# B Time dilation and 2 identical clocks

1. Dec 7, 2017

### Ibix

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

2. Dec 7, 2017

### 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 space ship is moving relative to arrangement 2 at v in the direction of v ...any questions ?

3. Dec 7, 2017

### Ross Arden

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

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4. Dec 7, 2017

### 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.

5. Dec 7, 2017

### Ross Arden

I have ignored length contraction

6. Dec 7, 2017

### Ibix

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

7. Dec 7, 2017

### Ross Arden

so wont the external observer perceive the egg timer in the space ship to be running faster than the egg timer at rest wrt the observer ?

8. Dec 7, 2017

Staff Emeritus
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.

9. Dec 7, 2017

### 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 space ship will have a higher pressure than the at rest one

10. Dec 7, 2017

### Ross Arden

okay

11. Dec 7, 2017

### jbriggs444

That is not correct.

12. Dec 7, 2017

### Ross Arden

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 ?

13. Dec 7, 2017

### Mister T

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?

14. Dec 7, 2017

### Ross Arden

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

15. Dec 7, 2017

### Mister T

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.

16. Dec 7, 2017

### Ross Arden

according to the external observer

the space ship 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 space ship minuses from the velocity of the grain?

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

Last edited: Dec 7, 2017
17. Dec 7, 2017

### Staff: Mentor

That is the proper time of the eggs.

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.

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.

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.

Last edited: Dec 8, 2017
18. Dec 7, 2017

### Mister T

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

19. Dec 7, 2017

### Mister T

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?

20. Dec 7, 2017

No.