# Time dilation when travelling

1. Oct 2, 2004

### vuxx

i understand that time slows down as compared to a stationary perspective when an object is travelling at a certain velocity, but by how much does time slows down for that object? whats the formula behind this time difference? also, i understand that Einstein has derived this idea when he was comparing the different perspective of an observer with how he observed the speed of light approaching him ie. the result is that the observer will measure the same speed of light from whatever perspective he was measuring. so is it right for me to say that when one is travelling at the speed of light (just an analogy although i know it is impossible), his time should stop as compared to our perspective??

2. Oct 2, 2004

### Staff: Mentor

Imagine two frames moving with respect to each other with some speed v. (Two rocketships, for example.) Each frame will measure the clocks in the other frame to be running slow compared to their own clocks. If a clock measures a time interval of $\Delta t_0$, then the other frame will measure the time that passed according to their clocks to be $\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2}$.

Of course, each frame views their own clocks as running normally.

3. Oct 2, 2004

### Fredrik

Staff Emeritus
The time dilation factor is

$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

Last edited: Oct 2, 2004
4. Oct 2, 2004

### vuxx

i dont understand. when u say 'a clock measures a time interval of...', which clock are u talking about?

5. Oct 3, 2004

### jdstokes

Either clock. Suppose I observe an oscillating pendulum whose centre of motion is at rest with respect to me. I measure the pendulum to execute one oscillation in an interval of time $\Delta t$. If you move away from me at some fraction of the speed of light $\beta$, you will measure the period of the pendulum as $\Delta t/\sqrt{1-\beta^2}$. The consequence is that intervals of time between events are measured as longer by observers in motion with respect to the events.

6. Oct 3, 2004

### vuxx

thanks a lot i understand what u mean. but what i want to know precisely is that when using the equation for the example given by Doc Al, the velocity 'v' is the velocity of the frame of the moving clock that has a time interval of $\Delta t_0$, correct?

Last edited: Oct 3, 2004
7. Oct 3, 2004

### jdstokes

Yes. In this case $v$ denotes the velocity of the clock wrt the observer. This is of course equal in magnitude to the velocity of the observer wrt the clock and opposite in sign. However, in this example the sign of the velocity is irrelevant as the equation involves only $v^2$.

8. Oct 4, 2004

### vuxx

thank you very much jdstokes. because im quite new to this topic so that is y i posted a lot of questions. another thing i want to verify is when measuring $\Delta t_0$ and $\Delta t$, are we comparing both intervals from the same frame? if yes, what is that frame?

9. Oct 4, 2004

### Staff: Mentor

different frames, different times

The two time intervals are measured in different frames, moving with speed v with respect to each other. Let's call the two frames frame #1 and frame #2. Let $\Delta t_0$ be the time interval measured by a clock (call it clock A) in frame #1, the rest frame frame of clock A. (The "rest frame" is the frame in which the clock is at rest, that is, its speed is zero. Clock A is stationary in frame #1, but moves with speed v in frame #2.) $\Delta t$ is a measurement made from frame #2. According to frame #2, clock A is running slow; the actual time elapsed according to frame #2 is $\Delta t$, while clock A only reads $\Delta t_0$.

10. Oct 5, 2004

### vuxx

Let $\Delta t_0$ be the time interval measured by a clock (call it clock A) in frame #1, the rest frame frame of clock A.

when u are saying time interval, whats the time difference measured from?

11. Oct 5, 2004

### Staff: Mentor

Between any two times measured by clock A there is an interval of time. For example, let $t'_1 = 12:00$ and $t'_2 = 12:05$ be two time measurements by clock A. Then $\Delta t_0 =5$ minutes is the time interval measured by clock A.

Of course, a different frame will measure different times ($t_1$ and $t_2$) and a different time interval $\Delta t$.

12. Oct 6, 2004

### vuxx

Thanks alot Doc Al, i understand now. To verify, i will give an example. Lets say we are going to conduct an experiment which is the same as the example given by you previously. I will start the experiment at 1pm and put a clock in the moving frame which is at first synchronised when at rest. I will then note down the time as the clock in the moving frame passes by me with velocity 'v' which is a significant fraction of 'c' (lets say i am able to see the clock). $\Delta t_0$ will then be the time interval of the difference in time between 1pm and the time i note down. Using $\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2}$, i will get the time interval which is at the rest frame where i am and that time interval is $\Delta t$. $\Delta t$ will be an interval greater than $\Delta t_0$. Am i correct?

13. Oct 6, 2004

### Staff: Mentor

I think you've got it. But since I don't completely understand your example, I will replace it with an equivalent one of my own. (One must be very precise when setting up these special relativity problems!) Let's say I have a huge laboratory with synchronized clocks at both ends of the lab (call them clocks A and B). A fast rocket with its own clock zooms by, first passing my clock A, then B. Let's pretend I am able to see what that moving clock is reading as it zooms past.

Let's also pretend that the rocket clock is set just right so that the moment it passes my clock A, both the rocket clock and clock A read 1 pm. Then I observe that the rocket clock reads a time $\Delta t'$ after 1pm exactly as it passes clock B, which reads a time of $\Delta t$ after 1pm. So the time interval measured in the rocket frame is $\Delta t'$, while the time interval according to my lab clocks is $\Delta t$. Special relativity tells us that the two intervals are related by $\Delta t = \Delta t' / \sqrt{1 - v^2/c^2}$. And, yes, $\Delta t > \Delta t'$, consistent with the time dilation maxim "moving clocks appear to run slow".

14. Oct 7, 2004

### vuxx

I have understand how the equation works and what it means now thanks! But i will like to ask you a paradox. From my perspective ( lets say a stationary frame on a train station platform on Earth), the clock in the moving frame (lets say a clock in a train) will appear slow to me. However, from the perspective of the person in the moving frame he will see me as the one moving and not him, hence he will see my clock to be slow and not his. Therefore, whos clock is slow?

15. Oct 8, 2004

Staff Emeritus

The answer to that is Yes. Both clocks are "right", because "right time" can only be defined relative to the inertial frame.

16. Oct 8, 2004

### Staff: Mentor

No paradox. Each observer views their own clocks operating normally and the other frame's clocks as being slow. (If it didn't work symmetrically, then there would be a problem as you would be able to tell which frame is "really" moving.)

To become comfortable with this, and realize that there is no contradiction or paradox, you need to learn a bit more about relativity. To fully understand how space and time are affected by relative motion, one must consider several relativistic effects operating together:
(1) time dilation
(2) length contraction
(3) the relativity of synchronization​
Check out any elementary book on special relativity.

17. Oct 8, 2004

### vuxx

If both clocks are 'right', doesnt that means both people in the stationary frame and the moving frame will see the same clock at one frame (lets say the stationary frame) to be 2 different times?

18. Oct 8, 2004

### Staff: Mentor

No. To remove any ambiguity, imagine that someone standing right next to a clock (clock A) in the "stationary" frame fires a gun. If that clock reads time t, then everyone will agree that according to the stationary clock, the gun went off at time t. Even observers in the "moving" frame, if they could see clock A as it whizzed by, would agree that clock A read time=t when the gun went off. But the "moving" frame will observe the gun to fire at a different time according to its clocks (clock B, for example).

19. Oct 9, 2004

### vuxx

But why should the person in the moving frame reads the same time 't'? Since from the perspective of the person in the moving frame, the stationary frame should be the one moving and not his, then he should read time 't' to be slower isnt it?

20. Oct 9, 2004

### Staff: Mentor

I'm not clear as to what you are talking about. Each frame's clocks read their own time. And each frame will observe the other's clocks to go slow. Do you have a concrete example in mind?

As I said in an earlier post, in order to analyze a particular situation you need to consider all three SR effects.