathcal{U} = \left{ (t,x,y,z) \quad | \quad \exists r, s \in \mathbb{R} \quad (t,x,y,z) = (s,r,10,0) \right} \\ \mathcal{V} = \left{ (t,x,y,z) \quad | \quad t = \frac{1}{c} \sqrt{x^2 + y^2 + z^2} \right} \\ \left< C -A , C -A \right> = \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \eta \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = -c^2 \\ \left< (1-r)A + rB - O, C -A \right> = \begin{pmatrix} 0 & r & 10 & 0 \end{pmatrix} \eta \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = 0 \\ \left< (1-r)A + rB - O, (1-r)A + rB - O \right> = \begin{pmatrix} 0 & r & 10 & 0 \end{pmatrix} \eta \begin{pmatrix} 0 \\ r \\ 10 \\ 0 \end{pmatrix} = 100 + r^2 \\ s = f(r) = \frac{1}{c} \sqrt{100 + r^2} \\ \mathcal{H} = \mathcal{U} \cap \mathcal{V} = \left{ (t,x,y,z) \quad | \quad \exists r \in \mathbb{R} \quad (t,x,y,z) = (\frac{1}{c} \sqrt{100 + r^2},r,10,0) \right} = \left{ (t,x,y,z) \quad | \quad \exists \theta \in \mathbb{R} \quad (t,x,y,z) = (\frac{10}{c} \, \cosh \theta , 10 \, \sinh \theta ,10,0) \right}$
Using $\theta = \sinh^{\tiny -1} \frac{r}{10}$ shows the two parameterizations of the hyperbola $\mathcal{H}$ are equivalent.

You are badly confused about the physics. $\mathcal{U}$ is the non-moving wire, $\mathcal{V}$ is the moving wave front, the spherical light wave, the forward light cone. $\mathcal{H}$ is the intersection of $\mathcal{U}$ and $\mathcal{V}$ but is not a thing and is definitely not a light wave. It is a succession of different parts of the spherical light wave $\mathcal{V}$ hitting different parts of the wire $\mathcal{U}$.
If you want to talk just about the part of spherical light wave $\mathcal{U}$ that runs through one part of the wire, you have to talk about a radial ray. $\mathcal{R} = \left{ (t,x,y,z) \quad | \quad \exists d \in \mathbb{R} \quad (t,x,y,z) = (\frac{d^2}{c} , 0 , d^2, 0) \right}$.
$\mathcal{R}$ describes one-dimensional motion with velocity:
$\vec{u}_{\mathcal{R}} = \begin{pmatrix} \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial t} \\ \frac{\partial z}{\partial t} \end{pmatrix} = \frac{\partial d}{\partial t} \begin{pmatrix} \frac{\partial x}{\partial d} \\ \frac{\partial y}{\partial d} \\ \frac{\partial z}{\partial d} \end{pmatrix} = \frac{c}{2d} \begin{pmatrix} 0 \\ 2d \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ c \\ 0 \end{pmatrix}$
By contrast, $\mathcal{H}$ describes the motion of the intersection of an expanding sphere intersecting a wire. Thus other than the initial time of tangency, it has TWO spots, both moving faster than the speed of light. This is entirely consistent with $\mathcal{H}$ describing an intersection of worldsheets, not a thing.
$\vec{u}_{\mathcal{H}} = \begin{pmatrix} \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial t} \\ \frac{\partial z}{\partial t} \end{pmatrix} = \frac{\partial \theta}{\partial t} \begin{pmatrix} \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial \theta} \end{pmatrix} = \frac{c}{10 \sinh \theta} \begin{pmatrix} 10 \cosh \theta \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} c \coth \theta \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{c^2 t}{x} \\ 0 \\ 0 \end{pmatrix}$
$\vec{u}_{\mathcal{H}} = \begin{pmatrix} \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial t} \\ \frac{\partial z}{\partial t} \end{pmatrix} = \frac{\partial r}{\partial t} \begin{pmatrix} \frac{\partial x}{\partial r} \\ \frac{\partial y}{\partial r} \\ \frac{\partial z}{\partial r} \end{pmatrix} = \frac{c \sqrt{100 + r^2}}{r} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} c \frac{\sqrt{100 + r^2}}{r} \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{c^2 t}{x} \\ 0 \\ 0 \end{pmatrix}$
As you see, $\vec{u}_{\mathcal{H}}$ is along the wire which extends in the X-direction and the sign of $\vec{u}_{\mathcal{H}}$ is equal to the sign of x and is away from the origin, and the magnitude of $\vec{u}_{\mathcal{H}}$ is always greater than c.

You are describing $\mathcal{H}$ not a spherical light wave. Also you should have written $c^2 t^2 = x^2 + 100$ which correctly gives $x = \pm \sqrt{c^2 t^2 - 100}$ such that $\frac{dx}{dt} = \frac{c^2 t}{x} = \pm \frac{c^2 t}{\sqrt{c^2 t^2 - 100}}$ which is everywhere greater in magnitude than c, which is further demonstration that at no point does this equation describe a spherical light pulse.

You can do this in a number of ways, all equivalent. You can transform the four events O, A, B and C and then compute the inner products and the equation of the hyperbola. You can use the result from my post [POST=3101532]Hard details part 1[/POST] to shortcut the calculation of the inner products in the primed frame. Or you can Lorentz transform the events on the hyperbola itself.
Starting with:
$\begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{10}{c} \, \cosh \theta \\ 10 \, \sinh \theta \\ 10 \\ 0 \end{pmatrix}$
we get:
$\begin{pmatrix} t' \\ x' \\ y' \\ z' \end{pmatrix} = { \Huge \Lambda } \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \cosh \, \rho \quad & c^{-1} \, \sinh \, \rho \quad & 0 & 0 \\ c \, \sinh \, \rho \quad & \cosh \, \rho \quad & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{10}{c} \, \cosh \theta \\ 10 \, \sinh \theta \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{10}{c} \, \left( \cosh \theta \cosh \rho + \sinh \theta \sinh \rho \right) \\ 10 \, \left( \sinh \theta \cosh \rho + \cosh \theta \sinh \rho \right) \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{10}{c} \, \cosh \left( \theta + \rho \right) \\ 10 \, \sinh \left( \theta + \rho \right) \\ 10 \\ 0 \end{pmatrix}$
which is a pure consequence of the Lorentz transform being a precise hyperbolic analogue of rotation.

Thus the description of $\mathcal{H}$ is identical in the primed coordinates and the unprimed coordinates. If there is an event in $\mathcal{H}$ with a particular description in primed coordinates, there is a different event with that identical description in unprimed coordinates. This property of identical coordinate descriptions of locii is shared with $\mathcal{U}$ (since the wire is parallel to the motion between the frames) and $\mathcal{V}$ (since the Lorentz transforms maps light-cones to light-cones and leaves event O untouched), but not $\mathcal{R}$ because the "place" $(x=0, y=10, z=0)$ is geometrically unlike the description of the "place" $(x'=0, y'=10, z'=0)$. These "places" are in relative motion.

That's not true because you are describing (as always) $\mathcal{H}$ not $\mathcal{R}$ or $\mathcal{V}$ and when interpreted as motion $\mathcal{H}$ is a superluminal not-a-thing and is not prevented from going backwards in time.

That's only part of the Lorentz transform which is a space-time transformation. If you neglect one you are breaking the rules of rational discourse. Breaking the rules of rational discourse invites banning even on this sub-forum.

That is an atrocious presentation. Only because you parrot the same bad math over and over do I have a chance of parsing your argument with broken assumptions and incorrect expressions that aren't even dimensional consistent.
First of all, you fix $\theta = 0, c \tanh \rho = v = -\frac{3}{5} c$ so the concrete event you are talking about has the following descriptions in the two coordinate systems:
$\begin{pmatrix} t_E \\ x_E \\ y_E \\ z_E \end{pmatrix} = \begin{pmatrix} \frac{10}{c} \\ 0 \\ 10 \\ 0 \end{pmatrix} \begin{pmatrix} t'_E \\ x'_E \\ y'_E \\ z'_E \end{pmatrix} = \begin{pmatrix} \frac{25}{2 c} \\ -\frac{15}{2} \\ 10 \\ 0 \end{pmatrix} \begin{pmatrix} \frac{25}{2 c} \\ -\frac{15}{2} \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{5}{4} & \frac{-3}{4c} & 0 & 0 \\ \frac{-3c}{4} & \frac{5}{4} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{10}{c} \\ 0 \\ 10 \\ 0 \end{pmatrix} \begin{pmatrix} \frac{10}{c} \\ 0 \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{5}{4} & \frac{3}{4c} & 0 & 0 \\ \frac{3c}{4} & \frac{5}{4} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{25}{2 c} \\ -\frac{15}{2} \\ 10 \\ 0 \end{pmatrix}$
These coordinates satisfy $c^2 t_E^2 - x_E^2 - y_E^2 - z_E^2 = 0 = c^2 t'_E^2 - x'_E^2 - y'_E^2 - z'_E^2$ as do all events in $\mathcal{V}$.
Likewise these coordinates satisfy $y_E = 10 = y'_E$ and $z_E = 0 = z'_E$ as do all events in $\mathcal{U}$.
Therefore these coordinates are in the intersection of $\mathcal{U}$ and $\mathcal{V}$ therefore this event is in $\mathcal{H}$.

Incorrect. If we are talking about the event at the instant the spherical light wave touches the wire in the unprimed frame, that event happens at the place (x,y,z) = (0,10,0) in the unprimed frame at time t = (10/c) and at the place (x',y',z') = (-7.5, 10, 0) in the primed coordinates at time t' = (12.5/c). It is incorrect to say t' > 10/c means anything about dx/dt' since every value of t' > 10/c happens for two values of x' and therefore two values of x. Specifically at the place (x',y',z') = (+7.5, 10, 0) in the primed coordinates at time t' = (12.5/c) which corresponds to place (x,y,z) = (18.75, 10, 0) and time t=(21.25/c). Also the ratio dx/dt' is incoherent and physically without meaning.

If you are close to unprimed (0,10,0), then the "speed" (dx/dt) of the non-thing $\mathcal{H}$ is many times the speed of light in the unprimed frame. But just because this event is the earliest event in the unprimed frame does not mean it is the earliest event on the hyperbola in the primed frame. So there is a lot of unsound geometry coming from you.

This is wholly false and without basis. You have badly misunderstood both the geometry and the Lorentz transform.

This is also many times false. What you call SLW is not spherical, not radially propagating, not moving at speed c. You ignore negative x for no reason. You ignore the "speed singularity" which happens AT (x,y,z) = (0,10,0) which would quickly demonstrate you don't understand the physical situation at all.

Rather than a contradiction in the geometry or physical theory, you have only demonstrated that you have made many mistakes in understanding and explaining either.

Grossly incoherent since the physical situation demands that at the instant the expanding light sphere touches the wire the intersection must propagate in both directions along the wire at infinite speed and this is true even if the sphere is not expanding at the speed of light because that misunderstanding of the geometry of the situation happens long before any Lorentzian properties of space-time are explored.

Last edited: Aug 24, 2013
8. ### AlphaNumericFully ionizedRegistered Senior Member

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Chinglu, rpenner has taken the time to gp through your attempt at a mathematical analysis. You show you still cannot do relativity properly, you show you still don't know what constitutes a good presentation of workings and you show you still bang on about the crap you were doing 2 years ago. Still cannot do Lorentz transforms properly, still don't get expanding light shells, still cannot work with fully general cases.

You seem forever stuck with very narrow horizons, incapable of opening your mind, unwilling to learn.

9. ### chingluValued Senior Member

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1,637
1) I am using c units and said so, therefore your concern there was unfounded for not using c in the LT equations.

2) You have not yet proven LT shows the SLW moves in the same direction along the y=10 line from the unprimed coordinate (0,10,0) in the calculations.

I proved dx/dt' is negative. You were unable to refute this fact. Yet, dx/dt>0. Since there is only one SLW, then t increases if and only if t' increases.

So, we have a contradiction since if both t and t' increase from unprimed (0,10,0), then dx/dt>0 proves the SLW travel along y=10 in a positive x direction. On the other hand, dx/dt'<0 proves the SLW can only move in a negative x direction.

Now, can you refute this yes or no?.

Stay on task this time.

10. ### chingluValued Senior Member

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1,637
Yea?

So, why don't you pick out his math that proves the SLW moves in a positive x direction from unprimed (0,10,0) in the calculations of both frames.

Let me know when you have completed this task.

Oh, don't forget, you only get one SLW.

11. ### AlphaNumericFully ionizedRegistered Senior Member

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All you can do is shift the burden of proof. Ylu asked for details and rpenner gave them. You didnt understand them and so just dismiised them in totality. Shows how dishonest you are when 5ou demand mathematics but then ignore it. You know I can do the mathematics of special relativity, we've geen through examples like this before. You have not shown you can do it and hence until you can engage in honest informed discussion you're a waste of time.

If you cannot respond to maths rpenner provides why should I provide more for you to ignore? You're dishonest and its very obvious.

12. ### rpennerFully WiredValued Senior Member

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This is incorrect.
The correct way to not use c in special relativity is to use units such that c=1. But you wrote both
and
where the undefined term "c units" cannot be interpreted to mean c=1 or the use of c in a velocity expression makes you look confused.

The valid choices were:
1. Use defined units where c has such-and-such a numerical value,
2. Use arbitrary units where c has such-and-such a numerical value, or
3. Use arbitrary units where c is a parameter standing for a fixed speed of light.
The third choice was the interpretation most consistent with your expression "v=3/5c" and so I ignored the question of just what exactly you meant by "c units" and used arbitrary units of length and since you were less than particular about the time coordinate, I used lengths divided by c to express time. This allows easy conversion to either of the two other conventions.

You again confuse the Spherical Light Wave, $\mathcal{V}$ that moves in every direction at exactly the speed c, with the locus of events as different parts of the Spherical Light Wave hit different parts of the wire, $\mathcal{H}$ that (to the extent a shadow or dot of laser illumination can be said to move) moves along the wire in both directions at speeds always greater than the speed of light.

Nowhere have you shown the least bit of geometric understanding of the nature of $\mathcal{H}$, yet understanding $\mathcal{H}$ is critical to any argument you want to make about the behavior of the system.
In every inertial coordinate system where the motion of the wire relative to the place of origin of the spherical light pulse $\mathcal{V}$ is parallel to the direction of the wire $\mathcal{U}$, the first place the light appears on the wire $\mathcal{H}$ is the spot on the wire closest to the place of origin of the spherical light pulse at the instant it touches the wire. The speed of the appearance of the light starts off initially as effectively infinite in both directions and an instant later there are two spots of light both receding at super-luminal speeds. In different frames this first event happens in different places because the Lorentz boost "rotates" the location of the bottom of the hyperbola.

This is not a valuable idea. You either misuse physics and impermissibly mix frame or you made a math mistake and wrote dx/dt' repeatedly when you mean dx'/dt'. You failed to identify the location of the first spot of light in each frame and thus miss the physical import. You probably miss the physical import because you are basing your argument on pre-Lorentzian preconceptions and thus the contradiction only exists in your head, not the physics of the Lorentz transform.

I demonstrated that for the numeric example you gave, $\rho = - \tanh^{\tiny -1} \frac{3}{5}$ and that all the events in $\mathcal{H}$ can be parametrized by $x = 10 \sinh (\theta), x' = 10 \sinh (\theta + \rho), t = \frac{10}{c} \cosh (\theta), t' = \frac{10}{c} \cosh (\theta + \rho)$ which give the relevant velocities of the same event in the two frames as $\frac{\partial x}{\partial t} = c \coth (\theta), \quad \left. \frac{\partial x'}{\partial t'} \right| _{\rho = - \tanh^{\tiny -1} \frac{3}{5}} = c \coth (\theta + \rho)$. Of course the signs differ when $\theta$ is between zero and $-\rho$, but there is no physical basis to expect the opposite. Velocities change signs all the time in coordinate transformations. What matters is the geometry.

When $\theta$ is between zero and $-\rho$, the event happens after the first event in both frames, but it happens on opposite sides of the first event in either frame. Thus the geometry requires you to expect the spot of light to run away from the direction of the first event in either case and therefore you should expect the Lorentz transform to change the direction of the velocity for the event. It is required for the Lorentz transform to be self-consistent.

Allow me to pick the values of $\theta$ that best illustrates this. Let $\theta_0 = 0, \theta_{\tiny 1/2} = \frac{1}{2} \tanh^{\tiny -1} \frac{3}{5} = \tanh^{\tiny -1} \frac{1}{3}, \theta_1 = \tanh^{\tiny -1} \frac{3}{5}$. Then we have:
$\begin{array}{cc|cccc|cc} & & x & x' & t & t' & \frac{\partial x}{\partial t} & \frac{\partial x'}{\partial t'} \theta & \theta + \rho & 10 \sinh (\theta) & 10 \sinh (\theta + \rho) & \frac{10}{c} \cosh (\theta) & \frac{10}{c} \cosh (\theta + \rho) & c \coth (\theta) & c \coth (\theta + \rho) \hline \hline - \frac{1}{4} \ln 2 = \tanh^{\tiny -1} \frac{1 - \sqrt{2}}{1 + \sqrt{2}} & - \frac{5}{4} \ln 2 & -5 \left( \sqrt[4]{2} - \frac{1}{\sqrt[4]{2}} \right) & -5 \left( \sqrt[4]{32} - \frac{1}{\sqrt[4]{32}} \right) & \frac{5}{c} \left( \sqrt[4]{2} + \frac{1}{\sqrt[4]{2}} \right) & \frac{5}{c} \left( \sqrt[4]{32} + \frac{1}{\sqrt[4]{32}} \right) & - \frac{\sqrt{2} + 1}{\sqrt{2} - 1} c & - \frac{\sqrt{32} + 1}{\sqrt{32} - 1} c 0 & - \ln 2 & 0 & - \frac{15}{2} & \frac{10}{c} & \frac{25}{2c} & \pm \infty & - \frac{5}{3} c \hline \frac{1}{4} \ln 2 = \tanh^{\tiny -1} \frac{\sqrt{2} - 1}{\sqrt{2} + 1} & - \frac{3}{4} \ln 2 & 5 \left( \sqrt[4]{2} - \frac{1}{\sqrt[4]{2}} \right) & -5 \left( \sqrt[4]{8} - \frac{1}{\sqrt[4]{8}} \right) & \frac{5}{c} \left( \sqrt[4]{2} + \frac{1}{\sqrt[4]{2}} \right) & \frac{5}{c} \left( \sqrt[4]{8} + \frac{1}{\sqrt[4]{8}} \right) & \frac{\sqrt{2} + 1}{\sqrt{2} - 1} c & - \frac{\sqrt{8} + 1}{\sqrt{8} - 1} c \frac{1}{2} \ln 2 = \tanh^{\tiny -1} \frac{1}{3} & - \frac{1}{2} \ln 2 & \frac{5 \sqrt{2}}{2} & -\frac{5 \sqrt{2}}{2} & \frac{15 \sqrt{2} }{2 c} & \frac{15 \sqrt{2} }{2 c} & 3 c & - 3 c \frac{3}{4} \ln 2 = \tanh^{\tiny -1} \frac{\sqrt{8} - 1}{\sqrt{8} + 1} & - \frac{1}{4} \ln 2 & 5 \left( \sqrt[4]{8} - \frac{1}{\sqrt[4]{8}} \right) & -5 \left( \sqrt[4]{2} - \frac{1}{\sqrt[4]{2}} \right) & \frac{5}{c} \left( \sqrt[4]{8} + \frac{1}{\sqrt[4]{8}} \right) & \frac{5}{c} \left( \sqrt[4]{2} + \frac{1}{\sqrt[4]{2}} \right) & \frac{\sqrt{8} + 1}{\sqrt{8} - 1} c & - \frac{\sqrt{2} + 1}{\sqrt{2} - 1} c \hline \ln 2 = \tanh^{\tiny -1} \frac{3}{5} & 0 & \frac{15}{2} & 0 & \frac{25}{2c} & \frac{10}{c} & \frac{5}{3} c & \pm \infty \frac{5}{4} \ln 2 = \tanh^{\tiny -1} \frac{\sqrt{32} - 1}{\sqrt{32} + 1} & \frac{1}{4} \ln 2 & 5 \left( \sqrt[4]{32} - \frac{1}{\sqrt[4]{32}} \right) & 5 \left( \sqrt[4]{2} - \frac{1}{\sqrt[4]{2}} \right) & \frac{5}{c} \left( \sqrt[4]{32} + \frac{1}{\sqrt[4]{32}} \right) & \frac{5}{c} \left( \sqrt[4]{2} + \frac{1}{\sqrt[4]{2}} \right) & \frac{\sqrt{32} + 1}{\sqrt{32} - 1} c & \frac{\sqrt{2} + 1}{\sqrt{2} - 1} c \end{array}$
And $y = 10 = y', \quad z = 0 = z'$.
For the sake of consistency, because the first event happens at x=0 in the unprimed frame and x' = 0 in the primed frame then it follows that the sign of x or x' is the sign of $\frac{\partial x}{\partial t}$ or $\frac{\partial x'}{\partial t'}$, respectively. Thus the signs of the velocities mismatching for the portion of $\mathcal{H}$ where the sign of the spatial coordinate changes signs is a demonstration that the predictions of special relativity are self-consistent. In the unprimed frame the earliest event in $\mathcal{H}$ is (t,x) = (10/c, 0) and in the primed fram the earliest event is (t',x') = (10/c, 0) which are different events corresponding to different values of $\theta$.

None of this demonstrates a contradiction because you don't understand the geometry and the meaning of velocity and that the places of one coordinate system are not the places of the other coordinate system.

People competent in geometry have seen that I have refuted your baseless assertions with calculations and logical arguments.

Last edited: Aug 25, 2013
13. ### chingluValued Senior Member

Messages:
1,637
1) My c unit usage is consistent. I do not need c=1. I can define v=3/5c and distance in light seconds.

2) You still have not proved that the SLW moves in 2 directions from unprimed (0,10,0) in the calculations of both frames. In the unprimed frame it is obvious.

But, the primed frame claims using the light postulate that the SLW moves only in the negative x direction from unprimed (0,10,0).

Here is your task since you seem to get all cornfused.

Prove the primed frame calculates that the SLW moves in the unprimed frame from (0,10,0) in the positive x direction along y=10.

You get the light postulate and you get LT.

So, now prove that.

14. ### chingluValued Senior Member

Messages:
1,637
No, that is not true.

I used simple calculus to prove the primed frame believes the SLW only moves in a negative x direction from unprimed (0,10,0).

I am waiting for rpenner and you to prove the primed frame calculates that the SLW moves in a positive x direction from unprimed (0,10,0). Neither of you have proven this.

What is taking you both so long?

Oh, I know, you cannot prove that.

15. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Yes, you can. That's why I interpreted #3 as the physics assumptions most compatible with your description.

You ignore the physical situation. If I could prove the Hyperbola $\mathcal{H}$ corresponds to movement in both direction at anywhere other than at the closest point to the origin of the light cone that would prove relativity inconsistant. So it is exactly correct in the unprimed frame that the position (x,y,z)=(0,10,0) is where $\mathcal{H}$ corresponds to movement in both directions (at infinite speed) while in the primed frame this happen at position (x',y',z') =(-7.5, 10,0) and therefore corresponds to movement in just one direction while a different event is closest to the origin of the light cone in the primed frame (x',y',z') = (0,10,0) and it is here that $\mathcal{H}$ corresponds to movement in both directions while this event has position (x,y,z)=(7.5,10,0) in the unprimed frame. Each frame fully obeys the laws of physics in a self-consistent manner and the Lorentz transform moves back and forth between them in a self-consistent manner. The only inconsistent thing is your baseless assumptions of nonsensical axioms that are nowhere part of physics or geometry.

The physics of the wire parallel to the X-axis are trivial, the physics of the light cone are trivial, the intersection between the two practically isn't even physics for the intersection is not a "thing" that force can be applied to or used to convey momentum and energy from one part of the wire to another part. But you assume absolutes that neither nature nor special relativity require and in these assumptions you create inconsistencies where there need be none.

The physical situation in the primed and unprimed frames are identical except that in the unprimed frame the wire is considered at rest, an assumption that nowhere enters the calculation of events on $\mathcal{H}$. Thus the location of the closest event of $\mathcal{H}$ to the origin in one frame has no bearing on the location of the closest event of $\mathcal{H}$ to the origin in another frame. The closest event in each frame also is the event in $\mathcal{H}$ which happens first, so the Lorentz transform also results in a reordering of history for the events that happen between $\theta = 0$ and $\theta = - \rho$. This should not be surprising to a student of relativity since Einstein pointed out that simultaneity is not absolute in section 2 of his famous 1905 paper while the Lorentz transform isn't introduced until section three.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Nowhere did you comment that the speeds associated with the intersection $\mathcal{H}$ are all superluminal, which means for any event there is always a coordinate frame where that speed is infinite and in both directions. This follows from the velocity addition law for parallel velocities:
If $u > c$ and $v = -c^2/u > -c$ then we have: $w = \frac{u + v}{1 + \frac{uv}{c^2}} = \frac{u - \frac{c^2}{u}}{1 - \frac{c^2}{c^2}} = \pm \infty$.
Likewise if $u = \frac{1}{h}$ and we compute $\lim_{h\to 0} w = \lim_{h\to 0} \frac{u + v}{1 + \frac{uv}{c^2}} = \lim_{h\to 0} \frac{\frac{1}{h} + v}{1 + \frac{v}{h c^2}} = \lim_{h\to 0} \frac{h v c^2 + c^2}{h c^2 + v} = \frac{c^2}{v}$
which is self-consistent.

I see that once again you have not addressed the algebraic or geometric or differential calculus aspects of any part of my four previous posts, so I take it you agree with them 100% and your only difficult is your preconceptions related to absolute time and space -- preconceptions which are categorically incompatible with special relativity. You are having your debate and losing badly because your position is incoherent and unphysical. Instead of actually demonstrating relativity is self-inconsistent, you instead now challenge me to prove something inconsistent with special relativity and geometry, which is not my burden. Where is there an audience member who rationally can advance a claim that you are not looking foolish?

It's insane to bulk-quote all of my arguments and calculations when you don't address their content unless you agree with them. In any case, you should agree with them because special relativity is self-consistent while your position is not.

Rather than developing your ideas mathematically to see that they are inconsistent, you focus on a tiny part of the math, run into problems and unfairly blame relativity rather than your misconceptions.

16. ### chingluValued Senior Member

Messages:
1,637
You have not demonstrated your position that the primed frame claims the SLW moves two directions from unprimed (0,10,0) in the coords of the unprimed frame.

Like I said, you get the light postulate in the primed frame and you get LT.

Now, from these only, prove the SLW moves in a positive x direction according to LT mapping the view of the unprimed frame from the view of the primed.

You have not yet done this task.

17. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Are you serious or did you make a typo?
Previously, you wrote:
which is called conceding the point. So if you are serious, then you are deeply dishonest.

If you made a typo and meant to write: "You have not demonstrated your position that the primed frame claims the SLW moves two directions from unprimed (0,10,0) in the coords of the primed frame." then you are dishonest and an idiot because that has never been my position. My position is that unprimed position (0,10,0) is nothing special at all in the primed coordinates frame, mostly because the world-line of unprimed position (0,10,0) isn't at rest in the primed coordinate system and therefore isn't event a place.

This has been trivially accepting since the days of Galileo, so you are only about 400 years behind the times.

Or are you trying to say I need to prove just from the primed frame that the physical situation in the unprimed frame is as the unprimed frame has always described it? That I have done a dozen times because I did it when I showed that Lorentz boosts of equal magnitude and opposite direction cancel each other out. I did it when I proved that the Lorentzian inner product gives the same result in every frame. And I did it when I showed the effect of the Lorentz transform on the hyperbola $\mathcal{H}$ was just a hyperbolic rotation and what was rotated can be rotated back.
$\begin{eqnarray} \theta & \rightarrow ^{\Lambda}& \theta + \rho \theta = \theta + \rho - \rho & \leftarrow ^{\Lambda^{\tiny -1}}& \theta + \rho \end{eqnarray}$

I see that you quoted without comment the part where I suggested in both bold and italic that you may be acting insane. Do you concede this point?

18. ### chingluValued Senior Member

Messages:
1,637
I have no idea why you think I am inconsistent.

I will explain it as simple as possible like I have several times.

Prove the primed frame claims using the LP and LT that the SLW moves in a positive x direction from unprimed (0,10,0) along y=10 in the coordinates of the unprimed frame.

I already proved it does not and you seem to be refuting my calculus proof.

So, prove your case or face the facts that I am right and you are wrong.

I have no idea why but you took on the wrong battle here. You are going to lose.

19. ### rpennerFully WiredValued Senior Member

Messages:
4,833
What is LP? This appears to be the first time in this thread you have ever used such a term so you aren't saving time by abbreviating.
The SLW is $\mathcal{V}$ but that which moves along the wire at y=10 and z=0 is $\mathcal{H}$ which is not spherical and does not travel at speed c and does not convey momentum and energy from one part of the wire to another part of the wire.

In post #44, I explicitly calculated the description of every event in $\mathcal{H}$ and every Lorentz transform that corresponds to relative movement along the X-axis.
So $\mathcal{H}$ only corresponds to movement in the +x direction on the parts of the wire with x > 0.

Likewise in the primed coordinate frame, the transformed events of $\mathcal{H}$ only correspond to movement in the +x' direction on the parts of the wire with x' > 0.

At this point it should be trivial to see in the primed frame we have:
$\vec{u}'_{\mathcal{H}} = \begin{pmatrix} \frac{\partial x'}{\partial t'} \\ \frac{\partial y'}{\partial t'} \\ \frac{\partial z'}{\partial t'} \end{pmatrix} = \frac{\partial \theta}{\partial t'} \begin{pmatrix} \frac{\partial x'}{\partial \theta} \\ \frac{\partial y'}{\partial \theta} \\ \frac{\partial z'}{\partial \theta} \end{pmatrix} = \frac{c}{10 \sinh (\theta + \rho)} \begin{pmatrix} 10 \cosh (\theta + \rho) \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} c \coth (\theta + \rho) \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{c^2 t'}{x'} \\ 0 \\ 0 \end{pmatrix}$

Later in post #49 I calculated the positions, times and velocities in both frames of seven different events in $\mathcal{H}$.

I agree. You would begin to get an idea by actually reading the posts you quote in entirety.
108 years of math, physics and geometry says different.

20. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
Since I've got a bank holiday Monday off I'll go through the scenario for chinglu. I'll also point out that the situation is pretty much identical to the one I refuted 2 years ago. Furthermore, since chinglu has used phrases such as "Let's settle SR once and for all" once this is done and he is unable to retort it we can all consider chinglu will be refuted 'once and for all' and any further repetition of these pointless threads will be considered trolling and dealt with as much.

Setup

* We ignore z directions as they are irrelevant. All units use c=1.
* Frame S has coordinates $x^{\mu} = (t,x,y)$ and frame S' has coordinates $(x')^{\mu} = (t',x',y')$
* In S there is an object moving at speed $v^{\mu} = (1,0.6,0)$ so that its space-time position is $(t,0.6t,0)$ at time t in S
* S' is defined by the object's rest frame, $(v')^{\mu} = (1,0,0,0)$
* At t=T=0 the object passes through x=X=0 and emits a pulse of light which spreads out at light speed in all directions.
* We wish to consider the times of arrival of the sphere at S spatial locations $A_{space}=(x_{A},y_{A}) = (0,10)$, $B_{space}=(x_{B},y_{B}) = (0,10+h)$ and their images in S'.

First I'll do the fully general case and then put in numbers. This will serve to illustrate that not only is there no contradiction chinglu claims, the result is necessary for Lorentz transforms to be viable. As such we are interested in the spatial location $A=(X,Y)$ and a neighbour $B=(X+p,Y+q)$ for p,q small

Frame S Calculations

The light sphere expands in time as the surface $x^{2}+y^{2}=t^{2}$. Therefore A and B are reached at times $t_{A} = \sqrt{X^{2}+Y^{2}}$ and $t_{B} = \sqrt{(X+p)^{2}+(Y+q)^{2}}$. On the assumption p,q are small we can expand $t_{B}$ in a series, dropping all non-linear terms,

$t_{B} = \sqrt{(X+p)^{2}+(Y+q)^{2}} \approx \sqrt{X^{2}+Y^{2}+2Xp+2Yq} = \sqrt{X^{2}+Y^{2}}\sqrt{1+\frac{2Xp+2Yq}{X^{2}+Y^{2}}} = t_{A}\left( 1+\frac{2Xp+2Yq}{X^{2}+Y^{2}} \right)^{\frac{1}{2}} \approx t_{A}\left( 1+ \frac{Xp+Yq}{X^{2}+Y^{2}} \right) = t_{A} + \frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}$

So if $Xp+Yq>0$ we have $t_{A}<t_{B}$ and the reverse if the term is negative. Therefore we have space-time events $A = (t_{A},x_{A},y_{A}) = \left( \sqrt{X^{2}+Y^{2}} , X , Y \right)$ and $B = (t_{B},x_{B},y_{B}) = \left( \sqrt{X^{2}+Y^{2}} + \frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}} , X+p , Y+q \right)$.

Frame S' Calculations

Without loss of generality we can take the motion of the object parallel to the x axis (as in the case described by chinglu) so we will. If we'd set q=0 this wouldn't be the case. Anyway, S' is constructed from S by a Lorentz transform of the form $(t',x',y') = (\gamma (t-vx) , \gamma (x-vt) , y )$ where $\gamma = \frac{1}{\sqrt{1-v^{2}}}$. The inverse of this is $(t,x,y) = (\gamma(t'+vx'),\gamma(x'+vt'),y')$.

To check this transform does as expected we apply it to the light sphere equation $-t^{2}+x^{2}+y^{2}=0$ defining the expanding light sphere. $-t^{2} + x^{2} + y^{2} = \gamma^{2}(-(t'+vx')^{2}+(x'+vt')^{2}) + y'^{2} = \gamma^{2} \left( -t'^{2} +2vt'x' - v^{2}x'^{2} + x'^{2} - 2vt'x' + v^{2}t'^{2} \right) + y'^{2} = \gamma^{2}(1-v^{2})(-t'^{2}+x'^{2})+y'^{2} = -t'^{2}+x'^{2}+y'^{2}$, the equation for an expanding light sphere centred on the origin in S', as required.

Now we compute the Lorentz transform images of A and B, giving A' and B'. To do this we compute the action of the Lorentz transform on A. If this is to be consistent we should then be able to put in resultant values and find they satisfy the same light sphere equation just worked out.

$A' = (t'_{A'},x'_{A'},y'_{A'}) = \left( \gamma (t_{A}-vx_{A}) , \gamma (x_{A}-vt_{A}) , y_{A} \right) = \left( \gamma (\sqrt{X^{2}+Y^{2}}-vX) , \gamma (X-v\sqrt{X^{2}+Y^{2}}) , Y \right)$

$= 0 -t'_{A'}^{2} + x'_{A'}^{2} + y'_{A'}^{2} = -\gamma^{2} (\sqrt{X^{2}+Y^{2}}-vX)^{2} + \gamma^{2} (X-v\sqrt{X^{2}+Y^{2}})^{2} + Y^{2} = -\gamma^{2}( X^{2}+Y^{2} - 2vX\sqrt{X^{2}+Y^{2}} + v^{2}X^{2} ) + \gamma^{2}( X^{2} - 2vX\sqrt{X^{2}+Y^{2}} + v^{2}(X^{2}+Y^{2}) ) + Y^{2} = \gamma^{2}( -(1-v^{2}(X^{2}+Y^{2}) + (1-v^{2})X^{2} ) + Y^{2} = -(X^{2}+Y^{2}) + X^{2} + Y^{2} = 0 = -t_{A}^{2} + x_{A}^{2} + y_{A}^{2}$

Unsurprisingly we indeed find everything cancels, since the original expression was equal to zero and Lorentz transforms do not alter that. Therefore A', the Lorentz image of A in Frame S', is at space-time location $A' = (t',x',y') = \left( \gamma (\sqrt{X^{2}+Y^{2}}-vX) , \gamma (X-v\sqrt{X^{2}+Y^{2}}) , Y \right)$

Repeating the same calculations for B will be identical except you replace all the X with X+p and all the Y with Y+q. Therefore $B' = (t',x',y') = \left( \gamma (\sqrt{(X+p)^{2}+(Y+q)^{2}}-v(X+p)) , \gamma ((X+p)-v\sqrt{(X+p)^{2}+(Y+q)^{2}}) , Y+p \right)$.

We have have the light sphere hitting A' at $t' = t'_{A'} = \gamma (\sqrt{X^{2}+Y^{2}}-vX)$ and hitting B' at $t' = t'_{B'} = \gamma (\sqrt{(X+p)^{2}+(Y+q)^{2}}-v(X+p))$. No now the question is what is the relation between there two times. As before let's do some linearisations,

$t'_{B'} = \gamma (\sqrt{(X+p)^{2}+(Y+q)^{2}}-v(X+p)) \approx \gamma (\sqrt{X^{2}+Y^{2} + Xp+Yq}-v(X+p)) = \gamma (\sqrt{X^{2}+Y^{2}}\sqrt{1 + \frac{Xp+Yq}{X^{2}+Y^{2}}}-v(X+p)) \approx \gamma \left( \sqrt{X^{2}+Y^{2}} + \frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}} \right) - v(X+p) ) = \gamma \left( \sqrt{X^{2}+Y^{2}} - vX \right) + \gamma( \frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}} - v p) = t'_{A'} + \gamma (\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}-vp)$

Frame Comparisons

Suppose p,q>0 so that B is to the up and right of A. So we have the following time values

$t_{A} = \sqrt{X^{2}+Y^{2}}$
$t_{B} = t_{A} + \frac{Xp+Yq}{X^{2}+Y^{2}}}$
$t'_{A'} = \gamma (\sqrt{X^{2}+Y^{2}}-vX)$
$t'_{B'} = t'_{A'} + \gamma (\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}-vp)$

Therefore we have $t_{A}<t_{B}$ if $Xp+Yq > 0$ in Frame S. In Frame S' we have $t'_{A'}<t'_{B'}$ if $\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}-vp > 0$.

The condition is not the same. If $Xp+Yq > 0$ B happens after A but if we make $vp>\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}$ then $\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}-vp<0$ and A' happens after B'. Chinglu considers this a contradiction, the order of times have changed, meaning that the spherical wave in S' expands out to hit B' first and then moves to the left to hit A'. Chinglu claims this is a contradiction.

There is no Contradiction

So why is there no contradiction? Well the resolution rests in something chinglu has failed to calculate, the x',y' locations of A' and B'. I would point out that this is EXACTLY the mistake he made 2 years ago which I wrote up multiple times and explained to him but he hasn't learnt from (thus proving he is either dishonest, stupid or both).

Since the Lorentz transformation doesn't alter the y values we know $y_{A}=y'_{A'}$ and $y_{B}=y'_{B'}$ so we just need to compute the other spatial component.

$x'_{A'} = \gamma (X-v\sqrt{X^{2}+Y^{2}})$
$x'_{B'} = \gamma ((X+p)-v\sqrt{(X+p)^{2}+(Y+q)^{2}}) \approx \gamma ((X+p)-v(\sqrt{X^{2}+Y^{2}} + \frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}})) = \gamma(X-v\sqrt{X^{2}+Y^{2}}) + \gamma (p-v\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}}) = x'_{A'} + \gamma (p-v\frac{Xp+Yq}{ \sqrt{X^{2}+Y^{2}}})$

We now set q=0 so that we restrict ourselves to a line parallel to the x axis, as in chinglu's example case.

$x'_{B'} = x'_{A'} + p\gamma (1-v\frac{X}{ \sqrt{X^{2}+Y^{2}}})$. With $p>0$ (so that B is to the right of A in the S frame) we find $x'_{A'}$ is always to the left of $x'_{B'}$ since $v\frac{X}{ \sqrt{X^{2}+Y^{2}}}<1$, making the difference always positive. But does this represent a contradiction? No, provided we work out $x'_{A'}$.

$x'_{A'} = \gamma (X-v\sqrt{X^{2}+Y^{2}})$ is not necessarily positive. The action of the Lorentz transform in the positive x direction is to have everything 'slide' along the x' axis to the left, as the S' frame moves to the right in the S frame. We have $x'_{A'}<0$ if $\gamma (X-v\sqrt{X^{2}+Y^{2}}) <0$, which amounts to $\frac{X}{\sqrt{X^{2}+Y^{2}}} < v$. But since we know $0\leq \frac{X}{\sqrt{X^{2}+Y^{2}}}<1$ it means that for sufficiently high values of v the Lorentz boost has the effect of sliding A and B so far to the left, once considering them in the S' frame, that now A' is further away from the origin (x',y')=(0,0)[/tex] than B'. In such an instance we'd DEMAND relativity says B' is hit first, which is what it does.

chinglu's mistake is equivalent to saying that since |3| < |4| if we apply a shift of -x to them both we should get |3-x| < |4-x| for all x. If x is small sure but if x=4 then we have |3-4| = |-1| and |4-4| = |0| and now |3-x| > |4-x|. More technically translations are not necessarily isometries of normed vector spaces.

Summary

In Frame S A is spatially further from the origin O than B and so A gets hit first by the light sphere. After a Lorentz transform of sufficiently high speed A' is now further from the origin O' than B' and gets hit second. Not only is there no contradiction, it would be a contradiction if that did NOT happen.

Chinglu, since I've shown the resolution of your complaint in generality and you have been throwing around comments like "Let's sort this out once and for all" if you wheel out another scenario like this again you'll be summarily banned. Nothing done here is any different from the methodology I used 2 years ago when you wheeled out an example of this kind by Andrew Banks. This level of calculus is expected of 1st year undergraduates in 'relativity and motion' courses I've taught so it isn't like I'm doing anything difficult. In the 2 years since I smacked down said example you could have learnt the relevant mathematics to understand special relativity. You didn't. This proves you are incapable of rational, honest discussion either because you're dishonest or stupid. I suspect it is both, in that you are sufficiently stupid to not grasp this material in any reasonable length of time and so you just don't bother. Your intuition is wrong, your understanding of relativity is non-existent, your misrepresentations of science reprehensible and your entire approach to scientific discourse pitiful.

You're a joke and a bad one at that.

Last edited: Aug 26, 2013
21. ### LakonValued Senior Member

Messages:
1,117
Bolded .. this seems most unfair. Did you forget this is the psuedoscience section ? By that measure, everything here in 'On The Fringe' should be banned.

PS - I see he HAS been banned ? Very unfair.

Last edited: Aug 27, 2013
22. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
I banned him for spouting nonsense in the main maths/physics forum, not for the 3 threads he currently has in this forum. I have no grounds to ban him for hackery in this subforum but he was warned about it for the main forum. In fact the threads currently in this fringe section he started in the main forum and I moved them. That was accompanied by a message not to post that crap in the main forum, keep it in the fringe section, it is why we have the fringe section. But chinglu couldn't help himself and had to take a thread in the main forum off topic so he could repeat the same nonsense from these threads in that one.

Specifically this post and the ones surrounding it. Despite me telling him just this weekend that publication does not imply proof and experiments which do not contradict special relativity are not viewed as proof of special relativity. I wasn't the only one saying this to him either. And even his own quotes of other sources backed up what I corrected him on and yet he continues. If he wants to be dishonest and ignore things in the fringe section fine, I didn't take any action because it isn't my moderating job to do this forum. However when he kept at it, repeatedly, in the main forum he's crossed a line he was warned repeatedly about. If he cannot present his case here, despite repeated requests, why should he be allowed to not make his case in more places.

He obviously doesn't know any special relativity, even to the level of someone taking an introductory course on Lorentz transforms often given to 1st year undergraduates fresh out of high school. I wrote my above post off the top of my head. I didn't sit down and crunch the algebra, I wrote it up as I thought it and it took easily less than 1 hour. That's what chinglu could do if he actually understood relativity to even the most basic level. But he doesn't, despite years of being retorted each and every time he tries this nonsense. I'd expect of a rational, honest person to spend the time the ban affords him to write up, in precise detail, the step by step algebraic derivation of the supposed inconsistency, so the moment they are unbanned they can provide what has been asked of them many times and go "There, as I stated". But we all know chinglu won't do that because he cannot.

Here, in this forum, he can avoid stepping up, provided he's happy to be seen as a dishonest hack. In the main forum I, nor anyone else who has any honest interest in physics and discussing physics, will not put up with that shit.

/edit

And as a side note can you edit your post so you just quote the relevant part of what I said please. Quoting a massive post to just respond to a small part clutters the thread. Thanks.

23. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Here is the x-t slice at y=10, z=0 in the coordinate systems S and S'. Green and red world-lines have been drawn in to illustrate that the place (x,y,z) = (0,10,0) in green is not the same thing as the place (x',y',z') = (0,10,0) in red.

Blue is the Hyperbola $\mathcal{H}$ where the SLW starting at the origin event (t,x,y,z) = (0,0,0,0) = (t', x', y', z') intersects this 2-dimensional world-sheet. $H_{-1}, \, H_0, \, H_1, \, H_2, \, H_3, \, H_4, \, H_5$ correspond to my table of points on the hyperbola. Notice that $H_0, C \; \textrm{and} \; A$ are all on the green line, while $H_4 \; \textrm{and} \; A$ are on the red line.

In frame S, the green line is the closest "place" to the origin event and thus it is the "place" which is hit first by the light. In frame S', this is the role of the red line. Thus in frame S, event $H_0$ happens before event $H_4$ while events $H_{-1}$ and $H_1$ happen at the same time. In frame S', event $H_4$ happens before event $H_0$ while events $H_3$ and $H_5$ happen at the same time. This is called relativity of simultaneity and is discussed in section 2 of Einstein's 1905 paper and every single textbook on the subject.

It is absurd for Andrew Banks or chinglu to think this is a mysterious part of special relativity only recently uncovered -- this is just a personal misunderstanding of Andrew Banks spread on vixra.org. On that note, I wrote the following comment on his recent paper:

http://vixra.org/abs/1303.0144

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Last edited: Aug 27, 2013