Question about time and measurement

  • #101
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[..] The correct distinction is between defined and un-defined conceptions of speed. For example, the "one-way speed of light" can have any value we choose, because it is an undefined concept, whereas the "one-way speed of light in terms of a system of coordinates in which the Newtonian equations of mechanics hold good (to the first approximation)" is unambiguously equal to c. This is objectively true for any pulse of light, whether that particular pulse's speed is measured or not. So the distinction isn't between measured and unmeasured quantities, it's between defined and un-defined quantities. Or, to put it another way, the distinction is between knowing what we are talking about, and not knowing what we are talking about.[..].
You seem to argue here that Newton and Lorentz proposed un-defined quantities and didn't know what they were talking about. No further comment. :devil: (anyway it's all off-topic).
 
  • #102
ghwellsjr
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My thought experiment:

We as a fourth observer are watching a spaceship move from left to right, viewed directly in front of us.

This spaceship launches two clocks out, at exactly the same speed.
The clock which it launched behind itself is seemingly at rest to our perspective as the fourth observer, because the exact rate at which the spaceship was moving was cancelled out.

To the spaceship, both clocks zoomed away from it in exactly the same way, and the spaceship would see both clocks ticking at exactly the same rate. which is slower than its own rate of time.

The spaceship already knows how fast each clock left its surface. The spaceship knows what distance is between the ship and the two clocks always.

after 10 minuits each clock sends a signal back to the ship how much time was accumulated during the 10 mins to the ship. lets say that the clocks both accumulated 8 seconds

What you guys (all of you) are trying to convey to me is that to the ship, both clocks have experienced less time, because to the ship both clocks zoomed away from it.

Fine.
Here are some spacetime diagrams to illustrate your scenario. First is the rest frame of the spaceship. The spaceship is shown in black with dots indicating the passage of each minute of time. The spaceship launched the red clock to the left and the blue clock to the right and the passage of minutes are also shown by their dots. The clocks are traveling at 0.6c:

attachment.php?attachmentid=62453&stc=1&d=1380909375.png

Note that at the Coordinate Time of 10 minutes, the two clocks send a signal to the spaceship saying that their own clocks were at 8 minutes. These are shown as the thin red and blue lines and they arrive simultaneously at the spaceship when its own clocks is at 16 minutes.

Does this make perfect sense so far?

Now we transform the scenario to our rest frame, which also happens to be the rest frame of the red clock:

attachment.php?attachmentid=62454&stc=1&d=1380909375.png

Now you will note that the red clock experiences no Time Dilation but the spaceship has the same Time Dilation that the clocks had in the previous frame and the blue clock has even more Time Dilation but still the signals sent at the speed of light arrive at the spaceship at its time of 16 minutes.

Are we clear on this?

Lets think of it this way.
Now lets say this ship was already travelling at 99.9%c. The ship does not know it is traveling at this speed.
Now you want to see what it looks like in a frame where the spaceship is traveling at 99.9%c. Unfortunately, this would require a very large screen or I would have to reduce the image to the point that the entire scenario would look like a bunch of lines along the 45 degree diagonal and you wouldn't be able to see any details. So how about we do it at 90%c, OK:

attachment.php?attachmentid=62455&stc=1&d=1380909375.png


Now you can see that both clocks and the spaceship have different amounts of Time Dilation (because they are traveling at different speeds) but still the signals travel at c and are sent and received just like they were before.

Are you getting this? If so, I think that you should be able to answer most of the rest of your questions in the rest of your post.
 

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  • #103
TumblingDice
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Plugging in c doesn't work and of course it's impossible to reach that speed, but taking the limit gives you zero clock frequency.
I'm not disagreeing with you, only trying to help the OP find a handle. You've jumped into the thread with facts that many of us already have a grasp on, and can digest with our own grain of salt. I don't think they're helpful to the OP. I understand limits - in this sense it's a convergence that can never be reached, like saying you can use X amount of energy to gain one-half of the difference between your relative speed and the speed of light. You can keep adding X, but halving the difference means you will never reach it. "Limit" is an acceptable mathematical term, and yet when you decided this had to be mentioned, I feel you detracted from helping the OP, choosing to add information that just blows smoke at this point.

I say this after embracing videos of Richard Feynman that I've watched that I learned of here on PF. One aspect I like is that Feynman was "all good" with keeping complications out of lectures when getting new students on board, freely explaining that "it's more complicated, but I'll tell you that later". To choose my own words, "A leader should never get too far ahead of their followers."

I'm more than game to discuss issues more deeply. Just don't think that's a good thing while we try to help questionator89.
 
  • #104
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I'm not disagreeing with you, only trying to help the OP find a handle. You've jumped into the thread with facts that many of us already have a grasp on, and can digest with our own grain of salt. I don't think they're helpful to the OP. [...] I feel you detracted from helping the OP [..]
The OP made clear (or so I understood it) on that same line that he knew that the speed of light cannot be reached; just ask him if your arguing that the speed of light cannot be reached so that he should not phrase it like Einstein did was any help to him.
 
  • #105
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So from the perspective of the spaceship both clocks leave and are zooming away.
But to the perspective of the fourth observer watching this all happen, the leading clock went faster than the speed of light, and the trailing clock is at rest and experiencing the same rate of time that we are?
No, velocities do not add that way. If your ship, which is moving at 99% c according to the 4th observer, sends out two probes at 99% c relative to itself, the one sent to the front is now moving at something like 99.99% c with respect to the 4th observer.

If light cannot break the speed of light, could you not at any velocity, throw a disco ball up and see which way the light goes slower?
Nope, from your frame the light goes equally as fast in all directions.

When we are driving at 80mph and we throw a baseball at 80mph we know when this ball leaves our hand it is traveling 160mph

Does light follow this rule? If not wouldn't time dilation also not follow this rule?
Actually the baseball is not traveling at 160mph. The velocity would be very very slightly under 160 thanks to the way velocities add together in SR. It's just that 160 mph is so slow compared to c that the relativistic effects aren't able to be seen without exceedingly accurate measuring devices. The differences between SR and Classical physics at this speed are so small that we don't bother using SR. Classical physics is accurate enough for practically everything at this scale.
 
  • #106
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Rephrasing: Maxwell thought that absolute speed can be measured.
Not true. Maxwell thought it was possible to measure speeds (well defined in terms of an inertial coordinate system) relative to the luminiferous aether, which he though was a substantial entity. He specifically denied any sense to the statement that the aether represented "absolute rest", just as he ridiculed the idea that we could determine the "absolute position" of a point. He was a thorough relativist.

You seem to argue here that Newton and Lorentz proposed un-defined quantities and didn't know what they were talking about.
Quite the contrary. As I said, Newton and Lorentz knew exactly what they were talking about. It's just some neo-Lorentzians that don't know (and don't even recognize the need to know) what they are talking about.

No further comment.
If I was in your position, I wouldn't have any further comment either.

(anyway it's all off-topic).
Not at all. The subject of the thread is "time and measurement", and my comments pertain explicitly to the issues of time (and space) and measurement, correcting mis-statements that have been made in the thread. This is hardly off-topic (unless the previous statements themselves were off topic).
 
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  • #107
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When we are driving at 80mph and we throw a baseball at 80mph we know when this ball leaves our hand it is traveling 160mph

Does light follow this rule?
Neither light nor the baseball follow that rule (although the baseball comes so close that we never notice the difference). The rule they both follow is:
[tex]
w=\frac{u+v}{1+\frac{uv}{c^2}}
[/tex]
where ##u## is the speed of the car relative to the ground, ##v## is the speed of the ball relative to the car, and ##w## is the speed of the ball relative to the ground, the thing that you're expecting to be equal to ##u+v##.

Just for grins, I tried calculating it for the 80 mph baseball thrown from an 80 mph car. The speed of the ball comes out to be 159.99999999998 mph. To make the difference interesting, you need to pick values of ##u## or ##v## which are an appreciable fraction of ##c##.
 
  • #108
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Well in the scenario the distance between them is shrinking. so either one is moving towards the other or they are both moving towards each other.
Sure. But the fact that they are moving relative to each other does not imply in any way that either of them ever accelerated. That is the point of my muon example.

Is what I do not understand is that time dilation is just an optical illusion?
It is certainly not just an optical illusion. The principle relativistic effects are what remain after properly accounting for the finite speed of light, etc.

Perhaps it will help a bit to look at what time dilation is mathematically. Mathematically it is dt/dτ, where t is the coordinate time in some coordinate system and τ is the proper time on some clock. The t is clearly coordinate dependent, and the τ is coordinate independent. So time dilation itself is coordinate dependent, since it contains a coordinate.

However, you can do things like take the ratio of two clocks which are compared in some coordinate-independent way (such as twins at departure and reunion), and the dt terms drop out, leaving only coordinate independent quantities.

How would you word it then?
I would say "object A is moving relative to reference frame S", making sure that A and S are unambiguous, e.g. by specifying that A is the first object and S is the second object's reference frame.

We are using a scenario where there are two objects. One is moving and one is not. How do we figure out which one is actually moving.
You don't figure out which one is actually moving. There is no such thing as "actually moving"; there is only "moving relative to".
 
  • #109
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I feel more insulted then helped anymore
I am not sure why you feel that. I looked back and didn't see any insults. I saw very factual responses and explanations.

I did see some frustration, but it didn't seem insulting. Of course, I could have missed it.
 
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  • #110
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Dalespam
all you do is argue about how I worded my posts. No helpful input
If you think back to your education you may realize that the bulk of any field of study consists of learning new concepts and the words used to describe and refer to those concepts. The same thing applies here. I am teaching you the words because they have meanings and understanding those meanings is how you learn any subject.

You consistently used the term "actually moving". I explained what that meant and why it is an incorrect concept in relativity. This is not just a literary critique, this is an explanation of both the concepts and the words used to describe the concepts. If you don't learn the words then you will never be able to communicate on this subject.
 
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  • #111
Ok I get what you are saying DaleSpam.

So if you read my other post about the spaceship and the clock and the fourth observer.
what this means is that:
The spaceship would read after 10 minutes that both the clocks have accumulated 8 minutes each since departure.
But if the spaceship were to choose a clock, accelerate for 10 minutes to retrieve the clock he chose. and find that He experienced a total of 20 minutes while the clock he chose experienced 30 minutes? Dont worry if my math is right Im more concerened with understanding the concept here.

So in the most normal scenario where the astronaut leaves earth, as he is leaving earth if he were to check after 10 minuits he would find that the earth has experienced 8 minuits.
even though we know damn well we just accelerated away from the earth.
But when he takes the 10 minutes to turn around and head back home, when he arrives all this time is realized and the earth has aged 30 minutes to his 20?


And in my scenario where the spaceship and the clocks (post#97)
What Harrylin is saying is that if your velocity is 90% of c, light leaving you would be viewed traveling at c but it is being shrunk basically into 2 dimensions?
 
  • #112
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What Harrylin is saying is that if your velocity is 90% of c, light leaving you would be viewed traveling at c but it is being shrunk basically into 2 dimensions?
Let's say traveller A is travelling at 90% of c relative to observer B. To B the light going in the same direction as A is moving 10% faster than A. B notices that A's measuring rods are length contracted and A's clocks are time dilated as well as being out of synchronisation, so B understands why A also measures the light to be travelling at c.

Now look at it the opposite way around. A is entitled to consider himself stationary. After all the speed of light in all directions as measured by a is c. To A, B is travelling at 90% of the speed of light and light going in the same direction as B is only moving 10% faster than B. A notices that B's measuring rods are length contracted and B's clocks are time dilated as well as being out of synchronisation, so A understands why B also measures the light to be travelling at c.

The situation is perfectly symmetrical. There is no sense in which either can claim themselves or the other to be 'actually moving' as everything they measure is identical. As Dalespam said. the term 'actually moving' has no meaning in relativity. On the other hand, the term 'actually accelerating' does have a physically significant meaning. When one of the observers changes speed or turns around, there is no question about who is actually accelerating and who is not. To compare the elapsed time on two clocks in an unequivocal way, they must start alongside each other and finish alongside each other. To do this, at least one of the clocks/observers has to accelerate and then the symmetry is broken, as in the twins paradox.
 
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  • #113
ghwellsjr
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Ok I get what you are saying DaleSpam.

So if you read my other post about the spaceship and the clock and the fourth observer.
what this means is that:
The spaceship would read after 10 minutes that both the clocks have accumulated 8 minutes each since departure.
Did you see the diagrams I made for you in post #102? Did you see that the spaceship does not receive the signals from the two clocks until 16 minutes after they were launched? The time of 10 minutes is not something that the spaceship can have any awareness of except by predicting that the times on the remote clocks will be at 8 minutes based on their speeds relative to the spaceship's rest frame, just like we are doing here. But if that was the issue you were concerned about, then what was the purpose of the signals the clocks sent to the spaceship?

But if the spaceship were to choose a clock, accelerate for 10 minutes to retrieve the clock he chose. and find that He experienced a total of 20 minutes while the clock he chose experienced 30 minutes? Dont worry if my math is right Im more concerened with understanding the concept here.
You're not quite right, if the spaceship took off when its was at 10 minutes and spent 10 more minutes reaching the blue clock, the blue clock at that point would read 25 minutes as depicted in this diagram showing just the black spaceship and the blue clock:

attachment.php?attachmentid=62464&stc=1&d=1380932694.png

But this is in fact nothing more than the most normal Twin Paradox scenario as depicted by this diagram showing the rest frame of the blue clock:

attachment.php?attachmentid=62465&stc=1&d=1380932694.png

So in the most normal scenario where the astronaut leaves earth, as he is leaving earth if he were to check after 10 minuits he would find that the earth has experienced 8 minuits.
even though we know damn well we just accelerated away from the earth.
But when he takes the 10 minutes to turn around and head back home, when he arrives all this time is realized and the earth has aged 30 minutes to his 20?
It's 25 minutes to his 20.

Consider this diagram and treat the astronaut as being in the black spaceship and the blue clock represents the time on earth:

attachment.php?attachmentid=62467&stc=1&d=1380934751.png

I have added signals to show when and how the black astronaut checks the earth time at his 10-minute mark. He sees the blue earth clock at 5 minutes (you have to count the dots along each worldline). He doesn't see the blue earth clock displaying 8 minutes until his own clock reaches 11.5 minutes. Do you see that?

Furthermore, the blue earth people see the astronaut's clock displaying 5 minutes when their own clock is at 10 minutes. It's reciprocal during this first part of the trip.

Does this all make sense to you?
 

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  • #114
ghwellsjr
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Here is a repeat of the last diagram but with only the two symmetrical light signals going from each observer's time mark of 5 minutes and going to the other observer's time mark of 10 minutes. Please note that in this and all subsequent diagrams, these two signals are present and always traveling at c although their trip times can vary and the times on the sending clock and receiving clocks is the same (based on the dots):

attachment.php?attachmentid=62472&stc=1&d=1380955052.png
Next, we see the rest frame for the black astronaut as he is leaving the blue earth:

attachment.php?attachmentid=62473&stc=1&d=1380955829.png
And the rest frame for the black astronaut as he is returning to the blue earth:

attachment.php?attachmentid=62474&stc=1&d=1380955829.png

Now we see the frame in which the black astronaut starts off traveling at 90%c:

attachment.php?attachmentid=62475&stc=1&d=1380955829.png

And finally a frame where both observers start off traveling away from each other at the same speed, 1/3 c.

And here's another one where the black astronaut starts off traveling at 50%c:

attachment.php?attachmentid=62479&stc=1&d=1380957016.png


And here's another one where the black astronaut starts off traveling at 50%c:

attachment.php?attachmentid=62480&stc=1&d=1380957016.png

Please notice that the speeds of the observers is different in each frame and therefore the Time Dilation is different in each frame which points out that Time Dilation cannot be directly observed.
 

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  • #115
ghwellsjr
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You lifted that quoted phrase from Einstein's 1905 paper (first article) where he proceeded to argue that the "one-way speed of light" requires an additional definition. Do you know something that Einstein didn't know?
I said the "one-way speed of light" (with no further specification) is undefined, and hence requires an additional condition to be well defined and meaningful. You point out that Einstein said the same thing, and then you ask if I know something Einstein didn't. That's a bit of a non-sequitur, isn't it?
No, it isn't, because you once again have misrepresented what I said (and you're misrepresenting what Einstein said in his 1905 paper). Maybe he said something different in subsequent writings but then I would think you should quote those later writings if you want to make your point.

As I previously pointed out, in the first article of his paper, he starts with a portion of your quote:

a system of coordinates in which the Newtonian equations of mechanics hold good (to the first approximation)
This system of coordinates has only three coordinates, all spatial, and he calls it the "stationary system". Then in the remainder of the article, proceeds to define time and the one-way speed of light and at the end of the article, he says:

It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it “the time of the stationary system.”
 
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  • #116
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questionator89, please try to be more concise in the future. This is exhausting to reply to. One good question that really gets at the heart of your confusion is better than a hundred repetitive small questions that nibble around the edges.
This spaceship launches two clocks out, at exactly the same speed.
I assume you mean the same speed relative to the spaceship's frame.

The clock which it launched behind itself is seemingly at rest to our perspective as the fourth observer, because the exact rate at which the spaceship was moving was cancelled out.
Not just seemingly. It is at rest in our frame.

To the spaceship, both clocks zoomed away from it in exactly the same way, and the spaceship would see both clocks ticking at exactly the same rate. which is slower than its own rate of time.
Yes. (in the spaceship's frame)

The spaceship already knows how fast each clock left its surface. The spaceship knows what distance is between the ship and the two clocks always.

after 10 minuits each clock sends a signal back to the ship how much time was accumulated during the 10 mins to the ship. lets say that the clocks both accumulated 8 seconds

What you guys (all of you) are trying to convey to me is that to the ship, both clocks have experienced less time, because to the ship both clocks zoomed away from it.
In the ship's frame, yes.

Lets think of it this way.
Now lets say this ship was already travelling at 99.9%c. The ship does not know it is traveling at this speed.
I assume that you mean it is travelling at .999 c in our frame.

So to us as the fourth observer we know that the propellants in the clocks cant break this speed. We would expect from watching that the clock launched in front, would not be able to be launched in front.
It will be able to be launched in front. Assuming that the ship is travelling at .999 c in our frame, and assuming that the clocks are travelling at ±.999 c in the ship's frame, then in our frame the clocks are travelling at 0. c and .999999 c. Velocities compose according to the relativistic velocity addition formula:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html (see the "projectile" section)

But what is conveyed to me is that, A) the ship has no idea that is travelling this fast, and cannot see anything around it including us as the fourth observer, except for the clocks.
Why wouldn't it be able to see us? Unless we are Klingons using our cloaking devices then the ship should be able to see us travelling at -.999 c in the ship's frame.

B) Both these clocks will zoom away from the perspective of the ship at the same speed and seem to experience less time
Yes.

So lets leave absolute speed completely out of this.
We know anything with invariant mass can never travel at the speed of light. Or our math tells us this?
Yes.

So from the perspective of the spaceship both clocks leave and are zooming away.
But to the perspective of the fourth observer watching this all happen, the leading clock went faster than the speed of light
No, the leading clock went .999999 c in our frame.

, and the trailing clock is at rest and experiencing the same rate of time that we are?
Yes, it is at rest in our frame and therefore it is not time dilated.

How is it that, because the spaceship has nothing to reference from and does not know that he is breaking the this law, this is able to happen?

Wouldnt the spaceship notice that the clock ahead of him never sent another signal again?
No law is being broken and the clock ahead functions normally (well, it is time dilated in both our frame and the ship's frame, but that is normal for moving clocks).

To us as the fourth observer the trailing clock is at rest and experiencing the same time rate as us, which is much faster than the ship and the forward clock.
Yes.

But to the ship this trailing clock is speeding away, and experiencing time dilation compared to the ships clock.
Yes. Sometimes this is called the reciprocity of time dilation. Also, in the ship's frame our clock is time dilated.

This means the the trailing clock sends a signal which would be identical to the leading ship, were it experiencing time.
Not sure what you mean here.

Us as the fourth observer know that this leading clock is traveling faster than light (if it left the spaceship at all)
No, it is travelling at .999999 c. The velocity addition formula ensures that it will never travel faster than c regardless of how fast it is launched from the ship.

Would the ship only receive one signal? or Because the ship has no frame of reference , would the clock go faster than c?
The ship certainly has a frame of reference, and you already asserted that the clock goes at .999 c in the ships frame, which is clearly not faster than c.

Lets say the spaceship didnt use clocks. Lets say he used lasers.
Would his leading laser leave his spacecraft?
Yes, it would leave at c in the ship's frame.

Would it appear to leave his spacecraft to the spaceship but not move to us as the fourth observer?
Yes, it would leave his spacecraft at c in our frame. He would be chasing right behind it at .999 c, but the laser would still outrun the ship in our frame.

If light cannot break the speed of light, could you not at any velocity, throw a disco ball up and see which way the light goes slower?
It doesn't go slower in any direction. That is what the Michelson Morely experiment measured, the isotropy of the speed of light.

So if you are already moving, you do not accelerate you do not slow down, you cannot know how fast(if at all) or in what direction you travel, you appear to be at rest.
Certainly you are at rest in your own frame.

And anything you launch away from yourself you assume is speeding up, because you believe you are at rest.
It isn't just a belief, it is a fact. You ARE at rest in your own frame (by definition). It is a frame-dependent fact since you are not at rest in other frames, but it is a fact nonetheless.

But isnt the speed of light completely independent from your speed/
Yes. There are other experiments that confirm this. I would recommend reading the sticky on the experimental basis of SR.

When we are driving at 80mph and we throw a baseball at 80mph we know when this ball leaves our hand it is traveling 160mph

Does light follow this rule? If not wouldn't time dilation also not follow this rule?
All objects, light, baseballs, bullets, clocks, follow the velocity addition rule I posted earlier.
 
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  • #117
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The spaceship would read after 10 minutes that both the clocks have accumulated 8 minutes each since departure.
But if the spaceship were to choose a clock, accelerate for 10 minutes to retrieve the clock he chose. and find that He experienced a total of 20 minutes while the clock he chose experienced 30 minutes? Dont worry if my math is right Im more concerened with understanding the concept here.
Yes. Note that the 10 min vs 8 min in the first part is a frame dependent fact since you are comparing two distant clocks. In different reference frames the accumulation will be different. However, the 20 min vs 30 min is frame independent since it is a comparison of co-located clocks.

So in the most normal scenario where the astronaut leaves earth, as he is leaving earth if he were to check after 10 minuits he would find that the earth has experienced 8 minuits.
In the astronaut's frame, yes. Please try to get in the habit of specifying the frame for frame-dependent quantities.

even though we know damn well we just accelerated away from the earth.
But when he takes the 10 minutes to turn around and head back home, when he arrives all this time is realized and the earth has aged 30 minutes to his 20?
Yes.

And in my scenario where the spaceship and the clocks (post#97)
I answered that in EXHAUSTIVE detail. Please let me know if you have questions, but please try to consolidate your questions.
 
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  • #118
well I think I understand now.

Thanks very much you guys, it was a lot of effort on your part to get me to understand this.

I'm sending the karma through the universe right now
 
  • #119
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...you're misrepresenting what Einstein said in his 1905 paper.
As I said, this particular aspect of Einstein's 1905 paper was not expressed as clearly as one could wish, and it is recognized as being at least slightly problematic (as you can already tell by Sommerfeld's footnote). In fact, the lack of clarity on this point was cited as one of "Einstein's Mistakes" in Ohanian's book of that name. I think Ohanian is wrong to call this a flat-out "mistake", since it IS possible, with some allowances, to construe Einstein's presentation in such a way as to make it correct. Nevertheless, it's undeniable that someone reading Einstein's 1905 paper without already having a good grasp of the subject can easily miss the essential point, which is the identity between (1) the measures of space and time in terms of which mechanical inertia is isotropic (so the equations of Newtonian mechanics hold good to the first approximation), and (2) the measures of space and time in terms of which the speed of light is isotropic. Einstein emphasized (2), and seems to pay scant attention to (1) after he first introduces it, but he never actually denies the identity, and of course the entire theory hinges on that identity.

The point is, it's perfectly correct (even obvious) to say that the "speed of light" can only be established by an additional stipulation, but it is incorrect to say that "the speed of light in terms of measures of space and time in which the equations of Newtonian mechanics hold good (to the first approximation) can only be established by an additional stipulation". Einstein asserted the former, which is correct, but his presentation could easily mislead someone into thinking he was asserting the latter. If he really was asserting the latter, then it was indeed a "mistake", as Ohanian claims. In that case, the answer to your question (Do we know something Einstein didn't?) would be yes. However, as I said, a careful (and charitable) reconstruction of the reasoning for the entire paper, with some allowances, permits (I think) the conclusion that Einstein understood it correctly and more or less expressed it correctly, but just wasn't as clear as he could have been. He certainly showed in subsequent writings that he understood it.

Maybe he said something different in subsequent writings but then I would think you should quote those later writings if you want to make your point.
He did indeed say things differently in later writings, and I did quote them. Of course, the issue of 'what Einstein said and when he said it' is of mainly historical interest. On the substantive issue, the intent here is not to argue by appealing to authority, but to explain the issue to make it conceptually clear. The main question you should try to answer is this: Does the requirement for Newton's equations of mechanics to hold good (to the first approximation) suffice to establish a definite simultaneity between spatially separate events? (Hint: Yes, it does.)
 
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