Understanding Time in a Moving Frame: Observer's Experiment with Egg Timers

[Ross Arden]

An observer standing on Earth has 4 egg timers (ET) ET1 to ET4. All the ETs are identical. Each ET only has one grain of sand in it and all the grains are identical.

When a grain of sand falls thru the neck of ET1 it emits a green flash of light. When the grain of sand hits the base of ET1 it emits a second flash of green light.

When a grain of sand falls thru the neck of ET2 it emits a blue flash of light. When the grain of sand hits the base of ET2 it emits a second flash of blue light.

Similarly ET3 emits red light and ET4 emits yellow light.

See the picture. The observer on Earth has ETs 3 and 4 and one ET is oriented upside down wrt the other, the control.

The observer places ETs 1 and 2 in a space ship. One upside down wrt the other as depicted. The spaceship blasts off and reaches a leisurely steady velocity v as depicted, say 100 km/hr. The observer wants to observe the period of time between the flashes from ETs 1 – 4. The period of time between to 2 blue flashes. The period of time between both yellow flashes etc etc .

The round dot in the ETs is the grain of sand.

The observer will observe the following.

The time between the flashes from ET1 will be less than the flashes from ET2 – ET4. As such the observer concludes that the time for a single grain of sand to pass thru ET1 will be less than the time for a single grain of sand to move thru the other egg timers.

As such time in the moving frame of reference ET1 is actually transpiring faster than ETs 2 – 4. The time in the moving frame of reference is transpiring faster than the stationary frame. Is that correct?

As one grain of sand is faster multiple grains of sand will also be faster. If there are 100 grains in each ET it will simply be the time of one grain multiplied by 100.

The observer conducts the same experiment millions of times and gets the same or roughly the same result ie it is reproducible.

Is the above correct ?

 

[sweet springs]

The spaceship blasts off and reaches a leisurely steady velocity v as depicted, say 100 km/hr.

After reaching the steady speed, no force work on sand grains so they do not fall.

 

[Ross Arden]

I thought someone might say that

we'll place an identical magnet on the base of each egg timer and replace the grain of sand with a grain of iron, and ignore any effects of gravity ...the result will be the same, the period of time between the blue flashes will be the smallest

 

[Ross Arden]

will the flash period be in this order

ET1 < ET3
ET1 < ET4
ET4 = ET3
ET2 > ET4

 

[Ross Arden]

so time in the moving frame of reference is simultaneously faster an slower than the stationary frame ?

 

[FactChecker]

Is the observer timing when the flashes reach him or is he using "clone" observers in his reference frame to record times at the position of the lights when they flash?

 

[Dale]

Is the above correct ?

No. Egg timers 1, 2, and 4 don’t work. Only egg timer 3 works.

 

[Ross Arden]

sorry ET1 emits green light, the green flashes will have the smallest period.

 

[Ross Arden]

Is the observer timing when the flashes reach him or is he using "clone" observers in his reference frame to record times at the position of the lights when they flash?

the observer is timing the time between like flashes. The amount of time between the two green flashes, the amount of time between the 2 blue etc etc

 

[Ross Arden]

No. Egg timers 1, 2, and 4 don’t work. Only egg timer 3 works.

why ?

 

[Ross Arden]

Is the observer timing when the flashes reach him or is he using "clone" observers in his reference frame to record times at the position of the lights when they flash?

when the flashes arrive at the observer

 

[Dale]

why ?

No gravity for 1 and 2. 4 has already run out.

 

[Nugatory]

The time between the flashes from ET1 will be less than the flashes from ET2 – ET4. As such the observer concludes that the time for a single grain of sand to pass thru ET1 will be less than the time for a single grain of sand to move thru the other egg timers.

Are we waiting until the ship has achieved its "leisurely constant velocity" before any of the egg timers do their flashing? That sounds like what you're describing but if so ET1 and ET2 will behave the same way: the same amount of time will elapse between the emission of the two green flashes as between the two blue flashes. This time will be greater than the time between the emission of the two red and two yellow flashes, so the earthbound observer will properly conclude that time is running slow on the spaceship. The earthbound observer will have to allow for the light travel time when calculating when all of these emission events happened: time the flash reaches Earth observer's eyes, minus the light travel time (distance divided by $c$) will be the emission time.

The spaceship observer looking at the red and yellow flashes will come to the conclusion that time is running slow back on earth. That is, he will find that after allowing for the light travel time, time between the emission of the two green and two blue flashes is less than the time between the emission of the two red and two yellow flashes.

Earth time is slower than spaceship time in a frame in which spaceship is at rest. Spaceship time is slower than Earth time in a frame in which Earth is at rest. To make sense of this scenario, you have to draw a spacetime diagram, or work through the Lorentz transforms to see how the relativity of simultaneity makes it all work, or (best!) do both.

 

[Ross Arden]

Is the observer timing when the flashes reach him or is he using "clone" observers in his reference frame to record times at the position of the lights when they flash?

Thanks for that I will do the sums and see how it works out

 

[Ross Arden]

Thanks for that I will do the sums and see how it works out

Hang on the distance of the observer to the spaceship it makes no difference

 

[Nugatory]

so time in the moving frame of reference is simultaneously faster an slower than the stationary frame ?

The time in the moving frame is always slower than the time frame in the stationary frame.

What you may be missing is that you can choose either frame to be the stationary frame. The physics is the same and there is no contradiction or paradox whether you choose a frame in which the spaceship is at rest (earth is moving, Earth time is dilated) or a frame in which the Earth is at rest (spaceship is moving, spaceship time is dilated).

 

[Ross Arden]

No gravity for 1 and 2. 4 has already run out.

see my post on using a magnet

 

[Ross Arden]

The time in the moving frame is always slower than the time frame in the stationary frame.

.

But doesn't time dilation say the observer who perceives the spaceship as moving should conclude time in the spaceship is dilated ?

 

[Dale]

see my post on using a magnet

This whole setup is needlessly complicated. You were not using any of the physics of egg timers to analyze the situation originally, and you are not using any of the physics of magnets to analyze them now, so why bother?. Just say that there are four accurate clocks emitting light flashes.

 

[Nugatory]

But doesn't time dilation say the observer who perceives the spaceship as moving should conclude time in the spaceship is dilated ?

Yes.

So if we choose a frame in which the spaceship is moving and the Earth is at rest then the spaceship time will be dilated. But we could just as well have chosen a frame in which the spaceship is at rest so the Earth is moving in the opposite direction; now the Earth is moving so it's Earth time that is dilated.

To make sense of this you should try a simpler scenario: the spaceship and the Earth are approaching one another at constant speed $v$. As the ship passes by the earth, both observers set their wristwatches to 12:00 noon. Sometime later, the spaceship observer looks at his wristwatch and sees that it reads 1:00 PM (and naturally the spaceship observer has aged one hour between these two events).

Using the frame in which the spaceship is at rest, what time appears on the Earth observer's wristwatch at the same time that the spaceship observer's wristwatch read 1:00 PM?

Using the frame in which the Earth is at rest, what time appears on the Earth observer's wristwatch at the same time that the spaceship observer's wristwatch read 1:00 PM?

Both questions are straightforward calculations using the Lorentz transformations.
Also, while you're at it, find any post by @Orodruin just so that you can see what's in his .signature.

 

[Ross Arden]

This whole setup is needlessly complicated. You were not using any of the physics of egg timers to analyze the situation originally, and you are not using any of the physics of magnets to analyze them now, so why bother?. Just say that there are four accurate clocks emitting light flashes.

too complicated

 

[Ross Arden]

Yes

So the observer on Earth should see time inside the spaceship as ticking slower?

 

[Ross Arden]

So if we choose a frame in which the spaceship is moving and the Earth is at rest then the spaceship time will be dilated. But we could just as well have chosen a frame in which the spaceship is at rest so the Earth is moving in the opposite direction; now the Earth is moving so it's Earth time that is dilated.

Im not choosing any frame I just want to know what the observer sees, not an observer observing the observer

 

[Ross Arden]

the answer is very simple will the observer observe
ET1 < ET3
ET1 < ET4
ET4 = ET3
ET2 > ET4

yes/no

 

[Nugatory]

Im not choosing any frame I just want to know what the observer sees, not an observer observing the observer

When you say something is moving, you are choosing a frame. The words "it is moving" are only meaningful if you specify a frame; when you say the ship is moving and the Earth is not you are specifying a frame in which the Earth is at rest and (because the spaceship and the Earth are moving relative to one another) the spaceship is not.

the answer is very simple will the observer observe
ET1 < ET3
ET1 < ET4
ET4 = ET3
ET2 > ET4

If those are supposed to be times between the emissions of the flashes and the spaceship is done accelerating so is moving at its leisurely constant speed, then ET1=ET2 and ET3=ET4, for everybody no matter where they are and what their state of motion relative to one another, the earth, and the spaceship. "Everybody" includes the guy on earth, of course.

Whether ET1 (and therefore also ET2 because it is equal to ET1) are less than, equal to, or greater than ET3 (and therefore also ET4 because it is equal to ET3) depends on the frame you choose. If you choose the frame in which Earth is at rest then ET1>ET3.

I really strongly suggest that you try the scenario in post #19 above. It is much much simpler to analyze, and until you understand that simpler case your egg timers will continue to be confusing.

 

[Ross Arden]

If those are supposed to be times between the emissions of the flashes and the spaceship is done accelerating so is moving at its leisurely constant speed, then ET1=ET2 and ET3=ET4, for everybody no matter where they are and what their state of motion relative to one another, the earth, and the spaceship. "Everybody" includes the guy on earth, of course.

I would like a second opinion. Do others agree with this ?

 

[Ross Arden]

Whether ET1 (and therefore also ET2 because it is equal to ET1) are less than, equal to, or greater than ET3 (and therefore also ET4 because it is equal to ET3) depends on the frame you choose. If you choose the frame in which Earth is at rest then ET1>ET3.

I have already told you which frame I choose. I choose the frame of the observer, what does the observer see!

 

[Ibix]

Assuming you are neglecting gravity and assuming you are correcting for light travel time then in the Earth frame ET1 = ET2 < ET3 = ET4 ET1 = ET2 > ET3 = ET4 (edit: I got confused about which timers were on the ship and which on Earth, so had the inequality the wrong way round - thanks to Nugatory for pointing that out).

If you aren't correcting for light travel time you need to specify positions of things because your diagram is not to scale.

 

[Nugatory]

I have already told you which frame I choose. I choose the frame of the observer, what does the observer see!

"See" is ambiguous here.

By "see" are we talking about when the two flashes of light of the same color reach the Earth observer's eyes (in which case we are not talking about time dilation or the rate at which time passes on the spaceship compared with earth)? Or are we talking about the amount of time that passes between when the two flashes of light are emitted?

Either way, the ET1=ET2, ET3=ET4, ET1>ET3 relationships hold in the situation you've set up. However, the analysis is way more complicated using the first meaning of "see"; we have to allow for Doppler effect, light travel time, and the changing distance between ship and earth. This is all assuming that:
- All flashes are emitted after the ship is done accelerating so the ship is moving at a constant speed relative to the earthbound observer.
- ET1 and ET2 are the ones on the ship.
- The ship is moving away from the earthbound observer. (This doesn't matter for the second, simpler, more common understanding of "see", but because of the Doppler effect it does matter for the first more complicated one).
- The physical dimensions of the egg timers are negligible. (Without this assumption, the problem becomes even more complicated without introducing any new physical insight, and trying to understand it before working through the simpler problem in post 19 is a complete waste of time, like trying to learn long division without learning subtraction first).

 

[Ibix]

Hm. I'm revising my answer. I was thinking of the egg timers as clocks, but they aren't clocks in the way they are being used here. The start and stop events aren't in the same spatial location in any relevant frame, so the relativity of simultaneity matters. So ET3 = ET4 < ET2 < ET1. (Edit: I'm assuming here that the various ETs are the times between flashes, measured in the Earth frame and with the different light travel times subtracted out, and that the average velocity of the sand grains in the rest frame of their "egg timers" is less than half the velocity of the rocket relative to Earth - assuming that the grain moves with a 1g acceleration, that last requires an egg timer length less than 40m).

The problem statement is rather misleading in referring to the devices as egg timers. They're just sand launchers and targets in this application. They don't time anything meaningful. Avoiding this issue is why Nugatory is assuming the egg timers are of negligible size in his last post, I suspect.

 

[Ibix]

We've had a very similar conversation before, by the way. https://www.physicsforums.com/threads/time-dilation-and-2-identical-clocks.932643/

 

[Dale]

the answer is very simple will the observer observe
ET1 < ET3
ET1 < ET4
ET4 = ET3
ET2 > ET4

yes/no

The answer is simple but the question isn’t well enough specified to get any answer. You have tons of unnecessary details (mechanism of clocks, colors of lights etc) and you omit some necessary details. For example: what are these ET numbers supposed to represent? Are they the times between the corresponding flashes being received according to an Earth observer or a rocket observer, or are they the times between the flashes being emitted according to an Earth observer or a rocket observer, or something else?

The reason you are not getting a straightforward answer is because your question does not allow a straightforward answer. Relativity does not depend on the clock mechanism, so don’t specify it, just use a generic “accurate clock”. Any experiment depends on exactly what is measured, so specify what ET1 is. There is a difference between coordinate and proper times, so if possible specify that as well or ask for help determining it. In relativity many quantities are frame variant, so specify which reference frame they are determined in.

 

[Mister T]

I would like a second opinion. Do others agree with this ?

One the things you'll notice is that people will correct each others' errors on this forum. If someone disagrees with an answer you receive, you will see that disagreement right away and it will get corrected.

 

[Dale]

A lack of disagreement could have other causes. In this case some people (e.g. me) don’t feel that the problem is well enough specified to even have a clear answer. The question is not well enough specified for me to either agree or disagree with @Nugatory (although on good questions we usually agree historically)