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

Mentospech

The passenger can always look only at one clock at a time - the one hes 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

Homework Helper
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 doesnt 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

Mentor
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 havent 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 dont you think?

Dale

Mentor
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 havent 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

Mentor
This book costs like 100 euro. Thats a little too steep for a book dont you think?
Might find one in a library.... and I recently saw a used copy advertised online for about 25 US dollars.

Dale

Mentor
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.

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Mentospech

He never discusses simultaneity in the inertial frame K'
He doesnt need to, there is no disparity in simultaneity involved at all. All the observations are made on objects at distance 0. There can be no difference in simultaneity.

He never makes the claim that by A stopping the time in the K frame would become the time in the other frame (K').
I dont know what you mean by "A stopping the time in the K frame" but I didnt talk about stopping time, but stopping the train. My point is that by stopping the train instanteniously in B you get the same reading as by just passing through B. So the frame of reference does not affect the reading. That is the essence of this thought experiment.

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.
Your logic here is this: effect X applies , therefore these equations hold ... and these equations hold because effect X applies. This doesnt make any sense. First you have to show why relativity of simultaneity should apply.
I have a very strong argument why it should not : relativity of simultaneity concerns only two spatially separated events. This is the point i contest.

Mentospech

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.
Equations by themselves are not proof of anything.

As was said in the paper and quoted here in one of the posts according to Einstein you can put points A = B, then put the rails in a circle and do this and each time A will lag more and more behind B. You can also have the circle arbitrarily short so you can watch the same clock at infinite number of instances lag more and more.

So either Einstein is wrong or you and your 'proof' is wrong and the equations you put forward do not actually relate to this experiment.

Dale

Mentor
He doesnt need to, there is no disparity in simultaneity involved at all. All the observations are made on objects at distance 0. There can be no difference in simultaneity.
He doesn't claim that, and the math doesn't support that.

I dont know what you mean by "A stopping the time in the K frame" but I didnt talk about stopping time, but stopping the train.
Oops, that should have been "by A stopping, the time in the K frame". Forgot the comma.

Your logic here is this: effect X applies , therefore these equations hold ... and these equations hold because effect X applies. This doesnt make any sense. First you have to show why relativity of simultaneity should apply.
I have a very strong argument why it should not : relativity of simultaneity concerns only two spatially separated events. This is the point i contest.
The Lorentz transform ALWAYS applies in special relativity. The equations hold, not because effect X applies but because they always hold. From the equations you can derived that effect X applies. My logic is sound and the derivation from the Lorentz transform shows it clearly.

Your assertion that the relativity of simultaneity concerns only two spatially separated events is incorrect as derived form the Lorentz transform and shown above.

So either Einstein is wrong or you and your 'proof' is wrong and the equations you put forward do not actually relate to this experiment.
So find the mistake. Use the Lorentz transform and do the math yourself. Show a full derivation directly from the Lorentz transform or point out in my first derivation exactly where I used it wrong.

I have pointed out explicitly where you are making your mistake and why it gives you the wrong answer, and I have backed it up with math three times three different ways. Once with the Lorentz transform, once with general considerations of the relativity of simultaneity, and once with the spacetime interval.

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Mentospech

So find the mistake.
I assume the mistake is that the original quoted formula holds true for all motion with constant velocity, it does not need to be in one inertial frame. Also it seems to describe an actual difference in time experienced. While the time dilation formula describes only observed difference in the speed of clock. I assume both of these effects take place, but only the one discussed here has actual casual impact on everyone. But im not sure yet i have to look more into this.

Dale

Mentor
I assume the mistake is that the original quoted formula holds true for all motion with constant velocity, it does not need to be in one inertial frame. Also it seems to describe an actual difference in time experienced. While the time dilation formula describes only observed difference in the speed of clock. I assume both of these effects take place, but only the one discussed here has actual casual impact on everyone. But im not sure yet i have to look more into this.
Sorry, I cross posted with you. Please see my revised response, I will try to avoid doing that in the future. However, instead of assuming the mistake, you need to show it explicitly as I showed explicitly yours. Use the Lorentz transform since it always applies in SR.

Mentospech

This is very similar to the twin paradox, and while the effect of time dilation is symmetrical to both of them, there is also the objective fact that one is aging faster, that fact is objective fact to all observers so it seems logical it would be propotional to the invariant- spacetime interval, and i believe that while time dilation occurs, this aging might be actualyl speeding up the clock for one of the observers

Mentospech

All the observations are made on objects at distance 0. There can be no difference in simultaneity.
He doesn't claim that, and the math doesn't support that.

The underlined portion implies distance 0 when comparing clocks.

Your assertion that the relativity of simultaneity concerns only two spatially separated events is incorrect
This is the exact opposite of what I found on the subject. Please consider the possibility that you are wrong and look it up. The terms concerning time of distant events always contain distance.

So find the mistake. Use the Lorentz transform and do the math yourself. Show a full derivation directly from the Lorentz transform or point out in my first derivation exactly where I used it wrong.
I cant. I dont have the capacity to do this well enough to be confident in the result yet. But contrary to what you seem to think, it doesnt mean that i cannot arrive at the right conclusions or spot the problems in your assessment.

George Jones

Staff Emeritus
Gold Member
Summary: Moving objects time dilation
View attachment 247722
Given the scenario of the attachment in the original post, the answer to the question of the thread's title is "Yes, according to B, the clock that moves from A to B appears to tick faster."

Mentospech

Given the scenario of the attachment in the original post, the answer to the question of the thread's title is "Yes, according to B, the clock that moves from A to B appears to tick faster."
Hi.
Thanks for responding to this question. But i think you reversed the order here. The moved clock (A) is lagging behind upon arrival according to the quote. So to me it seems that the clock B appears to tick faster according to the moving clock A. (disregading the doppler effect speeding it up, and despite the time dilation effect which is slowing it down).

George Jones

Staff Emeritus
Gold Member
Thanks for responding to this question. But i think you reversed the order here. The moved clock (A) is lagging behind upon arrival according to the quote. So to me it seems that the clock B appears to tick faster according to the moving clock A. (disregading the doppler effect speeding it up, and despite the time dilation effect which is slowing it down).
Then, it seems that we differ on the definition of the term "appear" in the question "Shouldn't a moving clock appear to be ticking faster instead of slower?"

Edits: 1) I have applied the question the to scenario in the attachment; 2) I do not question the validity of time dilation.

Mentospech

the answer to the question of the thread's title is "Yes, according to B, the clock that moves from A to B appears to tick faster."
But clock at B will have more time on it upon A's arrival, according to the
How could then A appear to tick faster, if then upon its arrival it is showing less time passed?

I do not question the validity of time dilation.
Me neither.

Dale

Mentor
The underlined portion implies distance 0 when comparing clocks.
Clearly. That doesn't imply what you want it to imply though.

I cant. I dont have the capacity to do this well enough to be confident in the result yet.
Then maybe pay attention to people who do have that capacity.

Since you are unwilling to listen to others, this thread is pointless. Please come back when you either:
a) can work through the math yourself
b) are willing to listen to those who can

The Lorentz transform always applies in SR, and I used it in my first derivation to show not only that you were wrong but exactly in careful detail why you were wrong. Until you can either accept that analysis or do the analysis on your own, this topic is closed.

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