Is time universal? NO (and its proof)

[Billy T]

...Lets call the launch point x=0 then after 10ship time units the photon is at x=10, but moving at 0.9c the mirror is now at x=19. I agree with you up to this point, by using your example of .9c. ...The mirror traveled another 9 meters while the photon was in flight, according to the distant observer. So, the photon will contact the mirror after it has traveled 19 meters. ....

No, you agree photon is at x=10 and mirror is then at x= 19. How will photon magically instantly jume to x +19? When the photon does get to x=19 (9 time units more) the mirror has moved 0.9times 9 = 8.1 units more to 27.1 as I originally stated. Please read more carefully.

Or understand that with a head strat distance that requires 10 time units for light to cover and light closing on the mirror by only 1 -0.9 = 0.1c instead of the stationary POV of the on board ship observer (and clock) full speed of light fro the distant observer it will take 100 time units to catch the moving mirror. (now you need to correct for the contraced separation that the distant obsrver see, so it is not really 100 instead of 10 time units but still much more than 10 for the photon to catch the mirror as the distant observer see it. Your idea that the phono at x =10 gets to x =19 in zero time is strange to say the least.

Look at the nice work, drawings Dale has made for you.

 

[Billy T]

...The problem with the diagrams you posted is the don't illustrate the proper POSITION of the distant (at rest) observer. ...

Dale's observer has what is sometimes called a "God's eye view" or you can think of him as a multitude of observers, one every where, so no perception delays, or only one observer who corrects for the perception delay correctly. Namely the V path Dales shows is the actual path in the "distant observer's frame" of the photon.

For example (to answer Geistkiesel's "not a V" objection) the photon was not fired at where the mirror is when the photon was "launched." It is like shoot a flying duck. it is aimed ahead of that point so that it and the mirror will both arrive at the same point together. Thus in the distant observer's frame, the photon does not strike the mirror perpendicular, but at half the V angle, just as a bullet would bounces off flat-sided "Super Duck"and make a V shaped trajectory, so does the photon.

As I said in prior post, because photon path is two times the length of the hypothese, not one 5m leg of the triangle twice, it takes longer to go the extra distance. In this "tranverse to speed" case, their is essentially no contraction to correct for. (Both frames agree that the distantance between source and mirror a constant 5m.)

 

[DaleSpam]

Thanks, DaleSpam, for your obvious effort in making this post.

You are quite welcome! That is what happens when I have spare time on a Sunday afternoon

The problem with the diagrams you posted is the don't illustrate the proper POSITION of the distant (at rest) observer. When a mathematician uses the Lorent transforms to calculate time dilation and length contraction, where is the observer located? He is at a rest point 'A' viewing a 'clock' (B) receed from him in a straight line, directly away from him. A photon from the light clock would appear to be traveling left-to-right for a light clock oriented perpendicular to the direction of travel. He is viewing the clock from the 'end' of the frame while the clock is receeding from his viewing location. The result would be the same as your first diagram. I don't think it is possible to illustrate this in a diagram.

I understand your concern. It makes perfect sense from an experimental or practical standpoint, but I don't think that it is a problem from a theoretical standpoint. True, I did not specify an observer position, but in a single frame you can theoretically put as many stationary observers as you need all around at known locations. You can syncronize all their clocks and they will all stay synchronized. They can then observe events at that location with no propagation delay and report the time back. So you can consider the observers to be located everywhere throughout the "rest" frame.

Of course, theory is nice, but experimental reality is considerably more difficult as you indicate. I think that to really do something like this experimentally you would need at least four observers with known relative locations and synchronized clocks. Then they could observe the time that the light for each event arrives. This would give one piece of data for each event, so four observers would give four pieces of data and you should be able to solve for the position and time. You may be able to get away with fewer observers if they can observe direction as well as time. In any case, I don't think it is a theoretical limitation.

The photon from the light clock oriented in the direction of travel would appear to receed from observer 'A' while traveling toward the mirror, and then travel back toward the observer after the photon was reflected from the mirror, both motions in-line with the observer's line-of-sight. Why do I insist on these orientations to illustrate a problem with the Lorentz transforms? Because these are the positions the Lorent transforms are calculated from. Your first diagram places the observer BESIDE the direction of travel of the light clock, time dilation will be evident, but NO length contraction will be seen from that perspective.

Yes, that is correct. Lorentz assumes no length contraction in the perpendicular direction which is illustrated in the first diagram. I am sure that you already noticed, but I just wanted to emphasize that the axes are different for the two diagrams. Both have x (the direction of motion) as the horizontal axis. The vertical axis of the first is y (a direction perpendicular to the motion) and the vertical axis of the second is time. That said, as I described above I don't think that the observer location is a theoretical limitation of the analysis.

This quote is in reference to the second diagram. No, we do not know time in the clock's frame is running slow. If this were the ONLY orientation of the light clock viewed, how would you determine that? Besides, the only thing shown in the first diagram was that a photon has to tranverse a bigger distance because of THE SAGNAC EFFECT. The laser/detector moved while the photon was in flight. Since the first diagram was using a PHOTON to measure time, if you state 'time slowed', then you are stating the speed of light is slower in that frame. Again, this is why I use a light clock. I'll stop at this point for now.

Here I disagree with your analysis fundamentally. You cannot analyze a single orientation and hope to derive a general transformation between frames. If the parallel orientation were the only orientation viewed then we would indeed have no reason to assume a length contraction. Instead we would probably follow Voigt's path and ascribe it purely to time dilation. In this case the time dilation would be 1.25². When we then tried to get a general transform by considering the perpendicular direction we, like Voigt, would find that we need length expansion to account for the perpendicular light clock. In both cases (Voigt and Lorentz), you must consider both parallel and perpendicular clocks to develop the general transform, from one you will get a pure time dilation with no length change and from the other you will get a length contraction or expansion with the time dilation found previously.

Are you trying to suggest a transform with no length changes but with anisotropic time dilation? If so, please elaborate, it sounds interesting.

-Dale

 

[DaleSpam]

Frankly I am at a loss for words. I didn't bring up colors. You did. You have specifically made them red and blue, which I thinks suggest doppler shift.

For the last time forget colors. Forget VIN's. The Mustang going 80 Mph (50 relative to you at 30 Mph) does not exist at all to the car going 50 Mph (another Mustang is going 100 Mph (50 Mph to it) and vice versa.

These Mustangs to not exist in but one frame and that is the frame where they are going 50ph relative to any observer.

Yes, I brought up the colors because I thought it strengthened the parallels between the Mustangs and light. And yes, the colors were meant to be suggestive of Doppler and time dilation effects. If you want to deliberately weaken your example you are certainly free to do that, but I like it better with colors.

For the 3rd or 4th time: I get your idea. You seem to think that this is some extremely difficult concept to grasp . The Mustangs in different frames are completely different vehicles. The only Mustangs that exist are going 50mph in the frame in which they exist. If you are going at an even slightly different velocity then a given Mustang does not exist in your frame.

False. Relativity is based on the assumption that these are the same mustangs and to have a constant relative veloicty to different observers that is where relativity comes in. Since these are not the same Mustangs there is no need for relativity.

Relativity makes no such assumption. From the Wikipedia entry (http://en.wikipedia.org/wiki/Special_relativity) the second postulate is: "Light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body". That says absolutely nothing about the same photons existing in different frames. It only describes the motion of the light, which you have not altered a bit with your proposal.

-Dale

 

[2inquisitive]

OK, the 'event' of the photon emission occurs at x=0, the time on the clock. The emission of the photon occured at the SAME location and time, the 'event' at x=0. At the speed of 'c' relative to the distant observer, the photon will travel 10 meters in 2.99792458 ns. It will only gain 1 meter on the mirror in that length of time, so it takes 10 times as long to gain the 10 meters, or 29.9792458 ns. However, on the return trip back to the mirror, it only has to travel 1 meter before it meets the oncoming mirror, or .299792458 ns. The total trip would take 30.2790038 ns if there were no length contraction of the moving frame. At .9c, the length contraction factor is .44. Multiply .44 times the 100 meters the photon has to travel, and the distance is now 44 meters. It takes a photon 13.190868152 ns to travel 44 meters. So the distant observer states it takes 13.190868152 ns for the photon to reach the mirror, and (.44 times .299792458) .13190868152 ns for the return path. The total time would be 13.3227768339 ns, according to the distant observer's clock. But wait, Special Theory states the gamma factor for the clock is 2.29! In the 'rest frame' of the clock we are speaking of, the photon reaches the mirror and returns in (20 meters times .299792458 ns per meter) 5.99584916 ns. Multiply 2.29 times 5.99584916 and the figure is 13.7304945764 ns. The time is off by .4077177425 ns. Shouldn't gamma be closer to 2.222 instead of 2.29 to transform between the rest frame and the moving frame? Or should the length contraction factor be closer to 45.3? Or is it that moving frames are not exactly equivalent to 'true' rest frames? Don't forget, the Devil is in the details. By the way, I still do not accept the premis of length contraction, just pointing out that the Lorentz transforms do not work out mathematically correct, unless I have made a mistake in my calculations.

 

[Billy T]

....At .9c, the length contraction factor is .44. Multiply .44 times the 100 meters the photon has to travel, and the distance is now 44 meters. It takes a photon 13.190868152 ns to travel 44 meters. So the distant observer states it takes 13.190868152 ns for the photon to reach the mirror, and (.44 times .299792458) .13190868152 ns for the return path....

I do not do the number any more (You got much more of this out of me than most.) I think you should not multiply the 100 meters, which you calculated by first neglecting contraction, but (assuming your .44 is corect) contract the original 10 meters first. (May work out to be the same end result, but is more logically correct - I mainly concern my self with ideas, not numbers nowdays.)

My whole point was to correct your false idea, at least very implicit is not explicitly stated) that the extra time going one way would be compensated for by the less time going the other. You now agree that is not the case.

I am not sure (because I am too lazy to read vey carefully severaltimes) exactly what you now mean. Have you accepted that the moving clock is in fact runing slow compared to the distant observer's or not? If yes, then can you also seen that MacM's "reciprocity is impossilbe" is not true? i.e the moving observer also measures that the distant clock is running slow?

 

[2inquisitive]

"My whole point was to correct your false idea, at least very implicit is not explicitly stated) that the extra time going one way would be compensated for by the less time going the other. You now agree that is not the case."
-------------------------------------------------------------------------------

Yes, but did you see the exercise about the airplane flying across country and back witn a wind vs. with no wind? Same type gedankin, and the times did not equal. THAT was my point.

"Have you accepted that the moving clock is in fact runing slow compared to the distant observer's or not? If yes, then can you also seen that MacM's "reciprocity is impossilbe" is not true? i.e the moving observer also measures that the distant clock is running slow?"
------------------------------------------------------------------------------

I think the moving clock MAY run slow, am I not yet shure, but gedankins based on how light bounces do not prove anything. The speed of light is constant in co-moving reference frames, it has not been shown it is constant between different reference frames. That is why the length contractions of STR are necessary for the theory to be correct. But clocks will run at different rates according to their position within a gravitational field, no relative velocity required. Will clocks slow when moving through a gravitational field due to the effect of the field, or because light bounces differently?

We have not even touched on 'MacM's reciprocity' in my examples. And, no, the moving clock running slower than the stay-at-home clock absolutely does not show reciprocity is true. Do you know what reciprocity refers to?

 

[MacM]

Have you accepted that the moving clock is in fact runing slow compared to the distant observer's or not? If yes, then can you also seen that MacM's "reciprocity is impossilbe" is not true? i.e the moving observer also measures that the distant clock is running slow?

How many times must you have your posts corrected before you stop posting your BS.?

I have made it clear each and every time this issue is discussed that we are not talking about illusions of motion but of actual physical accumulated time on the clocks.

Now once and for all demonstrate that two clocks brought together in a common frame after some period of relative velocity have each accumulated less time than the other.

THAT IS THE ONLY ISSUE EVER RAISED.

 

[Billy T]

How many times must {I tell} you... that we are not talking about illusions of motion but of actual physical accumulated time on the clocks....

Probably as often as I must tell you that "accumulated time" is measured between a "start event" and a "stop event."

No problem with having simultaneous "start events" if the clocks of the two frames are adjacent at the start of "accumulated time" , but the two observers do not agee that the other stopped his clock "correctly" (simultaneously with his) as their concepts of simultaneity are different. Each is sure the other stopped accumulation too late to be simultaneous with his stopping. If he had stopped it correctly (sooner than he did, when and it was showing less time accumulated) then it would have demonstrated that in the frame moving relative to him, time is passing more slowly. - the muons do get to Earth surface, etc. 2inquisitive's "light transit clock" are both running too slow, if the "stop events" were simultaneous in both frames, etc.

PS - Please also note that the thought experiment of this thread has no "illusions." - There are many clocks and only adjacent ones are used, no propagation or perception delays, no illusions, etc. at all for your duck and weaves diversions to drag up.

 

[Billy T]

...Now once and for all demonstrate that two clocks brought together in a common frame after some period of relative velocity have each accumulated less time than the other. THAT IS THE ONLY ISSUE EVER RAISED.

I know you admit that the GPS moving clock has a correcton applied to it for both the gravity difference and speed, so that it does run at the same time as some point on Earth (I don't care to hear again more about that point being Earth center, or some other point, and that the experts do it wrong, etc.)

I also do not know if any GPS clock has ever been bought back to the lab that made it etc. but if it has been agreeing with the Earth based clock WITH the SRT CORRECTION for many years, I strongly doubt the "hands" will suddenly rapid spin around thousands of times to disagree with Earth clocks by days. That is, the SRT correction is atrificailly added because the makers of GPS know moving clocks "run slow." - That SRT correction term must have been correct. I.e. forgetting about the larger (I think) gravity effect and its correction term, the orbiting clock was running slow and needs a speed correction term also to make it go at Earth clock rate.

That is about as close to your demand of "bring them back together" as we are likely to get. NASA can not a afford to send one away at high spped , turn it around, and bring it back just to please you. (They know how to correct for the moving clock's speed effect that by itself, makes it "go slow" and do not really care that you think it would "accumulate the same time" without the SRT correction for speed effect, if only the gravity effect correction were made and it were returned to sit next to the lab clock, again.) They think it would NOT agree with the lab clock if it did NOT have the speed correction. - that is why it has a speed correction. That is why it does agree. Tell them they are doing it wrong, not me.

 

[funkstar]

Now once and for all demonstrate that two clocks brought together in a common frame after some period of relative velocity have each accumulated less time than the other.

For the last fucking time: The only person claiming that relativity predicts this is you, and you alone! It has demonstrated to you repeatedly that this is in fact not the case, and is based on your rudimentary grasp of the subject involved; i.e. your use of the equation of time dilation where the full Lorentz transforms are needed. If you had EVER studied relativity theory in an educational context (i.e. at college or university), you would KNOW that this prediction is not that of relativity theory, because you cannot make a judgment of proper time by the time coordinates alone. It just doesn't work that way, and it's really fucking obvious when you sit down and look at the Lorentz transforms - something that I'm more and more convinced that you've actually never done.

It's an elementary error, and it's frankly embarrassing to watch you repeat the same mistake over and over.

Then there's the insulting part: You genuinely believe that you have somehow seen an inherint contradiction in the theory, whereby it predicts obvious nonsense (allowing two clocks to have each accumulated less time then the other), that everybody else who has ever studied relativity theory has somehow missed. Let me tell you something: Not only does everyone think of exactly this "paradox" when they first hear of time dilation, but most of us actually sit down with the Lorentz transforms and try to work it out - it's even usually given as an exercise, if it isn't explicitly described in the course material. It's a sobering experience, because, we find that we can't make the equations give the prediction! Do you think we're too stupid to see the obvious paradox that mutual time dilation seems to present? No, of course we aren't. Why, we even have a name for it: The Twin Fucking Paradox! And it's fucking wrong! (I don't want to hear a fucking peep out of you regarding GR on this count, because it is not needed and you fucking know it!)

Now, proper time for a given path between two spacetime events is provably invariant between frames, in relativity theory. That is: Transform the spacetime events and the path from one frame to another, and you'll find that the proper time in both frames, is the same. In other words, it's impossible for the theory to predict what you say it does.

You. Are. Wrong. How many times are we going to have to tell you?!?

 

[DaleSpam]

OK, the 'event' of the photon emission occurs at x=0, the time on the clock. The emission of the photon occured at the SAME location and time, the 'event' at x=0. At the speed of 'c' relative to the distant observer, the photon will travel 10 meters in 2.99792458 ns. It will only gain 1 meter on the mirror in that length of time, so it takes 10 times as long to gain the 10 meters, or 29.9792458 ns. However, on the return trip back to the mirror, it only has to travel 1 meter before it meets the oncoming mirror, or .299792458 ns. The total trip would take 30.2790038 ns if there were no length contraction of the moving frame. At .9c, the length contraction factor is .44. Multiply .44 times the 100 meters the photon has to travel, and the distance is now 44 meters. It takes a photon 13.190868152 ns to travel 44 meters. So the distant observer states it takes 13.190868152 ns for the photon to reach the mirror, and (.44 times .299792458) .13190868152 ns for the return path. The total time would be 13.3227768339 ns, according to the distant observer's clock. But wait, Special Theory states the gamma factor for the clock is 2.29! In the 'rest frame' of the clock we are speaking of, the photon reaches the mirror and returns in (20 meters times .299792458 ns per meter) 5.99584916 ns. Multiply 2.29 times 5.99584916 and the figure is 13.7304945764 ns. The time is off by .4077177425 ns. Shouldn't gamma be closer to 2.222 instead of 2.29 to transform between the rest frame and the moving frame? Or should the length contraction factor be closer to 45.3? Or is it that moving frames are not exactly equivalent to 'true' rest frames? Don't forget, the Devil is in the details. By the way, I still do not accept the premis of length contraction, just pointing out that the Lorentz transforms do not work out mathematically correct, unless I have made a mistake in my calculations.

Hi 2inquisitive,

All the numbers worked out for the Lorentz transform in my example with .6 c and light clocks measuring 4 sc (chosen to get a nice gamma of exactly 1.25 and to get nice 3-4-5 right triangles). I didn't understand why they weren't working for your example so I thought I would try crunching your more difficult numbers myself.

So here is your example as I understand it. You are looking at a light clock measuring 10 m in our frame and moving at .9 c in the parallel orientation. If we use the standard configuration then everything takes place on the x-x' axis. The unprimed coordinates are in our rest frame and the primed coordinates are in the clock's rest frame. We can start the origin (t=t'=0, x=x'=0) when the photon leaves the first end, and then calculate the times and positions in each frame for when the photon reflects off the second end and when it returns to the first end. According to my calculations:

The reflection is at:
(t, x) = (334 ns, 100 m) and (t', x') = (76.5 ns, 22.9 m)
which disagrees with your values of:
(t, x) = (30.0 ns, 100 m) and (t', x') = (13.2 ns, 44 m)

The return is at:
(t, x) = (351 ns, 94.7 m) and (t', x') = (153 ns, 0 m)
which disagrees with your values of:
(t, x) = (30.3 ns, ? m) and (t', x') = (13.3 ns, 0 m)

About the only thing that we agree on is that gamma is 2.29 for v = .9 c. You can see in my calculations that 153 ns gamma = 351 ns shows the correct time dilation and 22.9 m/gamma = 10 m shows the correct length contraction. In other words, the proper length of the clock is 22.9 m and it is shortened to 10 m in our frame. The proper "tick" of the clock is 153 ns which is dilated to 351 ns in our frame. Everything seems to work out correctly by my numbers.

I am particularly concerned that we do not agree about the values in our (unprimed) coordinates. Those values do not require any relativity or Lorentz transforms to calculate, so they should be straightforward. I did all my work in Mathematica, so I could go back and check:
(100 m)/(3E8 m/s) = 3.34E-7 s = 334 ns
I think you are off by a factor of about 10 for your c. Perhaps you misplaced a decimal?

-Dale

PS This kind of confusion is the reason I really like .6 c and 4 sc when you are just coming up with an example.

 

[MacM]

the muons do get to Earth surface,

Correct. NOW admit that this is a ONE WAY accumulation of time dilation, just as it is in particle accelerators. One way gamma functions can not be differentiated between "Relative Velocity" concepts and "Absolute Motion" (Energy) concepts.

PS - Please also note that the thought experiment of this thread has no "illusions." -

BULL! The moment you invoke SR you have invoked an illusion. This is not rhetoric it is fact. When you advance the concept of spatial contraction you have invoked an illusion. An illusion based on a false assertion.

Given 3E⁸ meter markers between a point in space and you on earth as measured by you and given a spacecraft moving along this path at 0.866c according to your earth measurement.

Your claim that the spacecraft only goes 1/2 half that distance (1.5E⁸ meters because d = vt based on proper times being equal and ignoring the dilated tick rate is only part of the problem.

The fact is simple ABSOLUTE physics shows that due to the dilated tick rate of the clock the pilot in motion would measure 5.18E⁸ m/s as he passed the meter markers. His calculated velocity would be 1.732c, not the same 0.866c.

There is no wiggle room for this error. Your pilot does not sense or measure the tick rate change. He cannot sense or measure any purported distance change. The simple fact is meter markers pass his window at the rate of 5.18E⁸ each second.

The fact is the craft goes 300,000,000 meters in 1.15473 seconds according to your clock, while going 259,800,000 m/s (0.866c).

But in the pilots frame 300,000,000 meter markers pass by in 0.577 seconds according to his clock. Hence his velocity is 1.732c, not 0.866c.

Distance did not and does not change. 0.577 seconds is correct at 1.732c to go 300,000,000 meters.

Hmmmm. Try again. Justify ignoring the meter rate as observed by the moving pilot.

Justify mixing frames using the rest frame velocity and the dilated tick rate clock in the moving frame. :bugeye:

Your spatial contraction is an illusion created by false stipulations. That is specifying an incorrect velocity for the moving pilot based on a frame other than his or his physics and his clock.

"Duck and Weave" your way out of these facts.

 

[MacM]

For the last fucking time:

It's an elementary error, and it's frankly embarrassing to watch you repeat the same mistake over and over.

Fuck you and the horse you rode in on.


http://itis.volta.alessandria.it/episteme/ep6/ep6-mull.htm

http://www.mathpages.com/home/kmath307/kmath307.htm

http://www.wbabin.net/babin/webdoc1.htm


Now smart ass. Suppose you justify the claim of spatial length contraction. You mix frames by stipulating the velocity in your frame for the moving frame when it is grade school arithmatic to show that is not the actual case.

Given 3E⁸ meter markers dispersed evenly between you on earth and a point in space and a spacecraft moving with a velocity of 0.866c (259,800,000 m/s as measured by you on earth) it is clear based on the gamma = 2.000 and the moving observers dilated clock, that he is measuring meter markers passing his window at the rate of 5.196E⁸ m/s.

That means he would calculate that he is going 1.732c, not 0.866c.!!

Where you would claim he took 1.154 seconds to traverse the distance, his clock would record only 0.577 seconds.

You want to claim that d = vt then means distance has been foreshortened but that is only based on your falsely stipulated velocity from your frame and ignores the physics of his frame.

The fact is the 0.577 seconds corresponds to traveling 300,000,000 meters at 1.732c.

Distance did not change. You don't have to like the > c figures but you do have to acknowledge that is the correct physics from the moving observers frame. Not the 0.866c figure you use to falsely create the illusion distance contracted.

Now stick your superority complex up your ass.

 

[2inquisitive]

DaleSpam, the speed of light is a defined value, 299,792,458 meters/second in a vacuum. A nanosecond is one billionth of a second. A photon will travel one meter in .299792458 nanosecond. You are correct, I screwed up. I multiplied 100 by the nanosecond value. That gave how far light would travel in 100 nanoseconds, instead of how long it takes to travel 100 meters. The shame, a grammer school error! I did the calculations as I typed in the quick reply block and didn't double check. Sorry.

 

[Pete]

Now once and for all demonstrate that two clocks brought together in a common frame after some period of relative velocity have each accumulated less time than the other.

Since that's not a prediction of special relativity, it's no surprise that nobody's tried to demonstrate it.

As we've discussed... over and over and over.

PS - Your three links are interesting, but is there a particular point you wanted to highlight from them?

 

[Pete]

Is this supposed to be supporting your contention that SR predicts that two clocks brought together in a common frame after some period of relative velocity have each accumulated less time than the other?
Oh... I see you're still stuck on the Muon experiment. Over and over and over again...

Given 3E8 meter markers dispersed evenly between you on earth and a point in space and a spacecraft moving with a velocity of 0.866c (259,800,000 m/s as measured by you on earth) it is clear based on the gamma = 2.000 and the moving observers dilated clock, that he is measuring meter markers passing his window at the rate of 5.196E8 m/s.

That means he would calculate that he is going 1.732c, not 0.866c.!!

Yes. If the traveler calculated velocity used Earth-frame meters and spaceship-frame clocks, he would indeed reach that conclusion. But is the calculation meaningful? If he prefers spaceship clocks to Earth clocks, then why would he prefer Earth meters to spaceship meters? Why would he mix Earth meters with spaceship clocks?

What velocity is he actually trying to measure? The ship's velocity in Earth's frame, or the markers' velocity in the ship's frame? Either way, he'll need to use clocks and rulers from the same frame.

If he calculates the velocity at which the markers pass using spaceship-frame meters and spaceship-frame clocks, he'll calculate that they pass at 0.866c, right?

 

[DaleSpam]

That gave how far light would travel in 100 nanoseconds, instead of how long it takes to travel 100 meters. ... Sorry.

No need to apologize. That explains why we were not off by exactly a factor of 10. Do you agree now that the Lorentz transform is at least self-consistent? (Something can be internally consistent but still wrong)

Let's use the .6 c example from now on, if you don't mind. I see no reason to write 9 digits when 3 will do

-Dale <-- lazy engineer

 

[MacM]

Is this supposed to be supporting your contention that SR predicts that two clocks brought together in a common frame after some period of relative velocity have each accumulated less time than the other?
Oh... I see you're still stuck on the Muon experiment. Over and over and over again...

I'll skip this since it simply isn't true.

Yes. If the traveler calculated velocity used Earth-frame meters and spaceship-frame clocks, he would indeed reach that conclusion. But is the calculation meaningful? If he prefers spaceship clocks to Earth clocks, then why would he prefer Earth meters to spaceship meters? Why would he mix Earth meters with spaceship clocks?

Surely you can do better than this.

I am not the one mixing frames SRT does. SRT has you specify a velocity based on one frame ignoring the physics of the frame in question.

What velocity is he actually trying to measure? The ship's velocity in Earth's frame, or the markers' velocity in the ship's frame? Either way, he'll need to use clocks and rulers from the same frame.

Correct. So why does SRT use the earth frame to assign a velocity to the moving frame when in the moving frame there exists a different velocity?

If he calculates the velocity at which the markers pass using spaceship-frame meters and spaceship-frame clocks, he'll calculate that they pass at 0.866c, right?

Incorrect. 3E⁸ meter markers being traversed in 0.577 seconds computes to 1.732c. Your assumption that the meters are no longer meter in length in that frame is nothing more than an assumption in advance of any calculation.

The actual calculation disputes that conclusion. In other words the only basis for length contraction is the deliberate stipulation of an assumed velocity in lieu of common sense and measurment in the moving frame.

 

[funkstar]

Fuck you and the horse you rode in on.

Would that be the dead horse you seem adamant in flogging?

http://itis.volta.alessandria.it/episteme/ep6/ep6-mull.htm

Philosophy, not physics.

http://www.mathpages.com/home/kmath307/kmath307.htm

Did you have a point with this? There's nothing there to support your view, you know...

http://www.wbabin.net/babin/webdoc1.htm

Philosophy, not physics.

Now smart ass. Suppose you justify the claim of spatial length contraction. You mix frames by stipulating the velocity in your frame for the moving frame when it is grade school arithmatic to show that is not the actual case.

First of all, I don't know what the sentence in red is supposed to mean. "Stipulating the velocity in my frame for the moving frame"? What's that supposed to mean? Yes, in my frame the moving frame has the velocity it has. There's an invariant, in that in the ship frame, my frame has the same speed. Is this what you're attacking? On what grounds?

Given 3E⁸ meter markers dispersed evenly between you on earth and a point in space and a spacecraft moving with a velocity of 0.866c (259,800,000 m/s as measured by you on earth) it is clear based on the gamma = 2.000 and the moving observers dilated clock, that he is measuring meter markers passing his window at the rate of 5.196E⁸ m/s.

That means he would calculate that he is going 1.732c, not 0.866c.!!

Not at all. He doesn't "calculate" that he is "going" at any speed at all. Because in his frame of reference, he isn't moving. It doesn't matter that he can see other things moving around him: In the coordinate system centered on him, which is an inertial frame, he isn't moving. He does, however, see markers passing at the rate of 5.196E8 pr. second, something he can simply measure. But he can also directly (and easily) measure the speed of the markers to be 0.866c, and is therefore able to measure the distance between the moving markers as half a meter.

Where you would claim he took 1.154 seconds to traverse the distance, his clock would record only 0.577 seconds.

Yes, that is correct.

You want to claim that d = vt then means distance has been foreshortened but that is only based on your falsely stipulated velocity from your frame and ignores the physics of his frame.

Absolutely not. Nowhere do the man on the ship use anything from my frame! The speed of the markers is measured directly in his frame (say, by timing them over a known distance such as the length of his ship). The length between them is measured directly from his frame (for instance, by measuring how long it takes for two markers to pass him, and using the speed he measured them to have.)

Absolutely nowhere does the man on the ship need to use any quantity from my frame to measure the speed of the markers nor their relative distance!

The fact is the 0.577 seconds corresponds to traveling 300,000,000 meters at 1.732c.

3E8 earth meters at 1.732 * 3E8 earth meters/ship second. Of course, this is definitinely not the same as saying that something travels at 1.732c, because the dimensions are wrong - speed has dimension meters/second. The meters and seconds has to be from the same frame. The conversion leads to the ship moving at 0.866c in the earth frame, and the markers moving at 0.866c in the ship frame.

Distance did not change. You don't have to like the > c figures but you do have to acknowledge that is the correct physics from the moving observers frame.

Absolutely not. In no way does the man on the ship measure the markers to be passing him with a speed of 1.732c! The only way he can see the 3E8 markers pass by in 0.577 seconds, with a speed of 0.866c, all of which are quantities measurable in his frame directly (he can be totally oblivous of the earth frame), is if there's half a meter between them (as he measures meters).

There's absolutely no reason at all for him to suddenly decide that his own meters aren't the real measure of distance.

Not the 0.866c figure you use to falsely create the illusion distance contracted.

So you are actually claiming that from the viewpoint of the ship the markers don't have a speed of 0.866c? What you're saying then, is that the relative velocity is not invariant between the two frames.

That's a truly extraordinary viewpoint! Do you have anything, anything at all to substantiate this claim, which goes against every bit of mechanical physics since Galileo?

Now stick your superority complex up your ass.

From a man who presumes himself a greater genius then virtually every physicist in the twentieth century (and apparently quite a few others), this is a somewhat comical statement.

In any case, I note that you did not, in fact, reply to my comment that you falsely and deliberately stated that relativity predicted something it patently does not. Now, will you acknowlegde that relativity theory does not predict that two clocks taken apart and brought together again will have to show less time than each other, or will you not?