Time Dilation Formula: Clarifying Confusion

[jtbell]

It depends on exactly what one means by $\Delta t$ and $\Delta t^{\prime}$.

 

[Rasalhague]

So would I be right in deducing that (2) is, in fact, the correct formula??

Sorry if I went off topic there...

As I said in post #52, I only used the prime symbol in those examples to indicate output, i.e. the value that we're using the formula to calculate. Here Einstein uses a different convention. In this particular formula, he labels his input t' (that's the given value, one second) and his output t (that's the value he's using the formula to compute). So the formula he uses is of type (1) in my scheme.

It's not a matter of either being correct or incorrect, as such, and (as we'd expect!) Einstein uses the correct formula for his purpose. They're each appropriate to a different task. The only disagreement is over which of them would be called time dilation (and which its inverse), and why... Does Einstein use the term "time dilation" (Zeitdilation)? If not, we can't be sure whether he'd label this particular formula "the time-dilation formula". That said, he does characterise his result as "a somewhat larger time", so like the majority of the writers I looked at (Adams, Freund, Lerner, Petkov, Schröder, Weinert, Taylor & Wheeler, Tipler & Mosca), presumably Einstein would regard time dilation as an operation that takes a smaller interval of time (a smaller total of seconds) and produces a larger interval of time (a larger total of seconds). The result of Einstein's calculation is a larger number. The above-named authors called the formula that produces a larger number "time dilation", and they called its result (the larger number) "dilated time".

This contasts with Lawden who presents the inverse of the formula Einstein uses here and calls that time dilation, and it contrasts with Anthony who called the smaller total produced by $$\Delta t' = \gamma^{-1} \; \Delta t$$ "dilated time".

 

[Grimble]

Should we not take Einstein's own usage as the convention in this case?

He referred to the stationary system as K and the moving system as K'; and his time t was that of the stationary system and t' that of the moving system as transformed by the Lorentz equations.

 

[Grimble]

...if the number of units can in one view increase and in the other decrease, in one sense the moving clock reads more time has passed and in the other that less time has passed, for do we not reckon time by the number of units of time passing rather than by the size of them?

So in what way does the clock slow?

Another interesting consideration is that however one measures it the total duration of whatever we are measuring is the same, moving or not.
The number of seconds multiplied by the length of one second gives the same total whether it is proper time or co-ordinate time. The difference is, that the unit of measurement changes: take the muon experiment referred to earlier where we have 2.2 microseconds proper time and approximately 65 microseconds co-ordinate time and the conversion is made applying the Lorentz factor which was 29.4

So in which way is it slowing?

And surely, whichever way we calculate it, the co-ordinate seconds have to be smaller than proper seconds, in the same way that co-ordinate metres are smaller than proper metres.

For how else can the speed of light in the moving system - measured in co-ordinate units from the stationary system - still be c?

If the transformed lengths are contracted and the times are dilated how can $$c = \frac{d}{t} = \frac {d^'}{t^'}$$

Grimble.

 

[Rasalhague]

Should we not take Einstein's own usage as the convention in this case?

He referred to the stationary system as K and the moving system as K'; and his time t was that of the stationary system and t' that of the moving system as transformed by the Lorentz equations.

In section 11 of the book you linked to, Einstein uses t' to denote a time which results from applying the Lorentz transformation to a time labelled t. But in section 12, he uses t to denote a time which results from applying the Lorentz transformation to a time labelled t'. So if there is a convention here, it's not defined relative to the Lorentz transformation, which fortunately does have a conventional form.

http://www.bartleby.com/173/11.html
http://www.bartleby.com/173/12.html

Einstein doesn't use the term "moving system" in either of these sections, as far as I can see. (I haven't read the whole book, so maybe it's used elsewhere.) But I don't know how the terms "moving system" and "stationary system" could be used to determine a convention for which frame to call K', since these terms are just as arbitrary, given that each system is moving relative to the other, and an observer at rest in either system will observe the same effect (slowness) in a clock at rest in the other system.

 

[Grimble]

In section 11 of the book you linked to, Einstein uses t' to denote a time which results from applying the Lorentz transformation to a time labelled t. But in section 12, he uses t to denote a time which results from applying the Lorentz transformation to a time labelled t'. So if there is a convention here, it's not defined relative to the Lorentz transformation, which fortunately does have a conventional form.

http://www.bartleby.com/173/11.html
http://www.bartleby.com/173/12.html

Einstein doesn't use the term "moving system" in either of these sections, as far as I can see. (I haven't read the whole book, so maybe it's used elsewhere.) But I don't know how the terms "moving system" and "stationary system" could be used to determine a convention for which frame to call K', since these terms are just as arbitrary, given that each system is moving relative to the other, and an observer at rest in either system will observe the same effect (slowness) in a clock at rest in the other system.

But in chapter 11 he writes:

A co-ordinate system K then corresponds to the embankment, and a co-ordinate system K' to the train.

and K' is the moving system in Figure 2.

Unfortunately it is confusing until we realize that what he is saying with relation to the Lorentz transformations is that the time in an inertial frame of reference (IFoR) is proper time and we apply the Lorentz transformation to convert it to co-ordinate time, the time perceived in another inertial frame of reference moving at a constant velocity with respect to the first.
If we then compare the resultant co-ordinate time with with the proper time in the 2nd IFoR we, not surprisingly, find that the relationship is the Lorentz Factor.
I believe we can change labels like system A and system B to either IFoR but, because they are fundamental to the Lorentz transformation equations, we need to have a solid convention for what they are ferring to.

 

[Rasalhague]

But in chapter 11 he writes:

"A co-ordinate system K then corresponds to the embankment, and a co-ordinate system K' to the train."

and K' is the moving system in Figure 2.

Okay, so K' is shown moving in Fig. 2, while K is at rest with respect to the page. This K' is what's known as the "rest frame" of the train (that's to say the frame where the train is at rest). In Chapter 12, K' is the label used for the rest frame of the clock. The equations of the Lorentz transformation in Chapter 11 tell us how to convert intervals of time and distances of space, as measured with respect to K, the embankment's rest frame, into the corresponding times and distances with respect to K', the train's rest frame.

It's natural for us to think of the embankment as stationary and the train as moving, but this is an artifact of the scenario. Taylor and Wheeler take a similar approach; they begin by defining one coordinate system as "the lab(oratory) frame" and another as "the rocket frame", meaning the rest frame of a laboratory and the rest frame of a rocket which they use in their thought experiments. Similarly with muons, it's natural for us to think of the muon's rest frame as moving relative to the earth. But in all of these scanarios, the effects due to special relativity are entirely symmetric. No frame is moving or stationary in an absolute sense, only relative to something.

Unfortunately it is confusing until we realize that what he is saying with relation to the Lorentz transformations is that the time in an inertial frame of reference (IFoR) is proper time and we apply the Lorentz transformation to convert it to co-ordinate time, the time perceived in another inertial frame of reference moving at a constant velocity with respect to the first.
If we then compare the resultant co-ordinate time with with the proper time in the 2nd IFoR we, not surprisingly, find that the relationship is the Lorentz Factor.
I believe we can change labels like system A and system B to either IFoR but, because they are fundamental to the Lorentz transformation equations, we need to have a solid convention for what they are ferring to.

I've been struggling with these issues myself as I've been learning about sepcial relativity, and I certainly don't have all the answers... I tend to prefer standard, general, abstract terms that will work in any scenario, and which don't hide the symmetry (rest frame, proper time, spacetime interval, etc.). But some people prefer to begin with more concrete sounding names like "laboratory frame" and "rocket frame" (those are Taylor and Wheeler's examples), or you could think of the train's rest frame and the embankment's rest frame in Einstein's example.

Proper time is a good term. It's traditionally represented by the Greek letter $$\tau$$ (tau). In special relativity, proper time is a timelike spacetime interval between two events. It's invariant, which means that it's the same in all inertial reference frames. It's equal to the coordinate time in the unique inertial frame of reference where the events happen in the same place as each other. It's unambiguous as long as we define which pair of events we're talking about the proper time between. In Einstein's example, both events lie on the worldline of a certain clock. The first event is the clock showing 0 seconds, and the other event is the clock showing 1 seconds. These events happen in the same place in the clock's rest frame (where, by definitio, the clock is at rest), so the coordinate time in that frame is equal to the proper time between these events. In any other frame, the coordinate time will be greater than the proper time. We find the coordinate time between these events using $$t_{coordinate} = \gamma \; \tau$$, which is just the time equation from the Lotentz transformation with x = 0.

 

[Rasalhague]

...if the number of units can in one view increase and in the other decrease, in one sense the moving clock reads more time has passed and in the other that less time has passed, for do we not reckon time by the number of units of time passing rather than by the size of them?

So in what way does the clock slow?

Yes, time is measured by the number of units, which is why it seemed more natural to me to call time dilation the process whereby a larger number of units is derived from a smaller one.

A "moving" clock runs slow in the sense that it shows less time passing (a smaller number of units) than a "stationary" clock. To an observer who considers themselves stationary, a clock moving relative to this observer will show a smaller number of units as having passed in any given interval than a clock which is stationary relative to the obsever.

The same scenario from a more symmetrical perspective: imagine two clocks, A and B, moving relative to each other at some constant velocity. The proper time between two events on clock A's worldline is always less than the proper time between two events on the worldline of clock B if the following two conditions hold. (1) The first event on clock A's worldline is simultaneous in clock B's rest frame with the first event on clock B's worldline; (2) the second event on clock A's worldline is simultaneous in clock B's rest frame with the second event on clock B's worldline.

Since A and B are arbitrary labels, this will still be true if you swap them.

Events are points in spacetime. In this example, they represent a clock showing a particular time. An object's worldline is its trajectory (the path it takes) through spacetime. In special relativity, which deals with objects moving constant velocity relative to intertial reference frames, worldlines are straight lines.

Another name for the longer of these two intervals of time (the proper time shown by clock B in this example) is the coordinate time in clock B's rest frame of the proper time between the stated pair of events on clock A's worldline.

Best read in with one eye on a spacetime diagram!

 

[Rasalhague]

Another interesting consideration is that however one measures it the total duration of whatever we are measuring is the same, moving or not.
The number of seconds multiplied by the length of one second gives the same total whether it is proper time or co-ordinate time. The difference is, that the unit of measurement changes: take the muon experiment referred to earlier where we have 2.2 microseconds proper time and approximately 65 microseconds co-ordinate time and the conversion is made applying the Lorentz factor which was 29.4

So in which way is it slowing?

I don't understand your first point here. We can conceive of the calculation as making the total bigger, or equivalently as making the units smaller, but not both, because that would be like multiplying by gamma squared, wouldn't it? Or if you made the total of one clock bigger, and the units of the clock you're comparing it to smaller, then that would be like making them both bigger, or both smaller, i.e. multiplying by gamma and dividing by gamma, i.e. no change. But there is a change, the calculation converts 2.2 into 65.

We're talking about the interval of time between two events on the muon's worldline, its creation and its annihilation. This interval is the muon's lifetime. The muon lasts 2.2 microseconds in its own rest frame and 65 microseconds in the rest frame of the lab. The time between events that happen in the same place is the proper time between them (a timelike spacetime interval). Since the muon's birth and death happen in the same place in the muon's rest frame, the proper time between them is 2.2 microseconds. In the lab's rest frame, the coordinate time of this interval is 2.2 * 29.4 = approx. 65 microseconds. If we think of the muon as a clock, we can say informally that the muon is running slow in comparison to a clock at rest in the lab's rest frame in the sense that the muon only counts to 2.2 microseconds while the lab clock counts to 65. From the perspective of an observer at rest with respect to the lab, a 2.2-microsecond muon is slow to disappear; it lasts for 65 seconds.

And surely, whichever way we calculate it, the co-ordinate seconds have to be smaller than proper seconds, in the same way that co-ordinate metres are smaller than proper metres.

For how else can the speed of light in the moving system - measured in co-ordinate units from the stationary system - still be c?

If the transformed lengths are contracted and the times are dilated how can $$c = \frac{d}{t} = \frac {d^'}{t^'}$$

Grimble.

You're right. If we conceptualise coordinate seconds as being smaller than proper seconds, and coordinate meters as smaller than proper meters, there will be more of either corresponding to a given proper time (timelike spacetime interval) or a given proper distance (spacelike spacetime interval). Time dilation, as most of the textbooks understand it, refers to the calculation of the coordinate time of the proper time between a given pair of events. To calculate this, we multiply the proper time by gamma. Likewise, to find the coordinate distance of the proper distance between two events, we multiply the proper distance by gamma. This latter procedure is not the same thing as "length contraction", for which the inverse equation is required.

 

[Grimble]

I don't understand your first point here. We can conceive of the calculation as making the total bigger, or equivalently as making the units smaller, but not both, because that would be like multiplying by gamma squared, wouldn't it? Or if you made the total of one clock bigger, and the units of the clock you're comparing it to smaller, then that would be like making them both bigger, or both smaller, i.e. multiplying by gamma and dividing by gamma, i.e. no change. But there is a change, the calculation converts 2.2 into 65.

I'm sorry, I haven't explained it well but in the following quote you say that Freund et al. take "dilated time" (co-ordinate units?) to mean an expanded total (of reduced units).

My point is just that Freund, like Adams, Lerner, Petkov and Schröder, takes "dilated time" to mean an expanded total (of reduced units). So for all of these authors, dilation seems to refer to the quantity of units, the total, rather than--as I thought you originally suggested--the size of individual units. If these authors had taken dilation to refer to the size of units, then surely they'd have used the label "dilated time" for the interval made up of a reduced quantity of these dilated units, wouldn't they?

And what I am saying is that if the co-ordinate time is an expanded number of reduced units, and if the 'expansion' and 'reduction' are both according to the Lorentz factor, then the only conclusion is that the total duration remains constant. It is precisely the number of units and the size of the units that change.

We're talking about the interval of time between two events on the muon's worldline, its creation and its annihilation. This interval is the muon's lifetime. The muon lasts 2.2 microseconds in its own rest frame and 65 microseconds in the rest frame of the lab. The time between events that happen in the same place is the proper time between them (a timelike spacetime interval). Since the muon's birth and death happen in the same place in the muon's rest frame, the proper time between them is 2.2 microseconds. In the lab's rest frame, the coordinate time of this interval is 2.2 * 29.4 = approx. 65 microseconds. If we think of the muon as a clock, we can say informally that the muon is running slow in comparison to a clock at rest in the lab's rest frame in the sense that the muon only counts to 2.2 microseconds while the lab clock counts to 65. From the perspective of an observer at rest with respect to the lab, a 2.2-microsecond muon is slow to disappear; it lasts for 65 seconds.

Yes, the 'clock' by which the muon's lifetime is measured is the muon's lifetime and that is occurring in the muon's rest frame(2.2 microseconds - proper time); the 'lab' time here (approx. 65 microseconds) is that time 'observed' from the "lab's" rest frame; and so it is in co-ordinate time; i.e. it is in co-ordinate microseconds not proper seconds.
But wherever and however it is measured, the duration of the muon's lifetime is and can only be 2.2 microseconds proper time, that is surely a physical constant. We are only discussing how that one fixed interval is measured in different circumstances.

So 2.2 proper microseconds are equal in duration to approx. 65 co-ordinate microseconds?


You have been a tremendous help so far, my friend,(if I may call you that?) and I see much clearer now and appreciate your patience, and following your suggestion I am putting all that I have learned into a new diagram that I hope will pull all these different threads together. I should be able to post it later today - it is certainly helping me to see things more clearly and I hope I am not too presumptious if I ask you to view it?

Thanks, once again for your help and guidance.

Grimble

 

[Rasalhague]

I'm sorry, I haven't explained it well but in the following quote you say that Freund et al. take "dilated time" (co-ordinate units?) to mean an expanded total (of reduced units).

I see what you mean. I got confused and contradicted myself. What I wrote above is one way of expressing the idea that these two time intervals (proper time and coordinate time) represent the same timelike spacetime interval. I suppose I was imagining proper time as being like one ruler, and the coordinate time as being like a previously identical ruler that's been squashed so that its units are closer together (smaller) which means that more of them will fit into the length of the original ruler representing proper time. Expanded (dilated) total of reduced (shrunk, contracted) units. When I objected to this idea in the later post, I think I must have had a different metaphor in mind. Sorry about that.

And what I am saying is that if the co-ordinate time is an expanded number of reduced units, and if the 'expansion' and 'reduction' are both according to the Lorentz factor, then the only conclusion is that the total duration remains constant. It is precisely the number of units and the size of the units that change.

That "sameness" is the way this particular view expresses the idea that the coordinate time here is the time component of the same spacetime separation (between two events) as the proper time is the invariant interval of.

Another common way of thinking about it is in terms of vectors in spacetime. The separation is a displacement vector in spacetime. Its interval is its magnitude. The magnitude ||Q - P|| of a vector in Euclidean space from point P to point Q is invariant when you rotate the coordinate axes, even though its x, y and z components change. Similarly, the interval of a displacement vector between two points (called events) in Minkowski spacetime is invariant when you switch between frames moving at a constant velocity relative to each other.

If there exists some frame in which the events happen in the same place (which is to say they happen along the possible worldline of some object), it's possible to represent the separation between them as a vector parallel to the time axis (vertical according to the usual convention). In this frame, the vector has only a time component and no spatial component, so its time component (its coordinate time) is equal to its proper time. This is like the way that a vector in a cartesian coordinate system, in Euclidean space, which is parallel to the y-axis has only a y component, and so its y component is equal to its magnitude (length). If we rotate our cartesian coordinate system in the xy plane, this vector will have some x component too, and so it's y component will be less than its magnitude so that the magnitude is still equal to sqrt(x^2 + y^2).

In Minkowski spacetime, the principle is similar, except that the "rotation" is hyperbolic: it takes the tip of the vector along a hyperbola rather than a circlular path. The faster the new frame is moving relative to the original frame where the events happened in the same place, the further the vector slopes, approaching the lightcone in the limit as the speed approaches c. This is because the invariant interval is equal to the absolute value of sqrt(t^2 - x^2), or in all 4 dimensions |sqrt(t^2 - x^2 - y^2 - z^2)|. A consequence of this is that, as the separation vector slopes and acquires a spatial component, its time component (coordinate time) must increase for the interval to remain invariant (in contrast to the y component of the cartesian vector which had to decrease). Hence coordinate time is always greater than proper time in any frame where the events happen in different locations.

Yes, the 'clock' by which the muon's lifetime is measured is the muon's lifetime and that is occurring in the muon's rest frame(2.2 microseconds - proper time); the 'lab' time here (approx. 65 microseconds) is that time 'observed' from the "lab's" rest frame; and so it is in co-ordinate time; i.e. it is in co-ordinate microseconds not proper seconds.
But wherever and however it is measured, the duration of the muon's lifetime is and can only be 2.2 microseconds proper time, that is surely a physical constant. We are only discussing how that one fixed interval is measured in different circumstances.

So 2.2 proper microseconds are equal in duration to approx. 65 co-ordinate microseconds?

That's right. Another scenario involving muons which is often used in introductions to special relativity is the example of muons created in the upper atmosphere by cosmic rays. A muon travels towards the earth. In the Earth's rest frame, the muon lasts a long time and travels a great distance, all the way from the upper atmosphere to the ground. In the muon's rest frame, the muon only lasts a couple of microseconds, but the Earth's atmosphere is length-contracted. So the speed with which the ground approaches in the muon in the muon's rest frame is the same as the speed with which the muon approaches the ground in the Earth's rest frame.

You have been a tremendous help so far, my friend,(if I may call you that?) and I see much clearer now and appreciate your patience, and following your suggestion I am putting all that I have learned into a new diagram that I hope will pull all these different threads together. I should be able to post it later today - it is certainly helping me to see things more clearly and I hope I am not too presumptious if I ask you to view it?

Thanks, once again for your help and guidance.

Grimble

You're welcome. I hope I haven't added to your confusion ;-)

 

[Grimble]

Hello again, my friend; well here it is:
http://img41.imageshack.us/img41/5448/specialrelativitydiagra.jpg
A Special Relativity diagram to demonstrate the relationship between two Inertial Frames of Reference moving with a constant relative velocity.
A and B represent a single axis (time or length) of Minkowski spacetime for each of the two IFoRs. They are drawn against a common background representing Proper Units
An observer at rest within each IFoR will be experiencing proper units (length and time) within that system; as shewn by the horizontal lines, labelled A and B.
But from each IFoR, the other frame's axis -- rotated according to their relative velocity -- will be reckoned in co-ordinate units, as shewn by the perpendicular projections from the coloured diagonals onto the observer's own axis.
The diagram is drawn to scale to represent two IFoRs with a constant relative velocity = 0.6c, giving $\gamma = 1.25$ and $\frac{1}{\gamma} = 0.8$

From this we can see exactly what Einstein was saying in http://www.bartleby.com/173/12.html" , when he writes:

But the metre-rod is moving with the velocity v relative to K. It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity v is

of a metre.

For the metre rod moving with the velocity 0.6c relative to B would be represented in the diagram by the number 1 on the red diagonal and we can see the projection onto B's x-axis (as it would be in this scenario) where it would be the green 1 co-ordinate unit.
And this agrees with $${x^'} = \frac {x}{\gamma}$$
i.e. ${x^'} = 0.8x$

Similarly, in the second part of that same chapter, Einstein writes:

**Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but

seconds, i.e. a somewhat larger time.

And again we can see just how this works, for this time he is converting the time from the observer's frame, $t$ (proper time units) into the observed time ${t^'}$ (co-ordinate time units) which would be to take the blue, 1 proper unit and project it upwards onto the red diagonal line or, indeed, one could read it off the green co-ordinate scale on B's axis.

Not surprisingly this agrees with Einstein's own equation:
$$t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ or
$$t = \gamma({t^'}) = 1.25$$ co-ordinate units

I am fairly certain that this diagram meets all the conditions, references and relationships between the elements that compose SR as described in this thread.

For instance, the co-ordinate units are greater in number but reduced in size.

The principal problem that is brought to light here is that there are far more elements than at first appear.
Consider if you will:

$1)$ We start with a solitary IFoR where the measurements are all, by definition, in proper units.

$2)$ We add a 2nd IFoR moving at a constant velocity with respect to the first: both are measured in proper units and their times are identical and synchronous.

$3)$ They then observe one another and upon doing so we find that the observed frames are rotated with respect to their observers, as shewn, but their units are still proper units.

$4)$ When observing the rotated frames, their proper units are projected vertically onto the observer's frame of reference, being there-by transformed into co-ordinate units.

$5)$ So we have the axis of each IFoR and its rotation, both measured in Proper units and its projection onto the other's axis measured in co-ordinate units.

$6)$ And as all this is matched reciprocally by the other IFoR we have this duplicated giving 4 measures in proper units and two in co-ordinate units.

Is it any wonder that we become confused when trying to deal with this using only primed and unprimed symbols?
And this is with it all reduced to two dimensions...


This exercise has certainly helped me to understand how it all fits together – I only hope that I have understood it correctly?

Many thanks, once again to Rasalhague, and Matheinste whose patience and explanations have been a great help.

 

[Rasalhague]

Looks good to me!

The conventional Minkowski spacetime diagram takes a slightly different approach to representing the same information. Here are a couple of examples.

The top diagram in my first attachment, the attachment labelled "parallels", shows which lines are parallel with which. The lower diagram in the same attachment shows the spacetime intervals between the origin and various events (represented by points). As you can see, in this kind of diagram, the intervals aren't exactly represented as the Euclidean distance between points since the hypotenuse of each of those right triangles corresponds to a shorter spacetime interval than the edge that lies along the axis, contrary to what we'd expect from Pythagoras. This is because the equation for the spacetime interval between two events is that of a hyperbola

$$||\mathbf{E}_{2} - \mathbf{E}_{1}|| = \sqrt[]{|\Delta t^{2} - \Delta x^{2}|}$$

rather than a circle like the equation that gives the distance between two points in Euclidean space:

$$||\mathbf{P}_{2} - \mathbf{P}_{1}|| = \sqrt[]{\Delta y^{2} + \Delta x^{2}}$$

One frame is represented with the vertical time axis and horizontal x axis. The axes of the other frame are inlined towards each other. The faster the frame shown with the inclined axes is traveling relative to the frame shown with orthogonal axes, the closer its axes will be to the dotted line at 45 degrees which represents the path of a photon travelling, of course, at c. The angle $$\beta$$ represents the speed of either frame relative to the other, as a fraction of the speed of light: v/c.

Any vertical line is the potential worldline of an object at rest in the frame with orthogonal axes; events on those lines happen in the same place in that frame. Any horizontal line is a line of simultaneity in the frame with orthogonal axes; all events on that line will be happen at the same time as each other in that frame.

Similarly, lines parallel to the time axis of the frame with oblique axes represent potential worldlines of objects at rest in the frame with oblique axes; events on those lines happen in the same place in that frame. Lines parallel to the x-axis of the frame with oblique axes are lines of simultaneity in that frame; all events on those lines happen at the same time as each other in that frame.

There's nothing special about either frame. The only difference is that they're moving in opposite directions relative to each other. So we could just as well change our diagram and show the oblique frame with orthogonal axes, and the orthogonal frame with oblique axes. The only difference would be that the axes of our new oblique frame would have a negative slope because its going in the opposite direction to the other frame. So the oblique axes would appear in the 2nd and 4th quadrants, rather than the 1st and 3rd.

My other diagram, the one labelled "hyperbola" is meant to show how this relates to hyperbolic geometry. If we draw a curve through all points on a Minkowski diagram which stand for events with a timelike interval of 1 from the origin, the curve is one branch of a unit hyperbola. Consider one of these events, the one that lies on the axis labelled tau in this diagram. The proper time between the origin and this event is 1. The coordinate time of this interval with respect to the frame represented by the axes labelled here t and x is

$$\gamma = \frac{1}{\sqrt[]{1 - \beta^{2}}} = cosh\left(\phi \right)$$

where $$\phi$$ is the "hyperbolic angle", the input of the hyperbolic trigonometric functions, which corresponds geometrically to the area shaded blue under the hyperbola here. Cosh, sinh and tanh are the hyperbolic cosine, sine and tangent respectively.

$$sinh\left(\phi \right) = \gamma \beta$$

and

$$tanh\left(\phi \right) = \frac{sinh\left(\phi \right)}{cosh\left(\phi \right)} = \beta$$

so $$\phi$$ is the value of the inverse hyperbolic tangent of $$\beta$$:

$$\phi = artanh \left(\beta \right)$$

I gather there are other ways of depicting the Lorentz transformation in two dimensions, but this is the one I'm familiar with, and seems to be the one most commonly used in textbooks. Some more examples here:

http://hubpages.com/author/lgsims96/hot/

So far every source I've looked at has followed this comvention for which way round to depict the time and space axes, with the exception of James Callahan's The Geometryu of Spacetime which has them the other way around:

http://www.google.co.uk/books?id=UM...:+Spacetime+and+Geometry#v=onepage&q=&f=false

 

[collectedsoul]

Is there a way to derive the time dilation formula without using light clocks? Specifically, without using Euclidean geometry, by simply using mechanics, maybe?

 

[Grimble]

Is there a way to derive the time dilation formula without using light clocks? Specifically, without using Euclidean geometry, by simply using mechanics, maybe?

Hello, I'm not sure that I can answer this, but I am assuming that you find the derivations using light clocks,http://www.answers.com/topic/time-d...ce_of_time_dilation_due_to_relative_velocity" are missing something -- and you would be quite right! For they are nothing more than plain straightforward GALILEAN TRANSFORMATIONS. The only reference they make to SR is that they use the constant speed of light and so let the time lengthen.

And, taking the $2^{nd}$ example; $${t_A} = {t_B}\sqrt{1 -\frac{v^2}{c^2}}$$ may be rewritten as $${t_B} = \frac{t_A}{\sqrt{1 -\frac{v^2}{c^2}}}$$

And this is the accepted formula for Time Dilation – BUT (and here lies the problem), none of it has anything to do with Special Relativity!
Everything done so far has been measured in Proper units; so all we have measured is a GALILEAN TRANSFORMATION! For in keeping the speed of light constant we have merely increased the time taken.

This is an example of what I would term the 'Slippery Slope of STR', by which I allude to the fact that, as space and time are no longer fixed scales, we have to be so very, very careful to be absolutely clear about what we are referring to at every step.

So let us now continue with the transformation into coordinate time:

1.The passage above brought us to the point where we had established the relationship between the longer path, seen when the clock is moving (relative to the observer), and that within the clock's own frame of reference. That is:
$${t_B} = \frac{t_A}{\sqrt{1 -\frac{v^2}{c^2}}}$$
where both sides of the equation are in common units of measurement.

2.And we know from Einstein's first Postulate (the Principle of Relativity), that time in the A's frame of reference is identical to that in B's frame of reference.

3.So $t_B$ in Proper units = $t_A$ in Proper units,


4.And $t_B$ in coordinate units $$= \frac{t_A}{\sqrt{1 -\frac{v^2}{c^2}}}$$ in coordinate units


4.Giving us the conversion factor, $$\frac{1}{\sqrt{1 -\frac{v^2}{c^2}}}$$ in coordinate units to convert from coordinate units to Proper units.

5.So giving us $${t_B} = {t_A}\sqrt{1 -\frac{v^2}{c^2}}$$

or $$t = {t^'}\sqrt{1 -\frac{v^2}{c^2}}$$ with the more usual terms, as the corrected formula for Time Dilation.

6.And $$\frac{1}{\sqrt{1 -\frac{v^2}{c^2}}}$$ for the Lorentz Transformation Factor.

The second common derivation that is often quoted is http://www.answers.com/topic/special-relativity#Time_dilation_and_length_contraction"

But the problem with this derivation is that we are using the clock in the unprimed system, but the moving clock is that in the primed system and the whole point of STR is to transform the Spacetime magnitudes of the moving clock into the unprimed system's frame of reference.

The clock in the unprimed system has no relevance to what we are doing.

No what we need to de here is to use the clock in the primed system. Where two consecutive ticks of this clock are characterized by
$\Delta{x^'} = 0$

If we want to know the relation between the times between these ticks as measured in both systems, we can use the third equation --
$$\Delta{t}=\gamma(\Delta{t^'} + \frac{v\Delta{x^'}}{c^2})$$

and find $\Delta {t} = \gamma\Delta{t^'}$ which is the correct formula for Time Dilation.

 

[collectedsoul]

Hi,

I've just recently learned about relativity by watching some videos and reading parts of some books and my understanding of it is very basic and quite shaky. It just seems so bizarre to me! So please bear with any errors on my part and correct me.

I think I understood what you said about the light clock derivation using Galilean relativity and simply keeping the speed of light constant and therefore stretching time to account for the increased distance. And this is where I had a doubt in the first place - isn't length being contracted in the moving frame wrt the stationary frame? If not the vertical length traveled by the light, then the horizontal length (is my reasoning in this regard valid)? If not at all, then in what cases are length contracted?

About your discussion of the coordinate units in comparison to proper units, are you referring in the first case to the relative measurements of a moving object wrt a stationary one, and in the second case to measurements within each frame itself? Then that makes sense to me.

I didn't follow you on the last bit though (not sure what an unprimed system is).

Coming back to the light clock derivation from your second link http://www.answers.com/topic/lorentz-factor#Derivation, I quote the premise:
Imagine two observers: the first, observer A, traveling at a constant speed v with respect to a second inertial reference frame in which observer B is stationary. A points a laser “upward” (perpendicular to the direction of travel). From B's perspective, the light is traveling at an angle.

When I asked the original qs., I was thinking of a case where A points the laser in the horizontal direction rather than vertical. Or simply the case where a stationary light source emits beams measured by two observers at a same distance x, one moving with velocity v and the other standing still. I was trying to find out the effect of light having the same measured speed by both observers using mechanics equations and getting really confused in the process. It'd be a big help if you could explain what is happening in this situation, and whether or not its possible to derive time dilation from this.

Thanks for your patience!

 

[Grimble]

Hi,

I've just recently learned about relativity by watching some videos and reading parts of some books and my understanding of it is very basic and quite shaky. It just seems so bizarre to me! So please bear with any errors on my part and correct me.

Hello, and welcome to the mysteries of relativity! it is great to find someone else, like me, that has learned about it by their own efforts! - It can be done! I just try to make sure I understand each step along the way - the mystery is more about how it is described, different uses of the same terminology, rather than relativity being difficult in itself.

I think I understood what you said about the light clock derivation using Galilean relativity and simply keeping the speed of light constant and therefore stretching time to account for the increased distance. And this is where I had a doubt in the first place - isn't length being contracted in the moving frame wrt the stationary frame?

Yes.

If not the vertical length traveled by the light, then the horizontal length (is my reasoning in this regard valid)? If not at all, then in what cases are length contracted?

Yes, that had me wondering, but you are quite right it is just the horizontal length.

About your discussion of the coordinate units in comparison to proper units, are you referring in the first case to the relative measurements of a moving object wrt a stationary one, and in the second case to measurements within each frame itself? Then that makes sense to me.

Yes, Proper time or length are the measurements within each frame, co-ordinate units are those proper units, converted by the Lorentz Transformation Equations into what a moving observer would see.

I didn't follow you on the last bit though (not sure what an unprimed system is).

Yes, that had me fooled for quite a time, but all it means is the use of the quote ('). It turns out that t' is primed and t is unprimed

Coming back to the light clock derivation from your second link http://www.answers.com/topic/lorentz-factor#Derivation, I quote the premise:
Imagine two observers: the first, observer A, traveling at a constant speed v with respect to a second inertial reference frame in which observer B is stationary. A points a laser “upward” (perpendicular to the direction of travel). From B's perspective, the light is traveling at an angle.

When I asked the original qs., I was thinking of a case where A points the laser in the horizontal direction rather than vertical. Or simply the case where a stationary light source emits beams measured by two observers at a same distance x, one moving with velocity v and the other standing still. I was trying to find out the effect of light having the same measured speed by both observers using mechanics equations and getting really confused in the process. It'd be a big help if you could explain what is happening in this situation, and whether or not its possible to derive time dilation from this.

Now I am sure there are many (if not most) of the other contributors out there that can give a more detailed answer than I can but essentially, the problem we have to overcome is thinking that time and space are absolute quantities.
In the situation you describe, each observer will measure the same speed for the light in his frame of reference, the tricky point to understand is that if either were to measure the speed of the light in the other frame of reference he would get the same result! It is the units themselves that are not constant.

The diagram in post 82 may help.

Thanks for your patience!

And don't give up it gets easier but remains fascinating, Grimble.

 

[collectedsoul]

Oh I'm not about to give up, its far too interesting! Thanks for answering my questions but a couple remain unanswered...any help from anyone else would be welcome.

In the light clock case, if length is contracted, then how is the derivation valid? Because it assumes that the moving clock travels a distance vt in the horizontal, when it should be less than that.

Secondly, can someone please explain why I can't derive time dilation from it? And what actually is happening in terms of each frame? Here is the case again:
Stationary light source sends out beams in the horizontal direction received by two observers. A is moving with horizontal velocity v and B is stationary. Both are same distance x from the light source. Since both measure the same speed of light, is there a way to derive time dilation from this case. And, like I asked before, what exactly is happening to spacetime in each frame, AND relative to each other?

 

[Rasalhague]

isn't length being contracted in the moving frame wrt the stationary frame? If not the vertical length traveled by the light, then the horizontal length (is my reasoning in this regard valid)? If not at all, then in what cases are length contracted?

Length is contracted in the direction of movement (and no other direction), but we don't need to worry about it in this derivation. That's because, using the notation of your link, vt_B and ct_B are distances measured wrt the frame where the light clock is moving, and ct_A is the same distance in either frame since length contraction only occurs in the direction of movement.

http://www.answers.com/topic/lorentz-factor#Derivation

So all distances in the derivation are consistent with respect to the frame where the clock is moving.

To see why distances must be the same in directions perpendicular to the direction of movement, imagine two vertical meter sticks moving past each other in a horizontal direction. Stick A has nails attached to either end. Stick B has balloons attached to either end. When they pass, the nails on stick A burst the balloons on stick B. If A was shorter or if B was shorter, the nails would miss. Each stick is moving relative to the other, so if one is shorter in the other's rest frame, the other should be shorter in the first one's rest frame. But the balloons either burst or don't burst; they can't do both. So neither stick can be shorter.

When I asked the original qs., I was thinking of a case where A points the laser in the horizontal direction rather than vertical. Or simply the case where a stationary light source emits beams measured by two observers at a same distance x, one moving with velocity v and the other standing still. I was trying to find out the effect of light having the same measured speed by both observers using mechanics equations and getting really confused in the process. It'd be a big help if you could explain what is happening in this situation, and whether or not its possible to derive time dilation from this.

If A emits light in the direction of movement, then length contraction does come into play. In fact this is a typical scenario used by textbooks to derive the length contraction formula.

http://www.pa.msu.edu/courses/2000spring/PHY232/lectures/relativity/contraction.html

 

[Rasalhague]

Another link with animations:

http://webphysics.davidson.edu/physlet_resources/special_relativity/illustration4.html

 

[Grimble]

If A emits light in the direction of movement, then length contraction does come into play. In fact this is a typical scenario used by textbooks to derive the length contraction formula.

http://www.pa.msu.edu/courses/2000spring/PHY232/lectures/relativity/contraction.html

I'm sorry, but surely this example is wrong: it is merely a GALILEAN transformation and has nothing to do with SR!

As I work it out if t_o = 2L_o/c
and t = 2L(c² - v²)^-1/2
and t = γt_o
then surely when we solve this we have

L(1 - v²c^-2)^-1/2 = γL_o

and as γ = (1 - v²c^-2)^-1/2

we arrive at L = L_o which is hardly surprising if, as I say this is merely a Galilean transformation.

Grimble:uhh:

 

[Ich]

and t = 2L(c2 - v2)-1/2

Where did you get that from?

 

[Grimble]

Where did you get that from?

Sorry my mistake the calculation is correct.

But as this uses light traveling at three different speeds c, c-v and c+v and we all know that light only travels at c, where does that leave one?

To use c+v and c-v it can only be describing a Galilean transformation, where either the speed of light inceases or the time increases.

 

[Ich]

But as this uses light traveling at three different speeds c, c-v and c+v

It doesn't. It calculates the time it takes for light moving at c to reach an object moving at v, that's where those denominators come from.
It is only your - wrong, btw. - interpretation that this somehow implies that the object sees the light moving at c+-v.

 

[Grimble]

Another link with animations:

http://webphysics.davidson.edu/physlet_resources/special_relativity/illustration4.html

Again purely Galilean transformations. With the addition of shortening the moving clock, which ticks slower although the text says it is the stationary clock that ticks slower.

The problem is that these derivations/demonstrations/proofs or whatever are only using Euclidean geometry so how can they have anything to do with SR? Merely adding a length contraction or time dilation proves nothing.

 

[collectedsoul]

Okay. Thanks Rasalhague for that excellent link to the animation.

Referring to the earlier derivation, http://www.pa.msu.edu/courses/2000sp...ntraction.html , isn't there something wrong with the formula? If I look at the final expression and substitute for c=L/t, I'm getting c=L₀/t₀gamma² So the speed of light is less in the moving frame? (btw how do I type symbols like the one for gamma?)

Besides, my original doubt was about what actually happens in the horizontal velocity scenario. If the receiver A is moving at velocity v, is both time dilation and length contraction occurring for him? I mean, in the above derivation they have just assumed the time dilation (I'm guessing from the previously derived light clock scenario where light moves in the vertical). But why? Why just assume this? Isn't it possible to derive time dilation in the second scenario?

I think part of my problem with these derivations is that they use Galilean transformations as Grimble says. I was looking for something not involving Euclidean geometry at all.

 

[JesseM]

Again purely Galilean transformations. With the addition of shortening the moving clock, which ticks slower although the text says it is the stationary clock that ticks slower.

No, the text says moving clocks tick slower as measured in the frame we label as "stationary":

Therefore it takes more clicks as measured by the stationary clock to measure a time interval of a moving clock. Observed from stationary frames, moving clocks run slower. This is called time dilation.

The problem is that these derivations/demonstrations/proofs or whatever are only using Euclidean geometry so how can they have anything to do with SR? Merely adding a length contraction or time dilation proves nothing.

The geometry of space at a given instant in any inertial frame is Euclidean, even if the geometry of spacetime is not. The spatial distance between points that are a distance dx apart along the x-axis of some inertial frame, dy along the y-axis and dz along the z-axis would just be $$\sqrt{dx^2 + dy^2 + dz^2}$$ in that frame for example.

 

[Grimble]

The geometry of space at a given instant in any inertial frame is Euclidean, even if the geometry of spacetime is not. The spatial distance between points that are a distance dx apart along the x-axis of some inertial frame, dy along the y-axis and dz along the z-axis would just be $$\sqrt{dx^2 + dy^2 + dz^2}$$ in that frame for example.

Yes of course, and I certainly agree with that, but one inertial frame observed from another is not, it is relativistic. Where in all these references is there one allusion to proper time, co-ordinate time or Lorentz transformations?

All I am seeing, as I say, are Galilean transformations, where the addition of an extra movement results in a longer path, and a longer time.

IF they hadn't complied with Einstein's 2nd postulate they could have complied with the 1st and kept the time the same but increased the velocity. THAT is the problem with Galilean transformations that Einstain was addressing - HOW to comply with the 1st and keep the time constant whilst AT THE SAME TIME complying with the 2nd and keep the speed of light constant. With these references they only do one at the expense of the other.

Einstein had to conceive of space and time not being absolute in order to comply with both postulates - thereby giving rise to transformations, and co-ordinate time.
(Which, not surprisingly has the same duration, in absolute terms, - unit length x number of units - as the proper time, for the same occurrence - (I would say event but that term has been appropriated already).)

Grimble

 

[Ich]

Don't you see that you are looking at very basic derivations of SR effects? There are no Galilean or Lorentzian transformations, they only rely on very general properties that you can derive from the postulates.
At the end of the process, one would find out how the transformations have to look like.
Your assertion "To use c+v and c-v it can only be describing a Galilean transformation" is plain nonsense. Just try to follow what they are saying, and try to understand.

 

[JesseM]

Yes of course, and I certainly agree with that, but one inertial frame observed from another is not, it is relativistic.

What do you mean it's not? The Euclidean formula for distance still works fine when dealing with objects that are moving at relativistic velocities in your frame. For example, if you have an object which in its own frame looks like a square 10 light-seconds on each side, and in your frame it's moving at 0.6c so the side moving parallel to the direction of motion is 8 light-seconds long in your frame while the side perpendicular to the direction of motion is 10 light-seconds long in your frame, then the distance between opposite corners of the square is just going to be given by the ordinary Euclidean formula $$\sqrt{8^2 + 10^2}$$. Relativity only plays a role in making one side shorter in your frame than it is in the object's own rest frame, but other than that nothing about the object's shape in your frame is any different than it would be if you were just dealing with an 8 by 10 rectangle in classical Newtonian physics.

Where in all these references is there one allusion to proper time, co-ordinate time or Lorentz transformations?

All I am seeing, as I say, are Galilean transformations, where the addition of an extra movement results in a longer path, and a longer time.

In relativity as in Galilean physics, velocity is defined as distance/time, so the time T for an object traveling at speed v to travel a given distance D must be given by T = D/v. And as I said, distances are given by the ordinary Euclidean formula, so if an object travels a horizontal distance D_h between points and a vertical distance D_v, it is just as true in an inertial SR frame as it is in Galilean physics that the total distance the object traveled must be $$\sqrt{{D_h}^2 + {D_v}^2}$$.

There is a uniquely relativistic aspect of the light clock thought-experiment though. Note that in Galilean physics, if the light bouncing between the two mirrors was moving at a speed c in the light clock's own rest frame, then it would not be moving at speed c in your frame where the light clock is in motion! In both Galilean physics and relativity, the total diagonal velocity of the light in your frame depends on both the vertical velocity in your frame (which according to the Galilean transformation would be c in your frame too if it was c in the light clock's rest frame and the light clock was moving horizontally in your frame) and the horizontal velocity v_h in your frame--the total velocity in Galilean physics would therefore be $$\sqrt{c^2 + {v_h}^2}$$, and you'd find that because the velocity was slightly greater, the time between ticks for the moving clock would not be slower than the time between ticks for your own light clock (assuming the light moves at c vertically for your clock too), despite the fact that the light has a longer path length to travel between ticks for the moving clock than for your clock. In relativity, on the other hand, it's a fundamental postulate that light travels at c in all inertial frames, so in the frame where the light clock is in motion the light must still travel at c along the diagonal path between mirrors.

IF they hadn't complied with Einstein's 2nd postulate they could have complied with the 1st and kept the time the same but increased the velocity. THAT is the problem with Galilean transformations that Einstain was addressing - HOW to comply with the 1st and keep the time constant whilst AT THE SAME TIME complying with the 2nd and keep the speed of light constant. With these references they only do one at the expense of the other.

I don't understand, why do you think the 1st postulate implies that the time should be constant? The 1st postulate implies that if you run the same experiment in different frames each frame will see the same result, so if you construct a light clock at rest in frame A and measures the time in frame A, it should give the same answer that you'd get if you constructed an identical light clock at rest in frame B and measured the time in frame B. But the 1st postulate does not say that if you construct a light clock at rest in frame A but in motion in frame B, and measure the time in frame B, that the result would be the same as if you construct a clock at rest in frame B and measure the time in frame B. In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B--nothing about the 1st postulate suggests that their ticking rates should be identical.

Einstein had to conceive of space and time not being absolute in order to comply with both postulates - thereby giving rise to transformations, and co-ordinate time.

I'm not sure what you mean by "space and time not being absolute"--certainly it's true that the distance and time between a pair of events will vary depending on what frame you use, but there is nothing in SR that implies we can't use the Euclidean formula for the distance between events in a single frame, or that we can't use the ordinary kinematical formula that velocity = distance/time in each frame. The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity, along with the uniquely relativistic notion that if the light is bouncing between mirrors at c in the light clock's own rest frame, it must also be bouncing between them at c in the frame where the light clock is in motion.

 

[Grimble]

Don't you see that you are looking at very basic derivations of SR effects? There are no Galilean or Lorentzian transformations, they only rely on very general properties that you can derive from the postulates.
At the end of the process, one would find out how the transformations have to look like.
Your assertion "To use c+v and c-v it can only be describing a Galilean transformation" is plain nonsense. Just try to follow what they are saying, and try to understand.

I'm sorry if I seem a little dim to you, but if you read http://www.bartleby.com/173/7.html" you should see why I am saying what I am saying.

 

[Grimble]

What do you mean it's not? The Euclidean formula for distance still works fine when dealing with objects that are moving at relativistic velocities in your frame. For example, if you have an object which in its own frame looks like a square 10 light-seconds on each side, and in your frame it's moving at 0.6c so the side moving parallel to the direction of motion is 8 light-seconds long in your frame while the side perpendicular to the direction of motion is 10 light-seconds long in your frame, then the distance between opposite corners of the square is just going to be given by the ordinary Euclidean formula $$\sqrt{8^2 + 10^2}$$. Relativity only plays a role in making one side shorter in your frame than it is in the object's own rest frame, but other than that nothing about the object's shape in your frame is any different than it would be if you were just dealing with an 8 by 10 rectangle in classical Newtonian physics.

Yes, I can see what you are saying BUT you have neglected to say that the 8 light-seconds is measured in co-ordinate seconds, not proper seconds and how do you apply Euclidean/Newtonian physics to a rectangle whose sides are measured in different units?

I don't understand, why do you think the 1st postulate implies that the time should be constant? The 1st postulate implies that if you run the same experiment in different frames each frame will see the same result, so if you construct a light clock at rest in frame A and measures the time in frame A, it should give the same answer that you'd get if you constructed an identical light clock at rest in frame B and measured the time in frame B. But the 1st postulate does not say that if you construct a light clock at rest in frame A but in motion in frame B, and measure the time in frame B, that the result would be the same as if you construct a clock at rest in frame B and measure the time in frame B. In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B--nothing about the 1st postulate suggests that their ticking rates should be identical.

You are quite right when you say that the two clocks should keep identical time; but consider; if the moving clock were to transmit a pulse of light each second, then allowing for the transmission time for the pulse of light, the stationary frame would still determine that the moving clock was 'ticking' at the same rate as its own clock.
It is only when the stationary frame observes the moving clock directly that the changes are seen, but will it really be 'ticking' at a different rate?

I'm not sure what you mean by "space and time not being absolute"--certainly it's true that the distance and time between a pair of events will vary depending on what frame you use, but there is nothing in SR that implies we can't use the Euclidean formula for the distance between events in a single frame, or that we can't use the ordinary kinematical formula that velocity = distance/time in each frame. The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity, along with the uniquely relativistic notion that if the light is bouncing between mirrors at c in the light clock's own rest frame, it must also be bouncing between them at c in the frame where the light clock is in motion.

By

space and time not being absolute

I am referring to Einstein's statement in http://www.bartleby.com/173/11.html" where he says

THE RESULTS of the last three sections show that the apparent incompatibility of the law of propagation of light with the principle of relativity (Section VII) has been derived by means of a consideration which borrowed two unjustifiable hypotheses from classical mechanics; these are as follows:

1. The time-interval (time) between two events is independent of the condition of motion of the body of reference.
2. The space-interval (distance) between two points of a rigid body is independent of the condition of motion of the body of reference.

Grimble

 

[Ich]

if you read this you should see why I am saying what I am saying.

I still don't see it. The problem Einstein refers to arises if you say that w=c-v is the speed of light as measured by the carriage. It isn't.
That does not mean that the equation w=c-v is evil and mustn't be used for whatever purposes. It's just that w is not the speed of light as measured by the carriage. It is the difference of the velocity of light and the velocity of the carriage as measured in the embankment system. That has nothing to do with any transformation laws, as it is a description in only one reference frame.

 

[Grimble]

I still don't see it. The problem Einstein refers to arises if you say that w=c-v is the speed of light as measured by the carriage. It isn't.
That does not mean that the equation w=c-v is evil and mustn't be used for whatever purposes. It's just that w is not the speed of light as measured by the carriage. It is the difference of the velocity of light and the velocity of the carriage as measured in the embankment system. That has nothing to do with any transformation laws, as it is a description in only one reference frame.

Taking the passage in question:

However to an observer who sees the clock pass at velocity v, the light takes more time to traverse the length of the clock when the pulse is traveling in the same direction as the clock, and it takes less time for the return trip.

Now I may be wrong, but I read that as saying: "to an observer of the moving clock, when observing the CLOCK'S FRAME OF REFERENCE, that the pulse of light moves slower one way and faster the other..."

For:

However to an observer who sees the clock pass at velocity v

- the moving clock

the light takes more time to traverse the length of the clock when the pulse is traveling in the same direction as the clock

Light takes longer - moves slower

and it takes less time for the return trip.

Light takes less time - moves faster

- whilst traveling the same distance.

Grimble

 

[Ich]

Now I may be wrong, but I read that as saying: "to an observer of the moving clock, when observing the CLOCK'S FRAME OF REFERENCE, that the pulse of light moves slower one way and faster the other..."

They didn't write what you read. There is no such thing as "observing the clock's frame of reference". You may observe the clock, recording time and date of such measurements as given by your frame of reference.
Why don't you read it as saying:
"However to an observer who sees the clock pass at velocity v, the light takes more time to traverse the length of the clock when the pulse is traveling in the same direction as the clock, and it takes less time for the return trip." ?

Apparently, you haven't understood the concept of a reference frame. For example

Light takes less time - moves faster

- whilst traveling the same distance.

(emphasis mine)
It isn't traveling the same distance.
Again: why don't you try to understand what they wrote instead of interpreting their every word? That has nothing to do with accusing you of being dim, but you definitely always skip the first step - reading and following their argument - and jump to the second: interpreting hat you have learned.

I think a reasonable first step would be if you draw a spacetime diagram of the described events; you can then easily read off distances, times, and velocities, and check with what the link is saying contrary to what you have assumed it was saying.