# B Time dilation and 2 identical clocks

1. 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
2. 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
3. Dec 7, 2017

### Mister T

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

4. 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?

5. Dec 7, 2017

No.

6. Dec 7, 2017

### Ross Arden

Okay thanks you have been very helpful

7. Dec 7, 2017

### Ross Arden

Oh Okay so the observer could fire a laser at the space ship 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 space ship 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 ?

8. Dec 8, 2017

### Ross Arden

wont the elapsed time be (L-(v(delta t))/x. if he knows L, x and v cant he calculate delta t ?

9. Dec 8, 2017

### Staff: Mentor

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

10. Dec 9, 2017

### Mister 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.)

Last edited: Dec 9, 2017
11. Dec 13, 2017

### Ross Arden

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 wont 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

12. Dec 13, 2017

### Mister T

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

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

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

13. Dec 13, 2017

### Staff: Mentor

Then does it really matter if the egg is undercooked?

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

14. Dec 13, 2017

### Ibix

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".

15. Dec 13, 2017

### 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.

16. Dec 15, 2017

### Mister T

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.