# The Naked Scientists Forum

### Author Topic: How long does 'C' take?  (Read 13442 times)

#### Farsight

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##### How long does 'C' take?
« Reply #25 on: 22/03/2010 12:55:55 »
I endorse that. If two twins A and B are moving apart in empty space with no external references, they can't say A is moving away from B, or vice versa. Their situation is symmetrical. It's only when twin B turns round and comes back that they can agree that it was twin B doing the moving. That turnaround requires a change in velocity - an acceleration. As you suggest, removing initial acceleration and final deceleration simplifies matters - start with the twin B passing twin A, as they pass they "tag" or touch each other to establish a synchronised event, and after B turns round they pass each other again and do the same to establish another synchronised event.

#### PhysBang

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##### How long does 'C' take?
« Reply #26 on: 22/03/2010 13:14:26 »
Relativity is correct, but people don't quite understand why. It's just because the local motion of light defines your time.
It is important to note at this point that this position is essentially the religious position of this poster. Farsight has no scientific evidence for this theory other than the literary criticism he performs on select quotations and he has no plans to actually demonstrate that his particular theories match what scientists investigate. He admits as much:
Quote from: Farsight
But actually, I don't want to cast this as a coherent mathematical model myself. That might sound odd, but think about it. If I locked myself away and came up with something that really flew, every theoretical physicist in the world would then be redundant. It's too late for them to get involved once it's finished. Moreover they'd look like crystal-sphere fools, and the public would feel betrayed. There would be a backlash, and the upshot would be a disaster. I'm trying to help physics, not destroy it.
Note the contempt for contemporary physicists.

#### Farsight

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##### How long does 'C' take?
« Reply #27 on: 22/03/2010 14:01:42 »
There's nothing religious or contemptuous about what I'm saying here. Here's the official definition of the second:

Since 1967, the second has been defined to be the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

An atomic clock employs microwave radiation, which is essentially light. We count 9,192,631,770 microwave peaks going past and call it a second. Hence we're defining the second using the motion of light. We then use the second to measure the motion of light. Hence we always measure a constant value. Physicists are saying this too. See Comments on "Note on varying speed of light theories" by Magueijo and Moffat at http://arxiv.org/abs/0705.4507, which includes:

"Following Ellis [1], let us first consider c as the speed of the photon. Can c vary? Could such a variation be
measured? As correctly pointed out by Ellis, within the current protocol for measuring time and space the answer
is no. The unit of time is defined by an oscillating system or the frequency of an atomic transition, and the unit of
space is defined in terms of the distance travelled by light in the unit of time. We therefore have a situation akin to
saying that the speed of light is “one light-year per year”, i.e. its constancy has become a tautology or a definition."

#### BenV

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##### How long does 'C' take?
« Reply #28 on: 22/03/2010 14:03:08 »
Physbang, two of your three posts in this thread have been nothing but personal attacks on Farsight.

If his science is wrong, kindly explain so without resulting to personal attacks.  If you have nothing to contribute but attacks, please don't contribute at all.

I think you've been warned for this before on a different thread.

#### Geezer

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##### How long does 'C' take?
« Reply #29 on: 22/03/2010 20:06:05 »

An atomic clock employs microwave radiation, which is essentially light. We count 9,192,631,770 microwave peaks going past and call it a second. Hence we're defining the second using the motion of light. We then use the second to measure the motion of light. Hence we always measure a constant value.

(I put the third sentence in bold)

This is quite erroneous. We might as well say something like:

"A guitar string employs acoustic radiation, which is really sound. We count n acoustic events and call it a second. Hence, we are defining the second using the motion of air. We then use the second to measure the motion of light. Hence we always measure a constant value."

Obviously, that makes no sense, but it is little different from Farsight's statement.

It might be possible to stretch things a bit and say that microwaves are a form of light, but that's not commonly accepted. It is however possible to say that they are both forms of electromagnetic radiation. However, the real problem lies in the third sentence.

We are absolutely not defining the second using the motion of light. What we are doing is defining the second in terms of atomic events. We are merely using the microwave energy released to detect and count the atomic events.

#### yor_on

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##### How long does 'C' take?
« Reply #30 on: 22/03/2010 21:07:54 »
There are some things here I need you to define Mr Fontwell.

By saying "constant relative motion with respect to each other they might pass each other at one point" are you thinking of them uniformly accelerating, or uniform moving? Also, do you mean that they are traveling in opposite directions meeting each other? To be able to pass each other it seems to me that this is what you are talking about?

If we assume that they are uniformly moving past each other in opposite directions and then say that 'A' do that 'turn around' and catches up to 'B' he will have to accelerate, right? And as you say an acceleration equals a 'compressed' timeframe for 'A' relative 'B'. But are you then saying that the uniform motion he will use after it to 'drift' up to 'B' and overtake him have nothing to do with the time dilation created? That's one of the things I'm really wondering about. If uniform motion also creates a time dilation. To me it seems that it should. That is if we assume that he does this turnaround after one years travel and then caches up to 'B' will give him a lesser time dilation relative 'B' than if he did it first after have traveled three years away from 'B' before doing the turn around.

And there is also the complication with it that he in both my examples will have to pace himself at a faster rate than 'B' to catch up with him, which sort of destroy the equivalence there, no matter which scenario you choose, at least I think so? If now not all 'uniform motion', no matter their velocity relative something else is equivalent of course. But thats the scenario that confuses me. To get it into perspective I actually have asked a physicist about it, but he wasn't sure himself of how to see it, there seems to be some heavy math involved in it.

As I said, to me it comes back to how to look at 'uniform motion' and whether all 'free falling' then could be said to be equivalent, no matter velocity? If it would be so that it is only the acceleration that creates a 'time dilation' and your uniform motion, for however long after the fact, won't have a bit to do with it then all free falling frames have to be equivalent (uniform moving). But if uniform moving do have something to do with the 'time dilation' observed then i can't say that any free fall will be equivalent to another 'free fall' as they will have to be absolutely the same velocity, and if so one starts to wonder about if their invariant mass also will play a role for it?

And if they (all free falling frames) on the other hand are equivalent then acceleration really becomes weird, at least to me. But probably I missed what you meant right :)

And I am interested in all views here. PhysBang, Farsight, Geezer, BEnV and all you  others that have a view on it. But first of all naturally you Mr Fontwell :) As it was your scenario that I, ah, manhandled here ::))

Hey, don't blame me. I'm just confused :)

« Last Edit: 22/03/2010 22:43:42 by yor_on »

#### Geezer

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##### How long does 'C' take?
« Reply #31 on: 22/03/2010 22:04:05 »
Fine, but let's try make it clear when we are within the bounds of supportable scientific theory rather than flying in the face of it.

#### fontwell

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##### How long does 'C' take?
« Reply #32 on: 23/03/2010 08:57:54 »
There are some things here I need you to define Mr Fontwell.

By saying "constant relative motion with respect to each other they might pass each other at one point" are you thinking of them uniformly accelerating, or uniform moving?

Neither party is accelerating. The effect is that the parties move at a constant velocity with respect to each other (really their frame of reference). It is understood in SR that if two parties are both in uniform motion (i.e. they don't experience personal acceleration) then they will move at a constant velocity with respect to each other.

Quote
Also, do you mean that they are traveling in opposite directions meeting each other? To be able to pass each other it seems to me that this is what you are talking about?

For uniform motion any direction is possible. If it is convenient to have them in the same place at t=0, then before that time they would have been moving together and after that time they move apart. Like trains on two parallel tracks which pass each other.

Quote
If we assume that they are uniformly moving past each other in opposite directions and then say that 'A' do that 'turn around' and catches up to 'B' he will have to accelerate, right? And as you say an acceleration equals a 'compressed' timeframe for 'A' relative 'B'.

Yes

Quote
But are you then saying that the uniform motion he will use after it to 'drift' up to 'B' and overtake him have nothing to do with the time dilation created? That's one of the things I'm really wondering about. If uniform motion also creates a time dilation. To me it seems that it should. That is if we assume that he does this turnaround after one years travel and then caches up to 'B' will give him a lesser time dilation relative 'B' than if he did it first after have traveled three years away from 'B' before doing the turn around.

The dilation due to uniform motion is symmetrical. This is an absolutely fundamental point. Two parties in constant motion both think the other has a slow clock and a short ruler. This is because they disagree about times and distances between events. This time dilation is what they each infer about the other through observation at a distance. One point to note is that in uniform motion they can only ever meet once, thus they can never get to see 'who is really slow'

Time dilation due to acceleration causes a difference between the two parties that is permanent and not symmetrical. this is because one party absolutely does the accelerating and the other doesn't.

When 'A' turns round and accelerates back to 'B' the total time 'A' has been drifting does affect the dilation between them, this is true. But up to that point it is still symmetrical, no one could say 'A' is actually slower than 'B'. But by accelerating back to 'B', 'A' fixes the dilation between them in favour of 'B' being static.

Also, the acceleration determines the new velocity between 'A' and 'B' and thus when they meet up again. So actually, by determining the return journey it already defines the next symmetrical time dilation of the next drift journey. They will each have a measurement of how long this journey takes. But at the end they meet at the same event. This means that the difference in their observations of the journey time has to be 'absorbed' by 'A' during the the turn around at the start of this portion.

Actually, assuming the return velocity is the same as the outgoing velocity, the entire journey is symmetrical in time. The start and end both has 'A' and 'B' at the same events. The drift away/toward from that event is seen the same way by 'A' and 'B'. Thus all the unbalanced dilation occurs during the turn around.

So the time dilation which causes the difference in their clocks when they finally meet again is all caused (or at least unbalanced) by the turn around acceleration. The duration of the drift does alter the amount of time by which the final dilation occurs, but it is the acceleration which swings this dilation round to put one party older than the other.

Quote
And there is also the complication with it that he in both my examples will have to pace himself at a faster rate than 'B' to catch up with him, which sort of destroy the equivalence there, no matter which scenario you choose, at least I think so? If now not all 'uniform motion', no matter their velocity relative something else is equivalent of course. But thats the scenario that confuses me. To get it into perspective I actually have asked a physicist about it, but he wasn't sure himself of how to see it, there seems to be some heavy math involved in it.

Are you talking about a light signal between them? I'm not sure what you are asking here.

Quote
As I said, to me it comes back to how to look at 'uniform motion' and whether all 'free falling' then could be said to be equivalent, no matter velocity? If it would be so that it is only the acceleration that creates a 'time dilation' and your uniform motion, for however long after the fact, won't have a bit to do with it then all free falling frames have to be equivalent (uniform moving). But if uniform moving do have something to do with the 'time dilation' observed then i can't say that any free fall will be equivalent to another 'free fall' as they will have to be absolutely the same velocity, and if so one starts to wonder about if their invariant mass also will play a role for it?

For parties in uniform motion the time dilation between them (symmetrical) is bigger as their relative velocity increases. Two parties will each observe the other's clock to be running slow and the faster they move with respect to each other, the slower that clock will seem to run. There can be many inertial observers and when they look at each other the time dilation they observe is a function of relative velocities.

Also, to me, 'free fall' is a phrase which implies falling toward a mass. That would be GR and something else. We have only been talking about SR here. The equivalent situation here is uniform motion = constant velocity = an inertial frame.
« Last Edit: 23/03/2010 10:10:25 by fontwell »

#### Farsight

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##### How long does 'C' take?
« Reply #33 on: 23/03/2010 12:29:06 »
An atomic clock employs microwave radiation, which is essentially light. We count 9,192,631,770 microwave peaks going past and call it a second. Hence we're defining the second using the motion of light. We then use the second to measure the motion of light. Hence we always measure a constant value.

(I put the third sentence in bold) This is quite erroneous.
It isn't, Geezer. This is what we do. We define our time using motion. In the old days we used the motion of the earth, later it was the motion of a pendulum, later still it was the motion within a mechanical clock. Nowadays our best clocks are atomic clocks, and they use the motion of microwave radiation.

We might as well say something like:

"A guitar string employs acoustic radiation, which is really sound. We count n acoustic events and call it a second. Hence, we are defining the second using the motion of air. We then use the second to measure the motion of light. Hence we always measure a constant value."

Obviously, that makes no sense, but it is little different from Farsight's statement.
That doesn't work because light and air are not the same thing.

It might be possible to stretch things a bit and say that microwaves are a form of light, but that's not commonly accepted. It is however possible to say that they are both forms of electromagnetic radiation.
Accepted. I was using the word "light" in the widest sense. Electromagnetic radiation is a better term.

However, the real problem lies in the third sentence. We are absolutely not defining the second using the motion of light. What we are doing is defining the second in terms of atomic events. We are merely using the microwave energy released to detect and count the atomic events.
We absolutely are. The atomic event concerned is the hyperfine transition. We aren't counting hyperfine transitions. That's like counting the number of plucks of the guitar string. We are counting the resultant EM wavepeaks going past. See http://en.wikipedia.org/wiki/Hyperfine_structure#Use_in_defining_the_SI_second_and_meter for more information.

#### Farsight

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##### How long does 'C' take?
« Reply #34 on: 23/03/2010 12:31:52 »
Yor-on, everything fontwell has said is IMHO bang on, but you seem to be thinking it's more complicated than it is. See the wikipedia article on time dilation, and look at this section:

http://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity

Don't worry about the arithmetic, just think about the parallel-mirror light clock. Now imagine you're twin A and you've got one of these light clocks in front of you. The light is moving up and down, so you can draw a light path like this , multiplied a zillion times whilst twin B is on his trip out and back. Now think of B's light clock, and draw the light path as you'd see it, something like this:

→ /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ →
↓
← /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ ←

The total light path length between your parallel mirrors is the same as his. He comes back time-dilated, and because of his motion through space, his light has bounced back and forth by less than a zillion times. When you get back together you can compare counters to prove it. But if that turnaround didn't happen, you can't, and twin B would assert that your light path was the one that was zigzag, not his.

#### yor_on

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##### How long does 'C' take?
« Reply #35 on: 23/03/2010 15:25:58 »
Okay Fontwell. If I got you right :) time dilation takes place both at uniform motion and acceleration, right? so assuming this, knowing that there are no 'preferred frames' defining a 'gold standard' in the universe, we can't really define the time dilation for any uniformly moving frame, can we?

It's all about comparison and definition via one 'inertial frame' relative the frame we observe (think Earth and a rocket). And our definition of that 'inertial frame' is arbitrarily made as we have no 'gold standard' for that either, right?

So, how would you define the circumstances describing two 'uniformly moving' frames as equivalent to each other?

#### yor_on

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##### How long does 'C' take?
« Reply #36 on: 23/03/2010 15:37:34 »
Farsight please, look at the thread before you answer :)

I do understand what you are saying, as you would see if you read it through. That all frames internally have the same time (more or less:) relative those/that inside that frame is 'self consistent' to me. That's not my discussion. I'm looking at how to define uniform motion relative acceleration, and time dilation.

It may be all self clear to you, but it's not to me.

And. .

#### Farsight

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##### How long does 'C' take?
« Reply #37 on: 23/03/2010 16:34:11 »
I've looked at the thread. I've read it all. Motion is relative, uniform motion is motion that is measurably occuring at a constant rate, relative acceleration is a change in relative motion, and time dilation is what you call it when your cumulative measure of local motion is occurring at a reduced rate compared to somebody else's. It always comes back to motion. Once you appreciate this, it's clear.

#### yor_on

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##### How long does 'C' take?
« Reply #38 on: 23/03/2010 16:42:04 »
Yes Farsight, that's your take on it and it might be mine too, but not yet. First I will ask my questions and then we will see if I understands it :)

#### fontwell

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##### How long does 'C' take?
« Reply #39 on: 23/03/2010 16:45:13 »
Okay Fontwell. If I got you right :) time dilation takes place both at uniform motion and acceleration, right?

yes...

Quote
so assuming this, knowing that there are no 'preferred frames' defining a 'gold standard' in the universe, we can't really define the time dilation for any uniformly moving frame, can we?

It depends on what you mean by gold standard, I'll come back to this...

I don't mean to sound rude but you don't seem to have understood what I mean't about symmetrical and unbalanced (non-symmetrical) time dilation.

For two observers in uniform motion, both think the other has a slow clock. There is no preferred position. The time dilation is purely what they infer about the other party by looking at their clock as it moves. They look toward the moving clock (also allowing for the time it takes light from the clock to arrive) and what they observe is that the moving clock runs too slow. This is one form of time dilation. It is symmetrical because for each party they both deduce the other to have a slow clock. This can work because they can only ever meet once, so they have no way to meet again to see who is correct (as long as they remain in uniform motion).

If you like, there is a gold standard of time dilation. For any particular velocity between to inertial parties, they both observe the same gold standard time dilation (slowness) in the other.

But this isn't an absolute gold standard because a third party in uniform motion will observe both of them and have a different time dilation with each one. And the simple sum of each dilation will not sum to the total between the first two parties.

Quote
It's all about comparison and definition via one 'inertial frame' relative the frame we observe (think Earth and a rocket). And our definition of that 'inertial frame' is arbitrarily made as we have no 'gold standard' for that either, right?

If I understand you correctly, not right. As I've said before, there is an easy test to see if you are in an inertial frame. You drop a stone. If it moves away from you then you are accelerating. On Earth we are not in an inertial frame because stones move away from us. Actually, on Earth we are feeling gravity and SR doesn't like to mention gravity.

In deep space our two parties can move away from each other in two inertial frames. They each drop a stone and it doesn't move away from either party. The stones move along side each party.

But for the parties to meet again one must turn round. It could be either but it has to be one. Whoever turns round will see his stone move away from him. It will track the path he would have taken.

This is the un-symmetrical time dilation which makes one clock absolutely ahead of the other when they meet again.

The gold standard is that any two parties with the same relative paths and acceleration would have the same time dilation experience.

But I could watch these two parties as I move at 0.99C past one of them and the whole situation would look different to me. So it is not a gold standard.

Quote
So, how would you define the circumstances describing two 'uniformly moving' frames as equivalent to each other?

Hopefully this is clear now. Each one experiences no acceleration (using the stone test). Each sees the other as having a slow clock. They can only meet once, so there is no way to get the two clocks together again to see 'who is right'. They can only meet twice and compare clocks twice if one of them ceases uniform motion.

#### yor_on

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##### How long does 'C' take?
« Reply #40 on: 23/03/2010 18:18:00 »
'It's all about comparison and definition via one 'inertial frame' relative the frame we observe (think Earth and a rocket). And our definition of that 'inertial frame' is arbitrarily made as we have no 'gold standard' for that either, right?'

"If I understand you correctly, not right. As I've said before, there is an easy test to see if you are in an inertial frame. You drop a stone. If it moves away from you then you are accelerating. On Earth we are not in an inertial frame because stones move away from us. Actually, on Earth we are feeling gravity and SR doesn't like to mention gravity."
==

What I was thinking is that when in uniform motion you have no definition of your velocity except when relating it to another frame of reference. So looking at it that way it seems to me that there is no absolute definition of the time dilation you might have in that frame other that relative another frame. Still there will be one, if I understands it right?

#### yor_on

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##### How long does 'C' take?
« Reply #41 on: 23/03/2010 18:27:27 »
You can also see it as you have a different time dilation, if so, against all other frames you measure your frame against? Which is a very weird idea. That as I don't see how you can define a uniform motion, except relative another frame of reference? And I know that this falls under SR not GR.

#### fontwell

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##### How long does 'C' take?
« Reply #42 on: 23/03/2010 18:35:24 »
What I was thinking is that when in uniform motion you have no definition of your velocity except when relating it to another frame of reference.

Yes!

Quote
So looking at it that way it seems to me that there is no absolute definition of the time dilation you might have in that frame other that relative another frame. Still there will be one, if I understands it right?

Yes! Relativity! It means observations are relative, not absolute. If you look at one moving object you see one time dilation, if you look at another moving object you see another time dilation, if they look at each other they see something else again.

There is equality between any two observers because they each observe each other to have the same dilation (slowness). But there is no gold standard dilation because other observers see both of them as time dilated to him.

Added
Quote
You can also see it as you have a different time dilation, if so, against all other frames you measure your frame against? Which is a very weird idea. That as I don't see how you can define a uniform motion, except relative another frame of reference? And I know that this falls under SR not GR.

You test for acceleration by dropping a stone. If it stays next to you you are in an inertial frame.

This is described as 'constant motion' but only with respect to another inertial frame. You always consider you are stationary. Why would you think you are moving? You drop a stone, it stays near to you - you are stationary.

The only thing is, every inertial frame considers that it is stationary too. This is fine because Relativity only concerns itself with relative motion. Relative motion between inertial frames, each one of which appears to be stationary (or not accelerating anyway).
« Last Edit: 23/03/2010 18:45:49 by fontwell »

#### yor_on

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##### How long does 'C' take?
« Reply #43 on: 23/03/2010 18:59:33 »
If I now was to compare my uniformly moving frame a against an accelerating one (Earth <-> accelerating rocket) then I understand that there is a generalized idea of how to differ between those two frames, by using the idea that motion isn't defined in relation to individual objects, such as our Earth <-> accelerating rocket system, but instead in relation to the distant stars. And yes, I know this one falls under GR, as I understands it.

But if you don't accept this definition then you could expect Earth to be the one aging relative the the accelerating rocket, again, as I understands it? And looking at it generally :) this one too brings up the question of how to define that time dilation. It seems easier to use your test to me and then say that acceleration, as shown by that stone moving towards its 'gravity well' is what defines the time dilation.

Which brings me back to my first wondering. How can we define the time dilation for an uniformly moving object like our Earth? And now I've given two examples that I believe to be correct? one under SR and one under GR (acceleration).

It do confuse me :) It would have been simpler with some 'gold standard' that we could define uniform motion against but not even the CBR can do that for us as I understands it?

#### yor_on

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##### How long does 'C' take?
« Reply #44 on: 23/03/2010 19:08:54 »
"This is described as 'constant motion' but only with respect to another inertial frame. You always consider you are stationary. Why would you think you are moving? You drop a stone, it stays near to you - you are stationary."

Well, it's there I don't agree. I can leave earth accelerating, when I then stop this acceleration I will be in a free fall. The stone will stay relative me but I know that I've accelerated first. If I now did this with two rockets that now both are in a free fall (uniform motion - coasting:) but firstly been accelerated at different velocities they should then be seen as equivalent? But their time dilation relative their origin (Earth) will differ although they are of the exact same mass and definition otherwise..

And that time dilation would then grow under the years of uniform motion - coasting too :) but unequally for our rockets versus Earth. And it's there I find it easier to see the acceleration in itself to become the 'time dilation' made by those rockets, but that wouldn't explain how uniform motion also have a hand in it. I hope you can see how I think :) As for being rude, explaining how one think is never rude, as long as one's not condescending of course. And I don't think you are Mr Fontwell :)

So to me there is a difference and that definition of the stone staying by your side doesn't cover it.
==

« Last Edit: 23/03/2010 19:21:29 by yor_on »

#### Geezer

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##### How long does 'C' take?
« Reply #45 on: 23/03/2010 19:12:11 »

However, the real problem lies in the third sentence. We are absolutely not defining the second using the motion of light. What we are doing is defining the second in terms of atomic events. We are merely using the microwave energy released to detect and count the atomic events.

We absolutely are. The atomic event concerned is the hyperfine transition. We aren't counting hyperfine transitions. That's like counting the number of plucks of the guitar string. We are counting the resultant EM wavepeaks going past. See http://en.wikipedia.org/wiki/Hyperfine_structure#Use_in_defining_the_SI_second_and_meter for more information.

Farsight, you fail to grasp the point, even although you actually stated it correctly.

The events are atomic as you say above (hint - do you think perhaps that's why it's called an "atomic" clock?). Only the method of detection of the atomic events is electromagnetic.

Your notion that atomic clocks measure time based on some characteristic of light is fundamentally wrong and liable to lead to lots of misconceptions and circular logic - QED. Atomic clocks measure time based on characteristics of atoms, not light.

Using your logic I could say that the tree outside my window is not really a tree, it's simply light because I use light to detect its presence.
« Last Edit: 23/03/2010 22:33:30 by Geezer »

#### yor_on

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##### How long does 'C' take?
« Reply #46 on: 23/03/2010 20:11:05 »
It all falls back on how you define your universe naturally. As a physicist/mathematician you might define the universe as being of two properties, one expressed through General relativity (acceleration) and the other being what happens under Special relativity (uniform motion). A little like we say we have a wave/particle duality.

But to me looking at it those definitions goes into each other, and I have severe difficulties defining where one starts and the other one ends, that as the universe in reality are a seamless experience to me, as to the rest of you too if you think about it.

And to my eyes it either have to make sense as one definition or we will have to start look at our definitions again. By that I do not mean that the theory of relativity is wrong in any way, rather that there should be an explanation that even I could understand. And it's trying to make sense of that definition I get stuck on this kind of things :) And I agree Mr Fontwell, I do blend GR and SR at times but it all goes back to how I see SpaceTime. As a 'whole', and that's why it at times might take some time to make my questions understandable, I guess ::))
==

To use another mans word, that physicist I talked with. And this is an GR question.
==

Would it make any difference for the time dilation, relative my point of origin (earth), if I took a second to reach 99% of the speed of light relative taking one year to reach that velocity?

He thought it wouldn't, just guessing there of course as we were just talking.
What do you think?
« Last Edit: 23/03/2010 20:28:36 by yor_on »

#### fontwell

• Jr. Member
• Posts: 39
##### How long does 'C' take?
« Reply #47 on: 23/03/2010 23:00:26 »
Geezer, your ideas about defining length belong in another ATM style thread, not this one.

#### fontwell

• Jr. Member
• Posts: 39
##### How long does 'C' take?
« Reply #48 on: 23/03/2010 23:32:04 »
yor_on, I think maybe you have such a fixed way of understanding what constitutes position in space that you will never be able to drop it. Unless you can do that you will always be looking at SR from a view point where it makes no sense. And it doesn't make much sense to begin with.

One last try.

For example, you do the stone test and see you are accelerating. Are you slowing down? Are you speeding up? when you stop accelerating are you now going really fast? are you stationary? Did you change direction completely?

There is no independent reference in SR, only many other inertial frames. Depending on which frames you look at, all of these answers can be true. Your relation to these frames is determined by the SR equations and is different for every frame. If this wasn't the case it wouldn't be called relativity.

When I say you will consider yourself stationary it might be a small liberty but it is basically true. Most of the time on Earth you consider yourself to be stationary (assuming you sitting at a desk or whatever and not in a car). Although you know you are travelling round the Sun and round the galaxy core and round the local galaxy cluster centre etc, you mostly think you are sitting in one spot.

Similarly, when you are not accelerating your situation is identical to that of being stationary. Since there is no independent frame of positions, you may as well call it stationary. If you decided to call it constant velocity the question arises, how fast? and in which direction? You cannot ever answer these questions. Except you can if you give them with respect to another inertial frame. But for every inertial frame you might use there is a different answer.

BTW, as you know, these 'fixed' stars are mainly rushing away from you in all directions (well, those in other galaxies).

Added
You mention about SR and GR and how the universe is really one thing so it gets difficult. But as I see it, even though they are related SR and GR explain different effects. When you sail on the sea you are blown by the wind and moved by the tide. These two things both affect you but you probably find it easier to calculate each one separately. SR covers issues purely down to relative velocities, GR covers issues related to mass/gravity.
« Last Edit: 24/03/2010 00:18:20 by fontwell »

#### fontwell

• Jr. Member
• Posts: 39
##### How long does 'C' take?
« Reply #49 on: 23/03/2010 23:54:25 »
Quote
Would it make any difference for the time dilation, relative my point of origin (earth), if I took a second to reach 99% of the speed of light relative taking one year to reach that velocity?

Yes. The time dilation depends on how far you have travelled and at what relative velocity. Assume you define the journey as being to a fixed location relative to Earth. If you travel along this path slowly there will be low dilation, if you travel fast there will be a lot. The total dilation will be a function of how much time was spent at every velocity during the journey. Since the quick acceleration means most of the journey is at the highest speed, this will cause the most dilation.

(comment removed - JB)
« Last Edit: 24/03/2010 00:54:34 by JimBob »

#### The Naked Scientists Forum

##### How long does 'C' take?
« Reply #49 on: 23/03/2010 23:54:25 »

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