# Relativity for the lay person

1. Feb 22, 2009

### mintparasol

Hi everyone,
The good Nick Summers from simplegravity.com directed me here as I'd been pestering him with questions regarding relativity and the structure of the cosmos.
I am neither a mathematician nor physicist. I have no formal training in the sciences beyond high school but have fond memories of physics and chemistry classes where the structure of reality was first outlined to us students about 13-14 years ago. I remember being taught about the structure of the atom, the uncertainty principle, the exclusion principle, valencies, newtonian mechanics, the gradated complexity of matter from the helium atom up to the heavy metals etc, etc. and thinking to myself at the time 'hey, this makes so much sense!'.
After leaving school, I kept alive my interest in the true nature of reality by trying to come to terms with the special theory of relativity, reading 'A Brief History of Time' and all sorts of physical and metaphysical books by the likes of Fritjof Capra and Robert Pirsig. I still often train my small telescope on planets and the moon and the like and am fascinated with the cosmos, the various explanations of the origins of the universe, the origins of life and so on.
My understanding of relativity needs a lot of work and I'm hoping someone here can answer the following question and perhaps direct me on to any source texts that are available online. This would be a big help in my understanding of the nature of the cosmos and any help would be greatly appreciated.
My question is this:-

Say I have 2 highly accurate clocks in perfect sync sitting side by side on the earths' surface. I take one of the clocks and go for a ride in my spaceship to the outer edge of the solar system or some other great distance from the earths' gravitational field. The clock I have with me on the spaceship should now be running faster than the clock I've left behind on earth, correct?
So now I return to earth and place the clocks side by side again. Will the clock I took on my space journey still be ahead of the clock I left behind or will both clocks be showing the exact same time?

I hope my question makes sense and look forward to any replies,
thanks in advance,
ad

2. Feb 25, 2009

### Naty1

You are asking about the twins paradox....do a search here for MANY discussions on the subject here or try Wikipedia:

http://en.wikipedia.org/wiki/Twin_paradox

....misunderstanding often arises because of subtlies involved....keep in mind its all RELATIVE: the moving observer and the "stationary" observer each see the OTHER clock running slower....yet each observer sees their own local clock tick as "NORMAL" ....what each observes about the OTHER clock is dependent on the relative velocity between them.

Make a mental note: when relative velocities are present between observer and observed, time slows and length contracts....for special relativity...

3. Feb 25, 2009

### Ich

It depends. If you made your journey slow enough, your clock will be ahead. If you went with high speed, you clock will lag behind.

4. Feb 25, 2009

### Saw

Ich, your answer doesn't look like the standard one. Can you explain how you obtain it?

I thought the standard answer was that the clock of the twin that flies away and accelerates to return would lag behind, this lack of symmetry being due to the fact that such clock has "changed frames".

5. Feb 25, 2009

### JesseM

In special relativity you're correct, but the original post specified that it was a scenario where the traveling twin moved a great distance from "earth's gravitational field" so it seems like mintparasol was interested in factoring in the effects of gravitational time dilation as well as velocity-based time dilation. Since clocks run slower deeper in Earth's gravity well, if you move away from Earth slowly enough, the gravitational time dilation should dominate and you will have aged more than your twin on Earth, not less as in the usual twin paradox scenario where gravity is ignored.

6. Feb 26, 2009

### mintparasol

Thanks folks.
Yes I was talking about a scenario that can actually be experienced by humans, velocities that our bodies can endure etc.
So if the clock I take on my space journey runs faster the farther it is from gravitational fields, I assume the opposite applies as I approach a strong gravitational field:- it should start running more slowly?
If this is the case, can we say that gravitational fields act to slow down the matter in any object?

7. Feb 26, 2009

### mintparasol

I'm very uncomfortable with this part of relativity. It's not clear how we can put observers into these pictures when the energies and velocities we're talking about would rip apart any lifeform capable of making such an observation. I mean, we're talking really high values before this is even observable. I know we have proof of this by induction but has it actually ever been directly proved experimentally?

8. Feb 26, 2009

### JesseM

Velocities make no difference to your body, only accelerations do. In relativity all velocity is relative anyway, there is no objective truth about how fast you are going. You could be traveling at 99.9999% the speed of light relative to the Earth through deep space, and as long as you were traveling at constant velocity with no acceleration, you would feel completely weightless, your experience would be indistinguishable from being in deep space at rest relative to the Earth. The idea that the non-accelerating observers moving at constant velocities relative to one another will all have identical experiences (at least if they don't look out the window to check their motion relative to the rest of the universe), and that any experiments they do on board their ship will yield identical results, is actually enshrined as one of the two basic postulates of relativity (the other being that such observers will all measure light to have the same speed relative to themselves).

9. Feb 26, 2009

### Staff: Mentor

No, relativistic time effects have been observed and studied with relative velocities that are achieved by airplanes in flight. See for example

http://tycho.usno.navy.mil/ptti/ptti2002/paper20.pdf [Broken]

in particular section 5, "Flight Tests."

Last edited by a moderator: May 4, 2017
10. Feb 26, 2009

### mintparasol

Please understand I'm not being deliberately difficult, it's just that my understanding is partially clouded by a slightly more 'mechanical' view of things.

A number of points here:-

1. How can the observer reach 99.99999% the speed of light from 'rest' without accelerating?

2. If the observer is travelling at 99.99999% the speed of light, the real universe is really whizzing by outside.

3. 'The other being that such observers will all measure light to have the same speed relative to themselves' - Does this mean that an observer travelling at, say, 87% the speed of light will observe light travelling at x m/s relative to themselves while an observer travelling at 99% the speed of light will also measure light travelling at the exact same speed relative to themselves?

11. Feb 26, 2009

### JesseM

You do have to accelerate to change velocities, but you're free to choose an acceleration as gentle as you like so no harm comes to your body, it'll just take longer to achieve a given change in velocity.

As a side note, keep in mind there is no such thing as absolute speed in relativity, so the phrase "99.99999% the speed of light" isn't meaningful unless you define what object (or what frame) the speed is being measured relative to. For example, if we are moving apart at 99.99999% the speed of light, then in my frame I am at rest and you are traveling at 99.99999% the speed of light, while in your frame it is you who are at rest and me who is traveling at 99.99999% the speed of light, and both perspectives are considered equally valid.
If you're moving at 99.99999% relative to the Earth, other stars will be moving very fast relative to you too (because although all stars don't share the same rest frame, their speed relative to one another tends to be small compared to light speed), but still the laws of physics are the same in your rest frame as in a frame where the Earth is at rest, so your body won't feel any different. The basic idea is that if two ships are moving at different velocities in deep space (far from sources of gravity) without accelerating, then as long as they don't look out their window all their local observations onboard the ship will be identical.
Yes, that's right, all observers moving at constant velocity measure the speed of light to be 299,792,458 meters/second relative to themselves. Keep in mind that each is assumed to use rulers and clocks at rest relative to themselves to measure speed = distance/time, and in relativity each observer sees the rulers of another observer in motion relative to themselves as shrunk by a factor of $$\sqrt{1 - v^2/c^2}$$ relative to their own ruler (length contraction), and the other observer's clocks to be running slow relative to their own clocks by the same factor (time dilation), plus there is something called the relativity of simultaneity which says that two clocks which are synchronized and a distance d apart in their own rest frame will be out-of-sync by vd/c^2 in another frame where they are moving with speed v along the axis joining them. On another thread I gave a numerical example of how these effects would come together to ensure that two observers both measure the same light beam to be traveling at the same speed, I'll repost it here:

Last edited: Feb 26, 2009
12. Feb 27, 2009

### mintparasol

WOW!! these guys are on the verge of calculating for absolute time. it kind of belittles things like war, conflict and the banking system when you see what we are capable of as a species.

Last edited by a moderator: May 4, 2017
13. Mar 3, 2009

### mintparasol

Quote from JesseM:-
"Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor (which determines the amount of length contraction and time dilation) is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.

Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.

If you want to also consider what happens if, after reaching the front end of the moving ruler at 100 seconds in my frame, the light then bounces back towards the back in the opposite direction towards the back end, then at 125 seconds in my frame the light will be at a position of 75 light-seconds on my ruler, and the back end of the moving ruler will be at that position as well. Since 125 seconds have passed in my frame, 125/1.25 = 100 seconds will have passed on the clock at the back of the moving ruler. Now remember that on the clock at the front read 50 seconds when the light reached it, and the ruler is 50 light-seconds long in its own rest frame, so an observer on the moving ruler will have measured the light to take an additional 50 seconds to travel the 50 light-seconds from front end to back end."

This is quite complicated for the layperson but I'm getting a handle on it.
Why is distance between the frames not a factor in this model?

14. Mar 3, 2009

### ZapperZ

Staff Emeritus
Maybe you want to start with this book:

http://www.oberlin.edu/physics/dstyer/Einstein/SRBook.pdf [Broken]

Zz.

Last edited by a moderator: May 4, 2017
15. Mar 3, 2009

### mintparasol

The numerical example I've quoted above makes sense but can you point me towards some actual experimental observations?

16. Mar 3, 2009

### Staff: Mentor

17. Mar 4, 2009

### mintparasol

Thanks,
ad

Last edited by a moderator: May 4, 2017
18. Mar 4, 2009

### mintparasol

19. Mar 18, 2009

### mintparasol

OK, special relativity aside, is it right or wrong to assert that the gravitational time dilation effect is a result of gravity slowing down the matter within an object?

20. Mar 19, 2009

### mintparasol

Anyone?

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