# B Time dilation understanding

1. Nov 11, 2017

### Staff: Mentor

These are not the same. You can have acceleration/deceleration in the absence of gravity. Also, in relativity, the proper definition of "acceleration" (and deceleration, which is just a form of acceleration) has nothing to do with gravity: it is proper acceleration, i.e., acceleration that is felt, and can be measured directly by an accelerometer.

2. Nov 11, 2017

### Edem

I thought that I read that Einstein wrote in SR that (and I paraphrase); the force of acceleration is equal to the force of gravity, no distinction can be made.

3. Nov 11, 2017

### Staff: Mentor

Where? Please give a reference.

Einstein certainly didn't say any such thing "in SR", because SR does not take gravity into account at all.

You might be misremembering or misunderstanding one of Einstein's statements of the equivalence principle; but unless you give a source for where you are getting this from, there's no way to tell for sure.

4. Nov 11, 2017

### Ibix

The words SR and gravity do not go together. If you want gravity you need GR.

The point about the equivalence principle is that the "natural" state of things in GR is free-fall. If you're standing on a surface and feeling weight you are being pushed out of your "natural" path by something. You cannot tell (by purely local measurements) if you are "trying" to free-fall to the centre of the Earth and being pushed out of that state by the floor being in the way, or if you are "trying" to move in a straight line in deep space and being pushed out of that state by a rocket hidden under the floor.

This does not say "gravity and acceleration are the same thing". It says "free-fall is free-fall whether you are near a planet or not", and "not being in free-fall is not being in free-fall, whatever the reason you are not in free-fall".

5. Nov 11, 2017

### Edem

Sorry, I guess it's symmetry. I thought it was a equivalency principal.
I thought that in an inertial state all frames of reference are equally valid.

6. Nov 11, 2017

### Edem

I will try to find one.

7. Nov 11, 2017

### laymanB

@Edem

I am still in the learning phase of SR and trying to get a grasp on it myself. Most of the math is still rather esoteric to me but hopefully not someday. Here is how I understand it today using a thought experiment.

There is one person on the surface of the Earth and another traveling in a spaceship near the speed of light at a constant velocity. When the spaceship traveler is perpendicular to the surface of the Earth where the other observer is, they synchronize their clocks and agree on the coordinate system they are going to use. Both of their clocks consist of two mirrors separated by 1/2 a light second in the Y direction (it is a large spaceship). The clocks send a light pulse from the bottom mirror to the top mirror where it is reflected and returns to the bottom mirror where a detector measures the incoming light pulse. This constitutes one second for each observer with their own clocks.

The spaceship is traveling with a velocity near the speed of light in the X direction relative to the Earth observer.

As the spaceship and the Earth are receding from each other let's say they have the ability to look at each other's clocks for a brief moment. From the perspective of each observer it appears that the other's clock is running slowly because each light pulse appears to take a longer path before it returns to the bottom mirror because the mirrors themselves appear to be moving along the X axis of relative motion.

So as each observer looks at the light clock of the other, they both come to the conclusion that the clocks of the other observer are running slowly. But in each of their own references frames, they will see their light clocks operating normally and counting off seconds.

For this thought experiment, let's assume that the size of the universe is small and has periodic boundaries so that the spaceship does not have to accelerate at all and comes back "around" to once again be perpendicular to the observer on the Earth after 10 light days, as measured by the spaceship. They compare their clocks again at the same point in space-time as before, and the clock in the spaceship has indeed recorded much less time than the observer on Earth. How is that possible?

The way I am (hopefully) seeing it, like the experts have posted, is that time is just another dimension. Like the spacial dimensions x,y, and z of familiar Cartesian coordinates we all know and love. Time can be measured in units of length by multiplying c (the speed of lights) by time. The Earth observer's change of position through space (orbiting around the Sun, the solar system orbiting around the galaxy) is negligible compared to the spaceship observer's change of position through space because the velocity of the spaceship dwarfs that of the Earth-solar system velocity. Here (I think) is the crux of the situation. Everyone is moving through space-time at the speed of light. The Earth observer is using most of his velocity to travel along the time axis and very little of the component velocity to travel through x, y, and z. So the Earth observer is moving very nearly to the speed of light along the time axis. The spaceship observer is traveling near the speed of light in the X direction so very little of his component velocity in traveling along the time axis. All the axis components of the velocity must add to c (the speed of light). So, you can either stay relativity still and travel through time quickly for your position in space-time, or you can travel about at velocities near the speed of light and come back to that same location in space-time without much time passing on your (spaceship) clock. You will be younger than the observer on Earth.

Hopefully that's close.

Last edited: Nov 11, 2017
8. Nov 11, 2017

### Edem

Thank you for the example laymanB. It does help some, I need to give it more thought.

9. Nov 11, 2017

### Edem

After doing a little research on points of error in my understanding, revealed by others in this thread. It is clear to me that I lack a basic understanding of some of the terms and words I was throwing around. My apologies.

I joined and posted here hoping to resolve my confusion on this topic (and others).
My initial post on this thread was meant to put out what I thought was right about the topic, mostly so that what was wrong would be pointed out.

I plan on more study of time dilation and related topics. If I learn more, I will be back to discuss. Thanks to all for your "time" and input.

10. Nov 11, 2017

### laymanB

You're welcome. Just don't go using my example in any doctoral dissertation, I'm sure it is full of loose language and probably errors. It could also be flat wrong, we will let the experts weigh in.

I think the first place to start is learning Galilean invariance. You talked several times about being in an inertial state. I'm not sure what that is but I am assuming you mean an inertial frame of reference. An inertial frame of reference is a frame of reference in which Newton's laws are true. This means someplace where you can do an activity like juggling and everything will feel "normal" to your everyday experience like on the surface of the Earth, even though the surface of the Earth is not technically an inertial frame, it is close enough. Once you pick an inertial frame of reference, then any other frame moving with constant velocity relative to that frame is also inertial. Here is a good YouTube video from 1960.

This should give you a good idea of why the light path looks different in the clock example above. If you already understand this fairly well, then disregard it. Once you understand that the Einstein's second postulate states that the speed of light will be measured with the same speed in any reference frame, you can begin to work out the implications how people will disagree about time and distance measurements in their frames of reference and neither is more correct. I'm still learning too.

Last edited: Nov 11, 2017
11. Nov 12, 2017

### vanhees71

That's the wrong approach. You have to learn the math first. You cannot even talk about physic without this math. Of course, muons are accelerated due to gravity of the Earth, but you can safely neglect the effects of gravity of the Earth in HEP physics. It's way too weak to have an important impact on the particles.

Of course, there are exceptions to this rule, as the beautiful example of the measurement of the energy levels of neutrons in the gravitational potential of the Earth (note that this is in the Newtonian approximation) above a reflecting surface, but that's of course physics of ultra-low energetic neutrons. Here is a diploma thesis on the subject:

http://www.pi.uni-hd.de/Publications/dipl_krantz.pdf

12. Nov 12, 2017

### Staff: Mentor

That is precisely the point. A measurement of time is a measurement of a kind of distance in spacetime, called the spacetime interval. Distance in normal Euclidean space is $ds^2=dx^2+dy^2+dz^2$ and the spacetime interval is $ds^2=-dt^2+dx^2+dy^2+dz^2$ so it is a distance in a spacetime with one timelike dimension and three spacelike dimensions.

The odometer analogy is intended to help you understand geometry in spacetime using mental experiences that you already have with geometry in space. You do yourself a great disservice by skipping it. It is one of the most powerful mental tools you have available.

13. Nov 12, 2017

### Staff: Mentor

Note that this assumption violates the principle of relativity. The geometry is called a four-torus. A four-torus may be locally flat, but there exists a preferred reference frame globally.

14. Nov 12, 2017

### 1977ub

If a twin can be understood to be an AI and its history can be transferred electronically between closely passing ships, then the outgoing twin has his history transferred to an incoming ship without deceleration per se. Then when the incoming ship returns home, his history is transferred to a computer at home without deceleration per se.

15. Nov 12, 2017

### PeroK

That's a neat idea. Slightly less fancifully you could transfer the clock reading to an identical clock moving in the opposite direction and thus measure the proper time of the out and return journeys without a physical turn around.

16. Nov 12, 2017

### Staff: Mentor

How did the "outgoing" twin come be "outgoing"?

17. Nov 12, 2017

### 1977ub

The initial outgoing ship was accelerated to speed before t=0 and then the AI was transferred from home to the passing ship at t=0

18. Nov 15, 2017

### Arkalius

I like this way of subverting the acceleration requirement, but everything still works out. The AI "twin" that arrives on Earth still did not occupy a single inertial reference frame, whereas the one on Earth did. This is actually a good way of illustrating why the acceleration itself is not what is important, but the fact that one twin did not occupy a single inertial reference frame while the other did.

19. Nov 16, 2017

### PAllen

Or even more simply, that time was not accumulated over a single inertial path; this statement does not even need to mention frames.