Looking to understand time dilation

[Grimble]

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.

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.

 

[Grimble]

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.

 

[Grimble]

Um... a much more sensible conclusion would be that not everyone contributing is an expert, rather than "none of the experts agree".

It can be good practice for a non-expert to try and identify the errors being made in so-called paradoxes or contradictions. The experts will generally provide helpful corrections, and I think THIS is what you are seeing.

Cheers -- sylas

Apologies to all

 

[Grimble]

cmb is cosmic background radiation; a useful reference against which velocities can be measured.



You can't just add the durations from two frames, because that is assuming and absolute notion of simultaneity. This is invariably behind the confusions people have with respect to so-called paradoxes of relativity.

Cheers -- sylas

I was not adding them but comparing them.

All I am saying here is that it appears to assume an absolute scale for measuring time. And that when Einstein says

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; ...

It implies that time passes at different rates in different frames.

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

 

[ghwellsjr]

I was not adding them but comparing them.

All I am saying here is that it appears to assume an absolute scale for measuring time. And that when Einstein says

It implies that time passes at different rates in different frames.

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

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? 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?

 

[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.