# Naked Science Forum

## Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: yor_on on 11/03/2010 18:56:16

Title: How long does 'C' take?
Post by: yor_on on 11/03/2010 18:56:16
simple Q.

Assume that I'm traveling extremely close to the light of speed. Watching a light signal from home (Frame at rest relative me) I see it coming at me, from behind, propagating at 'C' relative me, same as if I was standing still, right?

So, how does it do it?
And does it mean that the time it takes to finally reach me would be the exact same time it would take to reach that point in space where it finally reached me anyway?

From Earths 'frame of reference' it should be so, don't you agree?
But from mine? Near the light of speed?

You can assume one scenario where I steadily was accelerating the whole time, until the light reached me. In the other I had a short inertia-less acceleration of one second, as measured by Earth, to then go back to a 'free fall' (uniform motion) as observed from Earth.

I know that those two scenarios will get different length traveled, but the question isn't about the length but about the time when comparing when it reached me. Well, if I'm thinking right it do have to do with relative 'distance', but thats a later Q.

Hope the question is understandable.
Title: How long does 'C' take?
Post by: yor_on on 11/03/2010 20:58:31
The thing confounding me here is the realization that I will see that light coming at me at the light of speed, no matter what I do, standing still relative Earth or accelerating. You could look at it this way, as I accelerate away near light, still watching that light coming at 'C' from behind, all distances outside my frame of reference will 'compress' , except that lights, I will see it red shift but? I'm not even sure how to formulate it to make sense here :)
Title: How long does 'C' take?
Post by: Soul Surfer on 11/03/2010 23:35:12
You are having problems with your basic thinking here.  You talk about "seeing light coming"  this cannot be true. You only see light the moment it arrives and its frequency is red shifted according to the relative velocity of you and the light source.
Title: How long does 'C' take?
Post by: Farsight on 11/03/2010 23:41:04
Best to forget about acceleration for now, and about redshift/blueshift and the distance to the earth. Just think of yourself in a black box. How do you define a second? Probably using your atomic clock, like the official definition:

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.

That atomic clock employs microwave radiation, which is essentially light. You count 9,192,631,770 microwave peaks going past and call it a second. You're defining your second using the motion of light. And you'll define the metre likewise:

In 1983, the metre was redefined as the distance travelled by light in free space in 1⁄299,792,458 of a second

So your second and your metre is defined using the motion of light, so when you measure the rate of that motion, you're always going to get the same answer. Open a little window to let in some light that's come from the Earth, and you get the same answer again. You might try using a mechanical clock to define your second, but that doesn't help because mechanical clocks are made out of electromagnetic things like electrons, and they're affected like the light is affected. It's the same for rulers, and the same for you.

See http://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity for some arithmetic, though it isn't expressed as simply as it could be. Pythogoras' theorem applies because the hypotenuse is the light path with length of c=1 in natural units, and the base represents your speed as a fraction of c. Work out the height for a length contraction factor and take a reciprocal for a time dilation factor.
Title: How long does 'C' take?
Post by: yor_on on 12/03/2010 13:57:28
SoulSurfer, thanks for your considered response.
Deep and thought through, as always :)

And as alway I will point out that we're at TNS, I'm not trying to present a new theory, I'm asking a question, and you're nitpicking. If that's all you want to contribute to this thread? Don't bother please.
Title: How long does 'C' take?
Post by: yor_on on 12/03/2010 16:38:59
Farsight, thanks for your response, although it isn't the 'mainstream idea' that confounds me here, just some implications that I'm not even sure that I'm getting right, it may just as easily be that I need to think it through again.

As I said, I might need to rethink how the question should look. But if you feel you have two specific answers to give me, one for each sceniario I will read them with interest. F.ex, I'm more or less stating that the answers you'll get will be different, depending on time of acceleration, don't I :)

Is that right?
Why?

We spend the exact same amount of energy (Let's assume so for the Q.)
So why would the result differ, and how will it differ?

I will get two 'distances' but the same 'time displacement' relative the rest of the universe?

Or, I will get the same 'distance' and 'Time displacement'?

Or, I will the same 'distance but not 'time displacement'?

As I said, it's a remarkable subject.

Also. When I look at that signal it will according to relativity come at me at 'C' relative my own frame, no matter how you observe it, and me, from an 'inertial frame' like Earth f.ex if we assume that this were the origin for us both.

As I'm moving away relative Earth very near the speed of light (99.99999999~)what does that state about this light? If we now, as SoulSurfer pointed out, could have observed it. In reality the observation won't matter for my statement though, assuming that the concept of relativity is correct it have to be true.

Brain-gymnastics :)
And it do make me wonder about the concept of 'distance'?

So if you have ideas about the scenarios feel free to give them Farsight, and if you want to state them from another format than relativity, you're welcome, just make sure that I know you do :) So we're speaking the same language here, sort of.
Title: How long does 'C' take?
Post by: PhysBang on 13/03/2010 01:30:21
Farsight, thanks for your response, although it isn't the 'mainstream idea' that confounds me here, just some implications that I'm not even sure that I'm getting right, it may just as easily be that I need to think it through again.
Don't worry, nothing that Farsight wrote (aside from the link to something he is not correctly describing) has anything to do with the mainstream position.

What you probably need to remember in order to get everything to work out is that distances are not the same in every system of coordinates. If we use a system of coordinates in which (given your example) you are not moving, then the distance between you and home is a lot less than the same distance as determined in a system of coordinates in which home is not moving. The faster the difference in motion, the more dramatic the difference in distance.
Title: How long does 'C' take?
Post by: yor_on on 13/03/2010 14:33:20
Yes PhysBang, that is my thought too, and what made me wonder anew was the thought of that radiation coming to 'pick me up' :) from Earth, constantly moving at 'C'. We agree on one thing, don't we? That 'C' is a 'gold standard'. Well, at least I've understood it that way.

But remembering that light is defined as being at 'C' we suddenly gets three definitions it seems to me. One from the spectator at rest versus the accelerating/uniformly moving ship, the other from me in the ship observing that light reach me, the third via our statement that 'C' is a 'gold standard' so that all light will have that velocity in a vacuum.

And that only make sense to me when I start to question the concept of distance. If I stop looking at the observers and only look at the concept of 'C' in itself then I have a defined velocity related to some sort of idea of a defined 'distance'. Without that 'distance' C' won't make any sense, it needs it. Don't you agree?

So we assume a hidden 'gold standard' for 'distances' too, don't we? But, if that is true and, nota bene, assuming that light is a corpuscle moving at its own through a vacuum, how can we recombine those three concepts? One from the person at rest, one from the person moving, and one from the very concept we are assuming us to observe.

And that's where my questions come in, and why the answers interest me :) And they are tricky ones I think? Hard to answer, but they should be answerable, shouldn't they? And depending on how they differ I would get another glimpse into how to see it, well, I hope so :)
Title: How long does 'C' take?
Post by: PhysBang on 13/03/2010 17:27:06
how can we recombine those three concepts?
In Special Relativity, the Lorentz transformations. In General Relativity, it is the restriction of descriptions of physical laws to generally covariant form (and that's where it gets complicated). In contemporary gravitational research, the move is to what is called gauge invariance, which is a more generalized mathematical way of ensuring that our physical laws are properly described in any system of coordinates. And understanding that requires a commitment to some serious mathematics.
Title: How long does 'C' take?
Post by: yor_on on 13/03/2010 22:32:25
Okay fair enough, another simple question. Will your meter stick as observed from a 'far observer', being at rest relative that neutron star, shrink on the surface of a neutron star? We know it will shrink with motion as observed between two objects, passing each other near light speed. So do we observe the same effect when leaving our meter stick at that neutron stars surface?

Title: How long does 'C' take?
Post by: fontwell on 14/03/2010 22:47:16
Going back to the OP, both you and the guys back home observe the signal to be travelling at C with respect to yourselves, even though you are travelling near to C relative to each other, how come?

Apart from the initial event of you setting off from home, we need to identify two main events, the first in which the guys at home emit the light signal and the second in which the signal reaches you.

Since You are travelling in the same direction as the signal the situation works out to be this: You think the time and distance between the two events (sending and receiving the signal) are both less than the guys back home. Because the time and distance according to you are both scaled down by the same amount it still appears that the signal is travelling at C.

These things are almost impossible to understand but perhaps I can point out a few things to help it make more sense.

The first is that nobody has a privileged position. Home sending a signal to you is no different from you sending a signal back home. Everyone thinks all the signals travel at C. However, the particular situation you describe, home sending to you, is not symmetrical. Only home is present at the first event, so you have to deduce when/where it took place based on your observations. Only you are present at the second event, so home has to deduce when and where it took place based on their observations.

Two parties moving relative to each other who observe a series of events will not agree on the times or distances between events. They will disagree about which events are simultaneous and also the spatial separation. Crucially, distances that you used to think were particular lengths when you were previously sat at home in the other reference frame, like the distance to the next planet, will turn out to be much shorter than you were expecting. So that, for instance, if the signal reaches you just as you pass Mars, you will measure that you have travelled only 100,000,000km (say) whereas before you set off it appeared to be further away.

The result of these disagreements is that home think they sent the signal when you were still fairly close to them and that it took a long time and distance to reach you. But you will think they sent the signal quite late on and you didn't go too much further before it caught you up. In each case the ratio of distance between events to time between events results in a velocity between events of C.

This isn't really an explanation but I hope it helps anyway.
Title: How long does 'C' take?
Post by: yor_on on 15/03/2010 00:50:37
Yes I agree, it's frame dependent. But the question i started with was.

" 1. You can assume one scenario where I steadily was accelerating the whole time, until the light reached me.

In the other 2, I had a short inertia-less acceleration of one second, as measured by Earth, to then go back to a 'free fall' (uniform motion) as observed from Earth."

For the question we can also assume that I spent the same amount of energy.
Then my question become twofold.

For 1.
Will I get the same time & distance relative 2?
Or the same distance but a different time?
Or the same time but a different distance?
Or will both be different relative 2?

To me it's confusing :)

And when knowing that the light traveling towards me always, as seen from the frame I'm in, will do so at the same constant speed in a vacuum? What it really boils down to is what 'distance' should be seen as, and also how acceleration as compared to uniform moving transforms it. As I said I make the assumption that I will spend an equivalent amount of energy in both cases.
==

You can assume them to be exact twins, and also meet up after some equivalent spent acceleration/distance versus their point of origin (Earth) to compare later :)

It's a question I'm not sure on, but really would like to know.

Title: How long does 'C' take?
Post by: fontwell on 15/03/2010 07:44:29
The first 3/4 of your OP is the situation I covered. It assumes you go straight to your near C velocity. This addresses the question of how do you and home both observe the signal to be travelling at C relative to yourselves, despite travelling at near to C relative to each other. The summarised answer is that you see all the events as compressed into shorter times and distances compared to home.

I don't understand what the second part is asking or what you mean by inertialess acceleration or what energy has to do with it. But anyway...

If you keep accelerating, then the signal will take longer to reach you and the two events will be further apart in space and time. Also as you go faster your version of the events will seem to be even more compressed compared to what home thinks. So even though the events appear to be actually further apart to you when you continue to accelerate, home will see them as even more further apart.

I think a good place for you to start would be getting to grips with the Minkowski/Loedel diagrams for the uniform motion case.
Title: How long does 'C' take?
Post by: yor_on on 15/03/2010 13:51:08
Yes and no? I'm not sure what to make of it.
Are you telling me that I will get both different times, and distances between the two examples?

Why?

The only thing differing them is the time of acceleration, as I tried to set it up? And that's what I'm wondering about here? You can look at uniform motion several ways it seems. In one way it seems to me that all uniform motion are the same, as if you're in a black box you won't be able to define any motion at all, as I understands it, would you agree? you're in a 'free fall' no matter your uniform motion and inside that box your time will seem the same as always.

But then again, you will get different time dilations for my two cases when compared to the same frame of reference, depending on your acceleration / motion, again as I see it. And then we have non uniform acceleration expending energy, as well as uniform acceleration expending energy, and contrasting to that the acceleration you will get, as defined by an observer being at rest versus a gravity well, when 'free falling' into that same gravity well.

So I'm trying to see what differs them, telling me that it's all depending on the frames is how I see it too, but I'm trying to see if there is a difference and why it would be:) And there I don't expect that much understanding, one can always hope though. If you have an explanation to what and why makes those two scenarios differ I will be very interested. Or if you can show me that they are equivalent in form of 'time' and distance, or any of the combinations.

And that's where the 'energy' comes in. I've just tried to equalize my question so that the scenarios are as similar as possible. To be sure I'm not sure if that would work as I intuitively feel that the energy will differ :) But if one assume that they only have a limited energy one still would get to a point where that energy would be finished. Not necessarily the same point as when that light would reach my two cases. And that will then be another question of mine, but maybe not now. Assuming the same energy expenditure. Why should it differ, or not?

Hope I explained why a little clearer now.
Title: How long does 'C' take?
Post by: fontwell on 15/03/2010 21:16:21
This is has turned into a loser length post but I did it mainly to convince myself that I could explain what happens...

I think your questioning gets too complex too quickly and I don't really understand a lot of it. You really need to understand the uniform motion first, then maybe the rest is easier.

We both agree that no party is absolutely moving or stationary. All that matters is the relative motion. But even in this situation they always disagree about both when and where events happen. This is because the events will not be in positions which they view symmetrically. The only exception is events that are midway between them.

Actually, I might have got one part of my first post wrong, it was after midnight :) so here is an attempt to explain what happens.

The standard thing is that each party thinks the other has a slow clock and a short ruler. It helps if you just accept this as being how relativity works, each thinks the other is wrong in the same way.

TIMES

Home will measure the time the signal takes to travel to you as being between when they send the signal (which they can time directly) and when they think it gets to you. You will measure the time it takes the signal to travel as being between when you think they sent the signal to when it arrives at you (which you can time directly).

So what do we make of these times? Each thinks the other has a slow clock. So you think home sent the signal later than they do, because you think their clock next to that event was running slow. Home think you got the signal later than you do, because they see your clock when the signal arrives at you as being slow.

So if we get the ships logs and make a combined time line for what both parties think happened and when, it would look like this.

1) Home's clock reading when they send the signal
2) Time when you think they really sent the signal

signal travels

4) Time when home think you really got the signal

Note how both your times are squeezed between homes times. You definitely think it look less time than home do.

DISTANCES

Because the signal goes from Home to You an easy way to imagine this is that home have made a long corridor for you to move down in their frame of reference. They have cunningly built it so that the signal gets to you just as the corridor ends.

Home will see the distance the signal takes to travel to you as being the length of the corridor i.e. between them sending the signal (zero distance), and how far away they think you are when it gets to you. You will see the distance it takes the signal to travel as being between how far away you think home is when they send it, and you receiving it (zero distance from you). These are very different things.

Time to concentrate: Home will see the entire length of the corridor as being the distance between the two events. However, from your point of view, home sent a signal when you were at some particular distance (say, x) down the corridor. When it reaches you, the signal has only moved toward you by this distance. Your travel down the corridor doesn't affect this in any way. That's relativity. From your point of view the first event was x meters away from you and the second was at 0 meters away, total distance between events = x. You do not care about this corridor which the guys at home appear to be pulling past you. You just measure two events, one at distance x, one at distance 0.

In addition, you think home have made the corridor with short rulers and so you think its entire length is shorter than they do anyway.

So your version of the distance the signal travels definitely is much less than the guys at home. There is a combination of you thinking the whole corridor is shorter than home do, plus you perceiving that the signal only travels a distance equal to some fraction of the corridor.

So, because you think the distance between the two events was less than home do, this balances with your reduced times, allowing both you and home to measure the signal velocity as C.

As for acceleration, you and home will both continue to disagree about all times and distances and the events will all get farther apart.

[diagram=573_0]

This diagram kind of shows what happens and is mainly for my benefit(!) FYI, the diagonal grey axis lines represent light at velocity C, travelling past the reference event, which is you leaving home. E1 is home sending the signal, E2 is you receiving the signal. Note that the path between E1 and E2 is parallel to the light axis because it formed by the signal also moving at velocity C.
Title: How long does 'C' take?
Post by: yor_on on 16/03/2010 04:47:46
A very nice description Fontwell, and E1 and E2 lies on what kind of (arrows/axis), time? But yes, I see what you mean with 'C' and it was a lovely way of explaining how we both (home and traveler) can get the same answer for 'C'. There is one thing more I'm wondering a little over here. You write "because you think the distance between the two events was less than home do," it sounds as you are arguing that the distance observed by me then would be some sort of illusion?

If 'C' is a real 'gold standard' and according to your explanation rests on the relative distance as observed by me to home, then either that distance I measure is a 'real' one from my frame of observation, or it seems to me that one can't use that explanation to prove 'C:s' 'invariability' in SpaceTime?

But I loved your explanation fontwell, it was logical and understandable, (as I think:) anyway I hope I got the gist of it right. As for why I ask about uniform motion and acceleration, that's because I'm still unsure on where the differences between them lies. And to me it also goes back to how I should see 'uniformly moving' frames.

Title: How long does 'C' take?
Post by: fontwell on 16/03/2010 09:40:02
In the diagram your time axis is 't' and your distance axis is 'x'. Since you never move relative to yourself (!) you observe the world from along your 't' axis, where your 'x' is always zero. Similarly, home observe the world from along their 't' axis, where their 'x' is always zero.

E1 is in that position because it occurs at zero distance from home and also after home think some time has passed. Thus along homes 'x' axis it is at x=0, and along homes 't' axis is it some arbitrary distance.

The signal sets out from E1 moving at C, and by definition in the diagram, this means it is parallel to the grey 'speed of light' axis. E2 is the position when the signal reaches you, because at this point the signal is zero distance away from you i.e. it is at x=0 for you.

You write "because you think the distance between the two events was less than home do," it sounds as you are arguing that the distance observed by me then would be some sort of illusion?

'Illusion' might be too strong but it is the case that parties moving relative to each other will disagree about distances. Your can mark distances in your frame of reference and carry a 1 meter ruler but other moving observers will not agree with your version of the distances you mark or how long your ruler is.

How do you measure the distance between two objects? If the objects are both in line with you, and the objects are stationary to you it is easy, you can push out a ruler and stop when it reaches the furtherest object. Then you can see the distance between them along the ruler.

However, if the objects are moving towards you (but a fixed distance apart to each other) how do you do it? The only thing that makes any sense is to record where both objects are at the same point in time. If you think about measuring the length of a moving object you will see this is true.

So what you could do is put clocks all along your ruler, and at one point in time record where the two objects are, giving you the distance apart.

Now the tricky bit. The two objects have their own clocks (which agree with each other because they are stationary to each other) and they can detect when you record the position of each object. But you and moving objects don't agree on times. So, even though you think you measure where the two objects are at the same time, they observe that there was a time interval between your measurements. If the objects are moving toward you, they observe that you detected the front object first and then the back object. As a result, you think the objects are closer together than they do because during the interval (according to them) the back object moved closer to you, reducing what you think is the distance between the objects.

There are no 'gold standard' distances or times for objects in relative motion. Quite literally, the only thing all observers agree on is C.

Even though your frame of reference seems fixed and stable to you, any distances you actually measure are always distances between two events. Even if you measure a static distance, you are really checking the two end points at the same time. And because moving observers disagree about time and simultaneousness, they will disagree about distances too.
Title: How long does 'C' take?
Post by: yor_on on 16/03/2010 14:36:26
Okay, that explains the 't'. I was wondering a little there :) it's an arbitrarily representation where the relative placement of E! and E2 have to fit the description of 'C' propagating. But I'm still not sure if I'm getting how you think explaining relative distances? I see what you mean with needing clocks but if you look at a Lorentz contraction f.ex? There have to be some subtle truth you're trying to propose?

It's like you propose a 'gold standard' to distances, 'invincible' to us, with the explanation being our interpretation of the same depending on acceleration and mass (uniform motion too?). So if I say that distances actually contract with acceleration and not only in a illusionary way but really do? Would you agree to that?
===

Don't worry mr Fontwell, I'm having a very pleasant time reading you :)
And I'm learning a new way to look at it. It was a true pleasure reading you explaining how to fit 'C' to home and the traveler. I've sort of missed considering it from a 'gold standard' before, don't ask me why :)

Sheer laziness or?? Ah..
Well, whatever :)
Title: How long does 'C' take?
Post by: fontwell on 16/03/2010 16:44:41
I don't know the mathematical side of SR or how to deal properly with GR, all I have is a grasp of how to use a Minkowski/Loedel diagram (see above) as an aid to explaining what happens in a two observer situation when they are in relative motion.

The diagram is fully correct (allowing for drawing accuracy). One arbitrary part is the angle between the two 't' axis. This represents the relative velocity of the two parties, bigger angle = bigger velocity (max angle = 90 degrees = C). Once this angle is defined, all other axis are defined. The other slightly arbitrary part is where to put E1 on the 't' axis but this is only a matter of scaling. The path to E2, and all the other construction lines will always look the same after scaling. So really, seeing as there aren't any numbers on the scales anyway, the only random part is the relative velocities of the two parties.

Loedel diagrams are good because all axis have the same scales, allowing easy comparison between observers. It does appear that the observers axis are treated slightly differently. however, the observers can be swapped over and a new diagram constructed which will give identical times and distances for how each observer see the events.

Quote
It's like you propose a 'gold standard' to distances, 'invincible' to us,

No no no! Quite the opposite.

Quote
if I say that distances actually contract with acceleration and not only in a illusionary way but really do? Would you agree to that?

What does actually contract mean? To an observer it 'actually contracts' to another observer it 'actually contracts' less or more or not at all. There is no 'gold standard' distance. There is a gold standard space-time interval and that is how the Loedel diagram works.

Let me have another crack at distances.

You see, when people are trying understand GR I think that they don't have a big problem believing in time variation (in principal anyway) because time seems to be such a slippery substance anyway.

I think what really confuses the issue is distances.The distances we meet everyday all seem so solid and fixed. So there is one point I would like to address using the scenario of the OP.

There really is no fixed independent frame of reference into which anyone can place events. You think you know this already but it is really is very hard to think in a way which doesn't assume one.

So if two parties are relatively moving, they can only give positions of events in terms of how far away an event is from them. Nothing else has any meaning. When home send a signal they can only measure it as starting at zero distance from them. However, when you observe it, "zero distance from them" means nothing to you. You can only record it as being some particular distance from you. Does this mean you think it happened in a different place to each other? The question is not a proper question :-) There is no 'real place' where it happened. There may be other objects near the event when it happened, but that does not define a position in space.

So, for the event when home send a signal, they don't care how far away you are (that would be another event) they only care where the event is in relation to them, distance = zero. But you only care how far away you are. You don't care that they see it as zero distance away from them. There is no 'real place' where it happened.

At the next event it is the same thing. Home don't care that as the signal reaches you it is "zero distance from you" all they care about is how far away from them the event is. But from your point of view you only care that it is at zero distance from you. You do not care how far away home is at this point, as far as you are concerned they are running away from you anyway, so how would this affect the distance the signal travelled?

To repeat, there is is no 'real place' where events take place. All that observers can do is measure the distance to events from themselves. This means that to find the distance between two events (as distinct from absolute positions), they just subtract one distance away from the other.

Sorry this is so long again, I don't know if it helps but I hope it does.
Title: How long does 'C' take?
Post by: yor_on on 16/03/2010 20:38:11
I will have to ponder this. It's something to think about :)
And don't worry about writing long posts, I've always felt that it is better to explain properly what one think than just making statements. With statements nobody understands, most probably including the ones making them too :) You wouldn't happen to have a good link to where one can learn more about Minkowski/Loedel diagrams?
==

And yes, I agree to that view. I used my wording just to create one of those 'statements' to see your thought there. There is, as I see it, no 'set distances'. You need to use the concepts of clocks and rulers to make them 'connect' for two different frames. And there I'm still wondering what they really are:) Distance is such an 'normal' everyday concept to us, we expect them to be the same from day to day, and they are too :) at least inside our own frame.
Title: How long does 'C' take?
Post by: fontwell on 17/03/2010 14:35:45
Quote
There is, as I see it, no 'set distances'. You need to use the concepts of clocks and rulers to make them 'connect' for two different frames. And there I'm still wondering what they really are:) Distance is such an 'normal' everyday concept to us, we expect them to be the same from day to day, and they are too :) at least inside our own frame.

I had a look for links and all I can find (that look helpful) are these two...

http://wapedia.mobi/en/Minkowski_diagram

http://www.einsteins-theory-of-relativity-4engineers.com/loedel-diagrams.html

Both of them dive in quite quickly.

The way I think about the diagrams is like this (not why they work, only how to interpret)...

A normal space vs time diagram, which you know from school looks like this, except I have swapped x axis (distance) and time axis (t) over, because this is what Loedel diagrams do.

[BTW, I've put events on all these diagrams but they are not related, they are just examples of how to read coordinates from the axis.]

[diagram=575_0]

Note how there is a grid.

The horizontal lines represent things which all happen at the same time (from the origin). The vertical lines represent things which all happen at the same distance (from the origin). In relativity an observer makes all measurements relative to themselves i.e. they never move. So their 'world line' is the vertical 't' axis at x=0.

The units on the axis are defined in terms of C. Any unit can be used for distance, but the time unit used will then be "the time it takes light to travel one distance unit (in a vacuum)". Thus a light beam is always at 45 degrees on the diagram (to move one distance unit along, it takes one time unit up).

At low relative speeds two observers can use the same diagram. But at high relative speeds the effects of relativity can be modelled by 'compressing' the observers grid. It is just as if you had a trellis (like for plants) and squeeze it.

So for one observer we get a grid like this.

[diagram=576_0]

[Note how due to our choice of units for the time and distance axis, C is still at 45 degrees. The grey lines represent light signals which pass through the origin.]

And the other observer gets this.
[For some reason the events dots don't show on this diagram, just imagine :)]

[diagram=577_0]

Note that (apart from drawing accuracy) the two grids are mirror images of each other, both up-down and left-right. The only thing that doesn't reflect is the +/- on the axis.

So the Loedel diagram works by making a reference event where both observers were present and synchronised clocks, and puts this event at the origin. Then the two grids can then be overlaid. Note, I am not explaining why this works, just how to do it.

The resulting grids are not easy to draw on one diagram and it would just look a mess anyway. What usually happens is that only the axis are drawn, the events marked, and the lines from events to the axis to read off times and distances.

Note that for events remote to an observer, they see its time as being along their time grid line, and the distance as being along their distance grid lines.

[diagram=578_0]

So using this diagram as an example, there are two events marked and the construction lines to show what times and distances the observers measure.

Mr Green, from along his time line measures that there was some time before an event occurred, and an even longer time before the next one. However, Mr Red measures that he waited for some time in between what Mr Green thinks, and then both events happened at almost the same time, he even thinks they happened in reverse order.

Now, neither observer can define absolute positions where any event occurs, only the distance from themselves. But by measuring how far away from them each event is, they can determine the distance between events. By looking at the total distance between events on each observer's 'x' axis it can be seen that Mr Red think there was less distance between events than Mr Green, but not a huge difference compared to the time discrepancy.

There are good reasons why these diagrams work [to do with s^2 = (Ct)^2 - x^2] but I can't remember the full explanation any more. Suffice to say that if you try out any of the common 2 party scenarios, such as the OP, or why observers each think the other has a slow clock and a short ruler, or how causality is preserved even though observers can disagree on the order of events, then they always seem to give the correct explanation.

As a true exercise in relativity, you should do the following, which is actually quite tricky but worth the effort.

Place a straight edge (like a piece of paper, or a ruler) along one observers 'x' axis.

Mark the position of the 't' axis (x=0) on the straight edge. This mark represents the observer.

Slide the edge up the diagram, always keeping it parallel to that observers 'x' axis and always keeping the mark on that observers 't' axis (x=0). This takes a bit of practice.

As you move it up the page you can see what the observer (the mark) thinks is happening.

Events hit the straight edge in the time order he measures and at the distance he measures. It is his world view.

You can now do the same thing for the other observer using their axis and see their version of events.

Once you have done this enough times you can kind of do it in your head.

Note that for either observer, light beams either move toward them or away from them at velocity C. For example, the grey diagonals shown could represent light beams that coincided at the origin, along with our observers. Both observers measure that these beams move away from themselves in opposite directions, at velocity C.

It goes without saying that these diagrams only show one space dimension but for the armchair physicist that seems to be enough.

Have fun!
Title: How long does 'C' take?
Post by: yor_on on 17/03/2010 16:02:46
Wow, Fontwell, I will, in fact i will :)
Kind'a love this one.. But it will take me some time to 'melt' this. And I'm expecting to find good use of your explanation here while assimilating it.

It was very well done, and thought through. I'm sure I will have use for it as I look at your links. It's a pity when it comes to things like this that we don't have an area where we can put explanations like yours for others that are interested, at least I don't think we have? Do we? In an area of their own, sort of, search able for example on 'Minkowski diagram' 'Loedel diagram' etc. It couldn't hurt :) and would simplify for those not needing to make the same explanation twice.

As for why it works :) Oh yes, I'm wondering about that too, but I'm expecting to learn as I read. Otherwise I will have to pester you again I guess :) And thanks again Mr Fontwell, it's been a pleasure reading you.
Title: How long does 'C' take?
Post by: fontwell on 17/03/2010 17:23:54
You are most welcome yor_on, I'm glad you are interested enough to read my posts. Yes, it will take a while for this to 'melt' but once it does, it can be surprisingly easy to use.

The test of if you have understood it all will be if you can use my original diagram to help with the OP :)
Title: How long does 'C' take?
Post by: Farsight on 22/03/2010 00:42:08
Farsight, thanks for your response, although it isn't the 'mainstream idea' that confounds me here, just some implications that I'm not even sure that I'm getting right, it may just as easily be that I need to think it through again.
The real problem here is what time is. People rather think it's something that "flows", but there's no scientific evidence at all for this. Every measure of time relies on some form of motion, be it in a mechanical clock, an atomic clock, or a a light clock. If you're moving fast through the universe, the rate of local motion is of necessity reduced as per the parallel mirrors example. This affects all processes, so your thoughts and bodily functions slow down too. Hence you can't measure your time dilation locally.

As I said, I might need to rethink how the question should look. But if you feel you have two specific answers to give me, one for each sceniario I will read them with interest. F.ex, I'm more or less stating that the answers you'll get will be different, depending on time of acceleration, don't I? Is that right? Why?
I'm not quite clear what you mean, but the acceleration is only necessary to achieve a different velocity. The total time dilation experienced by say the travelling twin in the Twins Paradox is more if he coasts more with the same acceleration.

We spend the exact same amount of energy (Let's assume so for the Q.) So why would the result differ, and how will it differ? I will get two 'distances' but the same 'time displacement' relative the rest of the universe? Or, I will get the same 'distance' and 'Time displacement'? Or, I will the same 'distance but not 'time displacement'?
There is no reality to time displacement. In the Twins Paradox, the twins depart at the same "time", and at a later "time", meet up at the same time. They don't miss one another by six months! The interval is measuring how far the light has moved between their parallel mirrors, and it's the same distance irrespective of any travel. The time-dilated twin has merely experienced less local motion because of his motion through the universe. Hence he's time-dilated, and has to turn around and come back so that both twins agree on who was doing the moving.

Also. When I look at that signal it will according to relativity come at me at 'c' relative my own frame, no matter how you observe it, and me, from an 'inertial frame' like Earth f.ex if we assume that this were the origin for us both. As I'm moving away relative Earth very near the speed of light (99.99999999~) what does that state about this light? If we now, as SoulSurfer pointed out, could have observed it. In reality the observation won't matter for my statement though, assuming that the concept of relativity is correct it have to be true.
Relativity is correct, but people don't quite understand why. It's just because the local motion of light defines your time. If the rate of propogation of all electromagnetic phenomena in the universe was halved, your thinking and your bodily processes and your clocks etc would all run at half the previous rate, so you wouldn't measure a reduced c. Look at the definition of the second and the metre again.

And it do make me wonder about the concept of 'distance'?
That's defined by the motion of light too. But it's more fundamental than time. Hold your hands up a metre apart and you can see the distance between them. Waggle your hands and you can see motion. We use regular motion to mark out time, but you can't see time.

So if you have ideas about the scenarios feel free to give them Farsight, and if you want to state them from another format than relativity, you're welcome, just make sure that I know you do :) So we're speaking the same language here, sort of.
The things I talk about are relativity, because it's based on what Einstein said. Sadly people don't know about everything Einstein said. Things like the forgotten legacy of Godel and Einstein (http://www.amazon.co.uk/World-Without-Time-Forgotten-Einstein/dp/0713993871). However that means the correct fundamentals are space and motion, not space and time, and this is counter to Minkowski. Hence issues arise.
Title: How long does 'C' take?
Post by: fontwell on 22/03/2010 11:38:15
In the Twins Paradox...both twins agree on who was doing the moving.

"both twins agree on who was doing the moving"

There is something here which I'm not sure if people know and take for granted or if it is an area of confusion, so here goes...

If the two twins start in constant relative motion with respect to each other they might pass each other at one point and agree to synchronise clocks. Then they would naturally move apart. In this case their situation is exactly symmetrical. They each think the other has a slow clock and they are 'both correct'. Of course, since they are no longer both in the same place any more, this is because they both disagree about times and distances between any events.

However, when one of them turns round he really is changing direction and not the other. It could appear that everything is still symmetrical but it isn't. One of the twins will undergo acceleration as he changes direction. He will know this by, say, dropping a stone next to him which will continue to follow him as long as he moves at constant velocity. For the twins to meet again there is no avoiding that one twin must give up his inertial frame and he would see the dropped stone move away from him.

This unsymmetrical action is what results in the returning twin having experienced less time, rather than the situation being equal.

In fact as I understand it, if the twin always changes velocity instantaneously (including at the start) the part of the journey which causes a final difference in clocks is really the mid-point acceleration. During the constant motion part of the two journeys, the situation is identical for each twin.

For instance, (I believe this is true) when the twins have reached their furthest point apart, one twin could decide to change his motion so that he drifts at a constant distance to the first twin for a while i.e in the same inertial frame. Before he does this he will consider that he has moved a certain distance away from the first twin and that the first twin's clock is reading a certain (slow) time. As he changes his motion to match the other twin, he will observe the clock of his twin to suddenly race ahead (overtaking his own) and also the distance between them to suddenly grow.

After that, they would both agree about times and distances between subsequent events for as long as they both shared the same motion.

So the point is that even for two constant velocity journeys one twin will have to do the accelerating, and it is during the accelerating that the situation unbalances.

Title: How long does 'C' take?
Post by: Farsight 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.
Title: How long does 'C' take?
Post by: PhysBang 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 (http://forums.randi.org/showthread.php?p=5739464#post5739464) 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.
Title: How long does 'C' take?
Post by: Farsight 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."
Title: How long does 'C' take?
Post by: BenV 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.
Title: How long does 'C' take?
Post by: Geezer 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.
Title: How long does 'C' take?
Post by: yor_on 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 :)

Title: How long does 'C' take?
Post by: Geezer 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.
Title: How long does 'C' take?
Post by: fontwell 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.
Title: How long does 'C' take?
Post by: Farsight 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.
Title: How long does 'C' take?
Post by: Farsight 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.
Title: How long does 'C' take?
Post by: yor_on 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?
Title: How long does 'C' take?
Post by: yor_on on 23/03/2010 15:37:34

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. .
Title: How long does 'C' take?
Post by: Farsight 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.
Title: How long does 'C' take?
Post by: yor_on 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 :)
Title: How long does 'C' take?
Post by: fontwell 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.

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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.
Title: How long does 'C' take?
Post by: yor_on 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?
Title: How long does 'C' take?
Post by: yor_on 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.
Title: How long does 'C' take?
Post by: fontwell 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!

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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.

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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).
Title: How long does 'C' take?
Post by: yor_on 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?
Title: How long does 'C' take?
Post by: yor_on 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.
==

Title: How long does 'C' take?
Post by: Geezer 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.
Title: How long does 'C' take?
Post by: yor_on 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?
Title: How long does 'C' take?
Post by: fontwell on 23/03/2010 23:00:26
Title: How long does 'C' take?
Post by: fontwell 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).

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.
Title: How long does 'C' take?
Post by: fontwell on 23/03/2010 23:54:25
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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)
Title: How long does 'C' take?
Post by: Geezer on 24/03/2010 00:36:27

What do you mean by "length"?

What do you mean by "ATM Style"?

Stylish ATM machines? Asynchronous Transfer Mode? (although I don't think it's too stylish these days.)
Title: How long does 'C' take?
Post by: fontwell on 24/03/2010 06:40:22
What do you mean by "length"?

What do you mean by "ATM Style"?

Your posts to Farsight in this thread about how to define local time/distance. ATM = Against The Mainstream.

Although, apparently according to mods "Telling people where things belong is a moderator function" , not my function.

What with this and another mod message, I'm outta here.

Bye bye everybody.
Title: How long does 'C' take?
Post by: Geezer on 24/03/2010 07:13:24
Yikes! Please watch the door doesn't slam your ass on the way out.

BTW - ATM refers to Anything but The Mainstream.
Title: How long does 'C' take?
Post by: yor_on on 24/03/2010 18:57:42
Seems I missed something here?
What was it??

As for the the way you see it Mr Fontwell, I do see it.
But to me it doesn't explain the seamlessness. And I'm naturally inquisitive I'm afraid, never stops wondering, and I'm stubborn to a bone I admit. Don't worry though, it's just my nature to have that low imaginative threshold.

It always make me see to many positions to present an idea from. Not always the best thing, makes it hard to trust my own ideas too at times, And then I will argue. But I'm always learning, as we all are here, until I realize why my idea won't work. So me arguing from my position doesn't mean I can't see yours.
But I need to see where I go wrong to know why.
And .

And you're welcome back any time you like, we're not really that bad.
We're not actually, maybe a smite argumentative, some of us.
(Yep, that should be me.)

And we care about all that are here, no matter their level of physics.
I've been on sites where the decision was made that if you couldn't do the math you shouldn't open your mouth. No, they didn't throw me out, crazy as it was? :)
I threw myself out instead ::))

So when the mods react, it's because they are good mods, and try to set a good mode for all here. It's not 'personal', and it have nothing to do with your level of knowledge. We being here are quite cool with each other hopefully, just want all to feel that they have something to contribute with, as they do have, no matter where they are in their physics levels:)

Even stubborn, blind, ah, ** like me ::)) And I will study your loedel-diagram and try to wrap my mind around it, it was a lovely presentation Mr Fontwell. And risking repeating myself, you're welcome back. We all are I hope, the mods are actually the best mods I've seen anywhere.

Title: How long does 'C' take?
Post by: yor_on on 24/03/2010 19:37:35
You wrote "BTW, as you know, these 'fixed' stars are mainly rushing away from you in all directions (well, those in other galaxies)."
==

I quite agree Mr Fontwell. The reason I used it in my example was that it is a proposition from serious physicists (I hope?:)  LSU professor resolves Einstein's twin paradox. 2007 (http://www.physorg.com/news90697187.html) As for my position on this thread :) Well, I'm learning, nothing more. And I kind'a love that, as I think you might too? And I still think my arguments were good ones?
Title: How long does 'C' take?
Post by: yor_on on 08/04/2010 08:44:15
There's another thing confusing me too. I assume that when going near lights speed we will see the universe as 'contracted' any which way we look? Am I right there, that those 'distances' measured will, relative us traveling, shrink equivalently in all directions?

But light will behave differently depending on its vector relative us, won't it? The light 'catching up' to us will be redshifted and the infalling light meeting us as we travel towards it, will be blueshifted. Am I correct there?

So, although I could, if I now f.ex assumed that I was traveling at a constant one G uniformly, and therefore assume that it was equivalent to being on a planet, still observe a unequal distribution of energy, relative my frame of reference?

I know, that's peeking, but it still makes me wonder :)