Shouldn't a moving clock appear to be ticking faster instead of slower?

[Mentospech]

That is not time dilation either, but a related effect called differential aging.

So how does the universe know which should age faster.. the train or the rest of earth?

 

[Dale]

Im sorry but i don't know what primed and unprimed clock is.

A primed clock is a clock at rest in the primed reference frame and an unprimed clock is a clock at rest in an unprimed reference frame. Please go over the math again with that in mind.

So I do not understand why you try to explain this by saying the clocks are somehow badly setup.

It is nothing bad with your setup, it is just that your setup does not investigate time dilation. Your original question was about time dilation, but your setup simply does not address time dilation. Time dilation is about the rate of a single moving clock, not about the readings on an ensemble of moving clocks as they pass by a single stationary clock.

There is nothing wrong with the quantity you are calculating, but it is not time dilation, it is something else that I don't believe has a name. (It isn't differential aging either in the way that you have it set up). You are asking about apples and describing pears.

 

[Doc Al]

The passenger can always look only at one clock at a time - the one he's passing by. So that we can say they are at the same location, so we do not need to talk about their synchronization from frame K'

If all the passenger does is look at one clock, then what conclusion can he draw? Not much! But when observing multiple clocks as he passes them by, he'd better consider how they are synchronized in his frame if he is to make sense of their readings.

 

[Mentospech]

If all the passenger does is look at one clock, then what conclusion can he draw? Not much! But when observing multiple clocks as he passes them by, he'd better consider how they are synchronized in his frame if he is to make sense of their readings.

He looks at one clock at a time. Then he looks at another. What bad could happen there? Why does he have to consider anything about their synchronization, other than that they are synchronized in stationary frame? Does it not hold that all of their increments add up to the state of the final clock? Does it not hold that he could place arbitrary amount of them in arbitrarily small distances and measure their differences? Feels like the definition of differentiable function, so why not use their sampling to measure time-rate in the stationary frame ?

 

[Nugatory]

So how does the universe know which should age faster.. the train or the rest of earth?

Google for “twin paradox”, but the quick answer is:

The same way that a car odometer knows to count fewer kilometers when you drive a straight line between two cities than when you drive a circuitous path between them. We have two clocks traveling between the same two points in spacetime (the separation event and the reunion event) but on different paths through spacetime. A clock measures the “length” of its path through spacetime, and the two paths have different lengths so different elapsed time measured.

 

[Mentospech]

Google for “twin paradox”, but the quick answer is:

The same way that a car odometer knows to count fewer kilometers when you drive a straight line between two cities than when you drive a circuitous path between them. We have two clocks traveling between the same two points in spacetime (the separation event and the reunion event) but on different paths through spacetime. A clock measures the “length” of its path through spacetime, and the two paths have different lengths so different elapsed time measured.

And how do you tell which path through spacetime was longer?

 

[Mentospech]

I mean if an asteroid comes around Earth, its clock will not run at the same speed as ours. So how does the universe tell which clock should go faster and which slower

 

[Mentospech]

There is nothing wrong with the quantity you are calculating, but it is not time dilation, it is something else that I don't believe has a name. (It isn't differential aging either in the way that you have it set up). You are asking about apples and describing pears.

Well it should have a name, its from Einstein's paper. Its his setup, i think all i said is basically implied in the little quote i provided.

 

[Dale]

Its his setup, i think all i said is basically implied in the little quote i provided.

Sure, it is his setup in that specific part of the paper, but it is NOT what people refer to when they speak about "time dilation". He also did not call it "time dilation".

It is the difference between looking at x=0 and x'=0, as I set up the math above. Do you see that? Do you understand that x=0 and x'=0 are different lines in spacetime and represent different things.

 

[Mentospech]

Sure, it is his setup in that specific part of the paper, but it is NOT what people refer to when they speak about "time dilation". He also did not call it "time dilation".

It is the difference between looking at x=0 and x'=0, as I set up the math above. Do you see that? Do you understand that x=0 and x'=0 are different lines in spacetime and represent different things.

I understand you cannot use the one moving clock from the other frame, but i do not understand how you arrived at what you did, also it does not seem to correspond to what Albert arrived at. Your t / gamma is the time dilation equation, not like the one in OP.

 

[Dale]

i do not understand how you arrived at what you did

Are you familiar with the Lorentz transform? https://en.wikipedia.org/wiki/Lorentz_transformation

That is the transformation to transform coordinates in one reference frame to coordinates in another reference frame. The whole point of Einstein's paper was to derive this transformation from the two postulates.

If you have something moving at constant speed then in spacetime you can write it as a line. So, for example, $x=0.5 t+3$ would be something moving at a constant speed of 0.5 starting at an x position of 3. So, $x=0$ is something at rest (speed=0) at the origin. Does that make sense?

 

[Doc Al]

There's really nothing magical about Einstein's quoted formula for the difference in time on the moving and stationary clocks: a "lag" of $\frac{tv^2}{2c^2}$. According to the stationary frame, the travel time to get from A to B is $t$. The moving clock would only have recorded $t/\gamma = t(1- \frac{v^2}{c^2})^{1/2}$, due to time dilation. To find the "lag" just subtract the two, using a binomial expansion.

The above is doing the calculation from the stationary frame, which is easy. You can also calculate the lag from the moving frame, but then you'll need to consider synchronization and more. You'll get the same answer, of course.

 

[Mentospech]

There's really nothing magical about Einstein's quoted formula for the difference in time on the moving and stationary clocks: a "lag" of $\frac{tv^2}{2c^2}$. According to the stationary frame, the travel time to get from A to B is $t$. The moving clock would only have recorded $t/\gamma = t(1- \frac{v^2}{c^2})^{1/2}$, due to time dilation. To find the "lag" just subtract the two, using a binomial expansion.

The above is doing the calculation from the stationary frame, which is easy. You can also calculate the lag from the moving frame, but then you'll need to consider synchronization and more. You'll get the same answer, of course.

Thats not possible as your $t/\gamma$ could approach zero, whereas in the quoted equation the time perceived by the traveller could only go to about half of its original time

 

[Mentospech]

Are you familiar with the Lorentz transform? https://en.wikipedia.org/wiki/Lorentz_transformation

That is the transformation to transform coordinates in one reference frame to coordinates in another reference frame. The whole point of Einstein's paper was to derive this transformation from the two postulates.

If you have something moving at constant speed then in spacetime you can write it as a line. So, for example, $x=0.5 t+3$ would be something moving at a constant speed of 0.5 starting at an x position of 3. So, $x=0$ is something at rest (speed=0) at the origin. Does that make sense?

Yes it does, but look I can find this stuff on my own. What still bothers me is that the moving clock still appears to be going faster for one observer according to my logic.. oh well nevermind

 

[Doc Al]

Thats not possible as your $t/\gamma$ could approach zero, whereas in the quoted equation the time perceived by the traveller could only go to about half of its original time

The way to calculate the lag is to subtract those two times as indicated. That works for any speed. To get Einstein's formula, you need to ignore higher powers of $v^2/c^2$, which can only be done when that fraction is small enough. Einstein's formula is only an example calculation, not a fundamental principle of relativity.

 

[Nugatory]

And how do you tell which path through spacetime was longer?

You can calculate it.

Given two nearby events with coordinates $(x,t)$ and $(x+dt,t+dt)$ using the x and t coordinates assigned by whatever frame is convenient, the “length” or spacetime interval $ds$ between them is $ds=\sqrt{dt^2-dx^2}$ (note that that’s not quite the Pythagorean theorem, because the geometry of spacetime is not Euclidean). Integrate this quantity along the path to get its length.

We can simplify things a bit by considering the case in which one clock remains at rest in the chosen frame while the others one moves away at a constant speed v for time $\Delta{t}$ so traveling a distance $v\Delta{t}$, then turns around and returns at the same speed. In this case we don’t have to mess with the line integral to calculate the lengths: the stationary clock has followed a path of length $2\Delta{t}$ and the moving clock a path of length $2\sqrt{\Delta{t}^2-{v}\Delta{t}^2}$.

I very highly recommend “Spacetime Physics” by Taylor and Wheeler; a week or so of quality time with that book will clear up most of your confusions.

(Be aware that the discussion above is assuming that $dt\gt{dx}$ which will be the case for points on the path of a clock or any other object moving at less than the speed of light, and is taking $c=1$, as is the case if we measure distances in light-seconds and time in seconds)

 

[Dale]

Yes it does, but look I can find this stuff on my own.

OK, so then where are you getting lost with my derivation above? Once you have that, the rest is just algebra.

What still bothers me is that the moving clock still appears to be going faster for one observer according to my logic.

The reason that I did the math above was so that you could see exactly in detail what you are doing differently than what other people are doing when they talk about "time dilation". Other people are looking at a single moving clock as it passes multiple stationary clocks. You are looking at a single stationary clock as multiple moving clocks pass it. Those are different things, both physically and mathematically. It isn't that you are wrong with your logic, you are just talking about something different. What you are describing is not time dilation.

You can grumble about it, but the rest of the physicists in the world have been using this meaning for more than a century and are not going to change the terminology to suit you. I am afraid that you will have to simply accept that time dilation doesn't mean what you want it to mean here.

 

[Pencilvester]

What still bothers me is that the moving clock still appears to be going faster for one observer

To which moving clock are you referring? Because I was under the impression you were comparing the time of the clock on the train to the times on multiple clocks.

 

[Janus]

I mean if an asteroid comes around Earth, its clock will not run at the same speed as ours. So how does the universe tell which clock should go faster and which slower

The universe can't. Which clock runs faster depends on whether you are measuring from the asteroid or Earth.

let's start off with a simple example. We have four clocks A,B,C, and D. A and B are a pair, And C and D are a pair and these pairs are in relative motion with respect to each other. Each pair of clock is synchronized to each other according to each pair. In other words clocks A and B always consider themselves as reading the same at any given moment, and Clock C and D always consider themselves as reading the same at the same moment.

What the Relativity of simultaneity says is that Clocks A and B will not consider clocks C and D as being in sync with each other and clocks C and D will not consider clocks A and B as being in sync.

So first we consider things from the rest frame of A and B:

Here we assume that C and D are moving from right to left. We start with Clock A and C just passing each other while both read 0. At that same moment D is passing B. B, being in sync with A, also reads 0. However D not being in sync with C does not read 0 but already reads 2.5. An observer at clock B would record clock D reading 2.5 as it passed and his own clock read 0.

We then move forward to when D reaches A. In this particular example the relative speed of the pairs and the distance between clocks result in clock A reading 8.66 and Clock D reading 10 when they pass each other.

Both A and B have advanced by 8.66, while clocks C and D have only advanced by 7.5. Even though clock D shows a later time than A when they meet, D ticked slower.

Now we switch to the rest frame of C and D.
when we do this, we have to take a few things in consideration. In the above image the distances between A-B and C-D are equal. However, C-D is in motion relative to this frame, and is thus length contracted. In its own frame, the C-D distance will be measured as being greater than it was in the A-B frame. In addition, It si now A-B that is moving and will show length contraction. Also, in the A-B frame, it was clocks A and B that were in sync, while in the C-D frame, it is clocks C and D that will be measured as being in sync.
Again we start out with A and C passing each other when they read 0

You will note that when this occurs, due to the difference in A-B and C-D distances as measured in this frame, B has not yet aligned with D. In order for it to read 0 when this occurs, it now reads sometime before 0. Also note that clock D reads 0 and this moment and thus will advance some before B reaches it.

At the next stage we show when B and D meet. Both Clocks A and B have advanced by 2.165, while Clocks C and D have advanced by 2.5. As above, when B and D pass, B reads 0 and D reads 2.5. An observer at D would, when his clock read 2.5, be seeing B just passing by reading 0. Just like the observer at B would see his clock read 0 as D passes reading 2.5.

Lastly we get to when A and D meet. As in the first image, A reads 8.66 and D reads 10. In this case however, A advanced 8.66 while D advanced 10. clocks A and B tick slower than clocks C and D.

In this situation there is no "right" answer as to which clocks ticked slower. This relies on the fact that neither A-B or C-D never change their velocity during the exercise.

Things change if any of these pairs of clocks change velocity between any set of clock measurements. (any accelerations that might of taken place before the start of this setup don't matter). Then you can get a situation where both pairs of clocks will agree that one clock ended up accumulating more or less time than another( though even then, at any given moment they might not agree as to which clock was running fast.)

 

[Mentospech]

The universe can't. Which clock runs faster depends on whether you are measuring from the asteroid or Earth.

No. According to everything said here and also according to the twin paradox. The less time experienced is not an illusion or a matter of perspective. One HAS to be older and one HAS to be younger. If the asteroid hits the Earth it would have been of different age -> therefore the universe must somehow know which one should have been aging faster the whole time before.
Or am I wrong that relativity of simultaneity has no casual consequences? Meanwhile the difference in aging is witnessed by both observers and has casual consequences...

 

[Mentospech]

The passenger can always look only at one clock at a time - the one he's passing by

The passenger can do that, but doing so does not tell the passenger anything about the rate of a clock in the other frame as seen by his frame. See the math above.

Could you elaborate why such measurement would not tell anything about the rate of clock in the stationary time frame?
If you were to stop at that given point, all the clock would tell you exactly the same time, is that not the definition of time in the other time frame?

 

[Mentospech]

I found how to tell for which body the time runs faster - as Nugatory said it is given by the spacetime interval, which is invariant and therefore not relative value.

 

[jbriggs444]

I found how to tell for which body the time runs faster - as Nugatory said it is given by the spacetime interval, which is invariant and therefore not relative value.

A clock always records a space time interval of one second per second, moving or not.

 

[Mentospech]

A clock always records a space time interval of one second per second, moving or not.

Yes, but by being invariant doesn't this mean that both the stationary observer and the traveller on train will agree that it was the train who was moving and not the earth, or do I understand the concept of invariant wrongly?

 

[Dale]

If you were to stop at that given point

If you were to stop at that given point then you would be a non inertial observer and none of the standard formulas would apply to you anyway.

Could you elaborate why such measurement would not tell anything about the rate of clock in the stationary time frame?

Because of the relativity of simultaneity. Let’s say that the clocks that are moving in your frame all run at a rate $A$, and at your frame’s $t=0$ clock $i$ reads $B_i$. That means that the reading on any clock at any time is $r(t,i)=At+B_i$, and $A$ is the time dilation and $B_i$ is the relativity of simultaneity. Now, your goal is to find $A$ by observing clocks and reading the time.

Your approach is to observe at each time a different clock $i$. For example:
$r(1,1)=1A+B_1$
$r(2,2)=2A+B_2$

Note, you have two equations in three unknowns: $A$, $B_1$, and $B_2$. As you continue to look at different clocks you continue to get more unknowns because of the relativity of simultaneity. You can never gather enough data with this approach to solve for $A$.

Instead, the correct approach is to observe the same clock $i$ at multiple times. For example:
$r(1,1)=1A+B_1$
$r(2,1)=2A+B_1$

Now we have two equations in two unknowns which you can solve for $A$. If you want to determine $A$ then you must measure the same clock multiple times.

 

[Mentospech]

If you were to stop at that given point then you would be a non inertial observer and none of the standard formulas would apply to you anyway.

Not true.
The Einstein's formula i quoted clearly applies to that. Do you disagree with that ?

You still haven't demonstrated why you should not read different clocks. "relativity of simultaneity " applies to observing distant objects, this does not happen here at all. Introducing it here only confuses the issue.

On the other hand it is obviously demonstrable that the clock currently passed by shows exactly the time difference that the passenger experiences. It shows the exact amount of how much younger he would be compared to rest of the world due to his travel, if he stopped now. Yet you still maintain that you cannot use this reading to measure time rate in other frame.

So again: Why do you presume you cannot read different clocks for each reading if you know that this clock must provide the correct reading as it is by definition free of "relativity of simultaneity"

 

[Mentospech]

I very highly recommend “Spacetime Physics” by Taylor and Wheeler; a week or so of quality time with that book will clear up most of your confusions.

This book costs like 100 euro. Thats a little too steep for a book don't you think?

 

[Dale]

Not true.
The Einstein's formula i quoted clearly applies to that. Do you disagree with that ?

I disagree very strongly. In that section of Einsteins paper he only deals with simultaneity in the inertial frame K. He never discusses simultaneity in the inertial frame K' and he certainly never discusses simultaneity in A's non-inertial frame. He never makes the claim that by A stopping the time in the K frame would become the time in the other frame (K').

You still haven't demonstrated why you should not read different clocks. "relativity of simultaneity " applies to observing distant objects, this does not happen here at all. Introducing it here only confuses the issue.

I have demonstrated it twice now. You have failed to address the math at all. Relativity of simultaneity means what I wrote down, you are fooling yourself thinking it doesn't apply.

Instead of avoiding the issue, you need to address the math. If you are not willing to do that then you will not be able to make any progress.

So again: Why do you presume you cannot read different clocks for each reading if you know that this clock must provide the correct reading as it is by definition free of "relativity of simultaneity"

The math I have posted shows why. Again, you are wrong that the relativity of simultaneity doesn't apply here.

The Lorentz transform always applies. As I showed in the first derivation done directly from the Lorentz transform, what you are measuring is not what everyone else calls time dilation.

 

[Nugatory]

This book costs like 100 euro. Thats a little too steep for a book don't you think?

Might find one in a library... and I recently saw a used copy advertised online for about 25 US dollars.

 

[Dale]

I found how to tell for which body the time runs faster - as Nugatory said it is given by the spacetime interval, which is invariant and therefore not relative value.

This is another good approach. The spacetime interval of a clock moving arbitrarily in any inertial frame is given by $d\tau^2=dt^2-(dx^2+dy^2+dz^2)/c^2$. Note, while $d\tau$ is invariant $dt$ is not. Continuing with the brief derivation:$$\frac{d\tau}{dt}=\sqrt{1-\frac{1}{c^2}\left(\frac{dx^2}{dt^2}+\frac{dy^2}{dt^2}+\frac{dz^2}{dt^2}\right)}=\sqrt{1-\frac{v^2}{c^2}}=\frac{1}{\gamma}$$So again, any moving clock runs slow relative to any inertial frame by the $\gamma$ factor.

That pesky inconvenient math just keeps proving you wrong over and over again. Best just ignore it yet again, otherwise you might learn something.