# The Naked Scientists Forum

### Author Topic: How long does 'C' take?  (Read 13444 times)

#### yor_on

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##### How long does 'C' take?
« 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.

#### yor_on

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##### How long does 'C' take?
« Reply #1 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 :)

#### Soul Surfer

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##### How long does 'C' take?
« Reply #2 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.

#### Farsight

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##### How long does 'C' take?
« Reply #3 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.

#### yor_on

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##### How long does 'C' take?
« Reply #4 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.

#### yor_on

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##### How long does 'C' take?
« Reply #5 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.

#### PhysBang

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##### How long does 'C' take?
« Reply #6 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.

#### yor_on

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##### How long does 'C' take?
« Reply #7 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 :)
« Last Edit: 13/03/2010 14:35:50 by yor_on »

#### PhysBang

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##### How long does 'C' take?
« Reply #8 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.

#### yor_on

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##### How long does 'C' take?
« Reply #9 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?

« Last Edit: 13/03/2010 22:36:13 by yor_on »

#### fontwell

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##### How long does 'C' take?
« Reply #10 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.

#### yor_on

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##### How long does 'C' take?
« Reply #11 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.

« Last Edit: 15/03/2010 01:00:00 by yor_on »

#### fontwell

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##### How long does 'C' take?
« Reply #12 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.
« Last Edit: 15/03/2010 11:30:30 by fontwell »

#### yor_on

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##### How long does 'C' take?
« Reply #13 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.
« Last Edit: 15/03/2010 14:00:13 by yor_on »

#### fontwell

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##### How long does 'C' take?
« Reply #14 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.

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.
« Last Edit: 16/03/2010 00:55:10 by fontwell »

#### yor_on

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##### How long does 'C' take?
« Reply #15 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.

« Last Edit: 16/03/2010 04:56:52 by yor_on »

#### fontwell

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##### How long does 'C' take?
« Reply #16 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.
« Last Edit: 16/03/2010 09:54:06 by fontwell »

#### yor_on

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##### How long does 'C' take?
« Reply #17 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 :)
« Last Edit: 16/03/2010 14:45:53 by yor_on »

#### fontwell

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##### How long does 'C' take?
« Reply #18 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.
« Last Edit: 16/03/2010 18:40:09 by fontwell »

#### yor_on

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##### How long does 'C' take?
« Reply #19 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.
« Last Edit: 16/03/2010 21:15:19 by yor_on »

#### fontwell

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##### How long does 'C' take?
« Reply #20 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 [nofollow]

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

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.]

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.

[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 :)]

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.

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!

#### yor_on

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##### How long does 'C' take?
« Reply #21 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.

#### fontwell

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##### How long does 'C' take?
« Reply #22 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 :)
« Last Edit: 17/03/2010 17:37:22 by fontwell »

#### Farsight

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• Posts: 396
##### How long does 'C' take?
« Reply #23 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. However that means the correct fundamentals are space and motion, not space and time, and this is counter to Minkowski. Hence issues arise.
« Last Edit: 22/03/2010 00:51:52 by Farsight »

#### fontwell

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##### How long does 'C' take?
« Reply #24 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.

« Last Edit: 22/03/2010 11:47:25 by fontwell »

#### The Naked Scientists Forum

##### How long does 'C' take?
« Reply #24 on: 22/03/2010 11:38:15 »