Relativity of simultaneity and one-way speed of light

[david345]

In the train example the speed of light is the same in both directions for the person on the ground and the person on the train. Let us say for the person on the train the flashes happen at the same time. c is set to 1 and the person on the ground is moving at .5 c with respect to the person on the train.

 

person on train: flashes occur at (1,1) and (1,3) We are using (t,x) both flashes reach person on train at (2,2)

speed in direction 1= [math] d/t=c[/math]

speed in direction 2=c

 

Lorentz transformation from (t,x) to (t',x')

[math]\begin{vmatrix}t'\\x'\\\end{vmatrix}=\begin{vmatrix}\gamma&-\beta\gamma\\-\beta\gamma&\gamma\\\end{vmatrix}\begin{vmatrix}t\\x\\\end{vmatrix}[/math]

[math]\beta=.5[/math]

[math]\gamma=1.1547[/math]

 

person on ground: flashes occur at (.57735,.57735) and (-.57735,2.88675) both flashes reach person n train at (1.1547,1.1547)

speed in direction 1=c

speed in direction 2 =c

Edited September 4, 2015 by david345

[studiot]

 

I still think that this is consistent with Einstein's definition of simultaneity and with the convention used to define it.

I don't see how it shows an improvement over Einstein.

 

I'm not the one arguing against Einstein.

 

Apologies this line in post#66 should have read

 

I have used the capital delta to emphasise that all these are segment properties, not point properties.

 

 

I have emphasised that this all these are segment properties, not point properties.

Edited September 4, 2015 by studiot

[md65536]

I've said about all I have to offer on the topic, and do not seem to have done anyone a stitch of good. So I'm bailing. Apologies for acting out of frustration.

[studiot]

Here is an 1893 one way determination without a single clock.

 

 

 

[Andromacus]

Here is an 1893 one way determination without a single clock.

 

lightspeed2.jpg

 

 

That's nice. Of course if you consider spatial distance as absolute you don't need clocks, just measuring length and translating it to meters per second, but for this translation whether you believe it or not you are using the standard synchronization, so basically what you are saying is that this is the only possible synchronization. That's fine with me.

In the train example the speed of light is the same in both directions for the person on the ground and the person on the train. Let us say for the person on the train the flashes happen at the same time. c is set to 1 and the person on the ground is moving at .5 c with respect to the person on the train.

 

person on train: flashes occur at (1,1) and (1,3) We are using (t,x) both flashes reach person on train at (2,2)

speed in direction 1= [math] d/t=c[/math]

speed in direction 2=c

 

Lorentz transformation from (t,x) to (t',x')

[math]\begin{vmatrix}t'\\x'\\\end{vmatrix}=\begin{vmatrix}\gamma&-\beta\gamma\\-\beta\gamma&\gamma\\\end{vmatrix}\begin{vmatrix}t\\x\\\end{vmatrix}[/math]

[math]\beta=.5[/math]

[math]\gamma=1.1547[/math]

 

person on ground: flashes occur at (.57735,.57735) and (-.57735,2.88675) both flashes reach person n train at (1.1547,1.1547)

speed in direction 1=c

speed in direction 2 =c

Right, that is a Lorentz transformation from the train's frame to the ground's frame that is indeed linear, but that is not what is usually shown in a Minkowski diagram. Your example shows a transform from an observer at rest with the train frame to an observer at rest with the ground frame. It doesn't show the relativity of simultaneity. It is just a rotation of the axes that shows a velocity boost. While a Minkowski diagram shows a transformation from the oberver at rest in the train to an observer moving with respecto him in the ground as perceived by the former, or viceversa.

 

Here the relativity of simultaneity is evident, while in your transform we just see a change of speed, the relative velocity between train and ground, but time is absolute, and no time dilation-length contraction are shown, so the physical properties associated with SR are not shown. It's a transformation between stationary observers that are disconnected, with their velocity boosted.The fact that one sees flashes simultaneous and the other not is just the logical consequence of rotating the coordinates.

 

So the key here is acknowledging that the formal linear transformations can only be made between stationary observers. While a Minkowski diagram that is used to show the physical consequences of Lorentz transformations is done bewtween stationary and moving observers and requires the use of skew noninertial frames for the moving observer.

Edited September 4, 2015 by Andromacus

[studiot]

I don't consider anything as 'absolute'.

 

But what I did say a few posts back was that any 'correction' due to relativity would have been dwarfed by experimental errors

(note how accurate Blondlot was compared to the modern value.)

 

The same applies to Romer's value, I mentioned before.

 

The relative speeds of Jupiter and the Earth make for negligable corrections compared to the few seconds of time he could measure to.

 

However, I have never really thought about your synchronisation issue in the past and I agree that it is a ticklish one, which was why I separated out your two issues and, again as I said before, I am still thinking about it.

 

The main point about it I have made so far is how to avoid it - by choosing experimental methods that make it insignificant or by choosing an experimental method where another means of measurement is available for comparison.

 

Edited September 4, 2015 by studiot

[david345]

1. Right, that is a Lorentz transformation from the train's frame to the ground's frame

 

2. but that is not what is usually shown in a Minkowski diagram.

 

3. but time is absolute

 

4. in your transform we just see a change of speed, the relative velocity between train and ground, but time is absolute, and no time dilation-length contraction are shown, so the physical properties associated with SR are not shown.

1. I could have just as easily done a Lorentz transformation from the ground frame to the train frame.

 

2. The Minkowski diagram is just a simplified picture intended to make relativity easier for the amateur to understand. If you are using these pictures to perform your "calculations" then that is your problem.

 

3. What are you talking about? It is you who is using an absolute time. You claim we should only use the frame of the train and warp all other frames to fit this preferred frame.

 

4. Can you even count?

person on train: flashes occur at (1,1) and (1,3) both flashes reach person on train at (2,2)

person on ground: flashes occur at (.57735,.57735) and (-.57735,2.88675) both flashes reach person in train at (1.1547,1.1547)

Edited September 5, 2015 by david345

[pzkpfw]

... So the key here is acknowledging that the formal linear transformations can only be made between stationary observers. While a Minkowski diagram that is used to show the physical consequences of Lorentz transformations is done bewtween stationary and moving observers and requires the use of skew noninertial frames for the moving observer.

What do you mean by "stationary observers". It seems you mean stationary with respect to each other. In that case, what relevance are any "formal linear transformations"?

Edited September 5, 2015 by pzkpfw

[Andromacus]

1. I could have just as easily done a Lorentz transformation from the ground frame to the train frame.

 

2. The Minkowski diagram is just a simplified picture intended to make relativity easier for the amateur to understand. If you are using these pictures to perform your "calculations" then that is your problem.

 

3. What are you talking about? It is you who is using an absolute time. You claim we should only use the frame of the train and warp all other frames to fit this preferred frame.

 

4. Can you even count?

person on train: flashes occur at (1,1) and (1,3) both flashes reach person on train at (2,2)

person on ground: flashes occur at (.57735,.57735) and (-.57735,2.88675) both flashes reach person in train at (1.1547,1.1547)

1. Of course, I didn't imply otherwise, was just describing your setting. Axes can rotate clockwise or counterclockwise.

2.No, I'm not. I'm saying actually that they can be deceiving as to what is really going on. In fact they are not mathematically consistent. It's specially deceiving for the amateur, one needs to know some sophisticated math to know what they are really about. But it is a fact that they are very often used to justify the physics of SR in a sort of automatic fast and easy way not only by amateurs but by relativists in general.

3. Actually I'm not claiming that. I'm not suggesting a preferred frame. I admit I should have qualified better what I meant by saying that time is absolute in this context. I was referring to the time in each stationary frame separately as written by you d/t=c, this formula is preserved by axes d(x) and t remaining orthogonal, fixing an absolute simultaneity for light speed.

4. I think you should have quoted the next sentence in that paragraph too. There I tried to justify my claim by explaining that a passive transformation, i.e. rotating the axes while keeping the relative speed line fixed, leads to the numbers you calculated in relation with the new position of the speed line with respect to the rotated coordinate system. Why do you think the mathematically inconsistent Minkowski diagrams are used all the time to ilustrate the physical consequences of SR? Because a 2D diagram of a Lorentz transformation like the one you showed the calculation of, just shows a euclidean rotation of coodinates.

 

 

This diagrams are taken from wikipedia page about passive transformations vs active transformations.

The transformation you described correspond to the image on the right. A minkowski diagram would correspond to the image on the left of an active transformation if it wasn't because the axes are no longer perpendicular in the transformed frame of the M.diagram, and the transformation is no longer linear so the change in speed is purely coordinate just like in the passive transformation.

 

Now this is not easy to understand if one doesn't have some clear notion of geometry. The usual explanation is that one must abandon the Euclidean concepts and adopt the minkowski geometry. The problem is all this is happening in a flat plane(R2) and there is no way out of this.

 

Some reason as if picturing a circle in a plane should make the geometry of the plane "circular", and as if graphing a hyperbola in R2 should transform the geometry of the plane, Well, it doesn't.

What do you mean by "stationary observers". It seems you mean stationary with respect to each other. In that case, what relevance are any "formal linear transformations"?

Well, it's a bit tricky, but all it means is the notion that rest is equivalent to uniform motion if one is to talk about inertial frames. So a linear transformation in flat space is just a coordinate rotation, that in Euclidean terms leaves points invariant. But, and this is the tricky part in SR it is not interpreted like this, the special pseudo inner product of minkowski space leaves room to interpret the Euclidean rotation as a velocity boost like shown in Minkowski diagrams.

This thread is about how and why this is mathematically inconsistent. The problem has always been that two realms are maintained: the formal plane of the calculation posted by david345 which is formally consistent but that simply expresses a passive coordinate rotation in wich one of the axes happens to be called ct, and a physical plane that is not consistent(Minkowski diagram using a nonlinear transform to represent a linear transform) that is used to justify the relativity of simultaneity effects and that when questioned sends the questioner to the formal plane and shows how the numbers fit.

Edited September 5, 2015 by Andromacus

[david345]

Right, that is a Lorentz transformation from the train's frame to the ground's frame that is indeed linear, but that is not what is usually shown in a Minkowski diagram. Your example shows a transform from an observer at rest with the train frame to an observer at rest with the ground frame. It doesn't show the relativity of simultaneity. It is just a rotation of the axes that shows a velocity boost.

The boost is a rotation free lorentz transformation. The lorentz transformation as a hyperbolic rotation is:

 

[math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math]

This transformation is also a linear transformation.

 

 

This diagrams are taken from wikipedia page about passive transformations vs active transformations.

The transformation you described correspond to the image on the right. A minkowski diagram would correspond to the image on the left of an active transformation if it wasn't because the axes are no longer perpendicular in the transformed frame of the M.diagram, and the transformation is no longer linear so the change in speed is purely coordinate just like in the passive transformation.

 

 

Nope, the the transformation corrosponding to the image on the right is:

[math]\begin{vmatrix}x'\\t'\\\end{vmatrix}=\begin{vmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta\\\end{vmatrix}\begin{vmatrix}x\\t\\\end{vmatrix}[/math]

The Lorentz transformation is the transformation which is depicted in a Minkowski diagram. The Lorentz transformation is a linear transformation.

 

It is becomming clear that you are just throwing around big words to create the illusion you know what you are talking about.

Edited September 5, 2015 by david345

[Andromacus]

The boost is a rotation free lorentz transformation. The lorentz transformation as a hyperbolic rotation is:

 

[math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math]

This transformation is also a linear transformation.

So? These random assertions you have copied somewhere are fine but not contradicting the quoted paragraph.

 

Nope, the the transformation corrosponding to the image on the right is:

[math]\begin{vmatrix}x'\\t'\\\end{vmatrix}=\begin{vmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta\\\end{vmatrix}\begin{vmatrix}x\\t\\\end{vmatrix}[/math]

The Lorentz transformation is the transformation which is depicted in a Minkowski diagram. The Lorentz transformation is a linear transformation.

You are repeating what I wrote, so why the "Nope"? On the right there is a rotation, and on the left something similar to a Minkowski diagram. Previously you said what you wrote had nothing to do with a Minkowski diagram, and that they were of no use: "The Minkowski diagram is just a simplified picture intended to make relativity easier for the amateur to understand. If you are using these pictures to perform your "calculations" then that is your problem" ,now you have discovered that Minkowski diagrams are used to represent Lorentz transformations. Well, congratulations.

It is becomming clear that you are just throwing around big words to create the illusion you know what you are talking about.

I couldn't care lees if I tried about your illusions. But if you are here to troll I would kindly ask you not to post again in this thread. Thanks.

Edited September 5, 2015 by Andromacus

[david345]

 

The boost is a rotation free lorentz transformation. The lorentz transformation as a hyperbolic rotation is:

 

[math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math]

This transformation is also a linear transformation.

So? These random assertions you have copied somewhere are fine but not contradicting the quoted paragraph.

 

Nope, the the transformation corrosponding to the image on the right is:

[math]\begin{vmatrix}x'\\t'\\\end{vmatrix}=\begin{vmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta\\\end{vmatrix}\begin{vmatrix}x\\t\\\end{vmatrix}[/math]

The Lorentz transformation is the transformation which is depicted in a Minkowski diagram. The Lorentz transformation is a linear transformation.

You are repeating what I wrote, so why the "Nope"? On the right there is a rotation, and on the left something similar to a Minkowski diagram. Previously you said what you wrote had nothing to do with a Minkowski diagram, and that they were of no use: "The Minkowski diagram is just a simplified picture intended to make relativity easier for the amateur to understand. If you are using these pictures to perform your "calculations" then that is your problem" ,now you have discovered that Minkowski diagrams are used to represent Lorentz transformations. Well, congratulations.

It is becomming clear that you are just throwing around big words to create the illusion you know what you are talking about.

I couldn't care lees if I tried about your illusions. But if you are here to troll I would kindly ask you not to post again in this thread. Thanks.

 

 

Nope The transformation I wrote was:

 

Lorentz transformation from (t,x) to (t',x')

[math]\begin{vmatrix}t'\\x'\\\end{vmatrix}=\begin{vmatrix}\gamma&-\beta\gamma\\-\beta\gamma&\gamma\\\end{vmatrix}\begin{vmatrix}t\\x\\\end{vmatrix}[/math]

 

 

 

This is not a rotation. Contradicting your statement " It is just a rotation of the axes that shows a velocity boost. "

One can represent the lorentz transformation as a hyperbolic rotation by setting:

[math]\gamma = cosh\phi[/math]

[math]\beta = tanh\phi[/math]

 

You stated " The transformation you described correspond to the image on the right." The transformation I described does not correspond to the picture on the right. The picture is represented by the transformation:

[math]\begin{vmatrix}x'\\t'\\\end{vmatrix}=\begin{vmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta\\\end{vmatrix}\begin{vmatrix}x\\t\\\end{vmatrix}[/math]

 

I stated the Minkowski diagrams are simplified pictures used to help amateurs visualize relativity. They are no replacement for the mathematical equations. The diagrams you posted are not even the most accurate versions. Instead you should use something like this.

 

What's next? Will you argue that general relativity is wrong because the below image is contradictory.

It has become clear who the troll is.

[Andromacus]

To clarify, the 2 transformations shown above, the passive and the active are rotations, basically in flat space with cartesian/inertial coordinates active and passive transformations are equivalent, from this comes that the distinction between covariant and contravariant vectors is not necessary in this case. This should be also the case for Lorentz transformations that use inertial coordinates since Minkowski space is flat. However the indefinite signature of the pseudo inner product is supposed to justify the distinction between the active and the passive rotation. So what is called a boost or rotation free Lorentz transformation is just a passive rotation of the axes that includes the time axis(so not a real or active rotation).

 

Nope The transformation I wrote was:

This is not a rotation. Contradicting your statement " It is just a rotation of the axes that shows a velocity boost. "

You stated " The transformation you described correspond to the image on the right." The transformation I described does not correspond to the picture on the right. The picture is represented by the transformation:

[math]\begin{vmatrix}x'\\t'\\\end{vmatrix}=\begin{vmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta\\\end{vmatrix}\begin{vmatrix}x\\t\\\end{vmatrix}[/math]

 

You are deliberately misinterpreting my words, I was evidently not talking about the regular 3-space rotation when I say rotation.

 

There is no point discussing about science with someone with your attitude. I am asking you to leave and you come back insulting. I pity you but I'm not here to deal with people with such evident personal issues. Take them somewhere else.

I ask you again not to come back unless your attitude is constructive

[david345]

Fine I'm tired of responding to your crackpot garbage.

[Andromacus]

The rotations I've been talking about are obviously those including the time coordinate,either active or passive rotations by an imaginary angle. Since hyperbolic rotations have even been mentioned twice wich I took to mean it was understood I'm still a bit perplexed at the reaction.

[pzkpfw]

Well, it's a bit tricky, but all it means is the notion that rest is equivalent to uniform motion if one is to talk about inertial frames. So a linear transformation in flat space is just a coordinate rotation, that in Euclidean terms leaves points invariant. But, and this is the tricky part in SR it is not interpreted like this, the special pseudo inner product of minkowski space leaves room to interpret the Euclidean rotation as a velocity boost like shown in Minkowski diagrams.

This thread is about how and why this is mathematically inconsistent. The problem has always been that two realms are maintained: the formal plane of the calculation posted by david345 which is formally consistent but that simply expresses a passive coordinate rotation in wich one of the axes happens to be called ct, and a physical plane that is not consistent(Minkowski diagram using a nonlinear transform to represent a linear transform) that is used to justify the relativity of simultaneity effects and that when questioned sends the questioner to the formal plane and shows how the numbers fit.

That is such wordy salad, I can see it only as evasion. There are nuggets of truth in there, but you've not answered the question.

[david345]

It turns out this idea is quite popular among young earth creationists. No surprise here.

 

http://rationalwiki.org/wiki/Anisotropic_synchrony_convention

[david345]

Enough of this farce. Can you back up your crackpot claims with any citations at all?

Apparently you forgot to cite the paper you posted on vixra back in 2013.

http://www.scienceforums.net/topic/80668-sr-with-apparent-simultaneity/

This is just getting weird.

[Andromacus]

There are nuggets of truth in there, but you've not answered the question.

Ok, let me try again then.

1.It's what I'm calling a passive Lorentz transformation, in which we are just looking at the same event, just from different point of view, i.e. from different coordinate systems that are rotated by an imaginary angle. So we are alternating the rest train point of view coordinates and the rest embankment point of view coordinates, so both use perpendicular time and distance coordinate axes. We can do this because velocity is relative and in relativity we can arbitrarily consider each observer at rest with respect to the other. I hope we can agree so far. This is the sense in wich both observers can be considered stationary in the passive Lorentz trasformation.

 

2.There is a problem when trying to show such passive transformation with a diagram, because both coordinate systems are orthogonal and they would superpose unless they were represented like the right diagram in my analogy with active and passive real rotations, but then somebody could say we are trying to represent imaginary angle rotations, so why use real angle rotations of the axes. So it would seem we are stuck at representing either one of the observers perspective of the event but not both observers at once in a diagram.

 

3.What is usually done to represent a Lorentz transformation is using the regular Minkowski diagram wich in my analogy could be related to an active rotation (but please not literally, I'm just drawing analogies between imaginary and real angle rotations). To be able to do this with a hyperbolic rotation one must recurr to skewing the grid to express the different velocities of the moving observer with respect to the observer at rest. So we can represent the observer at rest with orthogonal axes and the moving one with oblique axes and the angle between them as the velocity boost.

 

4.There is a problem if one takes this representation literally, we are either not talking about inertial frames anymore or we are no longer representing a linear transformation. So you'll reply(if you have followed me this far): Don't take it literally. Well, ok, what's your alternative?

 

5. My point, (and I'm not asking you anymore to agree with me here, just presenting my view) is that Einstein did indeed find this skewing of the grid(although he only did it algebraically, not geometrically at the time) as his solution to what he presented as the apparent irreconcibility of his first and second postulates.

I'm saying that mathemaically it is incompatible with keeping observers inertial.

 

 

Please critizice each of the points specifically all you want, but don't just dismiss it globally as word salad.

Edited September 6, 2015 by Andromacus

[swansont]

Ok, let me try again then.

1.It's what I'm calling a passive Lorentz transformation, in which we are just looking at the same event, just from different point of view, i.e. from different coordinate systems that are rotated by an imaginary angle. So we are alternating the rest train point of view coordinates and the rest embankment point of view coordinates, so both use perpendicular time and distance coordinate axes. We can do this because velocity is relative and in relativity we can arbitrarily consider each observer at rest with respect to the other. I hope we can agree so far. This is the sense in wich both observers can be considered stationary in the passive Lorentz trasformation.

 

You'r making up new terminology, which is bad.

 

If it's a rotation, then call it a rotation. Rotating a coordinate system does not require a Lorentz transformation, so you're just inviting confusion by calling this a passive Lorentz transformation.

[Andromacus]

 

You'r making up new terminology, which is bad.

 

If it's a rotation, then call it a rotation. Rotating a coordinate system does not require a Lorentz transformation, so you're just inviting confusion by calling this a passive Lorentz transformation.

This is fair criticism to some extent, thanks. Everyone here talks as if if they were such relativity experts that I thought talking about lorentz transfromations as (hyperbolic)rotations wouldn't invite confusion.

However, I'm not making up new terminology, passive Lorentz transformations is terminology in use, a quick internet search gives this example: http://physics.stackexchange.com/questions/135112/a-few-questions-on-passive-vs-active-lorentz-transformations

 

I would also appreciate a more substantial criticism of the content besides the form(wich is fine)

Edited September 6, 2015 by Andromacus

[swansont]

Upon rereading, I see I missed where you said "imaginary".

 

I withdraw my comment.

[pzkpfw]

Ok, let me try again then.

1.It's what I'm calling a passive Lorentz transformation, in which we are just looking at the same event, just from different point of view, i.e. from different coordinate systems that are rotated by an imaginary angle. So we are alternating the rest train point of view coordinates and the rest embankment point of view coordinates, so both use perpendicular time and distance coordinate axes. We can do this because velocity is relative and in relativity we can arbitrarily consider each observer at rest with respect to the other. I hope we can agree so far. This is the sense in wich both observers can be considered stationary in the passive Lorentz trasformation.

 

...

Sure, it's a simple basic fact of relativity that any observer can consider themselves as at rest. e.g. The train observer sees themselves as at rest and the embankment observer is moving. And vice versa. But that still doesn't make sense out of the part of your post I was commenting on.

 

You wrote "... formal linear transformations can only be made between stationary observers ...".

 

{ Transformations } - { between } - { stationary observers }

 

Yes, both observers can consider themselves as stationary, but there's no outside view where both are stationary (if there's any relative motion), so it makes no sense to write "between stationary observers". You may as well have just written "... formal linear transformations can only be made between observers ..." ... in which case, I've lost the point you were originally trying to make back then. Why "can only"? By "stationary observers" here, do you really mean "inertial observers" (i.e. trying to keep acceleration out of it?)

Edited September 6, 2015 by pzkpfw

[Andromacus]

Sure, it's a simple basic fact of relativity that any observer can consider themselves as at rest. e.g. The train observer sees themselves as at rest and the embankment observer is moving. And vice versa. But that still doesn't make sense out of the part of your post I was commenting on.

 

You wrote "... formal linear transformations can only be made between stationary observers ...".

 

{ Transformations } - { between } - { stationary observers }

 

Yes, both observers can consider themselves as stationary, but there's no outside view where both are stationary (if there's any relative motion), so it makes no sense to write "between stationary observers". You may as well have just written "... formal linear transformations can only be made between observers ..." ... in which case, I've lost the point you were originally trying to make back then. Why "can only"? By "stationary observers" here, do you really mean "inertial observers" (i.e. trying to keep acceleration out of it?)

Yes I meant inertial observers and also I was trying to make a distinction between the passive transformation and the Minkowski diagrams where one observer is stationary while the other is moving. Probably not the most fortunate phrase to express what I was trying to convey.

 

Precisely the "there's no outside view where both are stationary" issue is discussed in the next points 2. and 3.

Edited September 6, 2015 by Andromacus

[Andromacus]

So my issue in 4. with the Minkowski diagram is a problem with the representation only, since the boost in one direction is obviously a linear transformation, because it may lead to the wrong physical conclusions in higher dimensions.