Newtonian vs Relativistic Mechanics

[Grimble]

The book from which I learned SR doesn't mention proper time until page 121 (at the same stage as four-vectors). By that time, it has already covered inertial reference frames, time dilation, lengths, simultaneity, paradoxes, the Lorentz Transformation and Spacetime Diagrams (in that order).

There seems to be no way I can refer to Proper Time and have what I say understood without someone or several misinterpretations of what I am trying to say - this may be due to my mis-phrasing, lack of understanding or to others reading between the lines and determining meanings that weren't there. Such is the baggage attatched to that term.

So may I plead with you, one and all, and change my terminology?

Allow me to separate, as least as far as this thread goes, measurements of time made by an observer at the origin of a frame on the clock that he is holding, which I will call Local clock time and measurements made by an observer in another frame as Remote clock time.

Now it seems to me that such Local clock time, measured on a standard universal clock that, obeying the same physical Laws in each and any Inertial Frame, ought to keep the same time - or what else is the first postulate for?

The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion.

All Inertial frames are the same, they all appear to be at rest in spacetime from their own perspective.

So how can the speed of their clock be affected by their speed? It will be affected by the relative speed of an observer, measuring from another frame. One frame will be moving at different speeds relative to different observers and each will measure the observed clock slow - but by different amounts depending on their relative speeds. That one clock that is observed cannot physically run at different rates - as measured by the local observer holding that clock it can only be ticking at one rate, surely?
The different speeds must be how the various observers calculate (Lorentz Transformations), yes, how they calculate the clock to have slowed, as they measure it.

Please believe me that I am not trying to rewrite anything but looking to understand how these different lines of thought and logic fit together.

Can you explain to me where reciprocity fits in - it seems such a very basic property that is at the very heart of how relativity works...

 

[Grimble]

Einstein's principle of velocity reciprocity (EPVR) reads:

We postulate that the relation between the coordinates of the two systems is linear. Then the inverse transformation is also linear and the complete non-preference of the one or the other system demands that the transformation shall be identical with the original one, except for a change of v to −v

Or: 'As B is to A, so A is to B'.

Now I know I found this under an the topic of Wigner Rotation - and I went no further down that route! Yet the principle seems to be a simple one concerning the fact that relative movement is reciprocal - only the sign of v changes.

I think this simple diagram is a good representation of reciprocity:
https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Chess%20Pieces.jpg?attachauth=ANoY7cpYvRnGrNbH-8vkwzdH0O_R6bwfd2drCpIqXJbLRBLEQRRaCtnjFA2GgiwgEIY6wWRkA_CpuhHnXS0YbGbnnfn33G0l7iThkJKAsPC0W7lpUxEha9-h1sStHOa9q5e5uKSbCsQPK98S8vutgwnznysYdxddjzhj1oSaW--avxYm4iAvPGEH7BX1LP-4N8-FbeTnDjl9vErRhNfFvsSULoj6pXXeRgOxnAqraZPoGLZbI-t-0PcigtAImjUeej6FBXfWJ11WmZYLRNAyBnGXJRSGxW51Eg%3D%3D&attredirects=0

 

[Simon Bridge]

Now it seems to me that such Local clock time, measured on a standard universal clock that, obeying the same physical Laws in each and any Inertial Frame, ought to keep the same time - or what else is the first postulate for?

... "same time" as what? Not each other - since they may have different relative velocities with respect to each other.
However - observers who are stationary wrt a clock will observe the clock tick off one second per second (i.e. the reference clock for that frame keeps the same time as all the other stationary clocks in that frame. Some care is needed to make sure of this due to the finite speed of light.
Generally it is not useful to make a local/distant distinction like this (local and non-local are technical terms that you need later anyway.)

All Inertial frames are the same, they all appear to be at rest in spacetime from their own perspective.

Everything is at rest in it's own frame. It does not "appear" to be at rest: it is at rest. "Appear" implies there is some true or absolute state motion - there is no such thing. Frames do not have their own perspective: that requires an observer. Frames are what observers use to make observations (of time and distance) in.
This is a kind-of mental discipline.

So how can the speed of their clock be affected by their speed? It will be affected by the relative speed of an observer, measuring from another frame. One frame will be moving at different speeds relative to different observers and each will measure the observed clock slow - but by different amounts depending on their relative speeds. That one clock that is observed cannot physically run at different rates - as measured by the local observer holding that clock it can only be ticking at one rate, surely?

That is correct - ones own reference frame is stationary but other's may be moving.
You will observe a moving clock to run slow, without any (lorentz) calculation involved ... you just use your clock to time events on the other clock just like you use it to time anything.

The clocks are not physically affected by relative speed: it is the observation of the clocks that are affected.

The observations are affected by relative speed in much the same way as lengths are affected by relative distance away ... "farther objects are smaller" would be the rule for perspective. The "proper length" of an object being what you measure when you are right next it. SR extends the rules of perspective to include relative velocity as well as relative distance... so now the proper length is measure right next to the object, while it is stationary.

[We generally think of distant objects as "appearing" smaller though... this is because we are used to thinking of the ground or something as providing an absolute reference frame.]

Can you explain to me where reciprocity fits in - it seems such a very basic property that is at the very heart of how relativity works...

It means that if someone is moving past you at speed v, then you are moving past them at speed -v. It's not just relativity.
[On the chessboard analogy, black see white move one square to the right, so white sees black move one square to the left... not happy with that analogy since one of them sees their own square change colour.]

In SR it means that if you see a passing clock tick slowly, then it's observer sees your clock tick just as slowly.
Since the twins can spend an arbitrarily long time when neither is accelerating, and they both notice the other's clock is slow, then how come the accelerated twin always ends up younger?
It's pretty easy to see how in space-time diagrams.

The toughest part is getting rid of ideas that rely on absolute motion.
I usually find that the following primer is accessible at HS level:
http://www.physicsguy.com/ftl/html/FTL_intro.html
... you want the bits about space time diagrams but it is probably worth plowing through the rest too.
There is a specific treatment of several solvable paradoxes in there ... tldr: different observers agree about the overall effect (like which twin ends up younger) but disagree about how it came about.

 

[Simon Bridge]

You originally asked about what is different...? Skimming - you seem to have got a lot of stuff about language and thought experiments to highlight the fun stuff.
It may be, however, that you may appreciate a different tack that does not need much beyond HS level understanding. vis:

To replace Galilean relativity, SR has to include it and then extend it to cover phenomena where Galilean relativity is unhelpful.
This gets checked by experiment - there is a list of experiments here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

The classic Mt Washington muon experiment is a teaching demo often used to show how time dilation looks in practise.


It is also possible to go through the HS physics course and show you the relativity relations next to all the ones you are used to.
ie. if we accelerate a particle, from rest, through potential U (i.e. you may release it from a spring so $U=\frac{1}{2}kx^2$) then it's final speed is given by $v=\sqrt{2U/m}$ (remember the equation for kinetic energy and conservation of energy?) ... but this suggests that a sufficiently big U will get v>c. You can work out how big that is: $U>\frac{1}{2}mc^2$ ... then do the experiment.
The improved relation is: $v = c\sqrt{1-1/(1+U/mc^2)^2}$ ... which is harder on the eyes, but is a good description of Nature.
It also gives answers close to the one you learned if the speed is a lot smaller than c.

... and so on.

 

[stevendaryl]

So how can the speed of their clock be affected by their speed? It will be affected by the relative speed of an observer, measuring from another frame. One frame will be moving at different speeds relative to different observers and each will measure the observed clock slow - but by different amounts depending on their relative speeds. That one clock that is observed cannot physically run at different rates - as measured by the local observer holding that clock it can only be ticking at one rate, surely?

You should keep in mind that the "rate" of a clock as it proceeds along a journey is a ratio of two numbers: $\frac{\delta \tau}{\delta t}$, where $\delta \tau$ is the elapsed time on the clock from the beginning of its journey to the end, and $\delta t$ is the difference in time coordinates between the beginning of the journey and the end. $\delta t$ is a physically meaningful quantity. In contrast, $\delta t$ is conventional: It depends on how we set up our coordinate system, and there are many different ways, and each of them has its own notion of $\delta t$.

I like to remind people of the analogies between the so-called paradoxes of Special Relativity and the good-old Euclidean geometry that people take for granted. Most paradoxes have a very close analogy.

Suppose we have a system of highways that cross each other, and bend around. Each highway has a system of "road markers", which are just signs beside the road with real numbers on them, and the numbers increase as you move down the highway (or decrease if you're going in the opposite direction). Those road markers give you a "local" view of your progress down a highway. You don't have to compare your progress with anyone else, you can just say: "I've traveled 50 units down highway A", and that uniquely describes where you are (well, assuming we know the starting point).

But now, if we want to compare two different journeys, this local view is not good enough. We have to set up a coordinate system. So here's a way that we can set up a coordinate system for our highways. We pick one highway, highway A, to be our standard, and we define the x-coordinate of any point P on any highway in this way: You move along highway A until you can get to a point P' such that the straight line between P and P' is perpendicular to highway A. Then you define the x-coordinate of P to be just equal to the value of the closest roadmarker for point P'. Now, in terms of this x-coordinate, we can define a "rate" for any highway:

rate = $\dfrac{\delta s}{\delta x}$

where $\delta s$ is the change in the roadmarker number as you move down the highway, and $\delta x$ is the change in the x-coordinate.

If you think about it, the rate of a highway will be different, depending on which highway you choose as your standard for setting up your coordinate system. (If the roadmarkers are evenly spaced, then the rate will be given by: rate = $\frac{1}{\sqrt{1 + m^2}}$, where $m$ is the slope of the highway, relative to the standard highway, where slope = $tan(\theta)$, where $\theta$ is the angle between them). This just means that this rate, while it might be useful for calculations, has no absolute physical meaning, because the physical meaning can't depend on the arbitrary choice of which highway is the standard.

 

[Zayl]

Allow me to ask a question here...
When I created this thread, the system insisted I insert a prefix at the start for the level. I put a B as when you start talking of:then I am lost - I had a 'high school education' I guess you would call it - I am from the UK.

I am thinking of this in a very basic way currently. I understand Newtonian Mechanics - that is what we were taught at school - very simple and straight forward. What I am looking for now, is some help in finding out which of Newton's laws are changed by Relativity.
Newtons Mechanics are represented in Cartesian coordinates with a constant time factor that is the same throughout.
In Relativistic Mechanics, I understand we are considering Time as the fourth dimension, rather than as a common standard, but how is this depicted in diagrams? In the Minkowski diagrams it seems to be the rotation of the moving frame's time access? OK but what of the other three axes? Are they not still orthogonal?

Newton's Laws are simply incorrect. It is nothing but an approximation which also fails to be an approximation at large enough relative motion. Then when the Lorentz factor is applied it becomes correct.

 

[Grimble]

... "same time" as what? Not each other - since they may have different relative velocities with respect to each other.
However - observers who are stationary wrt a clock will observe the clock tick off one second per second (i.e. the reference clock for that frame keeps the same time as all the other stationary clocks in that frame. Some care is needed to make sure of this due to the finite speed of light.

YES, every stationary clock in a frame will keep the same time, one second per second, because they are all at rest in that frame. They will tick at one second per second, measured internally with in that frame without any involvement of anything external to that frame. You said it - one second per second, the SAME as the time in every other frame measured internally without recourse to anything external to that frame. Each frame considered on its own which cannot be moving as we are making no reference to anything outside the individual frames. Each frame on its own. No reference to ANY other frame.

It is the relative speed between the observer and the clock that results in time dilation/clocks being measured to run slow. As you said:

The clocks are not physically affected by relative speed: it is the observation of the clocks that are affected.

[On the chessboard analogy, black see white move one square to the right, so white sees black move one square to the left... not happy with that analogy since one of them sees their own square change colour.]

Oh, for goodness sake! hehehe! The chess board is only there for reference! How about if we make the squares just lines, without colour... Why do you have to divert the topic all the time - not just you Simon, but so many of you 'experts' will divert away from the topic to go into the subtle meanings of how things are said! Sometimes it seems one can't make a statement without being told you are using the wrong words! Give us some room please - you know my level of science, for goodness sake, why expect me to use all the terminology as you would? That is unreasonable. It also means that at times I cannot say what I want for however I try - even inventing a phrase as I have done here - for me to use in explaining what I mean and even that is torn to pieces by scientific grammar n...

I am sorry, I apologise, but it is extremely stressful when anything I say is turned round to mean something different.

Another little point that confuses me in the twin paradox explained as a result of only one twin accelerating is that the total time difference when the traveller returns is dependent on the speed and the duration of the journey; not upon any factor related to the acceleration - neither to the rapidity of that acceleration nor to its duration.

Since the twins can spend an arbitrarily long time when neither is accelerating, and they both notice the other's clock is slow, then how come the accelerated twin always ends up younger?
It's pretty easy to see how in space-time diagrams.

It is because it is also the one that is moving relative to the observer who ends up younger. Remember - time dilation only affects the traveller

 

[Grimble]

Newton's Laws are simply incorrect. It is nothing but an approximation which also fails to be an approximation at large enough relative motion. Then when the Lorentz factor is applied it becomes correct.

Really? I understood that they were seen as a limiting case dealing solely with speeds much less than c; that they are limited in their application...

 

[Ibix]

If you take Zayl's view literally, relativity is also incorrect because it fails at singularities, and is presumably not giving exactly the same predictions as quantum gravity will elsewhere. A better view is that Newtonian mechanics is, as you say, an approximation to relativity. Where it's predictions are indistinguishable from those of relativity to the precision you are able to measure, it's as "right" as relativity. It does go completely off the handle with large velocities or displacements, as you are aware

 

[Grimble]

Oh, for goodness sake! hehehe! The chess board is only there for reference! How about if we make the squares just lines, without colour... Why do you have to divert the topic all the time - not just you Simon, but so many of you 'experts' will divert away from the topic to go into the subtle meanings of how things are said! Sometimes it seems one can't make a statement without being told you are using the wrong words! Give us some room please - you know my level of science, for goodness sake, why expect me to use all the terminology as you would? That is unreasonable. It also means that at times I cannot say what I want for however I try - even inventing a phrase as I have done here - for me to use in explaining what I mean and even that is torn to pieces by scientific grammar n...

I am sorry, I apologise, but it is extremely stressful when anything I say is turned round to mean something different.

May I apologise for the acerbic tone of that post. Ageing bones can be somewhat unrelenting when they stress the passing years...
which is no excuse! I ought to leave replies to my morning time when my patience is not tried by aches and pains.
On re-reading your post this morning, I find it very reasonable and very helpful, if I could I would remove the above critical passage - (Admins?)

 

[Grimble]

As for Time dilation and Relativistic Mechanics, the effect of the invariance of c is easily demonstrated; including exactly what it is, how and why it occurs - when I draw it with Occam's razor in hand! (that was a joke...)

https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Time%20diagrams%20%281%29.png?attachauth=ANoY7cp139iymu-8u-3t2YLYd-HgpGmmjme09LwHg3GlazYAWPN_3-FWURhbfgwzknY5Xs_4LuKGjQK9Xzeau2UEmEOObcbTtNoxyG2GRk1VTo7S6IeRSqnD5zp8gAyWcCuDkLl2rsh-_DVQDcrr7RD36Zp3rn0feheAA340kBquXZaX4kKKT3IXybolAz4XUgZnHFD8Let8yXJtJEWS5bPB4OIycOwOMIzhwf_vmn2rZyI8iWGH_hChGyjstAD7K77GlpFOOB0AcfNZBd_uYNYT7bhrHIkMjlLpUEtVvSs1NMff_z1R-nI%3D&attredirects=0

The Spacetime Interval as shewn in this link:

Armed with this insight, I recommend the following YouTube video:

is (ΔS)² = (Δt)² - (¹/_cΔx)²

When we draw this out in detail in the drawing on the right, we have the
Spacetime Interval measured, t = 0.8seconds
The time measured on the moving clock, t' = 1 second
The time it would take Zach and clock B to travel distance x = ¹/_cvt' = 0.6 seconds

This demonstrates that the Spacetime Interval measured in resting Alice's Frame, Δt, comprises the travel time of the light in the moving frame , Δt' less the time component of the translation of Zach and his Clock B which is ¹/_cΔvt' or making that calculation by means of Pythagoras:
(Δt)² = (Δt')² - (^v/_cΔt')²
or Δt = Δt'√1 - v²/c²

Let me say this is no new theory or new interpretation it is simply reading the spacetime diagram and reading what it is saying as a description of what is happening, using simple Euclidean geometry.

That the Spacetime Interval measured in the moving frame from the resting frame, comprises the total time passed in the moving frame, ct' less the time factor related to the translation of the moving frame. So the resting observer measures less time to have passed for the moving clock, it runs slow.

It seems to me that is due to simple Relativistic Mechanics because of the invariance of c. Once that is added into the equation the rest, time dilation, proper time, coordinate time and everything else falls out of it using no more than Euclidean Geometry.

After all, the 'principle of the invariance of the speed of light' is the postulate that changed everything.

 

[Dale]

Hmm, I don't know about this software. The lines should be hyperbolas.

 

[Grimble]

Yes, it seems to me that that is a fundamental conclusion developed in Minkowski's great concept of Spacetime, which as he states in his introduction is what this concept is built upon:

I would like to show you at first, how we can arrive – from mechanics as currently accepted – at the changed concepts about time and space, by purely mathematical considerations.

A natural line of reasoning coming from his equation c²t² - x² = 1, which plotted against cartesian coordinates produces a hyperbola.

Yet that is holding fast the Newtonian mechanics, where we can happily plot the time against the displacement with no qualms about exceeding the speed of light.
Indeed that leads directly (when we use seconds against light seconds to the familiar 45° limit for the speed of light. Yet when we add two speeds, as we do in the moving Light clock (the speed of the clock and the speed of the light in the clock, the result is >c.

This is the reason(?) that the plot becomes a hyperbola.

All absolutely right and correct.

Yet if we give due attention and gravity to the second postulate in Relativity, it seems to me that we should accept that we are given two facts. That light will always travel one light second per second and that the clock will have a horizontal displacement = vt and that the time axis for the moving clock (in this case) must pass through the point where the x coordinate crosses the 1 second point on the rotated time axis. (0.6,0.8 in the diagram).

Time dilation is shewn where the rotated, moving time scale, crosses the 1.0 coordinate of the observer (0.6c - Lorentz Factor = 1.25) where t' = 1.25 = γt and x' = 0.75, which by length contraction becomes 0.6 for the observer. x = ^x'/_γ

I first thought along these lines when considering the path of a fast moving body and I realized that after it had been traveling for time t, it would have traveled a distance of vt and the furthest a photon could travel being 1 light second per second would be along the x-axis where the time passed would be zero.

Again let me emphasise this is not changing anything, only how it is plotted. The only difference is that because light travel is the best measure of time as it comes mathematically from the second postulate, we measure the distance imaginary light would have traveled from the initial event, the departure of the moving body from a point of contact or crossing of paths and calculate the slowing of time measured in the observed body by the observer, i.e the moving clock slowing. (Literally in this case!).

https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Light%20clocks%20%281d%29small.png?attachauth=ANoY7cqN4iyplnt75M01m0saod0TBN71ldfM1KNUQ2GOj7uMWkrE1r75-RvifPuwOSf9mLYtNJcsXLhgGQNsGH9bukC4JDck5HFF170ijAOoGCfm4lC7uOQQQ8pE6UCTAsA-qolr-pHMMTZ3S_dSwHxvBO52GltUilvCJiFsWxXUOGVdacGap4EGdZTXV34BdUsVIAcEpArH1WpYi47JHfPhYTOd2Cn1eMiELWq_em2AQ9Cwm1hHZUroC47C860Ui5n-WtX0_r2tNsm2YSK4Utk7AlfYX2BSgEYPQ1-2fTOXvIZOm57TIOY%3D&attredirects=0

I am not trying to rewrite any mathematics. All I have done here is to stop adding two vectors to find a third; and instead subtracting one vector from the combined vector to determine the reduction in the third vector, Vector subtraction rather than addition.

As I say I am not trying to rewrite anything or change relativity or anything like that, but to check my understanding, and the way I visualise it.

 

[saim_]

Nobody explains relativity better than Einstein himself; at least for those not very mathematically sophisticated.

Relativity: The Special and General Theory by Einstein
http://www.schloss-sihlberg.ch/dl/6e29795ff0c56f00fb50a75e83a0eb47/relativity.pdf

I read this in eighth grade and it took me almost a month, it'll probably take you less than that. I would advise you to stop wrestling with four-vectors and try to understand the basics from this book first.

 

[Dale]

As I say I am not trying to rewrite anything or change relativity or anything like that, but to check my understanding, and the way I visualise it.

Your description here does not clarify what you are doing. Whatever this is it is not a spacetime diagram. It looks wrong to me and if this is your understanding then it seems wrong to me also.

Do you have a professional reference explaining what this type of diagram is and does. To me it looks flat out wrong, but I am willing to entertain the idea that it is simply unfamiliar if there is a good reference.

 

[the_emi_guy]

Your description here does not clarify what you are doing. Whatever this is it is not a spacetime diagram. It looks wrong to me and if this is your understanding then it seems wrong to me also.

Do you have a professional reference explaining what this type of diagram is and does. To me it looks flat out wrong, but I am willing to entertain the idea that it is simply unfamiliar if there is a good reference.

The OP is not intending for this to be a space-time diagram. Read posts 47 and 48. It is the very common pictorial used to show how the consistency of c leads to time dilation. If you want a reference: Feynman Lectures on Physics Vol 1, 15-4 "Transformation of Time" (or probably any other elementary text on SR).

Grimble,
Keep in mind that both axes in this pictorial are distance (there is no time axis) . We are intending to measure the time between two "events" by measuring the distance, horizontal and vertical, that light must travel between the events. (The first event is the flashlight being switched on, the second event is the light reaching the mirror).

On the other hand, the space-time diagram (video referenced in post 49), has time as the vertical axis, thus the hyperbolas and 45 degree "speed of light" line.

 

[Dale]

The OP is not intending for this to be a space-time diagram. Read posts 47 and 48. It is the very common pictorial used to show how the consistency of c leads to time dilation.

Maybe you are right. The vertical axis should be labeled "y" and not "ct" if that is the case. Let's see what he says.

 

[vanhees71]

The space-time diagram is obviously wrong. The lightcone is defined by $x=\pm ct$ in this (1+1)-dimensional diagrams. The units on the time-like axes of different inertial observers is determined by the hyperbola $(ct)^2-x^2=1 \text{unit}^2$ and not by the circle drawn in the diagram. It's a time-space plane in Minkowski rather than Euclidean space!

 

[the_emi_guy]

The space-time diagram is obviously wrong. The lightcone is defined by $x=\pm ct$ in this (1+1)-dimensional diagrams. The units on the time-like axes of different inertial observers is determined by the hyperbola $(ct)^2-x^2=1 \text{unit}^2$ and not by the circle drawn in the diagram. It's a time-space plane in Minkowski rather than Euclidean space!

What space-time diagram are you referring to? The only space-time diagram I see in this entire thread is in the video referenced in post 49.

 

[vanhees71]

I'm referring to the diagram in #83. According to the axes labels it should be a space-time (Minkowski) diagram, but at least the determination of the units on the time-like axes "clock A" and "clock B" is not correct, because you draw a circle rather than a hyperbola.

 

[Ibix]

What space-time diagram are you referring to? The only space-time diagram I see in this entire thread is in the video referenced in post 49.

The diagram in #83 has axes labelled as if it were a space-time diagram but looks more like a diagram in the x-y plane. Both Dale and vanhees71 are expressing confusion about exactly what is being shown. You seem to be interpreting it as the latter, but it isn't clear that that is what Grimble intends. You are assuming that you know which mistake Grimble has made - which may or may not help him correct whatever his misunderstanding is depending on whether your assumption is correct.

Edit: beaten to it by vanhees, I see.

 

[the_emi_guy]

The diagram in #83 has axes labelled as if it were a space-time diagram but looks more like a diagram in the x-y plane. Both Dale and vanhees71 are expressing confusion about exactly what is being shown. You seem to be interpreting it as the latter, but it isn't clear that that is what Grimble intends. You are assuming that you know which mistake Grimble has made - which may or may not help him correct whatever his misunderstanding is depending on whether your assumption is correct.

Edit: beaten to it by vanhees, I see.

The OP, who indicated a high school level of education, has derived the Lorentz transformation from first principals in a manner typically employed in college physics classes and this seems to have gone completely unnoticed.

Have you gone back and read posts 47 and 48? I suggested this just 5 posts ago to clarify where the OPs diagrams originated.

 

[Grimble]

OK. Let us examine the diagram in #83 that is causing such confusion and I will explain exactly what it is intended to shew.

First things first, this is not a Minkowski Spacetime diagram. It is no more than a diagram to demonstrate how, by making the invariance of 'c' central there is no need to use hyperbolae. That is NOT to say there is anything wrong with Minkowski’s Spacetime; it having stood the test of time very robustly.

In both traditional Newtonian and in spacetime diagrams - as I understand them - the time axis is vertical because time is being measured by the distance light would have traveled along the y axis.

In the clock diagrams time for each observer, in clock A and in clock B, is measured parallel to the y/ct axis vertically for each clock at rest. Which due to the construction of said clocks is also the path of the light in each clock, as observed by an observer at rest with each clock. So time measured by either observer is the distance light traveled in their own clock, it reaches the mirrors, 1 light second away in 1 second. Identical time for identical clocks, in their own frames.

Yet for time in each clock measured from the other clock, we know that in one second that light will have traveled for one light second, in a clock that has has traveled vt light seconds, at 0.6c that will be 0.6 light seconds.

So measured from the stationary clock, the light in the moving clock will have two components to its motion, the movement of the light in the clock and the movement of the clock itself. Hence at 0.6c the light in the moving clock must reach point (0.6,0.8) in the frame of the rest clock after 1 second. Less time will be measured by the resting observer as the speed of the light has two components, the speed of light in the clock and the speed of the clock. So measured from the resting clock the time measured must be less that in the resting clock(?).

Now to me that is plain simple Newtonian Mechanics, working from what is known rather than assumed, that the speed of the light will be traveling at c. As that is a constraint placed upon this scenario by relativity.

Time is measured by virtual light emitted at the initial event, traveling at c in every direction - so an expanding sphere of virtual light, centred on that initial event; any radius of that sphere will be a measure of the interval from that initial event. (in this two dimensional view it will, of course, be a circle)

For a body at rest - such as the observer whose frame we are drawing it will be vertical as there is no displacement.

For a moving body, the line along which time is measured will be rotated because of the lateral displacement. In Newtonian mechanics that displacement is measured after the interval measured on the observer's time axis - the vertical axis of the diagram. In the mechanics of relativity - because of the invariance of c - the displacement means that time in the moving clock has to be measured where the displacement (0.6 light seconds) intersect the 1 light second circle of time because those are the two constraints which we know are true: the displacement of the clock (0,6 light seconds in my diagrams) and the one light second the light must have travelled.

The time interval is measured on the moving body's rotated time axis. In the clock diagram the light in the clock will have traveled 1 light second - to the virtual time sphere while moving clock has traveled vt from the observer in the stationary clock. That is why the time axis for the moving clock is rotated through angle β (sin β = ^v/_c rather than angle α (tan α = ^v/_c because the 1 second coordinate constitutes the hypotenuse rather than the adjacent side of the triangle; it has to be that way because of the invariance of c.

This is no more than simple euclidean geometry.

We are intending to measure the time between two "events" by measuring the distance, horizontal and vertical, that light must travel between the events. (The first event is the flashlight being switched on, the second event is the light reaching the mirror).

On the other hand, the space-time diagram (video referenced in post 49), has time as the vertical axis, thus the hyperbolas and 45 degree "speed of light" line.

This is a diagram showing three views of the mechanics of relativity

1. First, classical Newtonian Mechanics with no account taken of the second postulate. Resulting in a speed greater than 'c'.
2. Second, Minkowski's great Spacetime where taking heed of the hyperbolic function caters for the invariance of 'c'. Yet it is truly a mathematicians solution that leaves those of us with a less mathematical background, difficulties in appreciation. There is nothing wrong with it - it has been accepted for more than 100 years!
3. Thirdly, is but a simple view of the mechanics, centred on the invariance of 'c' and trying with the best will in the world to apply Occam's razor. Keeping to the facts that we are certain of: that light will travel at 'c', that the moving clock will travel the distance vt, and how that rotation affects the measurement by the resting observer.

https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/3%20mechanics%20compared.png?attachauth=ANoY7coaMPUDTIO6hy2MrLS6UA0jDfN3g7ZAN_d_dYGxxFrLFD1F93wbzGyClgdYgDCPnXBpWs34Sj6gnWmMjSkd5bfAlV9Lj7jKMPV5opPLquyBEphh_th791WMA0n1AUE4SZxL33Lm-03eT2SI2nsHIXYMlMt5Y7BzfaILnrBs8eGMUiXDo3bI54IMJsB-9kDI1GAscn0uKYnRBjAnALgAxoinCixSOdGZdQHeSYWXx70nhmaL7j9pT5wUaMdey5By8-N87nu99VB0xO7wRrq8cCvEWDNi2h7iLWIGsz7OQ-4-_ci1rOk%3D&attredirects=0
The best thing that can be said for this last attempt to understand what happens in simple terms is that time dilation and the Lorentz factor fall out of it without any effort.

https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Lorentz%20factor.png?attachauth=ANoY7cp8JpFJWuip-BS5Zt2Bxq3gX06yEMV6SAXi_uGKqO3cv3bOgFtwPl9xMQBeFcuc7nXX5Fk4KZ49UpdrCKak0k-LvYZaSQQ0gY29LbmSRByGzGTFrkhNGjCtR5SCH_v-SP_PE1Y-hiuF-TgZA2NfIjGaVHVW3dGAFXbpBoS2Isknzam8bhAoocoEEszL4knxcneXi8ANdiVi9JWvQYR73UokJPY6dZt43u7fzQp_2htoojq4E6sbSo0Iw5WPtMMqmnaqYPWdWyIwF3GPBTdsJtVQalLdvw%3D%3D&attredirects=0

 

[vanhees71]

I don't understand your diagrams. At least the $(ct,x)$-plane cannot be a Euclidean plane. This wouldn't enable us to define a causality structure. The correct construction of the Lorentz transformation can be found in

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

pp. 8 and 9.

The Lorentz transformation implies all kinematic phenomena like time dilation and length contraction as is worked out in this manuscript too.

 

[bahamagreen]

"Nor must we overlook the fact that arguments which proceed from fundamental principles are different from arguments that lead up to them."
Aristotle

No doubt Einstein, Minkowski, et al knew this; and the difficulty when a change in the fundamental principles is itself the object of argument.

 

[Dale]

this is not a Minkowski Spacetime diagram

So then the vertical axis should not be labeled "ct", it should be labeled "y"

by making the invariance of 'c' central there is no need to use hyperbolae

There is still always going to be the need to use hyperbolas. The equation $\Delta s^2=-c^2 \Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2$ is a hyperboloid.

 

[Grimble]

I don't understand your diagrams. At least the $(ct,x)$-plane cannot be a Euclidean plane. This wouldn't enable us to define a causality structure. The correct construction of the Lorentz transformation can be found in

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

pp. 8 and 9.

The Lorentz transformation implies all kinematic phenomena like time dilation and length contraction as is worked out in this manuscript too.

Thank you, Vanhees71, I do understand that you are trying hard to help me, but with my level of High School Education (1960s) I am struggling with the first sentence of p8 and totally lost by sentence 2. Not by the wording but by the mathematical terms used, which I am unfamiliar with, in those terms at least.

On the other hand I am at a loss as to how you can fail to understand my diagrams!

I start with a simple diagram of two light clocks moving apart at 0.6c (to keep the diagram and any calculations simple. Perhaps if I take you through it again; starting with the difference between Newtonian mechanics with and without the invariance of 'c'.
https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Mechanics%20compared.png?attachauth=ANoY7cp9GaHYoo-aiMlFyVhXmgpAWLw7Y4ZXy-MbINOoEK7NI6A2GzRvClzDgb_ZLF0H9cw9RzRFyiqXrOLPuoLcvDGafLnGPtswJdjScrSzAjyeiZ2YxTnI_vNHz5uBfVDlONZZCaokzsr7MLFKjBSKJEcgYp2A-Z-dtLnkAZLt1Rw3XBHscpycCyeSJP-taYPoccpRQYImLwDqZ8QvofhyrXNW6doxkt4STBS-nk5nIkTz1Fde-dNpL8g4TNChWLGA9gR5peECVEhVa9AWxctK_KrdCDdpQg%3D%3D&attredirects=0
In both diagrams the vertical axis can be read as both/either a ct (time) axis or as a y (spatial) axis. It is the path of a flash of light normal to the x axis. Being the passage of light it is possible to use it as a time axis for the observer in the clock at 0,0.

In the first diagram speed is unlimited and the moving light travels 1.166 light seconds in one second measured on Clock A's time axis.

In the second diagram, observing the 2nd postulate, the moving light of clock A (in the frame of Clock B, which is at rest at rest) travels a rotated path, but still only 1 light second (The curving red line) from 0,0 while clock A also travels 0.6 light years in that same 1 second. Hence the light will arrive at (-0.6,0.8) after traveling for one second, measured along Clock A's rotated time axis, as measured by Observer B.

I drew the second diagram that way round to emphasise the reciprocality of relativity. Either clock can take either role.

The diagrams in Post #93 reduce those diagrams to the fundamental items, time measured on the observers time line, time measured on the moving clock's time line and the distance traveled by the moving clock.

The new part I introduced was the second part which I included to shew how the invariant spacetime interval function can be seen as a hyperbola.

Is there any point in particular that is causing you difficulty?

 

[vanhees71]

Well, perhaps I'm just too used to the usual Minkowski diagrams. I cannot make sense of your "Euclidean" space-time plane. It's contradicting any intuition we have about the relativistic space-time structure.

 

[DrGreg]

In both diagrams the vertical axis can be read as both/either a ct (time) axis or as a y (spatial) axis. It is the path of a flash of light normal to the x axis. Being the passage of light it is possible to use it as a time axis for the observer in the clock at 0,0.

I see two problems.

1. The vertical axis is the y axis only. You can't treat it as the ct axis as well, because, although $y = ct$ is true for one path in each diagram, it's not true for other paths in the same diagram.
2. In each diagram the red parts and the green parts refer to different observers, i.e. different coodinate systems. It's misleading to superimpose both in the same diagram, so really you ought to split each diagram into separate green and red diagrams. And use different names for the red and green coordinates. Traditionally the two coordinate systems are written as $(ct, x, y)$ and $(ct', x', y')$.

 

[Dale]

In both diagrams the vertical axis can be read as both/either a ct (time) axis or as a y (spatial) axis.

This is wrong. Time is orthogonal to space so it cannot be represented by the same axis.

It is the path of a flash of light normal to the x axis. Being the passage of light it is possible to use it as a time axis for the observer in the clock at 0,0.

This would be a different axis, neither time nor space. It would be called a null axis. Something like this is used in radar coordinates, but usually with two null axes.

 

[Grimble]

Thank you, Gentlemen. I can see what you mean and how my diagrams can be confusing. I have tried to include too much information in a (each) single diagram.
I will heed your comments and advice and redraw them, as separate diagrams. That makes a lot of sense. Thank you.

 

[the_emi_guy]

Thank you, Gentlemen. I can see what you mean and how my diagrams can be confusing. I have tried to include too much information in a (each) single diagram.
I will heed your comments and advice and redraw them, as separate diagrams. That makes a lot of sense. Thank you.

Grimble,
One thing to keep in mind is that one of the premises of this thought experiment is that each observer agrees about the vertical distance traveled by the light beam.
In relativity, observers will generally disagree about lengths (Lorentz contraction), but in this case, because there is no vertical motion, both will agree about this vertical distance. Since they are agreeing about this distance, we might as well make that distance something simple such as a meter. We can't have this distance indicated in terms of time because the observers disagree about elapsed time.

The next thing to consider is that this thought experiment shows that the two observers disagree about the elapsed time between two specific events (light switched on, light reaching mirror). Turns out that this implies that the two observers will disagree about the elapsed time between *any* two events, such as time time between beats of the travelers heart. Otherwise an observer would be able to detect if they were the one in motion by comparing the results of the "light clock" with their heartbeat. This is the other postulate of special relativity, that either observer can declare themself to be the one at rest.

I suggest taking a look at Feynman's very well written treatment of this thought experiment. Thanks to Caltech, it is freely available on the Web:
http://www.feynmanlectures.caltech.edu/I_15.html section 15.4
He describes why the two observers agree on vertical distance, and why the time dilation computed for the "light clock" applies to any other "clock".

Cheers

 

[Grimble]

https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Alice%20and%20Bobs%20light.png?attachauth=ANoY7coZsmLJQ0JAxFV72ZNxkbJBRzT6MCrPSCK-3J6r6q7Izraagl5byes4jkgMWSCX6HuegfgHK2T9WMjDRqFf4hiwqHDURP6iadByaTSGqqOQYVo6YMF5qgO27Bsr8136cganLCdkHQqxiFovfcGOXsUYCROHx5EmthiNTOzJ4cGRCeFtWBCgKdvX2gbuqm0ZWxrLfoiZJDt1VDrKloK8g7rMD7fa4nq766aEDc9T6Ygg0PEWufVX2U79EzbTBU7jnRGqgrycgQN2nox_xutMu9kDyjdx4HgjrbFyc0S49x0eXgZKP38%3D&attredirects=0

This is not an Minkowski Spacetime diagram. If anything it is a simple Newtonian diagram.
A simple diagram of the Frame of Reference of a resting observer, Alice; Bob is traveling along the x axis. He has a light clock with the light direction along his y axis. As he travels the path of the light in his clock is rotated by his motion. The speed of the light along the rotated path is 'c'. (The rotated path is only observed by Alice, Bob only sees his light travel at 'c' along his y axis).
Bob is traveling at 0.6c relative to Alice. After Bob has traveled 0.6 units, his light will have traveled 1 unit. along the rotated path, as measured by Alice.
Alice also has a light that emits a flash along her y-axis at Bob's departure.
Both lights travel in the y direction at the same rate, yet Bob's light also travels at 0.6c along the x-axis and has therefore traveled 1 unit along the rotated path, as measured by Alice, when Bob has traveled 0.6 units x-wards.
The increased duration of Bob's light journey (1 unit) is due to the relative speed between Alice and Bob and is only measured by Alice who measures Bob's light to have traveled 1 unit when her light has traveled 0.8 units; as Bob's light will be measured by Bob. Both lights are moving at 'c'; therefore the time passing for Bob, as measured by Alice, and only as measured by Alice, is measured to be 1 unit rather than 0.8 units.

That is purely a change in the time measured by Alice. Bob's clock has only run slow as measured by Alice, Not as measured by Bob. Indeed its rate is unchanged except as viewed by Alice due to her relative speed.

Now, before anyone complains that I am trying to rewrite anything, it seems to me to be crucial to this scenario, to relativity the experiments and all, to aver that Bob's unchanged passage of time is no more real than Alice's changed perception of time passing for Bob.

In Alice's Frame Bob's time does run slow and is measured to run slow. The Muon's time run's slow, because it is measured in the Earth's Frame of Reference in which it does indeed run slow.

Surely that is what is so important and fundamental to the whole concept and theory of relativity - it is all relative! No one observer's view is more correct than any other. Is that not what relativity is all about?

And if one swapped Alice and Bob in that thought experiment, would Bob not find that Alice's time runs slow? - And that both view are correct at the same time, whoever is considered to be the one moving? After all, all movement is and can only be relative...

 

[PeterDonis]

This is not an Minkowski Spacetime diagram. If anything it is a simple Newtonian diagram.

Yet you appear to be trying to demonstrate something about SR from it. I don't understand what you are doing or why you think it is valid. You can't use a "Newtonian diagram" to demonstrate something about SR. The two theories are inconsistent.

In fact, I'm not even sure your diagram correctly represents Newtonian physics. You appear to be assuming that in Newtonian physics, Alice's light travels 0.8 units in the same time that Bob's light travels 1 unit. I don't see anything in Newtonian physics that would lead to that result.

It seems to me that you are expending a lot of effort trying to invent new conceptual tools for something that you don't yet understand. That's not very likely to be a good strategy; so far it certainly hasn't appeared to work for you in this thread. I think you would be better served by cracking open a basic SR textbook, like Taylor & Wheeler, and trying to learn to use the conceptual tools that have already been invented by people who thoroughly understand the subject matter. Or you could try Einstein's own book for the layman, linked to in post #84.

 

[PeterDonis]

When we draw this out in detail in the drawing on the right, we have the
Spacetime Interval measured, t = 0.8seconds
The time measured on the moving clock, t' = 1 second
The time it would take Zach and clock B to travel distance x = 1/cvt' = 0.6 seconds

No, this is wrong--both your diagram (the one on the right) and your calculation as quoted just now. The equation for the interval that you gave, which is correct, is

$$
(\Delta S)^2 = (\Delta t)^2 - (\frac{1}{c} \Delta x)^2
$$

However, you are interpreting the terms wrong. The spacetime interval $\Delta S$ is the same as the time measured on the moving clock. The time interval $\Delta t$ is the coordinate time, i.e., the time according to the clock that stays at rest. So the correct calculation is: $\Delta t = 1$ (1 second elapsed on the clock that stays at rest); $\Delta x = v \Delta t = 0.6$ (the moving clock travels 0.6 light seconds in 1 second, both distance and time being measured according to the frame of the clock at rest); so $\Delta S = 0.8$ (0.8 seconds elapsed on the moving clock).

If you draw a proper spacetime diagram of the above, note that the triangle you draw will not obey the ordinary Pythagorean theorem of Euclidean geometry--the side of the triangle $\Delta S$ will be what looks like the "hypotenuse" of the triangle, and the side $\Delta t$ will be the vertical leg (the side $\Delta x$ will be the horizontal leg--your diagram did get that one thing right), even though $\Delta t$ is longer than $\Delta S$. That is because the geometry of spacetime is not Euclidean; it's Minkowskian. There is no way to draw an undistorted diagram of Minkowski spacetime on a Euclidean piece of paper; you have to accept that some things in the diagram will not work the way ordinary diagrams in Euclidean geometry work.