# The Light Speed Postulate and its Interpretation in Derivations of the Lorentz Transformation

Discussion in 'Pseudoscience' started by tsmid, Apr 24, 2016.

1. ### Confused2Registered Senior Member

Messages:
503
The link between the two frames is
x1²-c ²t1 ²=x2²-c ²t2 ²
If you don't get that by whatever means you use then you've gone wrong.

3. ### tsmidRegistered Senior Member

Messages:
368
There is no link between the class 1 and the class 2 events, as they are different events. See my previous post.

5. ### Neddy BateValued Senior Member

Messages:
1,749
The distinction between class 1 and class 2 events is your own terminology. What you are essentially saying is that when frame 1 measures the coordinates of an event, then that event is strictly a class 1 event, and not a class 2 event. What you don't seem to realise is that frame 2 can also measure the coordinates of that same event.

In fact the very purpose of the LT equations is to transform the coordinates of a single event from one frame to the other frame. Take a look at the following animation:

It is only for the sake of clarity that the x axis of system 1 is drawn higher than the x axis of system 2. In SR you are supposed to imagine that the x axes of both systems are superimposed on top of each other, so that the origin of both systems exist at the same point in space at time=0. Once you imagine that, then the events depicted in the above animation are as follows, in chronological order:

Event #0. A single ray of light is emitted from the origin of frame 1. This same event can be described as that same ray of light being emitted from the origin of frame 2. Given the spacial and temporal coordinates of this one event as measured by one frame, the LT could be used to find the spacial and temporal coordinates that would be measured by the other frame.

Event #1. That same ray of light reaches location x2 which is a coordinate in the system of frame 2. That same event can be described as that same ray of light reaching location x2' which is a coordinate in the system of frame 1. Given the spacial and temporal coordinates of this one event as measured by one frame, the LT could be used to find the spacial and temporal coordinates that would be measured by the other frame.

Event #2. That same ray of light reaches location x1 which is a coordinate in the system of frame 1. That same event can be described as that same ray of light reaching location x1' which is a coordinate in the system of frame 2. Given the spacial and temporal coordinates of this one event as measured by one frame, the LT could be used to find the spacial and temporal coordinates that would be measured by the other frame.

Last edited: Aug 22, 2016

7. ### tsmidRegistered Senior Member

Messages:
368
First of all, the light detectors in frame 2 are at rest in that frame, so by definition they are only associated with class 2 events. You could measure the coordinates of class 1 events in frame 2 only indirectly, by somehow communicating the detection events of the frame 1 detectors to some other (not light-) detectors in frame 2. This can of course in principle be done, but in any case, those secondary events would still be measured in terms of the system time t2. And assuming those events would propagate in frame 2 also with speed c would mean nothing but x1'=c*t2 (if we denote the location of those events with x1'). There is no clock rate t1' defined at all here (whatever you mean by that). The only clock rates that exist are t1 and t2 (and for good measure we have even made the convention that these are equal)

8. ### PhysBangValued Senior Member

Messages:
2,422
That sounds like crazy talk. We use detectors that are at rest to measure all kinds of things. Speedometers in cars, radar detectors...

Oh, wait, we could also think of those things as moving.

It must be that it doesn't matter if a detector is moving or not to use it to detect something.

9. ### Neddy BateValued Senior Member

Messages:
1,749
I'm glad you agree it can be done, even if only in principle. It is important to accept that an event, in theory, does not belong to any particular frame. All frames are equally entitled to measure the location of an event. Before you argue against this statement, let me give you a concrete example that you can try to argue against, if you wish.

Let's say that two cars have a head-on collision. Neither of the two cars are at rest with respect to the road frame at the time of the collision. But surely there must be some location on the road where the collision occurred. Even if you cannot identify where it happened, there still MUST be some location on the road where it occurred. To deny this is not only to deny SR, but also Galilean relativity, basic kinematics, mathematics of moving systems, etc.

10. ### rpennerFully WiredValued Senior Member

Messages:
4,833
A frame is in no way analogous to a physical box; It is never the case that one can say precisely that events happen "inside" a frame.

Instead, a frame is an inertial coordinate system, a methodology of assigning real numbers (at least 2 and in our experience not more than 4) to every point in flat space-time such that the coordinates associated with every inertial motion is described by the equations of a straight line:
$(x - A_x) = v_x ( t - A_t), \quad (y - A_y) = v_y ( t - A_t), \quad (z - A_z) = v_z ( t - A_t)$ is such a set of equations of a line characterized by an event on the line, $A$, and with frame-dependent velocity, $v$. Likewise
$\frac{x - A_x}{B_x - A_x} = \frac{y - A_y}{B_y - A_y} = \frac{z - A_z}{B_z - A_z} =\frac{t - A_t}{B_t - A_t}$ is also such a set of equations (provided none of the denominators are zero) for a line that goes through events A and B with frame-dependent velocity, $v = \left( \frac{B_x - A_x}{B_t - A_t}, \; \frac{B_y - A_y}{B_t - A_t}, \; \frac{B_z - A_z}{B_t - A_t} \right)$.​
Inertial motions through the event where all coordinates are zero are special in that the equation of their motion is a linear (as opposed to affine) relationship.
Inertial motions where the frame-dependent velocity is zero are special to that choice of frame in that we describe the inertial motion as "at rest."
But this special salience is entirely due to a choice of the coordinates.

So if we have two different frames, how may they be related?

1) Pure Translations.

Say frame $\Sigma$ is established where event A as the event which has all coordinates equal to zero while frame $\Sigma'$ has event B established for such a role, but they agree on the value of every frame-dependent velocity and measure of elapsed time. What is the rule that connects them? Since $A_x = 0, B_x' = 0, C_x - D_x = C_x' - D_x'$ we have: $x' = x' - B_x' = x - B_x = x + ( A_x - B_x)$ or generalizing:
$\begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} + \begin{pmatrix} A_x - B_x \\ A_y - B_y \\ A_z - B_z \\ A_t - B_t\end{pmatrix}$​

2) Pure Rotations.

Say frames $\Sigma$ and $\Sigma'$ share event A as the event where all coordinates are equal to zero. Likewise assume that the magnitude of all velocites are preserved. Likewise presume that all velocities parallel to $\vec{n}$ are unchanged. What is the rule that connects them?
$\begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} \quad + \quad \frac{\sin \theta}{| \vec{n} |} \begin{pmatrix} 0 & -n_z & n_y & 0 \\ n_z & 0 & -n_x & 0 \\ -n_y & n_x & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} \quad + \quad \frac{1 - \cos \theta}{\vec{n}^2} \begin{pmatrix} -n_y^2 - n_z^2 & n_x n_y & n_x n_z & 0 \\ n_x n_y & -n_x^2 - n_z^2 & n_y n_z & 0 \\ n_x n_z & n_y n_z & -n_x^2 -n_y^2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}$​

where $\theta$ is the angle which velocities rotate about the $\vec{n}$ axis.

3) Pure Boosts
Say frames $\Sigma$ and $\Sigma'$ share event A as the event where all coordinates are equal to zero. Likewise presume parallel motions separated by a offset in one of the 8 cardinal directions are still expressible as purely offset in that same direction. But assume motions at rest in one frame are not at rest in the other.

Then we have a pure boost, which is the point of the Galilean transform in Newtonian physical assumptions and the Lorentz transform in the Special Relativistic physical assumptions of what flat space-time is. Experiments readily favor the latter over the former when the accuracy of observations is sufficient to distinguish between them. So we stick with Lorentz transformations as our boost relation. From it comes a velocity transformation law.

4) Combinations. Two translastions combine to be a single translation. Two rotations combine (in complicated ways) to be a single rotation. A boost, rotation and boost combine to be the same as a rotation and boost. When you combine all three, you get what's known as a orthochronous, parity-preserving Poincaré transformation, the most general collection of transformations which preserve Newton's law of inertia, $c^2 (\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2$, the direction of time, and the sign of the triple product.

But if you can't understand how frames are not boxes and that every event has a representation in every frame, then you will never have something useful to say about frames.

11. ### tsmidRegistered Senior Member

Messages:
368
I don't argue against the statement. Obviously, an event occurs at a specific location and time in any reference frame. The question is only at which. The point I am making is that each frame uses its own clocks (and only its own clocks) to make the timings of these events. So in frame 1 the path of the events is tracked in terms of t1 i.e. we have a function x1(t1) , whereas in frame 2 the path of the same events is tracked in terms of t2 i.e. we have a function x1'(t2) (where x1' shall indicate that we are dealing with the events associated with the detectors at rest in frame 1, but with their location evaluated in frame 2).

Last edited: Aug 23, 2016
12. ### Neddy BateValued Senior Member

Messages:
1,749
It is true that x1' indicates that we are dealing with an event associated with a detector at rest in frame 1, but with its location evaluated in frame 2.

But x1' is not always going to be equal to x2 in general. And for the case that you specifically chose, (where x2 is numerically equal to x1), then x1' will certainly not be equal to x2 because the motion of the frames has to be considered (remember v?).

So you cannot assume the time that the x1' event happens as recorded by frame 2 clocks is going to be t2. In fact you did not even need to have x2 or t2 in the thought experiment at all, because the LT would transform x1 and t1 to x1' and t1' which are not the same thing as x2 and t2 at all, in your chosen case.

13. ### Neddy BateValued Senior Member

Messages:
1,749
Dear tsmid, please study this animation:

Do you see the brief flashes where x2' and x1' show up? Note that x2' shows up when the light reaches x2 (not when it reaches x1). Note also that x1' shows up when the light reaches x1 (not when it reaches x2). Yet x1 and x2 are numerically equal, as per your original request.

14. ### tsmidRegistered Senior Member

Messages:
368
This would contradict the light speed postulate as understood by Einstein, according to which the class 1 events must propagate also with speed c in frame 2. We all agree that the light speed postulate requires that the class 2 events propagate in frame 2 with speed c (i.e. x2=c*t2). Now if the class 1 events should also propagate with speed c in frame 2, this means that at each instant t2, the class 1 events must be co-located with the class 2 events i.e. x1'=x2=c*t2 (I am not saying this is in fact so, but this would be implied by Einstein's interpretation of the light speed postulate).

15. ### tsmidRegistered Senior Member

Messages:
368
If x1=x2, then t1=t2=t (assuming identical clocks and length units in both frames). And as per the argument in my previous post, Einstein's interpretation of the postulate would require the class 1 events also to travel the same distance x2 in frame 2 during that time t, i.e. x1'=x2=x1. So your animation could only be correct if the clock/length units or the speed of light in the two frames are different (which in any case would contradict the initial assumption).

16. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Events don't propagate. You need to know the difference between a point and a line.
You can't prove anything about speed until you consider both x1' and t1'; both x2 and t2; You can't just claim "contradiction" — you have to show it.

By assumption, event 1 has coordinates (x=0, t=0, x'=0, t' = 0)
By assumption, event 2 has coordinates (x = x2', t = x2'/c, x'=x2, t' = x2/c)
By assumption, event 3 has coordinates (x = x1, t = x1/c, x'=x1', t' = x1'/c)
By assumption, 0 < x1 = x2.

What does the Lorentz equation say? It says for every event: x' = (x + v t)/√(1 − v²/c²), t' = (t + v x/c²)/√(1 − v²/c²) where v is some constant.

So from the coordinates of event 2 we have x2 = (x2' + v x2'/c)/√(1 − v²/c²), x2/c = (x2'/c + v x2'/c²)/√(1 − v²/c²) or x2 = x2' ( √(1 + v/c) √(1 + v/c) )/ ( √(1 + v/c) √(1 − v/c) ) = x2' √((1 + v/c)/(1 − v/c))
And from the coordinates of event 3 we have x1' = (x1 + v x1/c)/√(1 − v²/c²), t1' = (x1/c + v x1/c²)/√(1 − v²/c²) or x1' = x1 √((1 + v/c)/(1 − v/c))

So when −c < v < c, the quantities x2', x1=x2, and x1' form a geometric progression.

Because, for every event, we have 2 equations, with four unknowns, if the speed of light postulate were incompatible with the Lorentz transformation, we would not have been able to solve the equations for all admissible velocities.

When v = 0, x2' = x2 = x1 = x1' (not surprising).
When v = +0.6c, x2' = (1/2) x2 = (1/2) x1, x1=x2, x1' = 2 x1 = 2 x2. All observers agree that event 1 happens before event 2 which happens before event 3 (0 < x2' < x1 = x2 < x1')
When v = +0.8c, x2' = (1/3) x2 = (1/3) x1, x1=x2, x1' = 3 x1 = 3 x2. All observers agree that event 1 happens before event 2 which happens before event 3
When v = +0.96c, x2' = (1/7) x2 = (1/7) x1, x1=x2, x1' = 7 x1 = 7 x2. All observers agree that event 1 happens before event 2 which happens before event 3

When v= −0.6c, x1' = (1/2) x1 = (1/2) x2, x1=x2, x2' = 2 x2 = 2 x1. All observers agree that event 1 happens before event 3 which happens before event 2 (0 < x1' < x2 = x1 < x2')
When v= −0.8c, x1' = (1/3) x1 = (1/3) x2, x1=x2, x2' = 3 x1 = 3 x1. All observers agree that event 1 happens before event 3 which happens before event 2
When v= −0.96c, x1' = (1/7) x1 = (1/7) x2, x1=x2, x2' = 7 x1 = 7 x1. All observers agree that event 1 happens before event 3 which happens before event 2

Last edited: Aug 25, 2016
17. ### Neddy BateValued Senior Member

Messages:
1,749
The light postulate is satisfied as long as:
x1/t1 = c ........ (where x1 and t1 are measured by system 1 using its own measuring rods and clocks)
x2/t2 = c ........ (where x2 and t2 are measured by system 2 using its own measuring rods and clocks)
x1'/t1' = c ........ (where x1' and t1' are measured by system 2 using its own measuring rods and clocks)
x2'/t2' = c ........ (where x2' and t2' are measured by system 1 using its own measuring rods and clocks)

Clearly x1' does not have to be equal to x2 for those equations to be satisfied. Simply accept that t1' might not be equal to t2, and you are all set. It is simply a matter of x1' and x2 being some distance apart on the x axis of system 2, which means the light will take some time to propagate from one to the other.

In fact, the only justification I can think of for you claiming t1' should be equal to t2 would be if you first claimed t1' must be equal to t1, which would be taking a time reading from a clock at rest in system 1 and trying to force it onto a clock at rest in system 2, a classic mistake.

18. ### tsmidRegistered Senior Member

Messages:
368
I suppose you would be even objecting then against the notion of propagating wave structures. It is indeed only localized structures that propagate. Lines don't propagate (at least not in the direction of the line)

19. ### tsmidRegistered Senior Member

Messages:
368
First of all, I would like to remind you that x1,x2,t1,t2 are all in principle continuous variables here (t1 and t2 the independent variables x1 and x2 the dependent variables). Assuming the light signal is sent from the origin of frame 1 at time t1=0, then for any system time t1 there will be exactly one detection event at this moment, namely at the coordinate x1(t1)=c*t1 (provided of course we have a detector positioned there). Likewise, in system 2, we have for any system time t2 exactly one detection event for the frame 2 detectors, namely at the coordinate x2(t2)=c*t2 (provided the light speed postulate holds, as the light source is not at rest in frame 2). This as such should already be sufficient to define the light speed postulate, but we can of course additionally ask how the location of the frame-1-detector events propagates in frame 2. For a given system time t2, there will be again exactly one such event, and assuming the speed of light postulate holds for these events as well, this must be at the location x1'(t2)=c*t2 =x2(t2). If you don't agree with this, then you are saying the light speed postulate does not hold in this case.

Last edited: Aug 25, 2016
20. ### Neddy BateValued Senior Member

Messages:
1,749
For a given system time t2, it is true that x2=c*t2 but that has nothing to do with x1'. You will need a transformation equation to determine what x1' is supposed to be, considering the two systems are moving relative to each other at velocity v, and also considering that there is no concept of "absolute time" in SR, despite your invalid assumption that t1'=t2. By the way, are you still claiming that t1' should not even be its own variable in this exercise?

21. ### tsmidRegistered Senior Member

Messages:
368
The events associated with x2 are obviously different events to those associated with x1', but the two must be co-located if the speed of light postulate applies to either of the events.
If you think differently, tell me where in system 2 you will observe the class 1 event (associated with the detectors at rest in frame 1) at time t2, if not at a location c*t2
We will get back to this after we agreed about the above.

22. ### rpennerFully WiredValued Senior Member

Messages:
4,833
What a ridiculous evasion of responding to a complete refutation of your baseless claim that the results of applying the Lorentz transform between the frames “would contradict the light speed postulate as understood by Einstein.”

Euclid taught a geometry of space without time. One particular space that Euclid covered was the Euclidean plane. Our concept of real numbers comes from exercises in plane geometry.

In 1637, René Descartes invented the notion of using pairs of numbers to identify points in the plane, what we call today a Cartesian coordinate system. Now geometry could be done by studying the algebra of the numbers that represented points on the plane. Typically, such numbers would be called x and y and (most) straight lines could be described in these equivalent forms:
$\begin{eqnarray} A x + B y + C & = & 0 \\ A (x - x_0) + B (y - y_0) &= & 0 \\ \frac{x - x_0}{B} & = & - \frac{y - y_0}{A} \\ (y - y_0) & = & m (x - x_0) \\ y & = & m x + b \end{eqnarray}$
where
$A = y_0 - y_1, B = x_1 - x_0, C = x_0 y_1 - x_1 y_0, m = - \frac{A}{B}, b = y_0 - m x_0$

Where Descartes described the geometry of the Euclidean plane, physicists use the same system to describe motion as the curves of trajectories in space-time. In a 1+1 representation of flat space-time, every event has a specific location, x, and specific instant, t, associated with it, giving 2 numbers to locate a point on the Cartesian space-time plane. Non-accelerating motion of a tiny object is represented by straight lines where each instant of time is associated with exactly one well-defined position such that in each identical interval of elapsed time, Δt, the line passes through the same measure of different positions, Δx, meaning the ratio of these quantities for the straight line, Δx/Δt, corresponds to the modeled velocity of the object, v.

Events cannot propagate because by definition they have Δx = 0 and Δt = 0. You need an extended 1-dimensional object to describe propagation in a 1+1 representation of flat space-time. You need a (possibly curved) line. The equation of a line in space-time through the event $\left( x_0, t_0 \right)$ with velocity v is: $(x - x_0) = v (t - t_0)$.
Then at time $t_1$ is known this can be used to solve for the corresponding position: $x(t_1) = x_1 = x_0 + v ( t_1 - t_0)$.

But what the Lorentz (and Poincaré) transforms say is that more than one Cartesian 1+1 representation of flat space-time is admissible. Your claim that the Lorentz transform is incompatible with the light speed postulate has been shown be baseless, unsupported and debunked. You lacked even the basic understanding of the geometric model of space-time being used so you couldn't understand the language of the claim, let alone have a chance of making an articulate statement about it.

Last edited: Aug 26, 2016
23. ### Confused2Registered Senior Member

Messages:
503
Just a drive-by post. I think tsmid might be confusing the detection of the event with the event itself.