# B Time dilation and 2 identical clocks

1. Dec 7, 2017

### Ross Arden

Okay thanks you have been very helpful

2. 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 ?

3. 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 ?

4. 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.

5. 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
6. 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

7. 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.

8. 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

9. 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".

10. 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.

11. 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.