Looking to understand time dilation

[ghwellsjr]

So you can no longer use the excuse that you have learned from Einstein that it makes a difference whether the traveling twin has stopped or not, or that it makes any difference in the reading on his clock whether he has stopped or not.

Please do not put words into my mouth! Read what I write not what you assume that I have written. That is a matter of simple courtesy.

I do read what you write. Here's what you have written:
1) The traveling twin, after he comes back to his starting point and stops so that he is at rest with the other twin, is the same age as the other twin.
2) While the twin is in motion, each twin sees the other one as getting younger than himself but this is just an illusion.
3) I learned all this from only Einstein's writings, not from any textbook, reference material, website, or teacher.

Is this an accurate portrayal of what you have written and what you believe?

(If you deny any of this, please tell me where you have posted that you have changed your mind and I will post exactly where you have made these kinds of statements.)

If you refuse to accept the fact that a traveling clock ends up losing time, then there is no point in trying to understand a theory that supports that fact because you will suspect the theory. That is what has been going on in this thread with you since the beginning.

No.No. No. One has to understand the theory to understand the results of the experiments or one arrives at something like phlogiston.

Let me repeat once more I HAVE NO PROBLEM WITH EINSTEIN'S THEORY OF SR
I have problems with conclusions that have been drawn from it. But that is not necessarily with those conclusions but the fact that NO ONE will answer my questions about them.

What are you saying "No. No. No." to? That I am misquoting you? That experimental evidence only makes sense when there is a theory to "explain" it?

Do you accept the experimental evidence of MMX that the measured round-trip speed of light always comes out the same in all directions, no matter what the speed of the experimental apparatus is relative to another time when the experiment was performed? This predates SR and there was no theoretical prediction at the time that said this would happen, in fact, quite the opposite. Existing theories had to be abandoned in favor of new theories that would co-incide with experimental evidence. Do you not agree that theories are driven by experiments and not the other way around?

 

[ghwellsjr]

I can't believe you are sincere in presenting these kinds of questions. Can you cite your reference to Einstein's quote? I'm sure we will see that he is in the process of developing an argument and I'm sure you could figure it out if you would just read the rest of his argument instead of trying to convince all the rest of us that you alone understand Einstein and his theories.

Reference: http://www.bartleby.com/173/M5.GIF"

And for goodness sake stop saying I am trying to say something that I am not.

I am raising a question about what appears to me to be an assumption.

Please explain why if time is seen to pass differently from different FoRs, then we should measure it on the same scale? It is surely a simple enough question.
Einstein is comparing two quantities. He says they are equal. I am merely querying in what way they are equal. That does not seem unreasonable to me.

To use the same scale of time on each side of the equation implies that time passes at the same RATE in every FoR and I don't believe that is what Einstein was saying. So show me where I am going wrong.

Your citation only displayed a graphic of the math. Can you provide the citation for Einstein's text, please?

Einstein is saying "that time passes at the same RATE in every FoR". What in his writings have led you to say "I don't believe that is what Einstein was saying"?

 

[Dale]

So if this causes so much affront to everyone will somebody EXPLAIN why.

I think that the best way to do that is to continue with the work on your spacetime diagram. As we clear that up I think that your questions will be cleared up also. I hope you do not stop the work on that.

 

[Grimble]

Your citation only displayed a graphic of the math. Can you provide the citation for Einstein's text, please?

Einstein is saying "that time passes at the same RATE in every FoR". What in his writings have led you to say "I don't believe that is what Einstein was saying"?

Yes, apologies: it is in chapter XII http://www.bartleby.com/173/12.html"

 

[ghwellsjr]

Yes, apologies: it is in chapter XII http://www.bartleby.com/173/12.html"

It's actually in chapter IX, The Relativity of Simultaneity, and here is the full paragraph:

"Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event."

You left off the first sentence of the paragraph in your quote. Do you have a problem with this paragraph? Do you understand it? Do you believe it?

And I would like a reply to my questions from post #251, please, I don't want to be accused of misquoting you.

 

[Grimble]

Grimble, listen to me: Pick anyone inertial reference frame. Time passes at some particular rate for all objects and clocks at rest in that frame, even imaginary clocks that don't exist but would be at rest if they did. Time is dilated for all other objects or clocks moving in that frame, the faster they move, the more the time dilation (which means the slower time passes for them).

Now you can pick any other inertial reference frame, say one in which one of those objects or clocks was moving in the first frame but in this new frame is at rest. In this new frame, time passes at exactly the same rate for all objects and clocks at rest in this new frame and more slowly for all other objects that are moving in this new frame.

This is the principle of relativity that you claim you have learned from Einstein. Do you agree or disagree with this concept?

Yes, that is basic relativity

I'm not asking you if you understand how it could be true, just if you agree that this is what Einstein and the theory of special relativity are saying?

Yes.

 

[Grimble]

I do read what you write. Here's what you have written:
1) The traveling twin, after he comes back to his starting point and stops so that he is at rest with the other twin, is the same age as the other twin.
2) While the twin is in motion, each twin sees the other one as getting younger than himself but this is just an illusion.
3) I learned all this from only Einstein's writings, not from any textbook, reference material, website, or teacher.

Is this an accurate portrayal of what you have written and what you believe?

No it is what I have found the theory to point to and what I want someone to tell me WHY it is wrong. I just happen to have said it as statements rather than questions as questions in the past have been answered by "go away and read about it".

(If you deny any of this, please tell me where you have posted that you have changed your mind and I will post exactly where you have made these kinds of statements.)


What are you saying "No. No. No." to? That I am misquoting you? That experimental evidence only makes sense when there is a theory to "explain" it?

What you wrote was

If you refuse to accept the fact that a traveling clock ends up losing time, then there is no point in trying to understand a theory that supports that fact because you will suspect the theory. That is what has been going on in this thread with you since the beginning.

But does the theory support the fact? It certainly supports the fact that it is seen to go slow in the 'reality' of another FoR, one in which it is moving; yet does it prove that it goes slows per se? In its own FoR? And if it does will someone explain it to me?

Do you accept the experimental evidence of MMX that the measured round-trip speed of light always comes out the same in all directions, no matter what the speed of the experimental apparatus is relative to another time when the experiment was performed? This predates SR and there was no theoretical prediction at the time that said this would happen, in fact, quite the opposite. Existing theories had to be abandoned in favor of new theories that would co-incide with experimental evidence. Do you not agree that theories are driven by experiments and not the other way around?

I'm sorry for my ignorance in recognising references but what is MMX.
AS for 'which drives which' I believe it happens both ways.

 

[Grimble]

It's actually in chapter IX, The Relativity of Simultaneity, and here is the full paragraph:

"Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event."

You left off the first sentence of the paragraph in your quote. Do you have a problem with this paragraph? Do you understand it? Do you believe it?

No, Yes, and yes. I also left out the rest of the chapter leading up to that point.

 

[sylas]

No it is what I have found the theory to point to and what I want someone to tell me WHY it is wrong. I just happen to have said it as statements rather than questions as questions in the past have been answered by "go away and read about it".

It's wrong because if you do the calculations correctly, as described by Einstein, you find that twin who went to a star and came back ends up younger than the twin who remains at Earth the whole time.

It is also incorrect to think of time dilation as "illusion". It is not illusion.

There is an ambiguity in your second point when you speak of "seeing" what the other clock is doing; we explained this previously. What you see is different from the time dilation because you also need to consider the changing distance and changing light travel time, and that makes a big difference to what you "see".

Your conclusions are wrong because they are different from the correct answer.

I suspect you are asking where specifically you go wrong; that I cannot be sure of. You haven't explained your own reasoning sufficiently clearly for me to see where you go wrong. Or possibly I haven't looked hard enough.

To get the correct answers for the age of EITHER twin, you just need to pick a frame of reference (any frame will do) and do all the calculations in that frame. The simplest frame to use is the Earth rest frame. But you can use other frames just as well, and you get the same answer.

Another useful approach is to consider what you do actually see another clock doing. For example, when a clock is moving away at velocity v, you see it moving more slowly by a factor \sqrt{\frac{c+v}{c-v}}. This is the doppler formula; it includes the effects of changing distance and also of time dilation. If v is negative, then a clock is approaching, and it is seen to be moving faster.

Consider a ship moving at 60% light speed to a star 6 light years distance, and then returning. For an observer on Earth, the ship takes 10 years to get there, and 6 years after that the light gets back for you to see the arrival. You thus see the ship moving away for 16 years, and all that time it is "seen" to tick more slowly by a factor of 2. You thus see 8 years pass on the clock. This is, of course, the same as what you get for 10 years travel with a clock dilated by the gamma factor 1.25.

For the next 4 years, you observe the ship returning, and over that time you see the clocks running twice as fast... you see 8 years recorded on the ship's clock; the same duration, of course.

So you see the traveling clock advance 16 years in total, during the 20 years it was away.

For the traveler, they see Earth moving away for 8 years (as they voyage to the star), and then then see Earth moving back again for 8 years (as they voyage back). On the first part, the Earth clock advances slowly, recording only 4 years. In the second part the clock advances more rapidly, recording 16 years. So in total, the traveling twin observes the Earth clock advance 20 years, during the 16 years of their trip recorded on the ship clock.

Both twins agree on what the clocks show. The twin who went to the star and returns is 16 years older, and the twin who stayed on Earth is 20 years older.

If you get anything different -- and your points 1 and 2 about no difference in age and about time changes being "illusion" are indeed different -- then you are incorrect.

Sorting out how you went astray may take a while.

Cheers -- sylas

 

[yuiop]

instead of a tape measure I like to use a
long line of stationary stations spaced 1 light sec apart and
each simultaneously sending out radio pulses at one sec intervals.

one can then create a second long line of stations' which are moving along with the moving twin.
from the traveling twins point of view these new stations' are spaced 1 light sec apart and
each is simultaneously sending out radio pulses at one sec intervals.

from the stationary twins point of view this new line of stations' is shrunk by a factor of gamma
and the time' between its pulses' is time dilated (increased) by a factor of gamma.

dont forget that the length of an object is the distance between front and back at one 'simultaneous' moment.

The attached drawings are based on a similar idea and illustrate time dilation and the relativity of simultaneity.

In the first drawing below (Blue Frame), the blue clocks are stationary and the red clocks are moving at v = sqrt(3/4) = 0.866c which was specially chosen because the time dilation factor works out as 2 which makes the numbers easier to handle. To read the drawing the initial situation is the lowest black rectangle, and subsequent rectangles working upwards illustrate how the situation unfolds with time. Each rectangle can be thought of as time slice in a given reference frame, or a frame in video recording.

In the next drawing (Red frame), the red clocks are stationary and the blue clocks are moving at 0.866c to the left.

When two clocks pass each other and are right alongside each other then this can called an event. Events are universal and different observers will agree on what the passing clocks show at a given event. For example call the event the B clock passing the F clock event BF. It can be seen that in either reference frame, that at event BF the blue B clock read +0.5 seconds and the red F clock read -2.0 seconds.

Consider the events AC and BC in the blue frame. It can be seen that 1 second elapses on the red C clock and this is a proper time measurement as it measured by a single clock present at both events. The coordinate time interval according to the blue reference frame is the time on the blue B clock minus the time on the blue A clock so the coordinate interval is 2 seconds and the blue observer considers the red C clock to be running slow because only 1 second elapses on the red clock. The Blue observer considers it valid to use two clocks because as far as they are concerned the clocks are synchronised.

Now consider events BG and BC in the second drawing (Red RF). 2 seconds elapse on the blue B clock. The red frame observer measures the interval between those two events as the time indicated on the red C clock (+1.0 seconds) minus the time indicated on the red G clock (-3.0 seconds) to give an elapsed coordinate time of 1-(-3) = 4 seconds. As far as the red observer is concerned 4 seconds has passed on his clocks when 2 seconds elapsed on the moving blue clock so the red observer considers the blue clock to be running slow. The red observer considers it equally valid to use two different clocks to measure the interval between the two events because they equally consider their clocks to be synchronised.

It can be easily seen in the diagrams that what one observer considers to be synchronised is not synchronised as far as another observer with relative velocity is concerned. It can be seen that all observers agree on the proper time interval between two events, recorded by a single clock that is present at both events. If the time interval between two events is made by two spatially separated but synchronised clocks, where neither clock is present at both events, then this is a coordinate measurement and is observer dependent.

I hope the diagrams are useful to anyone trying to understand time dilation and the relativity of simultaneity.

 

[granpa]

if the stations are 1 light sec apart and simultaneously sending out radio pulses at 1 sec intervals then
you can see that each sends out its signal when it receives the signal from the others.

If the line of stations is moving then you can easily see that they can't be synchronized from the point of view of a stationary observer

 

[yuiop]

if the stations are 1 light sec apart and simultaneously sending out radio pulses at 1 sec intervals then you can see that each sends out its signal when it receives the signal from the others. If the line of stations is moving then you can easily see that they can't be synchronized from the point of view of a stationary observer

I agree. It would be nice to do an animation of that sometime.

 

[Grimble]

It's wrong because if you do the calculations correctly, as described by Einstein, you find that twin who went to a star and came back ends up younger than the twin who remains at Earth the whole time.

It is also incorrect to think of time dilation as "illusion". It is not illusion.

There is an ambiguity in your second point when you speak of "seeing" what the other clock is doing; we explained this previously. What you see is different from the time dilation because you also need to consider the changing distance and changing light travel time, and that makes a big difference to what you "see".

Your conclusions are wrong because they are different from the correct answer.

I suspect you are asking where specifically you go wrong; that I cannot be sure of. You haven't explained your own reasoning sufficiently clearly for me to see where you go wrong. Or possibly I haven't looked hard enough.

Thank you Sylas, I appreciate your time and the time that others have spent in trying to answer my Confusion.

Several of you seem to be struggling to see where I am coming from and the points that are bothering me. It would seem prudent to me, therefore, to take a step back and tell you all just what I think and what I am asking.

Here goes: First of all I will repeat that the one document I have read and worked from is http://www.bartleby.com/173/" ; which according to his preface is:

intended, as far as possible, to give an exact insight into the theory of Relativity to those readers who, from a general scientific and philosophical point of view, are interested in the theory, but who are not conversant with the mathematical apparatus of theoretical physics.

which I have taken to be his 'everyman's guide'.

Now in Chapter VII: The Apparent Incompatibility of the Law of Propagation of Light with the Principle of Relativity; Einstein writes:

“At this juncture the theory of relativity entered the arena. As a result of an analysis of the physical conceptions of time and space, it became evident that in reality there is not the least incompatibility between the principle of relativity and the law of propagation of light,and that by systematically holding fast to both these laws a logically rigid theory could be arrived at. This theory has been called the special theory of relativity to distinguish it from the extended theory, with which we shall deal later. In the following pages we shall present the fundamental ideas of the special theory of relativity.”

Now when I first read this the question that was at the forefront of my mind was “and how does it do that?” So I continued reading but was disappointed. How Special Relativity addressed the question of holding fast to both postulates was not described.

My next realisation was that it had to be something simple, basic and straightforward, as he did not deem it necessary to spell it out.

Can anyone else explain what he was referring to and how exactly SR resolved the conundrum?

So I resolved to look at one of the most common starting points: the moving light clock with which I am sure you are all familiar.

So what is the problem with it?

Well if for an observer that is at rest with respect to the light clock he will measure 1 second for the light to hit the mirror and return.
Yet for an observer for whom the clock is moving the light takes a longer path; how then can it meet both of Einstein's postulates and both take the same time (relativity) and still travel at the speed of light (Constancy of 'c')?
For if it meets the first criterion of the constant time it must surely travel faster than 'c'; while on the contrary if it travels at 'c' it must take longer than 1 second to hit the mirror and return.

I found two things in solving this riddle:
1. I derived the Lorentz Transformation Equations as a natural mathematical outcome
2. As the speed of light has to be the same, then we are left with the fact that 1 second for the resting observer has the same duration as γ seconds does for the moving observer. That it is the rate of passage of time and the scale of the units of measurement that change.

In other words we are letting go of the concept of absolute time just as Einstein says we must in Chapter IX: The Relativity of Simultaneity.
Where he writes: “Now before the advent of the theory of relativity it had always tacitly been assumed in physics that the statement of time had an absolute significance, i.e. that it is independent of the state of motion of the body of reference. But we have just seen that this assumption is incompatible with the most natural definition of simultaneity; if we discard this assumption, then the conflict between the law of the propagation of light in vacuo and the principle of relativity (developed in Section VII) disappears.”

By letting go of that assumption we can see that it is the unit length * the number of units that is constant.

Very well, but could I find anything to support that contention?

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t' = γt'
which gives us c = x/t = x'γ²/t'

Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.

In order to address these concerns I turned to Chapter XII: the Behaviour of Measuring-Rods and Clocks in Motion.

Examining this all becomes clear.

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

And if the ratio of lengths is x/x' = 1/γ then that is also the ratio of unit sizes
while the ratio of times is t/t' = γ then that is the ratio of the number of units.
Which gives us the unit size * the number of units = γ/γ = 1.

Now to me that is all very simple and straightforward and is based solely upon that one paper and understanding what he wrote in it. So I hope everyone can see what I think, why I think it and that everything that follows should fit into that basic idea that the durations are equal. As that is how SR satisfies that original conundrum of the two apparently conflicting postulates.

Now can someone explain what is wrong with that reasoning/logic?

 

[Simplyh]

The attached drawings are based on a similar idea and illustrate time dilation and the relativity of simultaneity.

In the first drawing below (Blue Frame), the blue clocks are stationary .

I don't think your diagram is correct because you are assuming that two events at distance have the same time coordinate in the frame at rest, which is not correct.
SR definition of simultaneity of events at distance on a rest frame clearly states that the time coordinate at B must be equal to tA + (xB-xA)/c, being tA the time coordinate at A, xB and xA the space coordinates of B and A, respectively. In plain English: the time at B must be the time at A plus the time needed for light to travel from A to B.
There is no sense of claiming time to be absolute in the rest frame; otherwise it should also be absolute in all inertial frames.

 

[Simplyh]

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t' = γt'
which gives us c = x/t = x'γ²/t'

I think you made a mistake because of the ambiguous words of English. Time dilatation means that time flows slower, has a lesser value. So, mathematically it is the same as space contraction and t=t'/y, if y means a contraction factor. But I think you have realized it by yourself.

 

[Grimble]

I think you made a mistake because of the ambiguous words of English. Time dilatation means that time flows slower, has a lesser value. So, mathematically it is the same as space contraction and t=t'/y, if y means a contraction factor. But I think you have realized it by yourself.

I think you should read chapter XII and see what Einstein says!

 

[ghwellsjr]

If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t' = γt'
which gives us c = x/t = x'γ²/t'

Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.

You've got t' = γt' which is clearly a mistake. I'm going to guess that you meant t = γt' but that is also wrong. It should be t' = γt or t = t'/γ from which you can see that c = x'/t'.

Einstein derives this in chapter XI of your referenced book.

 

[Grimble]

You've got t' = γt' which is clearly a mistake. I'm going to guess that you meant t = γt'

thank you that is indeed the case.

but that is also wrong. It should be t' = γt or t = t'/γ from which you can see that c = x'/t'.

Einstein derives this in chapter XI of your referenced book.

Then I must be misreading Chapter XII.

In Chapter XI he derives the Lorentz Transformations whereas in Chapter XII he uses them viz

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.

Now to my understanding γ =

And Einstein has just written that using LT equations 1 and 4 that

where t' is set to 1 which I read as t = γt'

So either I am misreading it or are you saying that Einstein was wrong?

 

[Dale]

If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t' = γt'
which gives us c = x/t = x'γ²/t'

Avoid these formulas like the plague. Use only the full Lorentz transform. It will automatically simplify to these formulas when appropriate and you will not accidentally use them when inappropriate.

 

[ghwellsjr]

So either I am misreading it or are you saying that Einstein was wrong?

I would never say that Einstein was wrong so you must be misreading it.

Gamma is always a number greater than one, correct?

The last paragraph in chapter 12 says that the time in the moving frame is larger than in the rest frame, correct?

Therefore, if t is the time in the rest frame and t' is the time in the moving frame and you want to make an equation in which t' is larger than t, then it should be t' = γt, correct?

The equations that you have "quoted" from Einstein's book don't exist in chapter 12 but they are in chapter 11, correctly, that is, not as you have quoted them. In chapter 12, he is using them along with text and specific values applied to t and t' to illustrate the application of the equations from chapter 11. I will admit, it is confusing but I think you can see what he meant once you read carefully all the text.

 

[Grimble]

I would never say that Einstein was wrong so you must be misreading it.

Gamma is always a number greater than one, correct?

Yes of course, it has to be.

The last paragraph in chapter 12 says that the time in the moving frame is larger than in the rest frame, correct?

Well let us see what Einstein says viz

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.

Now read what Einstein is saying here. He is comparing 1 second of time, occurring in and 'judged' from the traveling frame (t') with the number of seconds (t) that are 'judged' to pass from the stationary frame. He says that 'judged' from the K frame more than 1 second passes between clicks in the traveling K' frame.

Therefore, if t is the time in the rest frame and t' is the time in the moving frame and you want to make an equation in which t' is larger than t, then it should be t' = γt, correct?

I think that you have to pay attention to what he is saying.

If you read further in what I wrote I EXPLAIN this and why these mistakes have been and are still being made.

The equations that you have "quoted" from Einstein's book don't exist in chapter 12 but they are in chapter 11, correctly, that is, not as you have quoted them

.
I am sorry but I fail to understand what you are trying to say here?
The quotes I made are direct quotes as anyone may see for themselves!

In chapter 12, he is using them along with text and specific values applied to t and t' to illustrate the application of the equations from chapter 11. I will admit, it is confusing

maybe to you but not I think to Einstein nor to anyone who reads what he wrote rather than what they think he should have written. I don't believe that after the time he spent writing a 'simple' guide that he left anything confusing.
The text of those Chapters is simple straightforward and easy to understand. and

I think you can see what he meant once you read carefully all the text.

 

[ghwellsjr]

Grimble, are still claiming that the correct equation for the speed of light in the moving frame has a gamma squared factor in it?

 

[nismaratwork]

Grimble, are still claiming that the correct equation for the speed of light in the moving frame has a gamma squared factor in it?

*nods*

 

[JesseM]

Now read what Einstein is saying here. He is comparing 1 second of time, occurring in and 'judged' from the traveling frame (t') with the number of seconds (t) that are 'judged' to pass from the stationary frame. He says that 'judged' from the K frame more than 1 second passes between clicks in the traveling K' frame.

To be clear, are you saying the clock is at rest in the "stationary frame" or the "travelling frame"? Einstein said the clock is permanently situated at the origin (x' = 0) of K' so I guess since you use a primed t' to denote the "travelling frame" you're saying the clock is at rest in the traveling frame, correct? In that case, if t' is the time between two ticks of the clock in the traveling frame, while t is the time between the same two ticks in the stationary frame, then t should be larger than t', meaning you're correct to say t = γt'. Nowadays the more common convention with primed vs. unprimed time intervals in the time dilation equation is to say unprimed is the time in the clock's rest frame while primed is the time in the frame where the clock is moving at speed v, the reverse of your (and Einstein's) notation, so probably that's why ghwellsjr thought you got it wrong.

 

[ghwellsjr]

To be clear, are you saying the clock is at rest in the "stationary frame" or the "travelling frame"? Einstein said the clock is permanently situated at the origin (x' = 0) of K' so I guess since you use a primed t' to denote the "travelling frame" you're saying the clock is at rest in the traveling frame, correct? In that case, if t' is the time between two ticks of the clock in the traveling frame, while t is the time between the same two ticks in the stationary frame, then t should be larger than t', meaning you're correct to say t = γt'. Nowadays the more common convention with primed vs. unprimed time intervals in the time dilation equation is to say unprimed is the time in the clock's rest frame while primed is the time in the frame where the clock is moving at speed v, the reverse of your (and Einstein's) notation, so probably that's why ghwellsjr thought you got it wrong.

Jesse, this is what Grimble is claiming (I've corrected his typo):

If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t = γt'
which gives us c = x/t = x'γ²/t'

 

[Dale]

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t' = γt'
which gives us c = x/t = x'γ²/t'

Hi Grimble, you haven't responded to my repeated suggestions to avoid these formulas (they are never necessary and often problematic) and stick to the Lorentz transform only. The reason you are getting a bad answer is because you are using formulas that do not apply. The time dilation formula applies when the clock is at rest in one of the frames, which is never the case for light. The length contraction formula only applies when you have a pair of events which are simultaneous in each frame and associated with the ends of a single object, which is also never the case for a single pulse of light.

Now, with that motivation, please try to re-do your line of reasoning using the Lorentz transform.

 

[yuiop]

...So I resolved to look at one of the most common starting points: the moving light clock with which I am sure you are all familiar.

So what is the problem with it?

Well if for an observer that is at rest with respect to the light clock he will measure 1 second for the light to hit the mirror and return.
Yet for an observer for whom the clock is moving the light takes a longer path;

Correct so far.

In the rest frame (S) of the light clock, the time taken is t = 1 s and the distance traveled by the photon is d = 1 ls.

In the frame in which the light clock is moving (S') the distance traveled by the photon is d' = γ ls and the time taken according to a clock moving relative to the light clock is t' = γ s.

... how then can it meet both of Einstein's postulates and both take the same time (relativity) and still travel at the speed of light (Constancy of 'c')?

This is where the confusion sets in. You say "both take the same time (relativity)" without specify according to what clocks or observers. If you are not always careful to specify the observer that makes the measurement in relativity you are doomed to be perpetually confused.

For if it meets the first criterion of the constant time it must surely travel faster than 'c'; while on the contrary if it travels at 'c' it must take longer than 1 second to hit the mirror and return.

Relativity states that the speed of light is always c in any inertial reference frame so you can rule out the first option. Since you have already stated that the photon takes one second in the rest frame of the light clock it can be be reasonably assumed you are talking about frame S' when you say "while on the contrary if it travels at 'c' it must take longer than 1 second to hit the mirror and return". This is correct. It DOES take longer than 1 second in frame S' for the photon to complete its round trip. In frame S' it takes t' = γ s > (1 s).

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …

Correct.

But x = x'/γ, while t' = γt'

x = x'/γ implies x' = xγ which is correct.
t' = γt' is obviously not correct. I assume you meant t' = tγ.
(Remember you said earlier " if it travels at 'c' it must take longer than 1 second to hit the mirror and return") .. so t' must be greater than t.

which gives us c = x/t = x'γ²/t'

when done correctly you should get c = x/t = (x'γ)/(t'γ) = x'/t',

Now can someone explain what is wrong with that reasoning/logic?

Yep, you are making simple algebraic errors.

 

[JesseM]

If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t = γt'
which gives us c = x/t = x'γ2/t'

OK, the problem here is that x = x'/γ and t = γt' don't work for any arbitrary pair of events, the time dilation formula t = γt' is only valid when you are talking about the time between a pair of events that both happened at the same position in the primed frame (like ticks of a clock at rest in the primed frame), while the length contraction formula is x = x'/γ only valid when you are talking about the length in both frames of an object at rest in the primed frame, or equivalently the distance in both frames between a pair of events simultaneous in the unprimed frame (if the events were on either end of an object at rest in the primed frame, then since they are simultaneous in the unprimed frame the distance between these events is the object's length in the unprimed frame, whereas even though they are non-simultaneous in the primed frame, since the object is at rest in the primed frame the distance between them still counts as the object's 'length' in the primed frame).

If you are talking about two events on the worldline of a light beam such that c=x/t and c=x'/t', neither of these conditions would be satisfied so you can't use the length contraction and time dilation formulas. As DaleSpam says, you should really use the general Lorentz contraction formulas to avoid this sort of confusion:

x' = γ(x - vt)
t' = γ(t - vx/c^2)

And

x = γ(x' + vt')
t = γ(t' + vx'/c^2)

You can see that in the special case where two events happened at the same position in the primed frame so x'=0, the equation t = γ(t' + vx'/c^2) reduces to the time dilation equation t = γt'. Likewise in the special case where the two events were simultaneous in the unprimed frame so t=0, the equation x' = γ(x - vt) reduces to x = x'/γ. But again, it's not valid to use the time dilation and length contraction equations in cases where the events you're considering don't satisfy the required conditions, whereas it's always valid to use the more general Lorentz transformation equations.

 

[yuiop]

And Einstein has just written that using LT equations 1 and 4 that

where t' is set to 1 which I read as t = γt'

So either I am misreading it or are you saying that Einstein was wrong?

It is more likely that he set t_0 to 1 so that t = \gamma t_0 where in this case t_0 is the proper time measured by a single clock according to an observer at rest with that clock and t is the coordinate time measured by an observer that is moving relative to that single clock.

The confusion comes about because in other posts and textbooks, t is used to represent proper time and t' is used to represent coordinate time (measured by spatially separated clocks) and sometimes the meanings of t' and t are reversed. What you can be sure o,f is that the coordinate time is always greater than the proper time.

 

[yuiop]

I don't think your diagram is correct because you are assuming that two events at distance have the same time coordinate in the frame at rest, which is not correct.
SR definition of simultaneity of events at distance on a rest frame clearly states that the time coordinate at B must be equal to tA + (xB-xA)/c, being tA the time coordinate at A, xB and xA the space coordinates of B and A, respectively. In plain English: the time at B must be the time at A plus the time needed for light to travel from A to B.
There is no sense of claiming time to be absolute in the rest frame; otherwise it should also be absolute in all inertial frames.

This is not correct. When synchronising the clocks a signal is sent from A at time tA. When B receives this signal he sets his clock to tA + (xB-xA)/c but in the meantime clock A's time has not stood still, but has advanced by (xB-xA)/c so that when B sets his clock to tA + (xB-xA)/c, clock A is also reading tA + (xB-xA)/c and both clocks are synchronised according to that frame. Other observers moving relative the frame in which clocks A and B are rest will not agree they are synchronised and will say they differ by a factor of (xB-xA)*v/c^2 with the trailing clocks showing more elapsed time than the trailing clocks.

 

[ghwellsjr]

...
So I resolved to look at one of the most common starting points: the moving light clock with which I am sure you are all familiar.
...
Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.
...

It's easy to misunderstand how length contraction combined with time dilation would result in the same speed (for light) being measured for a moving clock/observer as compared to a stationary one. Let me explain:

You seem to understand how a light clock works. In a rest frame where the speed of light is the same in all directions, it is easy to understand that if we place two mirrors a fixed distance apart in any orientation, the light will take the same time to make a round trip back and forth between the mirrors, no matter what the orientation of the two mirrors is.

Now we consider a light clock in motion. First we position the two mirrors so the light reflects back and forth at right angles to the direction of motion. You probably understand that the light takes longer to traverse the path between the two mirrors because it has to take diagonal paths and you can probably figure out that the total round-trip path is gamma multiplied by the previous path length (when the light clock was at rest) multiplied by 2 and that this demonstrates time dilation.

But what happens if we rotate the light clock 90 degrees so that the light is going back and forth along the direction of the motion of the mirrors? Well now if you follow the details (which I'm sure you can), you will need to move the mirrors closer together in order for the light to make the round trip in the same time as it did before we rotated the light clock. This demonstrates length contraction which is the original distance divided by gamma.

Now if we want to calculate the speed of light as demonstrated by this light clock, we might naively say that the speed is equal to the distance divided by the time and since the distance was the orginal distance divided by gamma and the time was the original time multiplied by gamma we should get the speed as distance divided by time divided by the square of gamma. This is what you calculated in a previous post, but can you see how this is wrong?

There are two ways to demonstrate that it is wrong. If we look at the first orientation of the light clock where length contraction is not involved, we use the actual distance that the light traveled along the diagonals, not the distance between the mirrors, correct? In other words, we make mental note of where the mirrors were when the light struck them and we use the actual distance traversed by the light when we calculate the total distance. This distance is gamma multiplied by the original distance multiplied by 2 (for the round trip).

And if we look at the second orientation of the light clock where length contraction is involved, we have to do the same thing as before: make mental note of where the mirrors were when the light struck them and again use the actual distance traversed by the light to calculate the total round trip distance. This distance is again gamma multiplied (not divided) by the original distance multiplied by 2.

If you actually do this exercise you will also see that in the last orientation, the two halves of the round trip are not equal in length--when the light is going in the same direction as the mirrors are moving it is longer, when the light is going in the opposite direction as the mirrors are moving it is shorter, and both these distances are different than the two halves of the distances in the first orientation. And in the same way we can say that the time intervals involved for light to traverse each path between the mirrors is different. It is this difference that is the reason for the relativity of simultaneity.

So you can see that even though we talk about length contraction for a moving object, for the light path, it is actually a length stretching because the light is striking the mirrors at different times between which the mirrors are moving to new locations. The Lorentz Transform takes care of the correct calculations.

 

[Grimble]

To be clear, are you saying the clock is at rest in the "stationary frame" or the "travelling frame"? Einstein said the clock is permanently situated at the origin (x' = 0) of K' so I guess since you use a primed t' to denote the "travelling frame" you're saying the clock is at rest in the traveling frame, correct? In that case, if t' is the time between two ticks of the clock in the traveling frame, while t is the time between the same two ticks in the stationary frame, then t should be larger than t', meaning you're correct to say t = γt'. Nowadays the more common convention with primed vs. unprimed time intervals in the time dilation equation is to say unprimed is the time in the clock's rest frame while primed is the time in the frame where the clock is moving at speed v, the reverse of your (and Einstein's) notation, so probably that's why ghwellsjr thought you got it wrong.

Yes absolutely! Thank you Jesse.

Let me refer back to my post 263 where I wrote

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t = γt'
which gives us c = x/t = x'γ²/t'

Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.

In order to address these concerns I turned to Chapter XII: the Behaviour of Measuring-Rods and Clocks in Motion.

Examining this all becomes clear.

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

And if the ratio of lengths is x/x' = 1/γ then that is also the ratio of unit sizes
while the ratio of times is t/t' = γ then that is the ratio of the number of units.
Which gives us the unit size * the number of units = γ/γ = 1.

In that post I was trying to explain where I was coming from and the logic behind some confusing and apparently conflicting aspects of SR, as I see it

And yes, if we change the time measurement to unit size instead of number of units, then the time formula becomes t = t'/γ and the speed of light become x'/t'.

As you can see I am not trying to be difficult.

Grimble

 

[Grimble]

Jesse, this is what Grimble is claiming (I've corrected his typo):

And if you read that passage in full you will see that I raise that as one of the puzzles that were niggling me and how it occurs and how to resolve it.

It is an APPARENT discrepancy that disappears upon investigation.

Stop taking it out of context. PLEASE

Grimble

 

[Grimble]

Hi Grimble, you haven't responded to my repeated suggestions to avoid these formulas (they are never necessary and often problematic) and stick to the Lorentz transform only. The reason you are getting a bad answer is because you are using formulas that do not apply. The time dilation formula applies when the clock is at rest in one of the frames, which is never the case for light. The length contraction formula only applies when you have a pair of events which are simultaneous in each frame and associated with the ends of a single object, which is also never the case for a single pulse of light.

Now, with that motivation, please try to re-do your line of reasoning using the Lorentz transform.

Once again, please read what I wrote in post 263. Not one part of it but all of it. I was explaining how I had arrived at my understanding.

I was saying that there was potential confusion with those formulae and why and how to understand it.

 

[Grimble]

Correct so far.

In the rest frame (S) of the light clock, the time taken is t = 1 s and the distance traveled by the photon is d = 1 ls.

In the frame in which the light clock is moving (S') the distance traveled by the photon is d' = γ ls and the time taken according to a clock moving relative to the light clock is t' = γ s.

This is where the confusion sets in. You say "both take the same time (relativity)" without specify according to what clocks or observers. If you are not always careful to specify the observer that makes the measurement in relativity you are doomed to be perpetually confused.

As I was discussing the longer path and it is only longer to the moving observer I think it reasonable to assume that that is the observer concerned.

Relativity states that the speed of light is always c in any inertial reference frame so you can rule out the first option. Since you have already stated that the photon takes one second in the rest frame of the light clock it can be be reasonably assumed you are talking about frame S' when you say "while on the contrary if it travels at 'c' it must take longer than 1 second to hit the mirror and return". This is correct. It DOES take longer than 1 second in frame S' for the photon to complete its round trip. In frame S' it takes t' = γ s > (1 s).

Which is what I am saying ...

x = x'/γ implies x' = xγ which is correct.
t' = γt' is obviously not correct. I assume you meant t' = tγ.
(Remember you said earlier " if it travels at 'c' it must take longer than 1 second to hit the mirror and return") .. so t' must be greater than t.

Again please read this passage in context...
[/QUOTE]

 

[Grimble]

OK, the problem here is that x = x'/γ and t = γt' don't work for any arbitrary pair of events, the time dilation formula t = γt' is only valid when you are talking about the time between a pair of events that both happened at the same position in the primed frame (like ticks of a clock at rest in the primed frame), while the length contraction formula is x = x'/γ only valid when you are talking about the length in both frames of an object at rest in the primed frame, or equivalently the distance in both frames between a pair of events simultaneous in the unprimed frame (if the events were on either end of an object at rest in the primed frame, then since they are simultaneous in the unprimed frame the distance between these events is the object's length in the unprimed frame, whereas even though they are non-simultaneous in the primed frame, since the object is at rest in the primed frame the distance between them still counts as the object's 'length' in the primed frame).

Yes exactly, that is what I am saying - one is measuring the size of the units and the other is counting the number of units.

If you are talking about two events on the worldline of a light beam such that c=x/t and c=x'/t', neither of these conditions would be satisfied so you can't use the length contraction and time dilation formulas. As DaleSpam says, you should really use the general Lorentz contraction formulas to avoid this sort of confusion:

x' = γ(x - vt)
t' = γ(t - vx/c^2)

Let me just point out here that in appendix 1, where Einstein shows a simple derivation of the Lorentz Transformations, in step 6 he writes viz

For the origin of k' we have permanently x' = 0, and hence according to the first of the equations (5)

If we call v the velocity with which the origin of k' is moving relative to K, we then have

i.e. substituting v with x/t

If we follow that substitution by replacing x with vt this gives us t' = γ(t - v²t/c²) or t' = γt(1 - v²/c²) = t/γ for t' = γ(t - vx/c^2)

And

x = γ(x' + vt')
t = γ(t' + vx'/c^2)

You can see that in the special case where two events happened at the same position in the primed frame so x'=0, the equation t = γ(t' + vx'/c^2) reduces to the time dilation equation t = γt'. Likewise in the special case where the two events were simultaneous in the unprimed frame so t=0, the equation x' = γ(x - vt) reduces to x = x'/γ. But again, it's not valid to use the time dilation and length contraction equations in cases where the events you're considering don't satisfy the required conditions, whereas it's always valid to use the more general Lorentz transformation equations.

 

[Dale]

And yes, if we change the time measurement to unit size instead of number of units, then the time formula becomes t = t'/³ and the speed of light become x'/t'.

As you can see I am not trying to be difficult.

No. The formulas DO NOT APPLY here. It is not simply a matter of switching unit size and number of units.

If you want to use those formulas (which I strongly recommend against) then you need to learn their domain of applicability.

 

[nismaratwork]

Grimble, you are of course free to argue this point until the end of time (har har), but if you actually TRY to take DaleSpam's advice and listen to JesseM and Yuiop, you'll learn something. I know I usually do when I that route.

 

[ghwellsjr]

Let me refer back to my post 263 where I wrote

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t = γt'
which gives us c = x/t = x'γ²/t'

Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.

In order to address these concerns I turned to Chapter XII: the Behaviour of Measuring-Rods and Clocks in Motion.

Examining this all becomes clear.

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

And if the ratio of lengths is x/x' = 1/γ then that is also the ratio of unit sizes
while the ratio of times is t/t' = γ then that is the ratio of the number of units.
Which gives us the unit size * the number of units = γ/γ = 1.

In that post I was trying to explain where I was coming from and the logic behind some confusing and apparently conflicting aspects of SR, as I see it

And yes, if we change the time measurement to unit size instead of number of units, then the time formula becomes t = t'/γ and the speed of light become x'/t'.

As you can see I am not trying to be difficult.

Grimble

So you started with t = γt' and you ended up with t = t'/γ? They're both true? Something to do with the first one uses number of units and the second one uses unit size? Can you show us where exactly in your referenced text Einstein explains this?

 

[JesseM]

Yes exactly, that is what I am saying - one is measuring the size of the units and the other is counting the number of units.

I have no idea what that means, both can be understood as the difference in coordinate position or coordinate time between some specific pair of events. Do you agree that the time dilation formula you wrote down is only applicable when you want to compare the time-interval in both frames between two events that happened at the same position in the primed frame, and that the length contraction formula you wrote down is only applicable when we're talking about the length in both frames of an object at rest in the primed frame, or equivalently when we're talking about the distance-interval in both frames between two events which are simultaneous in the unprimed frame? If you agree with that, then do you agree that we can not use these formulas when we want to know the separation in both frames between a pair of events which lie on the worldline of a light beam, i.e. a pair of events such that x/t = c (where x and t stand for the coordinate separation between the two events)?

Let me just point out here that in appendix 1, where Einstein shows a simple derivation of the Lorentz Transformations, in step 6 he writes viz

For the origin of k' we have permanently x' = 0, and hence according to the first of the equations (5)

If we call v the velocity with which the origin of k' is moving relative to K, we then have

i.e. substituting v with x/t

If we follow that substitution by replacing x with vt this gives us t' = γ(t - v²t/c²) or t' = γt(1 - v²/c²) = t/γ for t' = γ(t - vx/c^2)

No, that conclusion isn't valid in general, although it may be valid for some specific pairs of events. When reading Einstein's derivation in Appendix 1, you have to keep track of which equations are supposed to apply generally to the coordinates of arbitrary events like the Lorentz transformation, and which equations are only supposed to hold for events on a specific worldline, like the worldline of the light-signal which he introduces at the start of the derivation, or the worldline of the origin of k' for which that first equation x=(bc/a)*t above is supposed to hold. Note that at the start of the derivation he introduces the equation x=ct, hopefully it's obvious that this holds for events on the worldline of the light-signal sent from the origin at t=0, but that it is not a general equation that applies to the x,t coordinates of any arbitrary event in spacetime! You may want to take a look at this thread about Einstein's derivation in Appendix 1, where I wrote:

The confusing part may be that he wants a general transformation which can translate any event with coordinates x and t to corresponding coordinates x' and t', but he starts by considering the special case of a light beam emitted from the origin at t=0, so x=ct for any point on the light beam's path (this would obviously not be true for arbitrary events that don't lie on this path). He's pointing out that as long as the general transformation has the property (x'-ct') = λ(x-ct) (regardless of the value of x and t), that will guarantee that the light beam has the same speed of c in both coordinate systems , because if x=ct, that equation implies x'=ct' too. I don't know if it'd be possible to come up with a coordinate transformation where it was true that any (x,t) satisfying x=ct would also satisfy x'=ct', but it wasn't true that any arbitrary x,t would satisfy (x'-ct') = λ(x-ct). I guess you could come up with a coordinate transformation that did satisfy that equation but where λ was a function of x and t rather than being a constant, but then it wouldn't be a linear coordinate transformation...

Anyway, the specific step you mention, going from x=(bc/a)*t to v=(bc/a), only works because we are talking about the position as a function of time for a specific entity, namely the origin of k'. If you are talking about the position and time intervals between an arbitrary pair of events it doesn't work, because that "v" symbol specifically refers to the velocity of the k' frame relative to the K frame (note the line "If we call v the velocity with which the origin of k' is moving relative to K"), so for a pair of events on the worldline of an object not at rest in k' it wouldn't be true that x/t = v (of course if you have an arbitrary pair of events on the worldline of some object not at rest in k', you could use a different symbol like v₂ to refer to the object's velocity in the K frame, and then it would be true that x/t = v₂ for the pair of events on its worldline, but you can't use the same symbol v to denote both the velocity of the object and the velocity of the k' frame relative to K). If you are specifically comparing a pair of events on the worldline of an object at rest in k' (i.e. two events that have the same position-coordinate in k', so x'=0), then it would be true that x=vt and for those specific events your derivation t' = γ(t - v2t/c2) or t' = γt(1 - v2/c2) = t/γ would be valid, but of course I already specified that the time dilation equation t = γt' (which is just a rearrangement of t' = t/γ which you got there) only holds in the case where you're looking at two events which occurred at the same position in the t' frame.

 

[JesseM]

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

They are both "number of units". Suppose we have little lights on both the front and back end of an object which is moving relative to the K frame, and both lights flash simultaneously in the K frame. In this case, the distance between the positions of each flash qualifies as the "length" of the object in the K frame (since 'length' is always the distance between the positions of the front and back of an object at a single moment), so you're counting the number of distance-units between the flashes using a ruler at rest in the K frame and calling that the length x of the moving object in the K frame. Meanwhile the object is at rest in the k' frame, so although the flashes at either end of the object happen at different times in this frame (relativity of simultaneity), the distance between them in the k' frame is exactly the same as the distance you'd get if you set off two flashes on either end of the object which were simultaneous in the k' frame. Either way, you're counting the number of distance-units between the flashes using a ruler at rest in the k' frame and calling that the length x' of the object in the k' frame. Then you will find that x = x'/γ, so you can see that the length contraction equation is still talking about the "number of units" between a pair of events (the two flashes) as measured in each frame.

 

[Grimble]

So you started with t = γt' and you ended up with t = t'/γ? They're both true? Something to do with the first one uses number of units and the second one uses unit size? Can you show us where exactly in your referenced text Einstein explains this?

In chapter XII as I said back in post 263.

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.

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.

 

[Grimble]

I have no idea what that means, both can be understood as the difference in coordinate position or coordinate time between some specific pair of events. Do you agree that the time dilation formula you wrote down is only applicable when you want to compare the time-interval in both frames between two events that happened at the same position in the primed frame, and that the length contraction formula you wrote down is only applicable when we're talking about the length in both frames of an object at rest in the primed frame, or equivalently when we're talking about the distance-interval in both frames between two events which are simultaneous in the unprimed frame? If you agree with that, then do you agree that we can not use these formulas when we want to know the separation in both frames between a pair of events which lie on the worldline of a light beam, i.e. a pair of events such that x/t = c (where x and t stand for the coordinate separation between the two events)?

As I said (in that very quote) yes exactly.
All I am saying there in addition is that when Einstein says that the length contracts and the time gets larger it is because he is considering different aspects of the measurements. For the time he is measuring how the length is less and the number of seconds is greater. One may see this as when we calculate dime dilation we again use the size of the unit and t = t'/γ

No, that conclusion isn't valid in general, although it may be valid for some specific pairs of events. When reading Einstein's derivation in Appendix 1, you have to keep track of which equations are supposed to apply generally to the coordinates of arbitrary events like the Lorentz transformation, and which equations are only supposed to hold for events on a specific worldline, like the worldline of the light-signal which he introduces at the start of the derivation, or the worldline of the origin of k' for which that first equation x=(bc/a)*t above is supposed to hold. Note that at the start of the derivation he introduces the equation x=ct, hopefully it's obvious that this holds for events on the worldline of the light-signal sent from the origin at t=0, but that it is not a general equation that applies to the x,t coordinates of any arbitrary event in spacetime! You may want to take a look at this thread about Einstein's derivation in Appendix 1, where I wrote:

Anyway, the specific step you mention, going from x=(bc/a)*t to v=(bc/a), only works because we are talking about the position as a function of time for a specific entity, namely the origin of k'. If you are talking about the position and time intervals between an arbitrary pair of events it doesn't work, because that "v" symbol specifically refers to the velocity of the k' frame relative to the K frame (note the line "If we call v the velocity with which the origin of k' is moving relative to K"), so for a pair of events on the worldline of an object not at rest in k' it wouldn't be true that x/t = v (of course if you have an arbitrary pair of events on the worldline of some object not at rest in k', you could use a different symbol like v₂ to refer to the object's velocity in the K frame, and then it would be true that x/t = v₂ for the pair of events on its worldline, but you can't use the same symbol v to denote both the velocity of the object and the velocity of the k' frame relative to K). If you are specifically comparing a pair of events on the worldline of an object at rest in k' (i.e. two events that have the same position-coordinate in k', so x'=0), then it would be true that x=vt and for those specific events your derivation t' = γ(t - v2t/c2) or t' = γt(1 - v2/c2) = t/γ would be valid, but of course I already specified that the time dilation equation t = γt' (which is just a rearrangement of t' = t/γ which you got there) only holds in the case where you're looking at two events which occurred at the same position in the t' frame.

Thank you yes that does explain another little puzzle

 

[Grimble]

They are both "number of units". Suppose we have little lights on both the front and back end of an object which is moving relative to the K frame, and both lights flash simultaneously in the K frame. In this case, the distance between the positions of each flash qualifies as the "length" of the object in the K frame (since 'length' is always the distance between the positions of the front and back of an object at a single moment), so you're counting the number of distance-units between the flashes using a ruler at rest in the K frame and calling that the length x of the moving object in the K frame. Meanwhile the object is at rest in the k' frame, so although the flashes at either end of the object happen at different times in this frame (relativity of simultaneity), the distance between them in the k' frame is exactly the same as the distance you'd get if you set off two flashes on either end of the object which were simultaneous in the k' frame. Either way, you're counting the number of distance-units between the flashes using a ruler at rest in the k' frame and calling that the length x' of the object in the k' frame. Then you will find that x = x'/γ, so you can see that the length contraction equation is still talking about the "number of units" between a pair of events (the two flashes) as measured in each frame.

But are you not then assuming that the length unit is the same in both frames even though one is length contracted?

Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it - it is still the same clock face displaying the same time wherever it is seen from, whoever it is observed by.
Surely the difference in the time displayed is the length of the time units the ratio of t/t' is the ratio of the lengths of the time units.
A clock cannot tell different times to different observers, yet it may display different durations.

In the same way a ruler alongside an object will always shew the same length wherever it is observed from, it is the size of the units that will change not the number of them.

 

[JesseM]

But are you not then assuming that the length unit is the same in both frames even though one is length contracted?

What do you mean "the same in both frames"? They are both using markings on their own rulers to measure the distance between the two events, and each one would say that the markings on the other guy's ruler are shrunk down to a smaller distance apart than the markings on their own ruler. Of course their rulers are identically constructed so each one looks the same in its own rest frame, but the same is true of the clocks they use to measure the time between two events.

Part of the confusion here may be that the time dilation equation t = γt' and the length contraction equation x = x'/γ are usually seen as "going together" in a pair because the first compares the time in both frames between ticks of a clock at rest in the primed frame, while the second compares the length in both frames of an object at rest in the primed frame. But if you translate them into the language of events, then the first is dealing with the time between two events which occurred at the same spatial position in the primed frame, but the second is dealing with the distance between two events which occurred at the same time in the unprimed frame (I explained why this is equivalent to looking at the length in both frames of an object at rest in the primed frame in post #291). If instead you wanted to pair the time dilation equation with an equation giving the distance between two events which occurred at the same time in the primed frame (so in both cases, the pair of events have a 'special' property in the primed frame), then the pair would look like this:

t = γt'
x = γx'

This sort of confusing aspect of the meaning of the two equations (and why they are normally given as a matched pair) is why it may be a good idea to just take Dalespam's advice and use the more general Lorentz transformation equations, which always work for any pair of events...

Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it - it is still the same clock face displaying the same time wherever it is seen from, whoever it is observed by.

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event. But if you are asking a question like "what reading does the clock show 20 seconds after it passed by observer A", then different frames can disagree about this, because they disagree which event on the clock's worldline happened "20 seconds after it passed by observer A". For example, if the clock is moving at 0.6c relative to observer A, and the clock showed a reading of t'=0 at the time it passed observer A, then in the clock's own rest frame it shows a reading of t'=20 after 20 seconds of coordinate time in this frame, but in the rest frame of observer A it shows a reading of t'=16 after 20 seconds of coordinate time. It may help to realize that each frame is imagined to define the time-coordinates of events using clocks at rest in that frame (and synchronized in that frame according to the frame's definition of simultaneity). So in the clock's own rest frame, the clock itself is a valid way of measuring coordinate time, so 20 seconds passing on the clock means 20 seconds of coordinate time have passed. But in the rest frame of observer A, you'd need two clocks at rest and synchronized in this frame to measure the elapsed time on the moving clock. If a clock right next to observer A read t=0 at the moment the moving clock passed him reading t'=0, and then later the moving clock passed another clock at rest in A's frame (and synchronized with the one next to A) when that clock read t=20 and the moving clock read t'=16, then that's the physical meaning of the claim that the moving clock takes 20 seconds of coordinate time to tick forward by 16 seconds of its own time.

Surely the difference in the time displayed is the length of the time units the ratio of t/t' is the ratio of the lengths of the time units.

No, it's meaningless to talk about "the ratio of the lengths of the time units" because different frames disagree on the ratio--in my frame your time units are longer than mine (because your clock is ticking slower and thus takes longer to tick forward by a given amount), while in your frame your time units are shorter than mine, while in a third frame where we are both moving at the same speed in opposite directions, the ratio of my time units to your time units would be 1:1 (both our clocks are ticking at the same rate in this frame because they both have the same speed).

Again, the time dilation equation t = γt' is looking at some specific pair of events which occurred at the same position in the primed frame (like two readings on a clock at rest in the primed frame), and t is the amount of coordinate time between these specific events in the unprimed frame (as measured by clocks at rest and synchronized in the unprimed frame), while t' is the amount of coordinate time between these specific events in the primed frame (as measured by a clock at rest in the primed frame which is at the same position as each event when they occur).

A clock cannot tell different times to different observers

As I said I think that statement is overly vague. If you specify a specific event on the clock's worldline there can't be disagreement about what reading it shows at that event, but if you use some other type of specification like "the time on the clock 20 seconds of coordinate time after some event" then there can be disagreement on the reading.

 

[Grimble]

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event. But if you are asking a question like "what reading does the clock show 20 seconds after it passed by observer A", then different frames can disagree about this, because they disagree which event on the clock's worldline happened "20 seconds after it passed by observer A". For example, if the clock is moving at 0.6c relative to observer A, and the clock showed a reading of t'=0 at the time it passed observer A, then in the clock's own rest frame it shows a reading of t'=20 after 20 seconds of coordinate time in this frame, but in the rest frame of observer A it shows a reading of t'=16 after 20 seconds of coordinate time. It may help to realize that each frame is imagined to define the time-coordinates of events using clocks at rest in that frame (and synchronized in that frame according to the frame's definition of simultaneity). So in the clock's own rest frame, the clock itself is a valid way of measuring coordinate time, so 20 seconds passing on the clock means 20 seconds of coordinate time have passed. But in the rest frame of observer A, you'd need two clocks at rest and synchronized in this frame to measure the elapsed time on the moving clock. If a clock right next to observer A read t=0 at the moment the moving clock passed him reading t'=0, and then later the moving clock passed another clock at rest in A's frame (and synchronized with the one next to A) when that clock read t=20 and the moving clock read t'=16, then that's the physical meaning of the claim that the moving clock takes 20 seconds of coordinate time to tick forward by 16 seconds of its own time.

So the moving clock reads 20seconds in each frame? And in one 20 seconds has passed, but only 16 seconds in the other (measured on a clock in that frame?) But read from that frame the 20 seconds shown have taken only 16 seconds to pass? so those seconds are only 80% of the duration of 1 second in either frame measured from within that same frame?

Excuse me Jesse but are you not agreeing with me here?

No, it's meaningless to talk about "the ratio of the lengths of the time units" because different frames disagree on the ratio--in my frame your time units are longer than mine (because your clock is ticking slower and thus takes longer to tick forward by a given amount), while in your frame your time units are shorter than mine, while in a third frame where we are both moving at the same speed in opposite directions, the ratio of my time units to your time units would be 1:1 (both our clocks are ticking at the same rate in this frame because they both have the same speed).

No I am saying the ratio for the two times, (proper and coordinate) measured by the same observer.

Again, the time dilation equation t = γt' is looking at some specific pair of events which occurred at the same position in the primed frame (like two readings on a clock at rest in the primed frame), and t is the amount of coordinate time between these specific events in the unprimed frame (as measured by clocks at rest and synchronized in the unprimed frame), while t' is the amount of coordinate time between these specific events in the primed frame (as measured by a clock at rest in the primed frame which is at the same position as each event when they occur).

Yes, t is coordinate time (measured in one frame from the other frame) and t' is proper time(measured in the same frame it occurs in, local time as measured by an adjacent clock, time measured on that clocks worldline)

As I said I think that statement is overly vague. If you specify a specific event on the clock's worldline there can't be disagreement about what reading it shows at that event, but if you use some other type of specification like "the time on the clock 20 seconds of coordinate time after some event" then there can be disagreement on the reading.

I'm not quite sure what you are trying to say here?
The time a clock reads is the time on that clock at anyone instant An Event if you will.
Saying that it will read differently 20 seconds after an event as measured in two different frames means nothing, for then you are comparing two different events.

 

[JesseM]

So the moving clock reads 20seconds in each frame?

At some time in frame #1 the clock reads 20 seconds, and at some time in frame #1 it reads 19 seconds and 18 seconds and so forth, and likewise the same is true in frame #2. But there's no basis for saying it reads 20 seconds at the "same instant" in both frames or anything like that, each observer agrees it goes through a sequence of readings at different time-coordinates but they don't agree on the the time-coordinate of each reading. So if you were to say that you want to know the moving clock's reading when 20 seconds of coordinate time have elapsed since it passed observer A, then in observer A's frame the answer is that the moving clock's reading is 16 seconds, while in the clock's own frame the answer is that the reading is 20 seconds. In that sense one can disagree with your statement that "It cannot show a different time when it is moving can it - it is still the same clock face displaying the same time wherever it is seen from, whoever it is observed by", or at least say that the way you worded it was too vague.

And in one 20 seconds has passed, but only 16 seconds in the other (measured on a clock in that frame?)

No, you got it backwards. 20 and 16 were different opinions about the clock's reading after 20 seconds of coordinate time, not different opinions about the amount of coordinate time when the clock's reading is 20 seconds.

No I am saying the ratio for the two times, (proper and coordinate) measured by the same observer.

So you agree both are measuring the "number of units" of time that have passed between two events on a clock's worldline, with one measuring the number of seconds of proper time that passed between the events, and one measuring the number of seconds of coordinate time?

I'm not quite sure what you are trying to say here?
The time a clock reads is the time on that clock at anyone instant An Event if you will.

Because of the relativity of simultaneity "any one instant" isn't really meaningful. An "event" refers to a specific point in spacetime (like how we can talk about a specific geometrical point on a 2D plane even though that point may be assigned different coordinates by different cartesian coordinate systems on the plane), with there being some unique fact about what physical occurrences do or don't happen at each point (for example, we can specify that we are talking about the event of one clock passing next to another one, and all frames must agree on what reading each clock shows at that event).

Saying that it will read differently 20 seconds after an event as measured in two different frames means nothing, for then you are comparing two different events.

That's exactly why I said the following:

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event. But if you are asking a question like "what reading does the clock show 20 seconds after it passed by observer A", then different frames can disagree about this, because they disagree which event on the clock's worldline happened "20 seconds after it passed by observer A".

The point is just that your language was ambiguous, when you said "Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it" you didn't specify that you were asking if it could "show a different time" at the same event, you could instead have been talking about the reading the clock was showing at a specific coordinate time in each frame, in which case the different frames could disagree about the reading at that coordinate time.

 

[Grimble]

The point is just that your language was ambiguous, when you said "Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it" you didn't specify that you were asking if it could "show a different time" at the same event, you could instead have been talking about the reading the clock was showing at a specific coordinate time in each frame, in which case the different frames could disagree about the reading at that coordinate time.

I think, Jesse, that it would be prudent for me to start with a simple definition and see if you agree with it, for if you don't nothing built upon it will make any sense in your eyes.

So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical.

Anything wrong with that?

 

[Dale]

Are observer A and the clock co-moving?

 

[phyti]

grimble said:
"So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical."

Isn't he saying that the person and clock are moving together at some indeterminant speed?