Why does Einstein say that a clock slows when it is moving?

In summary: K0 shows time changing more slowly than in K.In summary, Einstein is discussing the concept of time dilation and how it applies to a clock that is moving with a velocity v in relation to a reference frame K. He explains that as judged from K, the moving clock appears to be ticking slower due to its motion, with t representing the time shown on the moving clock and t' representing the time shown on a clock at rest in K. By using the Lorentz Transformation equations, Einstein shows that 1 unit of proper time in K is equivalent to gamma units of coordinate time in the moving frame K0, indicating that the seconds in K are larger than the seconds in K0.
  • #1
Grimble
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In http://www.bartleby.com/173/12.html" [Broken] Einstein says the following:
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 [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
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 [tex] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest.

But what is he saying here?

That 1 proper second in the reference frame K is equal to gamma proper seconds in the moving frame K0, leading to the conclusion that the seconds in K are larger than the seconds in K0 and that therefore the moving clock slows?

No it isn't that, for the proper seconds in K0 have been transformed by the application of the Lorentz Transformation equations into coordinate time (as we term it), so he is saying that 1 proper second in the reference frame K is equal to gamma coordinate seconds in the moving frame K0.

So Einstein is actually describing a conversion from one set of units to another; that conversion being performed using the Lorentz Factor.

And does the moving clock slow? Well, no, for what we see is that 1 unit of time in the reference frame K is equal to gamma coordinate units, so if anything one would say that the moving clock goes faster, albeit in smaller seconds.

Grimble:uhh::confused::uhh:
 
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  • #2
What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.
 
  • #3
HallsofIvy said:
What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.

Yes but since the formula he is using is [tex]t = \frac{t^'}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] with [itex]t^'[/itex] set to 1, we have an equation where the dimensions are not the same on each side and, as I understand it, this is just plain wrong.
 
  • #4
t' = 1 second, the units remain
 
  • #5
Grimble said:
Yes but since the formula he is using is [tex]t = \frac{t^'}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] with [itex]t^'[/itex] set to 1, we have an equation where the dimensions are not the same on each side and, as I understand it, this is just plain wrong.
Why would you say the dimensions are wrong? v and c are both in "distance per time" units so [itex]v^2/c^2[/itex] is dimensionless and so is
[tex]\frac{1}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
Both sides have dimensions of time.

(Surely you are not thinking that "setting t= 1" removes the time units on the right? t is a time measure. Setting t= 1 means setting t equal to "1 second" or "1 minute" or "1 hour".)
 
  • #6
Pengwuino said:
t' = 1 second, the units remain

What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.


HallsofIvy said:
Why would you say the dimensions are wrong? v and c are both in "distance per time" units so [itex]v^2/c^2[/itex] is dimensionless and so is
[tex]\frac{1}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
Both sides have dimensions of time.

(Surely you are not thinking that "setting t= 1" removes the time units on the right? t is a time measure. Setting t= 1 means setting t equal to "1 second" or "1 minute" or "1 hour".)


Wait a minute there, you cannot be suggesting that an equation that has units of days on one side and milliseconds on the other has the same units, just because they both are units of time, can you? For that would be the equivalent of saying we could have feet on one side of an equation and centimetres on the other and that would be valid, as they are both dimensions of distance.

No, what the formula is saying is that 1 unit of proper time is equal to gamma seconds of coordinate time, that is coordinate time has units that are smaller than proper units by a factor of gamma.

Grimble
 
  • #7
Grimble said:
What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.

I don't know what you mean by "proper second" and "coordinate second", and what is the difference between them. As far as physics is concerned, there is only one type of "second" - the second shown by a well-designed clock. Einstein predicted that the "second" shown by a moving clock lasts longer in comparison to the "second" shown by the clock at rest. This is the relativistic time dilation.

Eugene.
 
  • #8
HallsofIvy said:
What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.

likewise by symmetry a person in frame K0 sees t=1 seconds tick off a clock in frame K while he sees [tex]t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] tick off his own. Since t' is larger than t, a clock in K shows time changing more slowly than in K0

Have I stated this correctly? Velocity is relative yes?
 
  • #9
edpell said:
likewise by symmetry a person in frame K0 sees t=1 seconds tick off a clock in frame K while he sees [tex]t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] tick off his own. Since t' is larger than t, a clock in K shows time changing more slowly than in K0

Have I stated this correctly? Velocity is relative yes?

If you and I are moving with respect to each other, then I say that your second (the second shown by your co-moving clock) is longer than mine. On the other hand, you say that my second is longer than your second. Both of us are correct. That's relativity.
 
  • #10
Hello Grimble

All this has already been addressed at length in a thread started by you entitled, as far as I can remember, Time Dilation Formula.
Matheinste
 
  • #11
Observation versus clock count. I see how both parties see the frequency of light flashes emitted by the other as slower due to their relative velocity away from each other. But suppose both observers agree to fire retro rockets after 1 million flashes (one flash per second) locally in their own frame and come to a mutual relative velocity of zero. And at that point they both agree to turn off the flasher. How many total flashes will each see before they see the flashing stop? I say both will count 1 million flashes. So what do you mean when you say a different amount of time passes in one frame versus the other?
 
  • #12
Let us say we have two identical clocks, one in system A, and one in system B.

Clock A records 10 seconds while clock B records only 8 seconds.

So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?
 
  • #13
Grimble said:
What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.
In this formula, a "second" is "a second as measured on the observer's clock". No, that does not convert between two different "kinds" of seconds.





Wait a minute there, you cannot be suggesting that an equation that has units of days on one side and milliseconds on the other has the same units, just because they both are units of time, can you? For that would be the equivalent of saying we could have feet on one side of an equation and centimetres on the other and that would be valid, as they are both dimensions of distance.
No, I am not suggesting nor did I say such a thing. There is only one "kind" of second here, one second as measured on the observers clock.

No, what the formula is saying is that 1 unit of proper time is equal to gamma seconds of coordinate time, that is coordinate time has units that are smaller than proper units by a factor of gamma.
No, that is NOT what the formula is saying.

Grimble
 
  • #14
edpell said:
Observation versus clock count. I see how both parties see the frequency of light flashes emitted by the other as slower due to their relative velocity away from each other. But suppose both observers agree to fire retro rockets after 1 million flashes (one flash per second) locally in their own frame and come to a mutual relative velocity of zero. And at that point they both agree to turn off the flasher. How many total flashes will each see before they see the flashing stop? I say both will count 1 million flashes. So what do you mean when you say a different amount of time passes in one frame versus the other?
Yes, of course, both will count 1 million flashes. But each will see the others flashes as coming slower than their own and will believe that he started his retro rockets before the other.
 
  • #15
Grimble said:
Let us say we have two identical clocks, one in system A, and one in system B.

Clock A records 10 seconds while clock B records only 8 seconds.

So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?
No, there is NOT "more time passing between two ticks" because it is those ticks that are measuring time. An observer in system A sees B's clock record 10 seconds only 2 seconds after his. He concludes that B's clock is running slower than his and that therefore time is running more slowly in B's system.
 
  • #16
Grimble said:
Let us say we have two identical clocks, one in system A, and one in system B.

Clock A records 10 seconds while clock B records only 8 seconds.

So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?

The rate of ticking of any clock IS the rate at which time is passing. There is no "real" rate of time passing that a clock's rate of ticking can be compared with. The rate of ticking is THE time for any clock. Time does not pass at some absolute rate, some clocks recording more of it and some less depending on their relative motion. All clocks are on an equal footing.

Matheinste.
 
  • #17
HallsofIvy said:
There is only one "kind" of second here, one second as measured on the observers clock.

OK, so if two clocks measure the same second, and it is longer for one clock than the other, but it is still the same second?

A measure, of any kind, is surely based upon some standard, and that standard is what measurements are relative to?

Then if the length of a second is different for two observers, yet is still equal to one second in each case, then the standard second for each observer has to be different?

And if the standard is different for each observer, then, surely, they must be measuring time on separate scales that are appliccable to each observer?

Or are you saying that there is, somehow a single "universal" second, or somekind of "Absolute" second that is measured differently by observer's moving at different relative velocities?

If a second has different durations, depending on the conditions under which it is measured the does this not imply defferent scales depending on those conditions?

And if all seconds are the same then where do the concepts of proper time and coordinate time come from?

Grimble
 
  • #18
Grimble said:
Or are you saying that there is, somehow a single "universal" second, or somekind of "Absolute" second that is measured differently by observer's moving at different relative velocities?

I would say there is only the local time in the observers inertia frame.

Let's say we made ten clocks all the same and tested them while they were all in the same inertia frame. We find they all run at the same rate. We then send all ten into unique inertia frames. In each frame we do various low speed physics experiments we will find that physics is working as expected when we use our local copy of the clock. Each local clock, each local time is as expected in terms of physical/chemical/biological processes.

If we want to translate between frames we can. The amount of local clock time in frame A (call it Ta) will be different than the amount of local clock time in frame B (Tb).

All our experience is in a slow Newtonian world we have no intuitive understanding of the relativistic mix of time and place. We have to use the math. We can always translate time and place in frame X into time and place in frame Y. It just feels wrong.
 
  • #19
Or to paraphrase "Give me a place to stand and a lever and I will move the world" we could say "Give me a place to stand and I will give you time (a clock to use)".
 
  • #20
edpell said:
I would say there is only the local time in the observers inertia frame.

Which is referred to as Proper time?

Let's say we made ten clocks all the same and tested them while they were all in the same inertia frame. We find they all run at the same rate. We then send all ten into unique inertia frames. In each frame we do various low speed physics experiments we will find that physics is working as expected when we use our local copy of the clock. Each local clock, each local time is as expected in terms of physical/chemical/biological processes.

And, as these clocks will all be subject to identical physical laws then all the clocks will be keeping identical time?

If we want to translate between frames we can. The amount of local clock time in frame A (call it Ta) will be different than the amount of local clock time in frame B (Tb).

I don't understand what you mean here. Is it that the local time in different inertial frames will be different?

Are you saying that, if we are in frame C, moving at the same relative velocity to frame A and to frame B, and we translate their local clock times, we will get different results?

So, are you saying that your identical clocks keep the same time, but that the time is different in each inertial frame?

Let me ask that another way; if we have identical clocks that keep identical time in one inertial frame, and we then distribute them among other inertial frames where they are subject to identical conditions (identical laws of physics), can you please explain to me how they can possibly experience different local times?

All our experience is in a slow Newtonian world we have no intuitive understanding of the relativistic mix of time and place. We have to use the math. We can always translate time and place in frame X into time and place in frame Y. It just feels wrong.

But it is only the difference between what is experienced/measured locally and what is experienced/measured when the subject and object are in relative motion.

"no intuitive understanding" (sometimes expressed as 'counter intuitive') - Is this not the one idea that is used to try and make relativity seem mysterious, as if it is some sort of magic that follows the laws of physics - yet doesn't? That enable statements to be made that otherwise would be seen as illogical?

e.g. 'seconds' seem to be some sort of magical unit, that is the same everywhere, yet can be different sizes in different places. It is like having two marks on a rod that define a metre and saying that that is a metre what ever the temperature of the rod.

One MUST define the conditions under which one is measuring.

Something is wrong somewhere, Grimble
 
  • #21
Grimble said:
Let me ask that another way; if we have identical clocks that keep identical time in one inertial frame, and we then distribute them among other inertial frames where they are subject to identical conditions (identical laws of physics), can you please explain to me how they can possibly experience different local times?
Because a different amount of time elapsed for each clock.

If the laws of physics are the same for two clocks, and they both keep proper time, then if more time elapses for one clock than the other, they will show different times.

In SR, the reason two clocks show different readings isn't because of anything physical about the clocks. The clocks are just measuring devices. It's the actual time elapsed being different that is the point.
 
  • #22
Al68 said:
Because a different amount of time elapsed for each clock.

If the laws of physics are the same for two clocks, and they both keep proper time, then if more time elapses for one clock than the other, they will show different times.

In SR, the reason two clocks show different readings isn't because of anything physical about the clocks. The clocks are just measuring devices. It's the actual time elapsed being different that is the point.

But that is just what is not happening is it?

Just step back a moment and consider what Einstein was doing.

He was trying to resolve the apparent dilemma of his first postulate and his second being mutually exclusive.

For if the first were true, and physical laws were the same in 'Galilean' ssytems (inertial frames) then the speed of light cannot be universal and vice versa.

And Einstein went on to show how that could be true by shewing how space and time varied when viewed by a moving observer.

Saying that the times in separate inertial frames are different contradicts the first postulate.

If that were true, there would be no need for special relativity! For the problem could be resolved by merely saying the time scales were different!

The difficulty that Einstein resolved was how the speed of light could be constant, while the clocks in different inertial frames did keep the same time.

The answer surely lies in the observing frame seeing the extra time required pass in the moving frame, whilst still taking the same (shorter) time to pass in the observer's own frame.
 
  • #23
Grimble said:
But that is just what is not happening is it?
Yes, that's exactly what is happening.
Saying that the times in separate inertial frames are different contradicts the first postulate.
Saying that the elapsed time between given events are different in different inertial frames certainly doesn't violate the first postulate. The first postulate says that the laws of physics are the same in each frame. There is no law of physics that says the time elapsed between given events is constant, or that the rate of a clock is constant. Time is a variable, and frame dependent.

The speed of light being a constant, not a variable, is a law of physics in SR, so must have the same value in each frame.

The only way that the laws of physics would require the clocks to read the same elapsed time between events is if the events were the same relative to each frame, ie a symmetrical situation.
The difficulty that Einstein resolved was how the speed of light could be constant, while the clocks in different inertial frames did keep the same time.
They keep the same time in their own respective frame, not relative to each other. And if each clock keeps good time in its own rest frame, and the actual elapsed time between events is different in different frames, the clocks must show different elapsed time between those events if the laws of physics are the same in each frame.
The answer surely lies in the observing frame seeing the extra time required pass in the moving frame, whilst still taking the same (shorter) time to pass in the observer's own frame.
I don't know what this means.

The bottom line is that a clock is a measuring device that measures (not determines) elapsed time. If a good clock measures an elapsed time of 10 hours, it's because 10 hours of proper time elapsed. Period. If not, the clock isn't a valid clock in SR/GR.
 
  • #24
Grimble said:
Saying that the times in separate inertial frames are different contradicts the first postulate.

Time is not a law of physics.

We measure time with things we call clocks which are mechanisms or systems which have a cyclic nature and whose workings are governed by the laws of physics. At the end of every cycle the mechanism returns to exactly the same state it was in at the end of the previous cycle. As the mechanisms of these clocks, whatever they may be, are governed by the laws of physics and as the first postulate says that these laws are the same in all inertial systems, the behaviour of clocks must be the same in all inertial systems. Therefore, if two identical clocks, moving inertially relative to each other, record different elapsed times it is because the elapsed times ARE different, whatever time thay are measuring, even if your reasoning or intuition says that these times should be the same. It is not because the clocks are behaving differently.

Think of a clock as some device that generatres ticks. The interval between these ticks is the unit of time. All identical clocks, by definition, whatever there inertial motion, generate the same unit interval between ticks. The observers role is to count these ticks, or by means of some sort of readout to read the number of ticks. The counted number of ticks is a measure of the time elapsed. It is a measure of the total number of EQUAL unit time intervals. If the count of ticks of the clocks is different it is because they have measured different numbers of equal unit time intervals.

Matheinste.
 
  • #25
Grimble said:
And, as these clocks will all be subject to identical physical laws then all the clocks will be keeping identical time?

I don't understand what you mean here. Is it that the local time in different inertial frames will be different?

In order to get to other inertial frames each clock has to experience a force and accelerate. In the process the mass of the clock and each of its parts increases as [tex] \gamma m_0 [/tex]. So how can you say that a clock that "now" (sorry) masses twice as much as another clock should tick tock in the same way?

They were only identical clocks before we started to boost (accelerate) them.

The problem is that in every frame except the local frame time and place are not orthogonal. So not only do observers in different frame have different times they measure different distances. For example one observer on Earth on observer in spaceship traveling with a gamma of 50 to a star 50 light-years away and stopping. At the end of the trip the Earth person says the ship traveled 50 light-years distance in a time of 50 years (approximately not going to do the math). The ship observer says I traveled one light-year distance in one years time. They do not agree on time nor place! Of course the traveler can say "now" that I am back in the initial inertia frame I know the distance in this frame was 50 light-years but the great thing is I was not doing my traveling in this frame! So my age is only one year older than when the trip started. So it would be fair for him or her to say in the English language effectively I was traveling at 50 times the speed of light or to say effectively time was going 50 times slower for me.

I think it has to do with the translation between an English language representation and a mathematical representation that gets people so angry. Please note I used the English words "effectively I was" I did not use the words "I was".
 
  • #26
Hello edpell,

When discussing concepts of this nature we usually use ideal clocks and rigid rods. Ideal clocks obey the clock hypothesis which says that they are unaffected by acceleration. In the real world this has been verified for certain forms of clock up to accelerations of [tex]10^{18}g[/tex].

Matheinste.
 
  • #27
Hi matheinste, I am happy with the clock hypothesis to quote wikipedia:

"The clock hypothesis is an assumption in special relativity. It states that the speed of clocks doesn't depend on their acceleration but only on their instantaneous velocity."

It is the instantaneous velocity part I am focused on. I am happy to say the acceleration per se had no effect. I would be happy to say the trip is over shot on both ends that is the acceleration part happens before the ship flies by Earth and we both start our clocks and the trip ending measurement is taken as the ship flies by the star (50 light-years distance from earth) at velocity and only latter bothers to stop. So no acceleration during the trip from Earth to the star just relative velocity.
 
  • #28
edpell said:
In order to get to other inertial frames each clock has to experience a force and accelerate. In the process the mass of the clock and each of its parts increases as [tex] \gamma m_0 [/tex]. So how can you say that a clock that "now" (sorry) masses twice as much as another clock should tick tock in the same way?

They were only identical clocks before we started to boost (accelerate) them.

I was only commenting on the above statements and pointing out that we can ignore such practicalities when discussing the scenario where various identical clocks are transported, to be at rest, in relatively moving inertial frames.

Matheinste.
 
  • #29
edpell said:
In order to get to other inertial frames each clock has to experience a force and accelerate. In the process the mass of the clock and each of its parts increases as [tex] \gamma m_0 [/tex]. So how can you say that a clock that "now" (sorry) masses twice as much as another clock should tick tock in the same way?
We don't make that claim, we only claim that a clock that does that isn't a valid clock in SR. If the clock ticks at a different rate than 1 second per second of proper time elapsed, it's not a valid clock.

A valid clock in SR runs at the same rate as a non-mechanical light clock it is at rest with.
 
  • #30
Al68 said:
A valid clock in SR runs at the same rate as a non-mechanical light clock it is at rest with.

Yes we all agree. In a frame clocks work as expected.

The confusion is between frames. The other guys clock always looks/appears/seems slow relative to the clock in my frame.
 
  • #31
edpell said:
Al68 said:
A valid clock in SR runs at the same rate as a non-mechanical light clock it is at rest with.
Yes we all agree. In a frame clocks work as expected.

The confusion is between frames. The other guys clock always looks/appears/seems slow relative to the clock in my frame.
I was referring to two clocks at rest with each other, not at rest in my frame. Both are in motion relative to me.

So, I'll rephrase: If a clock, at the same velocity relative to me as a non-mechanical light clock, runs at a different rate than that light clock, relative to my rest frame, it's not a valid clock in SR.
 
  • #32
Hello again Matheinste

matheinste said:
Time is not a law of physics.

We measure time with things we call clocks which are mechanisms or systems which have a cyclic nature and whose workings are governed by the laws of physics. At the end of every cycle the mechanism returns to exactly the same state it was in at the end of the previous cycle. As the mechanisms of these clocks, whatever they may be, are governed by the laws of physics and as the first postulate says that these laws are the same in all inertial systems, the behaviour of clocks must be the same in all inertial systems.
Yes, that is precisely how I understand it!
Therefore, if two identical clocks, moving inertially relative to each other, record different elapsed times
but do they? Surely, given what you have just said, their local (proper) times will be identical?
it is because the elapsed times ARE different,
but are they? is it not that it is the observed time that is different? If the clock is as you describe and it had a clock face, then the time that clock face displays must still be that it displays in proper time, it is the observer's clock that measures a different time, is it not?
whatever time thay are measuring, even if your reasoning or intuition says that these times should be the same. It is not because the clocks are behaving differently.

Think of a clock as some device that generatres ticks. The interval between these ticks is the unit of time. All identical clocks, by definition, whatever there inertial motion, generate the same unit interval between ticks. The observers role is to count these ticks, or by means of some sort of readout to read the number of ticks. The counted number of ticks is a measure of the time elapsed. It is a measure of the total number of EQUAL unit time intervals. If the count of ticks of the clocks is different it is because they have measured different numbers of equal unit time intervals.
So are you saying that the said clock is keeping, and displaying, two separate times, its proper local time and its coordinate time, the time measured by the remote observer?

That the conditions under which the observation is made causes the difference seems much more plausible to me.

Grimble:smile::smile::smile:
 
  • #33
I have been working on producing a diagram that pulls the multiplicity of elements constituting time dilation and length contraction together to see if I can find a visual representation that enables me to get a feel for just what is happening and this is what I have arrived at so far:

http://img41.imageshack.us/img41/287/specialrelativitydiagrar.jpg [Broken]

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 - see Note 1
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 γ = 1.25 and 1/γ = 0.8

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


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' = x/γ
i.e. x' = 0.8x

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


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:
E5.GIF


or t = γt' = 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 far as I have understood it.

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 see how it can all fit together.:)


Note 1 Proper time is that experienced within an Inertial Frame of Reference.
It has to be the same in any IFoR for the following reasons:
i.If two IFoRs are at rest with one another they are both effectively in the same IFoR and share the same proper time.
ii.If they are moving relative to one another, i.e. have a relative velocity, each can still be considered to be at rest and must, therefore, still measure the same proper time.
iii.Every IFoR obeys the same simple physical laws, therefore identical, synchronised, clocks situated in such frames must keep identical time.
iv.If the proper time in IFoRs COULD be anything other that identical, then the differences could negate the need for Special Relativity! As any conflicts between Einstein's first and second postulates would, possibly, be explained by the differences in the different measurements of time.
v.The proper times of two IFoRs can be calculated from a third IFoR by means of the Lorentz Transformation equations; and, if that third IFoR was permanently positioned at the midpoint between the two IFoRs in question, those calculated proper times would have to be identical. Otherwise we would be contravening the Special Principal of Relativity, Einstein's 1st Postulate, - The laws of physics are the same in all inertial frames of reference, in other words, there are no privileged inertial frames of reference.
vi.If two IFoRs are moving with a constant relative velocity with respect to one another, then the movement of one, being a combination of time and distance relative to the second, must be the reciprocal of the movement of that second one with respect to the first. Therefore they must be using the same proper units.

I don't know whether this will be considered to be ATM or not, but no doubt some will claim so, but to me it is drawing what Einstein described.o:)

Grimble:smile:
 
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