# Some Questions on Time Dilation.

1. Jul 15, 2009

### I_am_learning

I am not interested in knowing the detailed calculation but would be more than happy if you could just answer the simple questions I put.
Suppose in a Universe There are just two peoples A and B in their space ship. Now suppose they are initially at rest and both have a clock that now shows the same time. Now one of the ship starts up and moves away from the other at some speed.
Q. Will now both A and B see each other's clock tick slowly ?

2. Jul 15, 2009

### Tac-Tics

Yes.

3. Jul 15, 2009

### aaryan0077

"According to which frame? One wrt other, or both to some other, of none to single?"

"yes"

4. Jul 15, 2009

### I_am_learning

Answer:They are initially at rest with respect to each other, Since there are nothing except A and B in this universe.

OK You answered, Both will see each others time slow down in this scenario, Right ?. If they now see each other coming towards each other (perhaps one or both of the ship turned their propeller's direction) then what will they observe? Will they both see each others time slow down this time also?

(I have already learned the theories about time dilation so you don't need to provide links to "basics of time dilation" or the like. But I am still confused . You will help best by answering the sequential questions (which I have organized in a manner that will give me a clear view,) I will continue to post)

Last edited: Jul 15, 2009
5. Jul 15, 2009

### ZikZak

Any clock in motion relative to the observer will be observed to run slow. It makes no difference at all which direction the clock is going.

This "observation" is different from what they "see" through a telescope. In Relativity, an "observation" is made once one takes what one sees directly through a telescope and corrects it for the changes in light travel-time to the observer. The approaching clock will appear though a telescope to be running fast, but once light-travel times are corrected for, the observation will be that the clock is running too slow.

6. Jul 15, 2009

### I_am_learning

OK, I am also talking about the corrected time. So in this case also each others time appears to be slow, Right ? Now see, since from the beginning of the motion A and B see each others clock tick slowly now when they meet, Both of them should expect less time to have been passed to the other. What I mean is , A sees that only 5 minutes have passed for B when 10 minutes have passed for himself. So, when they meet A should expect B to be younger by 5 minutes. Similarly B should also Expect A to be younger by 5 minutes. But both can't be younger than the other, can they ? So where is the problem ?
(I have used 5 minutes and 10 minutes just as example)

7. Jul 16, 2009

### Staff: Mentor

In order for A and B to reunite, one of them must "turn around" by firing his rocket engines. Let's say that this one is B, for the sake of discussion. Suppose that B's rocket engines are powerful enough that he can reverse direction almost instantaneously. He will observe (after correcting for light-travel time of course) that A's clock has advanced by a much larger amount than his own, during his (B's) turnaround.

This does not mean than anything actually "happens" to A while B turns around. A feels or observes nothing in particular. This effect is analogous to the following:

Suppose that you have an x, y, z coordinate system attached to your nose, such that the x-axis points straight in front of you, the y-axis points to your left, and the z-axis points upward. Initially you are looking directly at an object that is 100 m in front of you. It has coordinates (x, y, z) = (100, 0, 0). You turn your head quickly so that after a fraction of a second the object is to the left of you. Its coordinates are now (0, 100, 0). In this coordinate system, the object has apparently moved a large distance, very quickly, starting from rest and ending at rest. Yet nothing has actually "happened" to the object.

In relativity, we use not an (x, y, z) coordinate system, but rather an (x, y, z, t) coordinate system. Changing your speed or direction of motion in relativity is analogous to rotating one's coordinate system in non-relativitistic physics. It has effects on your (x, y, z, t) coordinates of that object, even though nothing really happens to that object itself.

Last edited: Jul 16, 2009
8. Jul 16, 2009

### I_am_learning

Well jtbell, thanks for your time. But I am still not fully clear.
You answered, When now A and B approach towards each other, both will observe (ofcourse after light correction) each others clock to tick slowly. But depending on Who as ACTUALLY turned over, The person who as turned over will be very backward in time and when they meet he will be much younger. Right ?

Now I am confused by what it means by who has ACTUALLY turned over. How will one know if its A or B who is turning over by mere observation. For A, B would appear to turn Over For B, A would appear to turn Over.
And please don't tell me that-->"the one Who has really gone on acceleration would FEEL it". I question, When we are accelerated towards the earth when on free fall, do we feel it? Is the feeling any different than standing still in space ?

9. Jul 16, 2009

### sylas

There is an equivalence between gravity and the forces experienced by an accelerating observer. If you are standing still in the Earth's gravitational field, your clock runs slow also. So in fact, "feeling" the acceleration IS relevant, and it IS a real substantive difference between the two twins, and it DOES stand as a legitimate answer to the question of who has turned around. The one who experiences an acceleration is the one who turns around... given the constraints of the problem.

There's no amiguity as to who turns around. In Special Relativity, there is a special status for an "inertial observer", and every inertial observer can immediately identify who changes direction. There are also several different ways you can tell that you are NOT an inertial observer... one of these is observing a local gravitational field. Another way is that the observer who turns around can observe at the same time a change in the apparent size of the other twin; whereas the twin who remains inertial does not observe any size change.

Cheers -- sylas

10. Jul 16, 2009

### I_am_learning

So, you mean acceleration isn't relative, its ABSOLUTE. We can always say whether a body is at rest or its accelerating. Right ?
Then like velocity (which is relative) we don't need to say accelerating with respect to what. for eg.instead of saying the car is accelerating at 5m/s^2 w.r.t. the road, we can simply say accelerating at 5 m/s^2 . Do you mean this also ?
And also,
Can you tell me a very simple experiment I can carry out in a space to test whether I am accelerating or not.

11. Jul 16, 2009

### Staff: Mentor

You already know the answers and have already rejected them. I don't know what you expect from the rest of us, but it sounds like you have an agenda rather than a real inquiry.

12. Jul 16, 2009

### I_am_learning

I am sorry if I have appeared so rude like you have mentioned. But what I wanted was--- to ask the question -:
beforehand, if you would tell me we will feel the acceleration.

Last edited: Jul 16, 2009
13. Jul 16, 2009

### A.T.

For proper acceleration, yes.
No, this is coordinate acceleration (dv/dt), which is relative,
Sure, just drop something, and see if it moves away from you. Or get one of these:
http://en.wikipedia.org/wiki/Accelerometer
No, free fall means zero proper acceleration, or advancing straight in curved space-time. It is equivalent to moving inertially in space (tidal effects aside):
http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html [Broken]

Last edited by a moderator: May 4, 2017
14. Jul 16, 2009

### Staff: Mentor

Yes, you can directly measure proper acceleration using an accelerometer.
Yes, we can say that the norm of the proper acceleration is some value without reference to any coordinate system. Note that, unless the road is in free-fall the car's proper acceleration will be different than the coordinate acceleration wrt the road.
Yes, read the value of an accelerometer.
The proper acceleration of a freefalling observer is zero, it is only undergoing coordinate acceleration. Do you understand the difference between proper acceleration and coordinate acceleration?

Last edited: Jul 16, 2009
15. Jul 16, 2009

### I_am_learning

AH! Seems like here is my problem. Whats proper and co-ordinate acceleration? I would be more than happy if you could tell me in simple terms their difference rather than posting a link to a long and confusing topic. By the way I don't demand too simple language, I am a High-school pass student.

Last edited: Jul 16, 2009
16. Jul 16, 2009

### A.T.

Coordinate acceleration is the acceleration relative to an arbitrary frame of reference. Proper acceleration is the acceleration relative to a local free falling frame of reference.

17. Jul 16, 2009

### I_am_learning

Oops sorry, but I am afraid I have to ask, whats LOCAL FREE FALLING frame of reference?

18. Jul 16, 2009

### I_am_learning

ALSO Look at this scenario. Suppose A and B are at present moving towards each other at relativistic speed. Further suppose A and B sees each other (after light travel time correction) of same age at this instant. Now, since both see each others time pass slowly they should find each other younger when they meet. Where is the problem?

19. Jul 16, 2009

### A.T.

What don't you understand? What a "frame of reference" is? Or what "free fall" is?

In simple words: If you drop something, and measure it's coordinate acceleration (dv/dt) relative to you, then your proper acceleration is the negative of that value.

20. Jul 16, 2009

### I_am_learning

Then we can achieve proper acceleration only by the forces that will act on us but don't act on objects we drop; Right ?
So its never ever possible to accelerate (proper) by gravitational forces. But we can accelerate (proper) by cars and ships which accelerates us by providing forces through the seats, so that if we drop a coin, it will fly backwards.
Am i correct till now?

21. Jul 16, 2009

### A.T.

Yes, that's why there are no gravitational forces in General Relativity, because gravity doesn't produce any absolute acceleration. Proper acceleration is the deviation from free fall (or inertial motion).

22. Jul 16, 2009

### I_am_learning

Is it only gravity that can't produce proper acceleration?
I don't understand this. If you are accelerated by a high-current wind, then if you drop something it will also be driven by the wind so you observe it to be at rest. So, are you not accelerating (proper) this time also? Whats a sure-shot test ?

23. Jul 16, 2009

### A.T.

Any inertial forces which are proportional to the mass do not produce proper acceleration. They are called pseudo forces sometimes.
Drop something in an evacuated jar. Make sure it is not affected by electric or magnetic forces.

24. Jul 16, 2009

### I_am_learning

is it just the MASS or any force that is proportional to charge (magnetic, electric or the like) would also not produce proper acceleration.

If I am a charged particle and am accelerating (I think it should be proper acceleration) in an electric field then I drop a particle in an evacuated chamber which is also charged, so it also experience the force so would appear to be at rest. Then I conclude that I am not proper accelerating. Where am I wrong?

(If you like me to shield the chamber from electric and magnetic field then equivalently, I would like you to shield your chamber from Gravitational fields so that the scenario isn't any different from that of gravitational case )

(If you would say that, "You should Drop uncharged particle", but I further go onto supposition that Everything are charged, (not necessarily electrically charged), and I am being accelerated by the force (not known till today ) which accelerates this particular type of charge I am talking about)

25. Jul 16, 2009

### A.T.

No, just forces proportional to mass don't produce proper acceleration, because they accelerate everything at the same rate.
Drop two things with the same charge but different masses. Or just buy an http://en.wikipedia.org/wiki/Accelerometer" [Broken] if you cannot figure out the technicalities.
Physics deals with the observed reality so far, not with some potentially to come stuff.

Last edited by a moderator: May 4, 2017