B Time dilation and 2 identical clocks

[PeroK]

to restate my original post - the theorey is a clock inside a MFR should appears to tick slower if observed by an observer external to that MFR who is moving relative to the first mentioned MFR. My query is the timer inside the moving frame of reference will appear to be ticking FASTER , not slower, according to the external observer...It appears I have made a mistake but I don't know where

If your clock is an egg-timer, then the postulates of SR don't provide the rules to study this from first principles. The second postulate of SR states that the speed of light is invariant (the same in all inertial reference frames). As speed = diistance/time you can study the nature of time and space using the paths of light rays.

"Appears" may be a dangerous word, since the finiteness of the speed of light can delay the observation of measurements. And, the delay in receiving signals has nothing to do with SR or time dilation. It's the measurement of time that is required, which may be more than the observation of a moving clock.

 

[PeroK]

@Ross Arden For example. Suppose a clock is moving towards you at a speed of $\frac35 c$. In your reference frame that clock runs slow by a factor of $\frac45$. Suppose it reads $t' =0$ when it is $1$ light-second away from you - and emits a light signal at this point. This is all in your reference frame, by the way.

By the time you receive that signal, the clock has moved $\frac35$ of a light-second and is only $\frac25$ of a light-second away. And, due to time dilation, it reads only $t'=\frac45 s$ at this point. Assume it emits a second signal at this point.

That second signal will reach you only $\frac25 s$ after the first.

So, what you "observe" is:

At some time $t$ in your reference frame you observed the moving clock to read $0s$. At some time $t + \frac25 s$ you observed the clock to read $\frac45 s$. These raw observations appear to show the clock running fast.

Note that without time dilation (i.e. in classical physics) it would appear to run even faster.

But, these raw observations do not constitute a measurement of time. To get a measurement, you need (both in classical physics and SR) to take into account the travel time of light. Once you do that you get that the clock is running slow.

Note that this post doesn't explain time dilation itself (it assumes it), but it does show that you need to distinguish between raw observations and measurements of elapsed time between events. And that's why having local measurements and checking simultaneity is important in these thought experiments.

 

[Herman Trivilino]

I fail to see why a third clock is required, over an above the 2 egg timers, the counter and the egg.

Let's focus on one egg. If you have more than one egg ignore the others for now. Let's say that one timer is aboard the moving rocket ship with the egg and the other timer is at rest. Explain for us how the timer that's at rest can be used to time the cooking of the egg. If that timer is present when the egg starts to cook, it can't be present when the egg is finished cooking because the rocket will have separated it from the egg. You will need some kind of signalling device that will pass signals from the distant ship to the timer that's at rest.

 

[Dale]

elapsed time is measured by the egg. If the time is brief the egg is under cooked. if the time is long the egg is over cooked.

This is called the egg’s proper time. Here the word proper is in the sense of property, it is the eggs own time. The over/under cooking depends on the eggs proper time.

The other type of time is called coordinate time. This time requires some synchronization convention, and once you have chosen that synchronization convention then you can compare the time of distant events. This time does not belong to an object, but to a reference frame.

Is that clear? If so, then your main question appears to be “which clocks best measure the egg’s proper time”. Do I understand you right?

 

[Ross Arden]

Let's focus on one egg. If you have more than one egg ignore the others for now. Let's say that one timer is aboard the moving rocket ship with the egg and the other timer is at rest. Explain for us how the timer that's at rest can be used to time the cooking of the egg. If that timer is present when the egg starts to cook, it can't be present when the egg is finished cooking because the rocket will have separated it from the egg. You will need some kind of signalling device that will pass signals from the distant ship to the timer that's at rest.

sorry I don't understand what u are trying to say. there is 2 eggs, 2 timers, 2 counters, 2 stoves. Egg timer 1 counts the cooking of egg 1, Egg timer 2 is for egg 2...see the OP

 

[Ross Arden]

This is called the egg’s proper time. Here the word proper is in the sense of property, it is the eggs own time. The over/under cooking depends on the eggs proper time.

The other type of time is called coordinate time. This time requires some synchronization convention, and once you have chosen that synchronization convention then you can compare the time of distant events. This time does not belong to an object, but to a reference frame.

Is that clear? If so, then your main question appears to be “which clocks best measure the egg’s proper time”. Do I understand you right?

the time is compared by the degree of cooking of the egg, simply compare egg one with egg 2. Nothing need to be synchonised. If egg 1 starts to cook after or before or at the same time as egg 2 it makes no differnece the egg will still be over or under cooked

If u cook and egg for 1 minute and then an hour later u cook a second egg for an hour, and then compare them...u can figure the rest out

 

[Herman Trivilino]

Egg timer 1 counts the cooking of egg 1, Egg timer 2 is for egg 2...see the OP

Aren't you also interested in how Timer 1 counts the cooking of Egg 2?

Or how Timer 2 counts the cooking of Egg 1?

 

[Ibix]

the egg will still be over or under cooked

Why do you think so? Each egg's cooking is controlled by an egg-timer at rest with respect to it. When it's finished cooking, it's done (assuming the timer is an accurate indicator of how long you need to cook an egg). If you mean "at the same time as one egg finishes cooking, what is the state of the other egg?", then you need a simultaneity convention (i.e., you need to define "at the same time"), which is to say, you need to work out how you are synchronising clocks.

 

[Ross Arden]

Aren't you also interested in how Timer 1 counts the cooking of Egg 2?

Or how Timer 2 counts the cooking of Egg 1?

Im just interested in comparing how under,over, the saem the eggs are cooked timer 1 times egg 1 timer 2 times egg 2...so in short no

 

[Ross Arden]

Why do you think so? Each egg's cooking is controlled by an egg-timer at rest with respect to it. When it's finished cooking, it's done (assuming the timer is an accurate indicator of how long you need to cook an egg). If you mean "at the same time as one egg finishes cooking, what is the state of the other egg?", then you need a simultaneity convention (i.e., you need to define "at the same time"), which is to say, you need to work out how you are synchronising clocks.

If u cook and egg for 1 minute @ 100 degrees C and then 2 hours later u cook a second egg for 50 minutes @ 100 degrees C, in a moving spaceship wrt egg 1, and then compare the eggs...u can figure the rest out ...there is no requirement for synchronsiation...simply compare the eggs...both eggs cooking time is regulated by the egg timer associated with that egg. But if one egg timer is ticking slower than the other one then one egg will be undercooked compared to the other.

How is this a "telescope" scenario when the so called "telescope" is in the space ship? If that is the case isn't Einsteins original thought experiment with a light clock (light at one end, mirror at the other) in a railway carriage also just a "telescope"?

Maybe I am wrong as the heat transfer, to the egg, and heat transfer is a function of time, will be slower in the space ship, as time will be expiring slower in the spaceship so both eggs will be equally cooked ?

But the fact still remains the clock in the spaceship will appear to tick faster, than the external clock, according to the external observer

 

[Ibix]

If u cook and egg for 1 minute and then 2 hours later u cook a second egg for an 50 minutes, and then compare them...u can figure the rest out

But this has nothing to do with any scenario you have presented. I don't understand why you bring it up.

 

[Ross Arden]

Ill start again. See the pic
a = counter
c = light tube
b = egg timer
d is a light that pulses each time a grain falls thru the egg timer
L2 is the length of the light tube
L1 is the length of the egg timer
there are 2 identical arrangements, one in a space ship, one at rest wrt an observer external to the space. The spaceship is moving relative to arrangement 2 at v in the direction of v ...any questions ?

 

[Ross Arden]

the pic, the oval is an egg and the small red line under the egg is a heat source

 

[Ibix]

Your diagram doesn't appear to show length contraction of the apparatus in the moving ship. Assuming you understand that it should be length contracted, or that you've deliberately ignored length contraction for some reason, fine.

 

[Ross Arden]

Your diagram doesn't appear to show length contraction of the apparatus in the moving ship. Assuming you understand that it should be length contracted, or that you've deliberately ignored length contraction for some reason, fine.

I have ignored length contraction

 

[Ibix]

I have ignored length contraction

That may come back to bite, depending on where you're going with this, but fine for now. Go on.

 

[Ross Arden]

so won't the external observer perceive the egg timer in the spaceship to be running faster than the egg timer at rest wrt the observer ?

 

[Vanadium 50]

Ross, can you please use proper English and not text speak? People are already struggling to figure out what you mean; please don't make it any harder.

 

[Ross Arden]

BTW as there is no gravity the egg timer works of pressurized air. So if there is length contraction, as there will be less volume, the egg timer in the spaceship will have a higher pressure than the at rest one

 

[Ross Arden]

Ross, can you please use proper English and not text speak? People are already struggling to figure out what you mean; please don't make it any harder.

okay

 

[jbriggs444]

BTW as there is no gravity the egg timer works of pressurized air. So if there is length contraction, as there will be less volume, the egg timer in the spaceship will have a higher pressure than the at rest one

That is not correct.

 

[Ross Arden]

That is not correct.

if the 2 egg timers have a pressure chamber volume of 1000cc and are pressurised up to a pressure of 50 kg/cm^2
egg timer 2 is then put in the space ship, it blasts off and is moving relative to the observer
wont the observer observe length contraction so the amount of air is constant but the 1000 cc will now be observed to be less than 1000cc ergo same amount of air in a smaller volume = higher pressure ?

 

[Herman Trivilino]

Im just interested in comparing how under,over, the saem the eggs are cooked timer 1 times egg 1 timer 2 times egg 2...so in short no

Then all you have to do is decide how you like your eggs cooked. If a 3-minute egg is what you desire, then set Timer 1 to three minutes and cook Egg 1. Set Timer 2 to three minutes and cook Egg 2. Each egg will then be cooked just right, neither will be under-cooked or over-cooked. Of course, if you like your eggs cooked for a longer or shorter time you will adjust accordingly.

None of this has anything to do with your OP. In that post you show a person at rest viewing a clock in motion.

If u cook and egg for 1 minute @ 100 degrees C and then 2 hours later u cook a second egg for 50 minutes @ 100 degrees C, in a moving spaceship wrt egg 1, and then compare the eggs...u can figure the rest out

Yes, but you started this thread by saying there was something YOU couldn't figure out.

I know that an egg cooked for 1 minute will be underdone and an egg cooked for 50 minutes will be overdone. It makes no difference how fast those eggs are moving when I do it, as long as the clocks I use are moving alongside the eggs.

The theory of relativity enters our thoughts when we try to understand how a clock that's in motion relative to us behaves. If we try to use a moving clock to time the cooking of a stationary egg then the issue you raised in your OP will have to be tackled. Otherwise it won't!

But keep in mind that if you do want to tackle it you will need to address the issues I raised before, issues that you said you don't understand. If you want to understand them you can't just give up there. You will need to explain which parts of it you don't understand. Otherwise, how can we help you?

 

[Ross Arden]

Then all you have to do is decide how you like your eggs cooked. If a 3-minute egg is what you desire, then set Timer 1 to three minutes and cook Egg 1. Set Timer 2 to three minutes and cook Egg 2. Each egg will then be cooked just right, neither will be under-cooked or over-cooked. Of course, if you like your eggs cooked for a longer or shorter time you will adjust accordingly.

None of this has anything to do with your OP. In that post you show a person at rest viewing a clock in motion.
Yes, but you started this thread by saying there was something YOU couldn't figure out.

I know that an egg cooked for 1 minute will be underdone and an egg cooked for 50 minutes will be overdone. It makes no difference how fast those eggs are moving when I do it, as long as the clocks I use are moving alongside the eggs.

The theory of relativity enters our thoughts when we try to understand how a clock that's in motion relative to us behaves. If we try to use a moving clock to time the cooking of a stationary egg then the issue you raised in your OP will have to be tackled. Otherwise it won't!

But keep in mind that if you do want to tackle it you will need to address the issues I raised before, issues that you said you don't understand. If you want to understand them you can't just give up there. You will need to explain which parts of it you don't understand. Otherwise, how can we help you?

I repostsed the OP and made it simpler, as simple as I could, as it appears people had no idea what I was talking about

 

[Herman Trivilino]

so won't the external observer perceive the egg timer in the spaceship to be running faster than the egg timer at rest wrt the observer ?

No. Slower. If the light clock ticks 1000 times during the time that the egg cooks, then that's what it will do regardless of its state of motion. If it's in motion relative to some observer then those 1000 ticks will take longer than 1000 ticks on an identical clock at rest with respect to the observer, but the egg will still take 1000 ticks to cook.

 

[Ross Arden]

No. Slower. If the light clock ticks 1000 times during the time that the egg cooks, then that's what it will do regardless of its state of motion. If it's in motion relative to some observer then those 1000 ticks will take longer than 1000 ticks on an identical clock at rest with respect to the observer, but the egg will still take 1000 ticks to cook.

according to the external observer

the spaceship with the egg timer inside is moving to the right at v
a grain from the egg timer leaves the neck of the egg timer at t1
it arrives at the base of the egg timer at time t2
in the time it takes for the grain to leave the neck and arrive at the base the spaceship has moved a distance of v x (t2-t1) to the right
if the length from the neck to the base of the egg timer is L
the distance traveled by the grain will be L - (v x (t2-t1))
the grain moves at a velocity x (this is constant for both egg timers)

the time taken for the grain in the space ship, according to the external observer, will be distance/velocity = (L - (v x (t2-t1))) /x

The observer then looks at the egg timer next to him and measures the time for a grain to hit the base to be
L/x

but L/x > (L - (v x (t2-t1))) /x

is that right?

is the error the velocity of the spaceship minuses from the velocity of the grain?

if that was the case what if the grains were replaced by tunneling photons

 

[Dale]

the time is compared by the degree of cooking of the egg, simply compare egg one with egg 2. Nothing need to be synchonised.

That is the proper time of the eggs.

Nothing need to be synchonised.

If you are not interested in comparing spatially separated events then there is no need for synchronization or coordinate time. In your opening post it sure seemed like you wanted to do that kind of comparison.

If not, then I am not sure what your question is. If you cook an egg according to an egg timer that is comoving with the egg then it will be cooked to the same level of doneness. That is proper time.

If u cook

Please don’t use “u” as shorthand for “you”. It is distracting here since there are many physics formulas that traditionally use “u” as a variable.

if one egg timer is ticking slower than the other one

FYI, here is where you are comparing spatially separated events and you need a synchronization convention. This is coordinate time.

Time dilation is what happens when you compare the proper time of a clock to the coordinate time in a reference frame where it is moving.

 

[Herman Trivilino]

the grain moves at a velocity x (this is constant for both egg timers)

Not possible. The timers are moving relative to each other.

 

[Herman Trivilino]

the time taken for the grain in the space ship, according to the external observer, will be

1. Can the external observer measure that elapsed time without knowing the reading on his clock when the grain begins to fall and the reading on his clock when the grain lands?

2. Because to take those two readings he will have a difficulty. If he's next to the grain when it begins to fall he can take a clock reading, but when the grain lands his clock will be far away from the grain, so how can he determine that clock reading?

3. He will need some kind of signalling device to send a signal from the place where the grain lands to the place where his clock is located. And he will have to subtract off the travel time of the signal to find out what time it was at the place where the grain landed at the time that it landed. In other words, he will need to know what time it was at that place when the grain landed. This is the process we're referring to when we speak of synchronizing distant clocks. It's what you have to do if you want to use a clock located here to find out what time it is over there.

If you don't understand the above please explain where it was you first started to not understand. Was it the paragraph labelled 1, 2, or 3?

 

[jbriggs444]

if the 2 egg timers have a pressure chamber volume of 1000cc and are pressurised up to a pressure of 50 kg/cm^2
egg timer 2 is then put in the space ship, it blasts off and is moving relative to the observer
wont the observer observe length contraction so the amount of air is constant but the 1000 cc will now be observed to be less than 1000cc ergo same amount of air in a smaller volume = higher pressure ?

No.

 

[Ross Arden]

No.

Okay thanks you have been very helpful

 

[Ross Arden]

1. Can the external observer measure that elapsed time without knowing the reading on his clock when the grain begins to fall and the reading on his clock when the grain lands?

2. Because to take those two readings he will have a difficulty. If he's next to the grain when it begins to fall he can take a clock reading, but when the grain lands his clock will be far away from the grain, so how can he determine that clock reading?

3. He will need some kind of signalling device to send a signal from the place where the grain lands to the place where his clock is located. And he will have to subtract off the travel time of the signal to find out what time it was at the place where the grain landed at the time that it landed. In other words, he will need to know what time it was at that place when the grain landed. This is the process we're referring to when we speak of synchronizing distant clocks. It's what you have to do if you want to use a clock located here to find out what time it is over there.

If you don't understand the above please explain where it was you first started to not understand. Was it the paragraph labelled 1, 2, or 3?

Oh Okay so the observer could fire a laser at the spaceship to determine how far away it was. A second laser pulse now the observer has direction, distance and speed of the space ship. Then he could start his stop watch when the first pulse arrives from the spaceship and stop his stop watch when the last pulse arrives, allow for distance, speed and direction of the space ship, and compare that to his local egg timer ?

 

[Ross Arden]

1. Can the external observer measure that elapsed time without knowing the reading on his clock when the grain begins to fall and the reading on his clock when the grain lands?

wont the elapsed time be (L-(v(delta t))/x. if he knows L, x and v can't he calculate delta t ?

 

[Dale]

Oh Okay so the observer could fire a laser at the spaceship to determine how far away it was. A second laser pulse now the observer has direction, distance and speed of the space ship. Then he could start his stop watch when the first pulse arrives from the spaceship and stop his stop watch when the last pulse arrives, allow for distance, speed and direction of the space ship, and compare that to his local egg timer ?

Yes, this would be coordinate time. He is using a synchronization convention to determine the time at a distant location.

 

[Herman Trivilino]

wont the elapsed time be (L-(v(delta t))/x. if he knows L, x and v can't he calculate delta t ?

If $\Delta t$ is the proper time, that is the time as measured by the observer aboard the rocket ship, the elapsed time measured by your external observer would be $\frac{\Delta t}{\sqrt{1-(v/c)^2}}$.

But regardless of the nature of the relationship between them, the answer is yes, given one you can calculate the other.

Note that to do so your external observer will need to measure the speed $v$ of the rocket ship, so again he needs to know what time it is on his clock when the rocket ship is at some distant location to make that measurement.


(By the way, use LaTeX to express your mathematical expressions so that things like (L-(v(delta t))/x become $\frac{L-v \Delta t}{x}$. To see how just quote this message in your response and you'll see the LaTeX coding. Otherwise people will not be reading those expressions.)

 

[Ross Arden]

If $\Delta t$ is the proper time, that is the time as measured by the observer aboard the rocket ship, the elapsed time measured by your external observer would be $\frac{\Delta t}{\sqrt{1-(v/c)^2}}$.

But regardless of the nature of the relationship between them, the answer is yes, given one you can calculate the other.

Note that to do so your external observer will need to measure the speed $v$ of the rocket ship, so again he needs to know what time it is on his clock when the rocket ship is at some distant location to make that measurement.


(By the way, use LaTeX to express your mathematical expressions so that things like (L-(v(delta t))/x become $\frac{L-v \Delta t}{x}$. To see how just quote this message in your response and you'll see the LaTeX coding. Otherwise people will not be reading those expressions.)

there is no observer on the ship

The observer external to the ship will percieve the egg timer on the ship as ticking faster than the egg timer at rest wrt the observer

there is not "diagonal" of a right triangle in my example so there won't be a square root of anything $\frac{\Delta t}{\sqrt{1-(v/c)^2}}$.it is all straight lines

there is no "triangle" full stop so how you can have a triangle calculation ${\sqrt{1-(v/c)^2}}$.is beyond me

 

[Herman Trivilino]

there is no observer on the ship

It's the time that elapses on the clock that's aboard the ship. Even when there's no observer present.

The observer external to the ship will percieve the egg timer on the ship as ticking faster than the egg timer at rest wrt the observer

Do you have a reference to support that claim? Because theory predicts, and experiments confirm, that it runs slower.

there is not "diagonal" of a right triangle in my example so there won't be a square root of anything $\frac{\Delta t}{\sqrt{1-(v/c)^2}}$.it is all straight lines

You don't need diagonals to get square roots. That's the expression predicted from theory and confirmed by experiment.

 

[Dale]

there is no observer on the ship

Then does it really matter if the egg is undercooked?

The observer external to the ship will percieve the egg timer on the ship as ticking faster than the egg timer at rest wrt the observer

No. The moving egg timer will tick slower than the resting egg timer in the observers frame

 

[Ibix]

there is not "diagonal" of a right triangle in my example

Note that it's a trivial matter to add light clocks to your experiment wherever we want them, thereby introducing "diagonal paths" wherever we want them. Not trying to measure something doesn't mean it's not there. Unless you are trying to argue that your experiment works differently depending on whether a clock is present or not, your quoted statement boils down to "if I don't look at a clock it won't tick normally".

 

[hutchphd]

I am afraid that some of the responses here are overly complicated. When you make a light clock on your lab bench you can sit next to it a regular clock and choose the length of the arms of the light clock so that it ticks once a second as does the clock next to it. All observers will agree about that synchrony.

If the light clock is in motion relative to another Observer, then that Observer will see the pulse of light traveling in the triangular path that has been discussed. . He will conclude that the ticking of that light clock takes longer than one second. He will also comclude that the regular clock in synchrony with the light clock is also running slow . His only consistent conclusion is it all of your clocks in the lab are running slow.

That's it.

 

[Herman Trivilino]

there is no observer on the ship

The observer external to the ship will percieve the egg timer on the ship as ticking faster than the egg timer at rest wrt the observer

It occurs to me that you may be misunderstanding what we mean when we say that the external observer will observe that the clock on the ship is running slow compared to the time-keeping devices that are at rest relative to him.

Let's say that three minutes pass on the clock aboard the ship, and during that time an egg is cooked aboard the ship. It will be a perfectly-cooked three minute egg, and all observers, regardless of their state of motion relative to it, will observe that it did indeed take three minutes to cook it. They will all agree that three minutes went by on the timer and that that is the reason the egg came out perfectly cooked.

This is an example of what we call a relativistic invariant. It simply means that all observers, regardless of their motion relative to a device used to make a measurement, will agree on the result of the measurement. In this case the measurement is 3 minutes. But the measurement could also be of the air pressure inside the ship. If a pressure gauge inside the ships reads 101 kPa, all observers will agree that the gauge reads 101 kPa.

The reason we say that the external observer will conclude that the egg timer is running slow is because of the results of measurements that he takes with other clocks. Clocks that are at rest with respect to that external observer, and synchronized in the rest frame of that external observer, are carefully placed so that one is next to the ship when the timer starts and the other is next to the ship when the timer stops. They might show that, for example, five minutes of time passed while the egg was cooking. This would be the case if the ship were moving at a speed of $\frac{4}{5}c$ relative to your external observer.