Sorry guys, Lorentz's Contraction "Paradox" again!

There is a ship moving away from earth at velocity V1=0.6c relative to earth. In V1 reference frame, the ship accelerates at acceleration a=1E6 m/s² for an elapse time of <font face=symbol>D</font>t = 1 second and reaches velocity V2 relative to earth.

Velocity u gained in V1 reference frame is approximately 1E6*1=1E6 m/s and V2:

V2 = (0.6c+1E6)/(1+0.6c*1E6/c²) = 180,638,722 m/s (based on c = 3E8 m/s)

Gamma (<font face=symbol>g</font>) factor for V1 and V2 is respectively 1/0.8 and 1/0.7984

Along the direction of motion, there are (distance are as seen by earth observer):

1) Moon, 400,000 km away
2) Sun, 150,000,000 km away
3) Alpha Centaury, 4.3 light years away

What is the change of distance for those object as seen by the ship observer when its velocity increase from V1 to V2, aka change of Lorentz's contraction? The general formula for computing this is:

<font face=symbol>D</font>L = L (1/<font face=symbol>g</font>₁ - 1/<font face=symbol>g</font>₂), where L is the object distance as listed above. Since, only L is variable, we can write:

<font face=symbol>D</font>L = L (0.8 - 0.7984) = 0.0016*L

Having this simple equation, we can now easily compute <font face=symbol>D</font>L and the respective V_M defined as
<font face=symbol>D</font>L/<font face=symbol>D</font>t, which results:

1) With respect to Moon, <font face=symbol>D</font>L=640 km and V_M = 640 km/s
2) With respect to Sun, <font face=symbol>D</font>L=240,000 km and V_M = 240,000 km/s
3) With respect to Alpha Centaury, <font face=symbol>D</font>L=6.51E10 km and V_M = 6.51E10 km/s (over 2 millions times the speed of light in vacuum, or is it?)


It should be noted that the above V_M velocities are only last for one second, that is only exist during the acceleration period. This is one of the oddest properties of this "velocity". Clearly, this "velocity" is not normal velocity. It is also useless to compare this "velocity" with, for example, the speed of light. So, when its value exceed 300,000 km/s, it doesn't mean it is faster than light...just not that kind of velocity.

It should be also clear from the above that this velocity V_M is not well defined as it could be any value; for every different L we would get different V_M. Is there any paradox? Ahhh, I think this is too much!

Paul T,
Sorry guys, Lorentz's Contraction "Paradox" again!

There is a ship moving away from earth at velocity V1=0.6c relative to earth. In V1 reference frame, the ship accelerates at acceleration a=1E6 m/s2 for an elapse time of Dt = 1 second and reaches velocity V2 relative to earth.

Velocity u gained in V1 reference frame is approximately 1E6*1=1E6 m/s and V2:

V2 = (0.6c+1E6)/(1+0.6c*1E6/c2) = 180,638,722 m/s (based on c = 3E6 m/s)

Gamma (g) factor for V1 and V2 is respectively 1/0.8 and 1/0.7984

Along the direction of motion, there are (distance are as seen by earth observer):

1) Moon, 400,000 km away
2) Sun, 150,000,000 km away
3) Alpha Centaury, 4.3 light years away

What is the change of distance for those object as seen by the ship observer when its velocity increase from V1 to V2, aka change of Lorentz's contraction? The general formula for computing this is:

DL = L (1/g1 - 1/g2), where L is the object distance as listed above. Since, only L is variable, we can write:

DL = L (0.8 - 0.7984) = 0.0016*L

Having this simple equation, we can now easily compute DL and the respective VM defined as
DL/Dt, which results:

1) With respect to Moon, DL=640 km and VM = 640 km/s
2) With respect to Sun, DL=240,000 km and VM = 240,000 km/s
3) With respect to Alpha Centaury, DL=6.51E10 km and VM = 6.51E10 km/s (over 2 millions times the speed of light in vacuum, or is it?)
It should be noted that the above VM velocities are only last for one second, that is only exist during the acceleration period. This is one of the oddest properties of this "velocity". Clearly, this "velocity" is not normal velocity. It is also useless to compare this "velocity" with, for example, the speed of light. So, when its value exceed 300,000 km/s, it doesn't mean it is faster than light...just not that kind of velocity.

It should be also clear from the above that this velocity VM is not well defined as it could be any value; for every different L we would get different VM. Is there any paradox? Ahhh, I think this is too much!
Thank you. Finally your tone has become appropriate and hence I choose to continue this dialog.

(based on c = 3E6 m/s)

What is this?

Posted by MacM in original Lorentz Contraction Paradox (LCP) introduction post: "(for general example without computing the gamma function) would appear to have moved 10 B Ly in a matter of an hour.

Posted by MacM in LCP 5/30 @ 4:40 PM: "Therefore the spatial contraction must according to that collapse many times the speed of light as I approach v = c. As the spatial dimension contracts I am getting closer, hence the faster I fly away the faster I would appear to be getting closer.

Posted by MacM in LCP 6/1 @ 8:28 PM: "Now I already know you will scream "Time Dilation". Go ahead, at V2 time dilation is such that 1 sec is actually 7,071 seconds earth time. So while I shouldn't I'll give you time dilation and still claim that if spatial contraction in Relativity is as you claim then in one second (or 7,071 seconds) moving AWAY from earth I would see earth move CLOSER by 967,009 Lyr!!!"

AND

"Note this is Light YEARS per Second. "

I could post many more examples of my statements but I think this should suffice. Now:

Posted by Paul T in LCP 6/4 @ 4:34 PM: "You watch too much cartoon, MacM. ............................."

AND "Now, give me a good reason why addition of velocities is not applicable in this situation!"

Posted by Paul T in LCP 6/6 @ 10:51 AM: " 2). You claimed that addition of velocities was wrongly applied for a rocket moving at velocity V1 and then accelerate to V2 (both V1 and V2 are velocities relative to earth). You are narrow minded in this case.

Posted in "MacM's Exercise"6/7 @ 7:27 AM: " Posted by MacM to James R:“ Can you comment yet on the velocity addition situation? ”

Posted by James R in response: "I don't see how it is relevant. I will wait for Paul T to explain, if he wishes."

Posted by Paul T from above: "Velocity u gained in V1 reference frame is approximately 1E6*1=1E6 m/s and V2:

First I'll note that you have now tried to qualify your work by adding the "approximately" term. Because you now realize you are attempting to perform this calculation in an accelerating frame.

Posted by Paul T above: "V2 = (0.6c+1E6)/(1+0.6c*1E6/c2) = 180,638,722 m/s (based on c = 3E6 m/s)"

I'll note that you are calculating V2, where in my example I approximated an acceleration and specified V2. This is not a complaint, just a note about the changing of criteria of the example. I also note the continued value of c = 3E6 m/s. It is repeated throughout so I don't see that as a typo.

Posted by Paul T above: "3) With respect to Alpha Centaury, DL=6.51E10 km and VM = 6.51E10 km/s (over 2 millions times the speed of light in vacuum, or is it?)

It should be noted that the above V_M velocities are only last for one second, that is only exist during the acceleration period. This is one of the oddest properties of this "velocity". Clearly, this "velocity" is not normal velocity. It is also useless to compare this "velocity" with, for example, the speed of light. So, when its value exceed 300,000 km/s, it doesn't mean it is faster than light...just not that kind of velocity.

It should be also clear from the above that this velocity V_M is not well defined as it could be any value; for every different L we would get different V_M. Is there any paradox? Ahhh, I think this is too much!"

Hmmmmmm.

I find this whole thing a bit curious and amusing.

You start your own thread of the same topic and lead off with "Sorry", like you are proving something important that contridicts conclusions of my original thread which you opposed.

Even altered slightly, at your option and perogative, you ultimately come to the same conclusion. But you note that the affect is only for the duration of the acceleration. Do you not think that it would only exist during the closure (acceleration) of my example also. The ultimate distance becomes fixed once V2 in my example was achieved but during the acceleration is where the FTL observation and paradox occurs. Good job - Thanks.

But because you proved it to yourself and the rest of us after all, you now want to phoo-phoo the result and say lets ignore it, it can't be real and it only last 1 second.

Do you not realize your 2 million c velocity for one second could also be > c for just under 2 million seconds? 555 hours or 23 + days!

As you like to say Paul T, "You are a funny man".

MacM,


I'll note that you are calculating V2, where in my example I approximated an acceleration and specified V2. This is not a complaint, just a note about the changing of criteria of the example. I also note the continued value of c = 3E6 m/s. It is repeated throughout so I don't see that as a typo.


It was just a typo, Mr MacM. It didn't affect the whole calculation.

So Paul T
Do you feel that you have found a paradox or just a mathematical anomally?

Paul T,

MacM,
It was just a typo, Mr MacM. It didn't affect the whole calculation.

I'll accept your claim but please try to remember these threads and your own error, typos and such, in the future before calling others ignorant or worse.

Have a nice day.

QQ,

So Paul T
Do you feel that you have found a paradox or just a mathematical anomally?

He found??? Oh the burden of fleeting fame. :D

MacM,


I'll accept your claim but please try to remember these threads and your own error, typos and such, in the future before calling others ignorant or worse.

Have a nice day.

This error was made by me and I admit it. Have you admitted any of your errors that I pointed out. I have seen none. Rather than admitting your errors, you went on covering them up. I remember when I was a kid, I tried one or two times to put the dead rat into plastic bag, sealed it tightly so that I could throw them into the rubbish bin...but, it still smell just the same as your errors. Should I remind you what are your errors that you still have not found as errors (at least, that what's you tried to convince us)? I don't know about you, but for me, I would respect you more if you did admit your error, corrected them and move on to the nest topic.

To agree with something some one else has found is to find that which others have found as founded. :D

In other words to truely learn we have to find it ( again ) for ourselves.

Quantum Quack,

So Paul T
Do you feel that you have found a paradox or just a mathematical anomally?

No, I did not found paradox. MacM did, hehe. The velocity associated to Lorentz's Contraction was named MacM's Velocity, denoted by V_M. But, of course, MacM's Velocity is a mere product of hallucination.

Velocity (the real one) that we know goes something like this. To determine velocity, say from a moving rocket relative to a star, the observer first measure the distance to that star and after an elapse time of <font face=symbol>D</font>t, he measure again and obtain <font face=symbol>D</font>s. He then determine the relative velocity of his rocket to the star using v=<font face=symbol>D</font>s/<font face=symbol>D</font>t. Assuming that the velocity is constant. If another observer at the star measures the velocity of the rocket, she would obtain v the same (numerically) as that ontained by the rocket observer. The real velocity behaves this way.

MacM's Velocity behaves very differently. As I have shown before, it has no specific value (ir can be any value) and it last during the rocket acceleration only. MacM's Velocity is therefore not a real velocity, it cannot be compared to, say the speed of light.

MacM thought there is a paradox because he found from his exercise that the rocket accelerating away from earth could observe the earth-rocket distance to decrease instead of increase. I don't see this as paradox, as the the rocket is in the non-inertial reference frame.

So,
Paul T
Can you tell me what your thread starter was all about?
Was it a demonstration of your own approach or a demonstration of someone elses approach?

V=2000000*'c' striked me as strange

So,
Paul T
Can you tell me what your thread starter was all about?
Was it a demonstration of your own approach or a demonstration of someone elses approach?


I would say, my approach and someone else problem.


V=2000000*'c' striked me as strange

It strange to you, it's unbelievable to me. The "velocity" is not of the kind that we know for the speed of light and therefore is not comparable to normal velocity. I presented the computation to show that MacM's Velocity has no specific value.

That's all I have to say.

Paul T,

This error was made by me and I admit it.

Fantastic. We probably won't become best of buddies but I think we might be able to have a mutually respectful communication in the future. I see your mathematical skills exceed my own but that doesn't make me enept of the processes. Like you said to another poster. "I can bite" :D

Have you admitted any of your errors that I pointed out. I have seen none.

Actually I have but not in a point by point case. Se my post to James R where I admitted to some errors, etc.

Rather than admitting your errors, you went on covering them up. I remember when I was a kid, I tried one or two times to put the dead rat into plastic bag, sealed it tightly so that I could throw them into the rubbish bin...but, it still smell just the same as your errors. Should I remind you what are your errors that you still have not found as errors (at least, that what's you tried to convince us)? I don't know about you, but for me, I would respect you more if you did admit your error, corrected them and move on to the nest topic.

You got it. If you would bother to list what you think are errors I will be pleased to post a point by point reply. If they are errors I will gladly admit it. I have in the past here (SciF) in the past. However, be prepared for me to also qualify certain posts.

For example I was aware that computing the acceleration at relavistic velocities was going to be beyond me, so I qualified that post and gave a general (approximation) answer and said (1 second or 7,071 seconds). So while mathematically my number may not be correct I knew it wasn't but also commented that in any case it didn't alter the ultimate claim. But list it and I will admit it is numerically incorrect.

Also I can do the algebra that you have done. I am good at algebra. I don't do calculus but understand its principles. My problem would be not knowing if the correct algebra is being used in some cases.

PaulT
Please excuse my confusion and I know you said you wanted to say no more on the subject but from my perspective I woudl like to ask:

You state that it is a velocity that can not be compared to normal velocity?

If it can't be a velocity comparable to normal velocity what sort of velocity can it be compared to?

Obviously the figure is so extreme that it brings up the question of validity.

But are we just excersising a convenience by stating that
The "velocity" is not of the kind that we know for the speed of light and therefore is not comparable to normal velocity

Could it be the theoretical speed limit may impose a restriction on having this sort of result.

I am sure that the result is well formed but unable to be explained using current thinking about the nature of light.
Am I right in feeling that your extreme figure is correctly arrived at using the matematical methods that are currently accepted as valid?

Paul T,

Posted by Paul T to Quantum Quack: "It strange to you, it's unbelievable to me."

Maybe we did learn something here. :D

Just wait for my next topic. You will love it too I'm sure.


PS: I am not adding this closing statement to piss you off but as a statement of fact.

MacM's velcocity is Relativity's velocity and it MUST be just as real as Relativity's purported contraction of space. Perhaps in time you will learn why I hold some of the views I do. I think different than most of you. Part of my training I guess. :o

You state that it is a velocity that can not be compared to normal velocity?

If it can't be a velocity comparable to normal velocity what sort of velocity can it be compared to?


It was just not properly defined as "velocity" and therefore is not velocity at all. See below.


Obviously the figure is so extreme that it brings up the question of validity.

But are we just excersising a convenience by stating that


It's invalid not because it gives "velocity" too extreme, but because it was not correctly defined.


I am sure that the result is well formed but unable to be explained using current thinking about the nature of light.
Am I right in feeling that your extreme figure is correctly arrived at using the matematical methods that are currently accepted as valid?

Strickly speaking, no. There is no problem in getting velocity (an instant velocity; don't remember if this term correct) for an accelerating object or from an accelerating reference frame. What we need is, take <font face=symbol>D</font>t as small as possible (ussually denoted as dt), so that velocity during that infinetesimal time can be considered as constant. This was not done in the first post under this thread and obviously something MacM did not consider. By not taking infinitesimal time, we are unable to determine the so-called instant velocity and also differential Lorentz's contraction (which can not be considered to exist if we use dt instead) enter into the picture.

Paul T,

Strickly speaking, no. There is no problem in getting velocity (an instant velocity; don't remember if this term correct) for an accelerating object or from an accelerating reference frame. What we need is, take Dt as small as possible (ussually denoted as dt), so that velocity during that infinetesimal time can be considered as constant.

This was not done in the first post under this thread and obviously something MacM did not consider. By not taking infinitesimal time, we are unable to determine the so-called instant velocity and also differential Lorentz's contraction (which can not be considered to exist if we use dt instead) enter into the picture.

Two points.

1 - Only to clarify: I did not use the delta t process but I am aware of it. Rather than have you suggest otherwise, if you are not aware, then I'll inform you that James R., and I have had that discussion on this forum before in dealing with rockets under acceleration, etc.

2 - Have you given any thought to re-doing your presentation using the delta t process?. I think it would be informative to the readers.

It does seem to me that the distance stipulated for V1 and the relavistically adjusted distance for V2 are terminal velocities. That is they are constant velocities and hence the distance change during the interval of acceleration time still leaves you with the observed paradox. Using dt can only change how one gets from point A to point B, not the final point otherwise the relavistic contraction formula would be invalid.

MacM,


Two points.

1 - Only to clarify: I did not use the delta t process but I am aware of it. Rather than have you suggest otherwise, if you are not aware, then I'll inform you that James R., and I have had that discussion on this forum before in dealing with rockets under acceleration, etc.

2 - Have you given any thought to re-doing your presentation using the delta t process?. I think it would be informative to the readers.


No. Not necessary. If you use dt instead, you will get instant velocity starting from V1 to V2 and you won't see differential Lorentz's contraction (because of the infinitesimal dt). When you measure length/distance when the rocket at velocity V1 and then at V2 (while V1 and V2 are substantially difference), you would of course get differential length attributed to Lorentz's contraction. However, that would mean you are deducting length measured from two different inertial reference frame (you can't do this). When you use dt instead, you assume that your system is still under an inertial reference frame.

O, I just want to remind you that you were wrong about SR cannot handle acceleration; with some trick, it can. Check Baez's site about relativistic rocket, it was derived using SR and there is acceleration.


It does seem to me that the distance stipulated for V1 and the relavistically adjusted distance for V2 are terminal velocities. That is they are constant velocities and hence the distance change during the interval of acceleration time still leaves you with the observed paradox. Using dt can only change how one gets from point A to point B, not the final point otherwise the relavistic contraction formula would be invalid.

You can consider velocity at V1 and V2 constant, but they are two different constant velocities. You can't mix (length) measurement obtained for V1 and V2 reference frames and use them to compute some sort of "velocity".

Paul T,

MacM,
No. Not necessary. If you use dt instead, you will get instant velocity starting from V1 to V2 and you won't see differential Lorentz's contraction (because of the infinitesimal dt). When you measure length/distance when the rocket at velocity V1 and then at V2 (while V1 and V2 are substantially difference), you would of course get differential length attributed to Lorentz's contraction. However, that would mean you are deducting length measured from two different inertial reference frame (you can't do this). When you use dt instead, you assume that your system is still under an inertial reference frame.

My question is what is the ultimate difference in the end when you have two velocities, over a given time, with two distances linked by relativity to those velocities. Your process is different but the physical reality is the same.

I understand you are trying to set up a condition where you can claim ds/dt is not a velocity. I'm looking for what you want to replace it with?

O, I just want to remind you that you were wrong about SR cannot handle acceleration; with some trick, it can. Check Baez's site about relativistic rocket, it was derived using SR and there is acceleration.

Not necessary. I looked back and indeed I saw where I said something to the affect that you can't compute that in an accelerating frame. So I'll accept your statement but want to qualify that I actually meant the statement to apply to the simple application of the velocity addition formula in the accelerting frame. Before you object I ask you look at the many other comments I made on that issue. They were directed to your direct use of the VAF. I did not mean it was impossible to calculate via SR. Indeed as I have said the dt process in accelerating frames has been discussed here before.

You can consider velocity at V1 and V2 constant, but they are two different constant velocities. You can't mix (length) measurement obtained for V1 and V2 reference frames and use them to compute some sort of "velocity".

Why not? Because it gives results that you don't want? v = ds/dt. I dare say that if you were in the captains chair you would sound the abandon ship alarm because of the velocity you saw the earth coming at you!

Paul T,

Velocity (the real one) that we know goes something like this. To determine velocity, say from a moving rocket relative to a star, the observer first measure the distance to that star and after an elapse time of Dt, he measure again and obtain Ds. He then determine the relative velocity of his rocket to the star using v=Ds/Dt. Assuming that the velocity is constant. If another observer at the star measures the velocity of the rocket, she would obtain v the same (numerically) as that ontained by the rocket observer. The real velocity behaves this way.

I also meant to comment on this. Rather than me put my foot in my mouth by making the statement I would like to make I'll pose it in the form of a question.

Can you explain to us how "she" would see the ds/dt measurements any differently than the rocket pilot? If not then does not the behavior mimic your "real" velocity in every respect?

Your arguements seem to require that things vanish or cease to exist during acceleration since velocity can't be observed as ds/dt. I suggest that I very much can see a Mac truck when it is accelerating my way.

MacM,


My question is what is the ultimate difference in the end when you have two velocities, over a given time, with two distances linked by relativity to those velocities. Your process is different but the physical reality is the same.

I understand you are trying to set up a condition where you can claim ds/dt is not a velocity. I'm looking for what you want to replace it with?


I did not suggest that you need to replace v=ds/dt, which is the correct representation for velocity, where ds and dt are infinitesimal displacement and time.


You can consider velocity at V1 and V2 constant, but they are two different constant velocities. You can't mix (length) measurement obtained for V1 and V2 reference frames and use them to compute some sort of "velocity".



Why not? Because it gives results that you don't want? v = ds/dt. I dare say that if you were in the captains chair you would sound the abandon ship alarm because of the velocity you saw the earth coming at you!

Differential Lorentz's contraction only exist if you have two velocities, V1 and V2. If you able to measure V1 and V2, your ds is no longer infinitesimal as for infinitesimal ds, you will get only one velocity, not two. In English, you should take ds (and dt) limit to zero or very very close to zero.

MacM,


Velocity (the real one) that we know goes something like this. To determine velocity, say from a moving rocket relative to a star, the observer first measure the distance to that star and after an elapse time of Dt, he measure again and obtain Ds. He then determine the relative velocity of his rocket to the star using v=Ds/Dt. Assuming that the velocity is constant. If another observer at the star measures the velocity of the rocket, she would obtain v the same (numerically) as that ontained by the rocket observer. The real velocity behaves this way.



Can you explain to us how "she" would see the ds/dt measurements any differently than the rocket pilot?


She would obtain ds and dt numerically different then those obtained by rocket observer, but ds/dt would be the same for both.


If not then does not the behavior mimic your "real" velocity in every respect?


Please explain.


Your arguements seem to require that things vanish or cease to exist during acceleration since velocity can't be observed as ds/dt. I suggest that I very much can see a Mac truck when it is accelerating my way.

I do not understand. Please explain. Velocity can be observed as ds/dt even in accelerated reference frame. For case not involving acceleration, you have liberty to pick ds and dt not infinitesimal (become <font face=symbol>D</font>s and <font face=symbol>D</font>t), but for system involving acceleration, you have no such choice.

Paul T,

I did not suggest that you need to replace v=ds/dt, which is the correct representation for velocity, where ds and dt are infinitesimal displacement and time.

I understand that but it appears to me when you resolve ds/dt toward "zero" you don't destroy the claim of contraction. You merely now have a specific instantaneous velocity, which according to Relativity has associated with it a specific spatial dimension. Each increment therefor of ds/dt will have an incremental changing dimension. You are merely quantisizing the process. Both processes must ultimately yield the same result at the terminal velocities. You can destroy the calculation but not the reality.

Differential Lorentz's contraction only exist if you have two velocities, V1 and V2. If you able to measure V1 and V2, your ds is no longer infinitesimal as for infinitesimal ds, you will get only one velocity, not two. In English, you should take ds (and dt) limit to zero or very very close to zero.

I understand that process but I do not see differential Lorentz contraction as an issue. As said above while you are dealing with one velocity in this process and hence ignore any contraction calculation, the fact remains that at any and every relative velocity along the path from V1 to V2 you have a fixed spatial dimension which is changing.

It seems the process is more of a dodge of the issue than it is a resolution.

Paul T,

She would obtain ds and dt numerically different then those obtained by rocket observer, but ds/dt would be the same for both.

We agree but what I fail to see is:

1 - How getting the same result i.e. - velocity in any way alters the underlying correlation between velocity and spatial dimension. I understand you claim the process makes differential contraction equal to zero and yet we know the spatial contraction is linked to incremental velocity change.

The problem seems to be using a process where you can arbitrarily claim one function and deny the other. That is if ds/dt were actually 0/0 what do you think you have?

Not being actually 0/0 you want to claim an instantaneous velocity but want to deny any contraction. Not actually being "Actually 0/0" means you still have differential velocity and hence contraction.

Do you follow my point? 0d/0t is incalcuable and meaningless, just as you claim for MacM's velocity Vm.

Please explain.

I have tried, don't know if you get the signifigance or not.

I do not understand. Please explain. Velocity can be observed as ds/dt even in accelerated reference frame. For case not involving acceleration, you have liberty to pick ds and dt not infinitesimal (become Ds and Dt), but for system involving acceleration, you have no such choice.

I question the merits of the claim of "no choice" for one observer and having a choice for the other. That is as my point has been from the start either both observers see spatial contraction or neither see spatial contraction.

Otherwise velocity is not truely relative.

My personal belief is there is no spatial contraction per Relativity and since there is no empirical evidence or data to support the view that there is, I say it is one of a number of such issues that distorts our reality in current physics.

Paul T,

I have been thinking about this issue and how to present it for discussion vs making a specific claim which only induces arguement.

Where 's' is distance as in linear formulations:

Your use of ds/dt = 0 seems to me to be nothing more than a cop out. You create a situation where you express only V = x and declare there is no dt hence no ds and contraction is not at issue.

The fact is ds/dt does no such thing. When you state ds/dt = 0 all you are saying is V' = Vo + at/2, where t has been set to t = 0.

It seems equally valid (when 'a' is linear) to using the conventional V' = Vo + at/2 for any 't' during the acceleration. In such case you now also have a valid V', a valid 't' and a valid ds from which you can get to points of valid V's and t's to compute contraction which will equal the same value regardless of where along the 't' string you take your data.

That is I can compute individual lorentz values for any given valid V' at any selected 't' and simply subtract one from the other to see the change in ds/dt or I can calculate it using individual V1 and V2 data points and get the same result.

What is the purpose of not doing so? Other than trying to avoid the results which you prefer to deny.

MacM,

In general, your three above posts make things all fall back to square. I thought you started to see the point, but when you say ds/dt is 0/0 or ds/dt=0 or "instantenuos velocity" (hope it is the same as what I have in mind), you seem to really missed the point.

ds/dt is not 0/0. ds and dt are limit zero but NOT ZERO! Limit zero and zero are totally difference.

I never say ds/dt=0, if it does it mean the velocity is zero. We didn't talk about zero velocity.

You don't seem to understand why v must be ds/dt in general (including for system with acceleration) and <font face=symbol>D</font>s/<font face=symbol>D</font>t (which you adopted in your exercise) is only applicable for constant velocity. Velocity (particularly the one involving acceleration) must be strickly determined for time interval as small as possible so that assumption that within that time interval velocity remain constant hold true. To understand this you need to have some knowledge about limit concept or calculus.

When you use dt very very very close to zero (but not zero), you have no differential Lorentz's contraction to compute. During that small time interval you only have one velocity (that is v) that I called it instant velocity, which I mean velocity at one point (mathematically, two very very close points).

Paul T,

ds/dt is not 0/0. ds and dt are limit zero but NOT ZERO! Limit zero and zero are totally difference.

Actually I didn't miss it at all. In fact you missed the point where I said if you only approach ds/dt = 0 that you still have both ds and dt and that contraction can be computed.

I think I see your arguement however, in that as the value of ds/dt changes due to the non-linear response of contraction to dv that you get varible answers as a function the magnitude of ds/dt that you use.

The bottom line however would still seem to be the fact that at V1 you have a claimed spatial contraction and that at V2 that contraction has increased. These are absolute numbers which exist at the V's and going from V1 to V2 in a given period still means you have a change in distance over time which computes as FTL.

I personally do not accept spatil contraction as occurring since as has been made clear that all observers in a relative velocity situation cannot see the same affect. I believe it is a gross error to suggest it occurs, ecspecially since there is no empericl evidence that it does.

The example given by James R., was a rocket between earth and Alpha C. Earth and Alpha C. clearly cannot see their spatial seperation change because the rocket is in motion relative to them but HE & YOU both still claimed in the presentation I made which was simply a rocket and the earth that the rocket sees the distance between earth and the rocket as changing but that earth did not.

That is unacceptable and not advocated by Relativity at all.

What is consistant is 'A', 'B' & 'C' where 'B' is in motion between 'A' & 'C':

A & B see contraction between them vice-versa. B & C see contraction between themselves and vice-versa but A & C do not see contraction between themselves.

MacM,


Actually I didn't miss it at all. In fact you missed the point where I said if you only approach ds/dt = 0 that you still have both ds and dt and that contraction can be computed.


Certainly I missed your point. Indeed, I don't know what do you mean by approaching ds/dt=0? It must mean, zero velocity, no relative motion. If there is no relative motion, why do we care about relativistic issue such as Lorentz's contraction? Can you explain your argument, please?


I think I see your arguement however, in that as the value of ds/dt changes due to the non-linear response of contraction to dv that you get varible answers as a function the magnitude of ds/dt that you use.

The bottom line however would still seem to be the fact that at V1 you have a claimed spatial contraction and that at V2 that contraction has increased. These are absolute numbers which exist at the V's and going from V1 to V2 in a given period still means you have a change in distance over time which computes as FTL.


Let me summarize the following:

I hope you understand my example in the first post under this thread. I have shown that your MacM's Velocity as defined (incorrectly) by you has no specific value (as you could derive any value of velocity, depending on to what distance you compute the Lorentz's contraction) and therefore it is not a kind of velocity that we know.

Further, I stated that velocity is defined as ds/dt, which is applicable for either constant velocity motion and also accelerated motion (as in your "Lorentz's contraction paradox" exercise). For accelerated motion, ds and dt must be limit zero in order to avoid the acceleration effect to the velocity determination.

You determine your MacM's velocity like this. Take for example, James' example about a rocket moving at v=0.6c toward Alpha Centaury 4 Ly away from earth. According to rocket observer, the distance is 0.8 x 4 Ly = 3.2 Ly. You take note this distance. Then the rocket accelerate to v=0.8c, assuming it doesn't move forward very much so that we don't have to take into account the displacement during the acceleration. Now, the distance to AC is 0.6 x 4 Ly = 2.4 Ly. You take the balance, 3.2 Ly - 2.4 Ly = 0.8 Ly. If the rocket achieves velocity 0.8c from 0.6c in 1 hour (rocket time, well... in between 0.6c and 0.8 c reference frames), your MacM's velocity is 0.8 Ly/3600 second, of course a very large value in km/s. And you claimed, this is FTL. It is not FTL. In fact, the "velocity" is not velocity at all because you compute it based on information gathered from two difference inertial reference frames. The key word is INERTIAL REFERENCE FRAME. You can't compute velocity bridging two different reference frames. STOP, don't argue that since your are in the same rocket you must be in the same reference frame; yes if you consider you are in a non-inertial reference frame, but if you want to evaluate the problem using SR, you have to break the non-inertial reference frame into inertial reference frames. During the acceleration from v=0.6c to v=0.8c, your reference frame is continuously changing. At v=0.6c you are in one inertial reference frame and at v=0.8c you are in another.


I personally do not accept spatil contraction as occurring since as has been made clear that all observers in a relative velocity situation cannot see the same affect. I believe it is a gross error to suggest it occurs, ecspecially since there is no empericl evidence that it does.


Your personal view is incorrect. SR cannot survive without Lorentz's contraction. If Lorentz's contraction doesn't exist, neither time dilation as both are two faces of the same coin. Don't you know about experiment done by flying an atomic clock in commercial jet? Sure you do, don't you? That was one of large scale test on time dilation that also, indirectly, proving Lorentz's contraction.

If you don't understand my point, let me explain. Back to James' example. If the rocket observer measure the distance to AC as 4 Ly, the same as earth observer, but its clock move slower, this rocket observer must say that the rocket move as velocity 1.25 x 0.6c = 0.75c. SR falls apart, if this is true. Or, you want to say, clock in that rocket would run as the same rate as that on earth? How do you going to explain many many experiments verifying time dilation? Please, those experiments verifying SR quantitatively, not qualitatively.


The example given by James R., was a rocket between earth and Alpha C. Earth and Alpha C. clearly cannot see their spatial seperation change because the rocket is in motion relative to them but then still claims that the rocket sees the distance between earth and Aljpha C. as changing.

That is unacceptable and not advocated by Relativity at all.

What do you mean? Such result against relativity? What relativity? Certainly not SR.

MacM, please explain why do you think earth and rocket observers in James R's example must see the earth-Alpha Centaury distance as the same numerically? You don't have to say, SR says so since SR doesn't say so. Use your own argument.

Paul T,

Note: I posted some corrections and clarifications to my last post while you were posting this.

Certainly I missed your point. Indeed, I don't know what do you mean by approaching ds/dt=0? It must mean, zero velocity, no relative motion. If there is no relative motion, why do we care about relativistic issue such as Lorentz's contraction? Can you explain your argument, please?

Approaching means approaching i.e. close but not = zero. In which case you still have both ds/dt and contraction could be computed for that increment. But I also made the point in agreement with you that it can give a variable answer as to what dt and hence ds you use. I don't particularily care about computing values along the line of acceleration. My point is that you would see earth getting closer even though you had your rocket pointing away from earth and pouring on the coal (assuming that the result didn't become FTL as I have stated).

Let me summarize the following:

I hope you understand my example in the first post under this thread. I have shown that your MacM's Velocity as defined (incorrectly) by you has no specific value (as you could derive any value of velocity, depending on to what distance you compute the Lorentz's contraction) and therefore it is not a kind of velocity that we know.[quote]

I had just said that in the paragraph above this post.

[quote]Further, I stated that velocity is defined as ds/dt, which is applicable for either constant velocity motion and also accelerated motion (as in your "Lorentz's contraction paradox" exercise). For accelerated motion, ds and dt must be limit zero in order to avoid the acceleration effect to the velocity determination.

In which case you never increase speed since dt = 0 at any acceleration yields ds = 0.

You determine your MacM's velocity like this. Take for example, James' example about a rocket moving at v=0.6c toward Alpha Centaury 4 Ly away from earth. According to rocket observer, the distance is 0.8 x 4 Ly = 3.2 Ly. You take note this distance. Then the rocket accelerate to v=0.8c, assuming it doesn't move forward very much so that we don't have to take into account the displacement during the acceleration. Now, the distance to AC is 0.6 x 4 Ly = 2.4 Ly. You take the balance, 3.2 Ly - 2.4 Ly = 0.8 Ly. If the rocket achieves velocity 0.8c from 0.6c in 1 hour (rocket time, well... in between 0.6c and 0.8 c reference frames), your MacM's velocity is 0.8 Ly/3600 second, of course a very large value in km/s. And you claimed, this is FTL. It is not FTL. In fact, the "velocity" is not velocity at all because you compute it based on information gathered from two difference inertial reference frames. The key word is INERTIAL REFERENCE FRAME. You can't compute velocity bridging two different reference frames. STOP, don't argue that since your are in the same rocket you must be in the same reference frame; yes if you consider you are in a non-inertial reference frame, but if you want to evaluate the problem using SR, you have to break the non-inertial reference frame into inertial reference frames. During the acceleration from v=0.6c to v=0.8c, your reference frame is continuously changing. At v=0.6c you are in one inertial reference frame and at v=0.8c you are in another.

I have no arguement with that. My arguement is that using SR and your logic you seem to suggest that the pilot no longer sees the universe changing around him (assuming such change is subluminal) but the reality is that if you continue to see objects in space during acceleration (which you do) then if your rocket is pointed away from earth pouring on the coal, you will still see earth getting closer. Yes or No? The FTL issue is argumentative I agree but I think it is short sighted to say it isn't FTL when the distance change is purported to be real and it is achieved in a specified time frame.

You may not have mathematics to clearly define it as standard velocity but it is "observed" as velocity none the less. That is observing the change in distance over time. Yes or No?

Your personal view is incorrect. SR cannot survive without Lorentz's contraction. If Lorentz's contraction doesn't exist, neither time dilation as both are two faces of the same coin.

Survival of SR is neither my goal nor is to overturn it. You should be careful stating a view is wrong when there is no emperical evidence that your view is valid. It is inappropriate to abdocate common sense however in support of SR. If its application creates unacceptable results then it should be shit canned. Or as a minimum clarified and limited.

Don't you know about experiment done by flying an atomic clock in commercial jet? Sure you do, don't you? That was one of large scale test on time dilation that also, indirectly, proving Lorentz's contraction.

Of course I do. I also have posted a number of times the following:

1 - Clocks to not measure time. They measure a process. If I use a pot with marking up the side and evaporation rate of the water as a clock, I have a clock. Not a good one granted but I have a clock. The reason it is not a good clock is that certain things such as humidity, temperature and air currents will alter the process of evaporation and hence my measure of time.

If I light a fire under my clock I could see time go by rather fastly.

Point being that motion may infact alter the clocks process but that has nothing to do with time dilation or even proves time exists for that matter.

2 - In the H&K clock experiment detailed counter studies found that the variation between the clocks and the number of other issues which could alter the function of the clocks and the clock rates before they were even transported, etc had such wide variation (noise to signal ratio) that such data would have been virtually impossible to get.

They also found on investigation that the results were actually published after massaging data on a statistical basis. These were not clean and clear results. That doesn't make them wrong but it taints them to a point that you should never attach the term "proof" anywhere near them. There have infact been alegations that they published outright fraud. I don't know either way but I do know that the published results are in serious question.

If you don't understand my point, let me explain. Back to James' example. If the rocket observer measure the distance to AC as 4 Ly, the same as earth observer, but its clock move slower, this rocket observer must say that the rocket move as velocity 0.8 x 0.6c = 0.48c. SR falls apart, if this is true. Or, you want to say, clock in that rocket would run as the same rate as that on earth? How do you going to explain many many experiments verifying time dilation? Please, those experiments verifying SR quantitatively, not qualitatively.

Asked and anwered above. Muon life time is also a process.

What do you mean? Such result against relativity? What relativity? Certainly not SR.

Yes SR. If you still agree with James that in the case "I" presented which was only two observers, the rocket and the earth, that one sees contraction and the other doesn't, then SR is dead in the water. See my corrections and clarification in the above post for this paragraph.

MacM, please explain why do you think earth and rocket observers in James R's example must see the earth-Alpha Centaury distance as the same numerically? You don't have to say, SR says so since SR doesn't say so. Use your own argument.

I don't and haven't. My arguement has been as stated above and to which You and James said was not true and that is that the rocket and earth both see the contraction for the relative velocity between them. My case had nothing to do with Alpha C. as a third observer and the distance between earth and A.C. Crisp seems to share my view on this.

You in fact placed a string from earth into space and pointed out that you would not see it change dimension as your arguement. I would agree. But that has nothing to do with the issue. The string (like Alpha C., is not in motion to earth but the rocket is.

So what must happen (if it happens at all) is that earth would see the rocket at a different position along the string than the rocket would. Both must see the spactial contraction between them or neither do.

MacM,


Further, I stated that velocity is defined as ds/dt, which is applicable for either constant velocity motion and also accelerated motion (as in your "Lorentz's contraction paradox" exercise). For accelerated motion, ds and dt must be limit zero in order to avoid the acceleration effect to the velocity determination.


In which case you never increase speed since dt = 0 at any acceleration yields ds = 0.


No, MacM. Given v=0.6c, for example, as the velocity of your rocket relative to earth, you measure the difference of displacement ds for an interval of dt very very small but NOT ZERO. The ratio of ds and dt should give you v=0.6c for that particular time. If your rocket accelerates and you make another measurement, you should obtain another set of ds and dt that should give you v larger than 0.6c. Your mistake is, you take dt as zero while infinitesimal dt is not equal to dt=0!


My arguement is that using SR and your logic you seem to suggest that the pilot no longer sees the universe changing around him (assuming such change is subluminal) but the reality is that if you continue to see objects in space during acceleration (which you do) then if your rocket is pointed away from earth pouring on the coal, you will still see earth getting closer. Yes or No? The FTL issue is argumentative I agree but I think it is short sighted to say it isn't FTL when the distance change is purported to be real and it is achieved in a specified time frame.


I have never said that you would not see earth as getting closer. I thought that was very clear. Your definition of velocity just simply incorrect.


You may not have mathematics to clearly define it as standard velocity but it is "observed" as velocity none the less. That is observing the change in distance over time. Yes or No?


Why do you need another kind of definition for velocity? If you really want to, just keep that new definition separate from the normal definition and stop comparing your incorrectly defined velocity to normal velocity. Also, do not forget, while your rocket is accelerating you are in a non-inertial reference frame; you feel the acceleration. This should be enough for you to let go your uneasiness feeling that earth get closer to you the faster you move.


Survival of SR is neither my goal nor is to overturn it. You should be careful stating a view is wrong when there is no emperical evidence that your view is valid. It is inappropriate to abdocate common sense however in support of SR. If its application creates unacceptable results then it should be shit canned. Or as a minimum clarified and limited.


What unacceptable results are you referring to here?


Of course I do. I also have posted a number of times the following:

1 - Clocks to not measure time. They measure a process. If I use a pot with marking up the side and evaporation rate of the water as a clock, I have a clock. Not a good one granted but I have a clock. The reason it is not a good clock is that certain things such as humidity, temperature and air currents will alter the process of evaporation and hence my measure of time.

If I light a fire under my clock I could see time go by rather fastly.

Point being that motion may infact alter the clocks process but that has nothing to do with time dilation or even proves time exists for that matter.

2 - In the H&K clock experiment detailed counter studies found that the variation between the clocks and the number of other issues which could alter the function of the clocks and the clock rates before they were even transported, etc had such wide variation (noise to signal ratio) that such data would have been virtually impossible to get.

They also found on investigation that the results were actually published after massaging data on a statistical basis. These were not clean and clear results. That doesn't make them wrong but it taints them to a point that you should never attach the term "proof" anywhere near them. There have infact been alegations that they published outright fraud. I don't know either way but I do know that the published results are in serious question.


I am not interested to comment on that and receive a crackpot link from your later on.

But, I do want to say one thing. If you have a stop watch and it ticks 10 times, you say 10 seconds have passed. This is applicable on earth as well as in your rocket. That's all matter to me. Whether that stop watch measure "time" or "process" is immaterial.


Yes SR. If you still agree with James that in the case "I" presented which was only two observers, the rocket and the earth, that one sees contraction and the other doesn't, then SR is dead in the water. See my corrections and clarification in the above post for this paragraph.


I have never seen your answer as to why earth and rocket observers must see the distance (whatever, distance between earth and Alpha Centaury -- while the rocket move toward this star -- or distance between earth and the rocket) as the same. SR doesn't say so. Why do you think SR requires that?

MacM,

No, MacM. Given v=0.6c, for example, as the velocity of your rocket relative to earth, you measure the difference of displacement ds for an interval of dt very very small but NOT ZERO. The ratio of ds and dt should give you v=0.6c for that particular time. If your rocket accelerates and you make another measurement, you should obtain another set of ds and dt that should give you v larger than 0.6c. Your mistake is, you take dt as zero while infinitesimal dt is not equal to dt=0!

I have never said that you would not see earth as getting closer. I thought that was very clear. Your definition of velocity just simply incorrect.

Nor have I said you did. I have said you denied earth seeing the rocket get closer.

Why do you need another kind of definition for velocity? If you really want to, just keep that new definition separate from the normal definition and stop comparing your incorrectly defined velocity to normal velocity. Also, do not forget, while your rocket is accelerating you are in a non-inertial reference frame; you feel the acceleration. This should be enough for you to let go your uneasiness feeling that earth get closer to you the faster you move.

I have no uneasyness. It is Relativity and you that are misrepresenting its functions that seem to be uneasy. I have clearly agreed (assuming spatial contraction occurs at all) that the rocket sees the earth getting closer. My arguement is that the earth must also see the rocket getting closer and you have disagreed on that point. Please address that issue and not claim disagreement where there is none.

What unacceptable results are you referring to here?

I am not interested to comment on that and receive a crackpot link from your later on.

Another dodge. Refusal to answer.

But, I do want to say one thing. If you have a stop watch and it ticks 10 times, you say 10 seconds have passed. This is applicable on earth as well as in your rocket. That's all matter to me. Whether that stop watch measure "time" or "process" is immaterial.

Good then you have said clocks and their time dilation may not be actual time dilation. We can agree on that issue.

I have never seen your answer as to why earth and rocket observers must see the distance (whatever, distance between earth and Alpha Centaury -- while the rocket move toward this star -- or distance between earth and the rocket) as the same. SR doesn't say so. Why do you think SR requires that?

Here we go again with Alpha C.

1 - I have never said A.C. was involved. You introduced that (or James R.) as a confusion factor. And I have indeed answered that question as clearly stated that since there is no relative velocity between A.C. and earth there are no relavistic affects.


2 - -- or distance between earth and the rocket) as the same. SR doesn't say so. Why do you think SR requires that?

It is this portion of your responses that are in error.

Please explain how it is that 'A' has relative velocity to 'B' and sees lorentz contraction but you think 'B' (which MUST also have relative velocity to 'A' does not see relavistic affect to 'A'?

In plain english please. Your concept violates Relativity.

A common resolution to this string may be found at:

http://www.sciforums.com/showthread.php?t=37305&page=3

My god, do you enjoy spamming?

"I should drag as many topics down as possible, then point to an example of me being ignorant"

My god, do you enjoy spamming?

"I should drag as many topics down as possible, then point to an example of me being ignorant"

Those topics had closed down and were left with a lot of BS and no resolution. Readers have a right to an answer. A common answer has finally evolved (Like pulling molars but none the less an important distinction or clarification has been made).

I can understand that you might prefer that it hadn't but your wishes don't rule the universe.

To find examples of you being ignorant shouldn't take that much effort.You spam us with that all the time. :D

Those topics had closed down and were left with a lot of BS and no resolution.

That's true, MacM. But, most of the BS were written by you.:D

In case, you wonder why there are no resolution...I think you should check the answer in yourself. I have seen many solutions given by many bright posters. The problem is on you who unable to see them and keep claiming that there are no resolution, as if there is really a paradox as you have suggested. Since you said you are a bright (not so young) star...brighter than me or at least that what you want us to think, it seems logical that you are the only one (with a few rare exceptions, I would say) who see the "paradox".

Paul T,

I have stated before that I consider you are superior (overall) both mathematically and perhaps generally on the subject of Relativity. But that certainly doesn't mean you know it all or that you are right on every issue (which I think has now been shown).

I have no particular interest in pursueing your little pissing contest. But you want to attack my intelligence then I will respond in kind.

Perhaps it is your obvious lack of a complete mastery of the english language that you find rapidly getting closer to something while accelerating away is not a "Paradox".

In case you missed it I'll re-post the definition:

From Webster
"Paradox":

2 - A statement that seems contridictory, unbelievable,, or absurd but that may be true in fact.

Now if the issue I have brought forth here is not a "Paradox", I would like to see a better definition.

MacM,


I have stated before that I consider you are superior (overall) both mathematically and perhaps generally on the subject of Relativity. But that certainly doesn't mean you know it all or that you are right on every issue (which I think has now been shown).


I have no worry about who is more superior, but anyway, thanks. Have I ever said that I know everything about relativity? In fact, I said in other thread that I wanted to learn from you since you said you have a deeper understanding on relativity. Now, I have to ask you again, what is right here? Okay, we are talking about whether there is paradox as you claim or there isn't? I said no paradox and you keep telling us, there is paradox. Does that make you right and I am wrong? I would be happy to admit it if that really the case.


I have no particular interest in pursueing your little pissing contest. But you want to attack my intelligence then I will respond in kind.


Attacking you intelligence? Did I? You are so sensitive about your intelligence, I don't know why. I just said that most of the BS were written by you. You have said similar things to me many times and I have never accused you attacking my intelligence.


Perhaps it is your obvious lack of a complete mastery of the english language that you find rapidly getting closer to something while accelerating away is not a "Paradox".

In case you missed it I'll re-post the definition:

From Webster
"Paradox":

2 - A statement that seems contridictory, unbelievable,, or absurd but that may be true in fact.

Now if the issue I have brought forth here is not a "Paradox", I would like to see a better definition.

Hey, are we talking about English literature or something? As far as I know, paradox with respect of a theory like SR indicates that there is something wrong with the theory. A theory cannot live happily with a paradox. It must show that the paradox is just a misconception view or else the theory falls. What is your stand here, MacM? You keep claiming that the paradox is real but do not want to give us a firm statement whether SR is correct or wrong with respect to this "paradox', at least.

When you travel away from earth in an accelerating rocket, you see that your distance to earth could decrease instead of increase. Did you also think that your distance to a destination point also decrease faster? Your thought is correct to this point. But, when you go on try to compute your MacM's velocity as we have discussed, then you are going too far.There is only one type of velocity, which for an accelerating system must be obtained based on ds/dt where both ds and dt are very very small. You did not use ds/dt, instead you picked a distance while your rocket was at one velocity, deduct it from another distance obtained while your rocket was at another velocity and then compute your MacM's velocity. That's invalid!

MacM,

I have no worry about who is more superior, but anyway, thanks. Have I ever said that I know everything about relativity? In fact, I said in other thread that I wanted to learn from you since you said you have a deeper understanding on relativity. Now, I have to ask you again, what is right here? Okay, we are talking about whether there is paradox as you claim or there isn't? I said no paradox and you keep telling us, there is paradox. Does that make you right and I am wrong? I would be happy to admit it if that really the case.

It does indeed appear that part of the problem here is uderstanding "English". That is not meant as a put down. Re-read the definition of "Paradox". The answer to your question is the close. "................... actually being true". Claiming a "paradox" does not claim SR is invalid in any way.

Attacking you intelligence? Did I? You are so sensitive about your intelligence, I don't know why. I just said that most of the BS were written by you. You have said similar things to me many times and I have never accused you attacking my intelligence.

My comment was very general. That is covers your posts in general, not just your last statement. My respnses to you were just that "Responses" to attacks by you. I have never once attacked somebody first on this forum.

Hey, are we talking about English literature or something? As far as I know, paradox with respect of a theory like SR indicates that there is something wrong with the theory. A theory cannot live happily with a paradox. It must show that the paradox is just a misconception view or else the theory falls. What is your stand here, MacM? You keep claiming that the paradox is real but do not want to give us a firm statement whether SR is correct or wrong with respect to this "paradox', at least.

Re-read my statetment above and the definition of what "Paradox" means. You are in error that a theory must be invalid to also have a paradox.

When you travel away from earth in an accelerating rocket, you see that your distance to earth could decrease instead of increase. Did you also think that your distance to a destination point also decrease faster? Your thought is correct to this point. But, when you go on try to compute your MacM's velocity as we have discussed, then you are going too far.There is only one type of velocity, which for an accelerating system must be obtained based on ds/dt where both ds and dt are very very small. You did not use ds/dt, instead you picked a distance while your rocket was at one velocity, deduct it from another distance obtained while your rocket was at another velocity and then compute your MacM's velocity. That's invalid!

You again are hung up on a technicality. You can argue your mathematical nuances all you want but the fact remains that there is a change in distance over time which is (as you yourself calculated) can be millions of times the speed of light. I am not claiming a velocity during the period of acceleration, in that case your ds/dt = 0 is the correct approach. However, my ds/dt = actual ds/dt viewed from a non-accelerating V1 and V2 in a given period is also a valid statement of average velocity (V'). V' = Vo + at/2.

Since you don't like the result you choose to deny its reality. If you don't want to call it velocity and want to name it MacM's velocity, I have no real objection to that, other than you want to imply it is somehow false and unreal.

I would like to see your explanation of a spatial distance change which could be 1 B LYr in 1 hour. Distance at V1 is real. Distance at V2 is real and that real distance change occurs in real time. How else do you suggest we write the relationship other than ds/dt = V'?

You don't find this an interesting aspect of accepting Relativity? You don't see the humor of a theory that given a real ds/dt, results in a violation of one of its own principles - i.e. a faster than light result? I do.

MacM,

It does indeed appear that part of the problem here is uderstanding "English". That is not meant as a put down. Re-read the definition of "Paradox". The answer to your question is the close. "................... actually being true". Claiming a "paradox" does not claim SR is invalid in any way.


Hahaha. If you talk like this in an English literature forum, you are forgivable. But,...let me see, this forum is called "Physics and Math" forum, therefore....tetttt, wrong answer!


Re-read my statetment above and the definition of what "Paradox" means. You are in error that a theory must be invalid to also have a paradox.


If "twin paradox" is a real paradox, then SR sucks. However, as "twin paradox" is not a paradox, SR survive. What kind of paradox do you have here that doesn't prove anything wrong on SR? Why do you still call it a paradox, then? Just for fancy reason?


You again are hung up on a technicality. You can argue your mathematical nuances all you want but the fact remains that there is a change in distance over time which is (as you yourself calculated) can be millions of times the speed of light. I am not claiming a velocity during the period of acceleration, in that case your ds/dt = 0 is the correct approach. However, my ds/dt = actual ds/dt viewed from a non-accelerating V1 and V2 in a given period is also a valid statement of average velocity (V'). V' = Vo + at/2.


You are wrong again, MacM. My argument about the correct definition for velocity is not mathematical nuances, but a conceptual approach. Apparently you still don't understand. I gave you computation for MacM's velocity only to show you that your defined velocity is absurd. Since you still saying ds/dt = 0, it is obvious you really have no clue on what I have said. There is no ds/dt = 0 in our discussion as it means v=0. The rocket is moving, its velocity is therefore non-zero. I think you should consider reading something about velocity, MacM. On second thought, I think you will say that you don't need to as you know very well what velocity really mean, hahahaha.

And what do you mean by your actual ds/dt? Velocity defined as ds/dt is applicable for any velocity, V1 or V2 or any other velocities. You can only apply ds/dt for system under acceleration for dt very-very small (I am about to get very bored saying this over and over again), like while the rocket velocity is V1 or V2. So, you see, while you are at V1, you get ds/dt=V1 and so on.

I have seen your equation: V' = Vo + at/2, for quite a few times. I first thought, it was just a typo, but since you repeatedly write it that way, I am now convince you don't even know what the correct equation for velocity is. May be you will tell me: "Well, it just a minor error. It will not change the fact that there is a PARADOX!" But, I would not agree with you. Since you don't even know this simple velocity equation, I can't let myself believe on your claim for other more important matter. Please, do not think that I am attacking you. If I want to attack you, I have done it the very first moment I catch such error.


Since you don't like the result you choose to deny its reality. If you don't want to call it velocity and want to name it MacM's velocity, I have no real objection to that, other than you want to imply it is somehow false and unreal.


It's not a matter of like or dislike, but if something is wrong, it's wrong...whether you like it or you don't like it. I named it MacM's velocity, because it is not velocity as defined in physics. I think you still don't get it.


I would like to see your explanation of a spatial distance change which could be 1 B LYr in 1 hour. Distance at V1 is real. Distance at V2 is real and that real distance change occurs in real time. How else do you suggest we write the relationship other than ds/dt = V'?


Yes, your MacM's velocity could assume any value; it could be 1 billions Ly/hour as you said and at the same instance 1000m/s. Don't you see the absurdity of your MacM's velocity? There is a conceptual error in your defined velocity!


You don't find this an interesting aspect of accepting Relativity? You don't see the humor of a theory that given a real ds/dt, results in a violation of one of its own principles - i.e. a faster than light result? I do.

No, I don't see humor on SR with respect to your "paradox". The result probably only violates your "common sense". You think, a rocket accelerating away from earth must see that its distance to earth increasing with time. After applying SR, you found that, for certain condition, it is not the case and you therefore claim there is a "PARADOX". It is a paradox to your "common sense", but not to the real physics. SR violates many other "common senses". Do you see all of them as paradoxes or humors??? I must tell you, the paradox is not in SR, but it is in your own argument.

Paul T,

Hahaha. If you talk like this in an English literature forum, you are forgivable. But,...let me see, this forum is called "Physics and Math" forum, therefore....tetttt, wrong answer!

I guess that makes sense. Physicists don't need to know how to read and understand the meaning of words. :bugeye:

If "twin paradox" is a real paradox, then SR sucks. However, as "twin paradox" is not a paradox, SR survive. What kind of paradox do you have here that doesn't prove anything wrong on SR? Why do you still call it a paradox, then? Just for fancy reason?

Nobody has referred to the "twin Paradox" here. Since it is theoretically resolved by Relativity, the paradox doesn't exist. HOwever, it seems in this case the paradox is infact mandated by Relativity.

To understand why it is called a paradox go learn the meaning of the word before you condon its application.

You are wrong again, MacM. My argument about the correct definition for velocity is not mathematical nuances, but a conceptual approach. Apparently you still don't understand. I gave you computation for MacM's velocity only to show you that your defined velocity is absurd. Since you still saying ds/dt = 0, it is obvious you really have no clue on what I have said. There is no ds/dt = 0 in our discussion as it means v=0. The rocket is moving, its velocity is therefore non-zero. I think you should consider reading something about velocity, MacM. On second thought, I think you will say that you don't need to as you know very well what velocity really mean, hahahaha.

You are a fool. It has been said many times here that the expression ds/dt = 0 means approaching '0' not absolute zero. We have already discussed this point, don't continue to babble about people not understanding.

Instantaneous Velocity = lim ds/dt or V = Vo + at.
dt->0

and is actually only an average velocity during the time interval.

I explained to you the validity of what you want to dub MacM's velocity; which is nothing more that the valid mathematical calculation for standard velocity calculation when acceleration is involved. V' = ds/dt. You choose to not recognize standard physics formulas that is your perogative. But it does shed considerable doubt on your physics comphrension.

And what do you mean by your actual ds/dt? Velocity defined as ds/dt is applicable for any velocity, V1 or V2 or any other velocities. You can only apply ds/dt for system under acceleration for dt very-very small (I am about to get very bored saying this over and over again), like while the rocket velocity is V1 or V2. So, you see, while you are at V1, you get ds/dt=V1 and so on.

You should really get bored saying it over and over. You are boring the rest of us as well. We all know about acceleration and instantaneous velocity. That is not and has not been the issue. The issue is one you choose to deny but is actually very standard (highschool level) physics.

V1 and V2 are given starting and ending velocities. At V1 you have a given gamma (hence distance adjusted for relavistic affects). At V2 you have a different gamma and distance. The differential distances calculated by gamma at each V becomes ds. Pretty damn simple actually. This change in distance occurs in a [b]given] amount of time, which is ds.

NOthing funny here than you ineptitude. V' = ds/dt. This is not MacM's velocity it is standard physics velocity. You are beating a dead horse.

I have seen your equation: V' = Vo + at/2, for quite a few times. I first thought, it was just a typo, but since you repeatedly write it that way, I am now convince you don't even know what the correct equation for velocity is. May be you will tell me: "Well, it just a minor error.

Actually you are correct. I did miss type it it is not at/2 which ahs to do with distance but it has been corrected above before I got to this paragra. Need I remind you of your repeated c = 3E6 m/s posts, which I pointed out to you.

It will not change the fact that there is a PARADOX!" But, I would not agree with you. Since you don't even know this simple velocity equation,

Kiss my arse - you blithering idiot. Several have now conceeded that one sees the earth rapidly getting closer as one simultaneously accelerates away from it. You don't want to lable that a paradox. I really care less. It only shows your stupidity and not even knowing the meaning of the word paradox.

I can't let myself believe on your claim for other more important matter. Please, do not think that I am attacking you. If I want to attack you, I have done it the very first moment I catch such error.

Whoppie! You dumb SOB. c = 3E6 m/s???? HeHeHe. :bugeye:

It's not a matter of like or dislike, but if something is wrong, it's wrong...whether you like it or you don't like it. I named it MacM's velocity, because it is not velocity as defined in physics. I think you still don't get it.

Oh, I get it alright. You don't even know highschool physics.

Yes, your MacM's velocity could assume any value; it could be 1 billions Ly/hour as you said and at the same instance 1000m/s. Don't you see the absurdity of your MacM's velocity? There is a conceptual error in your defined velocity!

No error what so ever. Just a unique fact if you accept Relativity. To bad you don't understand it well enough to appreciate it. Bumping your gums doesn't change the physical reality as calculated by Relativity. ds/dt can be counter intuitive (backwards) and FTL. Sorry you lose. Live with it.

No, I don't see humor on SR with respect to your "paradox". The result probably only violates your "common sense". You think, a rocket accelerating away from earth must see that its distance to earth increasing with time. After applying SR, you found that, for certain condition, it is not the case and you therefore claim there is a "PARADOX". It is a paradox to your "common sense", but not to the real physics. SR violates many other "common senses". Do you see all of them as paradoxes or humors??? I must tell you, the paradox is not in SR, but it is in your own argument.

Only because you are to stupid to understand the enaing of the word paradox. In the beginning you tried to argue it didn't occur at all now you want to jpretend you knew it all along and that it really isn't a paradox. Pathetic loser.

MacM,


I guess that makes sense. Physicists don't need to know how to read and understand the meaning of words. :bugeye:


Save your English course for kid at school, MacM. Talk about physics or math here, since this is "Physics & Math".


Nobody has referred to the "twin Paradox" here.


Silly! I did, hahahaha.


To understand why it is called a paradox go learn the meaning of the word before you condon its application.


Again, I said...save the crap for school kid. On second thought, you better keep a distance from kid, hehe. You are a dengeruos fellow, who would destroy kid's future with all your BS preaching, hahaha.


You are a fool. It has been said many times here that the expression ds/dt = 0 means approaching '0' not absolute zero. We have already discussed this point, don't continue to babble about people not understanding.


Let's see where the fool is hidding. The name is MacM. Where did you learn that ds/dt = 0 means aproaching '0'??? What's approaching zero? It's not ds/dt which is velocity. What's a silly thought you have there, MacM.


Instantaneous Velocity = lim ds/dt or V = Vo + at.
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;dt->0


Hm, improvement. At least you are trying to give us some impression on your math ability. But, let see....it's wrong MacM. In math, ds/dt is calculus notation. You've gotten it all mix up with limit concept. Also, instantaneous velocity is not Vo + at, hahaha. You are a real dengeruous fellow for kid MacM.


and is actually only an average velocity during the time interval.


Wrong again. Instantenous velocity is not the same as average velocity. MacM, I suggest you to pay a visit to your nearby high school and talk to some of kids there to learn something about the difference of the two. But, please...don't try to teach them physics, you'll ruin their future.


I explained to you the validity of what you want to dub MacM's velocity; which is nothing more that the valid mathematical calculation for standard velocity calculation when acceleration is involved. V' = ds/dt. You choose to not recognize standard physics formulas that is your perogative. But it does shed considerable doubt on your physics comphrension.


Hahaha. The truth is, you had it all screwed-up, MacM. Instantenuous velocity is actually average velocity? Hahahaha, this is an obvious indication that you need some basic physics guidance from highschool kid. Take my advise, don't be that stuborn.


You should really get bored saying it over and over. You are boring the rest of us as well. We all know about acceleration and instantaneous velocity. That is not and has not been the issue. The issue is one you choose to deny but is actually very standard (highschool level) physics.


Hahaha. You know what you problem is? You have serious problem with highschool physics. What's happen, MacM. Did you skip school a lot when you were young? How did you manage to get all the "high physics" knowledge without going through the basic? Ooops, you'd found a short cut, apparently.


V1 and V2 are given starting and ending velocities. At V1 you have a given gamma (hence distance adjusted for relavistic affects). At V2 you have a different gamma and distance. The differential distances calculated by gamma at each V becomes ds. Pretty damn simple actually. This change in distance occurs in a [b]given] amount of time, which is ds.


Yep, pretty damn simple truth...and all wrong. Go and read some calculus and understand what ds means. It's infinitesimal, MacM. The differential distance of 1 billions Ly is not infinitesimal, hahahaha. You are a funny guy, MacM.


NOthing funny here than you ineptitude. V' = ds/dt. This is not MacM's velocity it is standard physics velocity. You are beating a dead horse.


Who said nothing funny here, you are funny. You are having problem with the meaning of infinitesimal s and t. I am not talking about the english meaning of the word, the math meaning; afterall we are in this "Physics & Math" forum.


Actually you are correct. I did miss type it it is not at/2 which ahs to do with distance but it has been corrected above before I got to this paragra. Need I remind you of your repeated c = 3E6 m/s posts, which I pointed out to you.


Just pointing out to you. Good, if you know the mistake.


Kiss my arse - you blithering idiot. Several have now conceeded that one sees the earth rapidly getting closer as one simultaneously accelerates away from it. You don't want to lable that a paradox. I really care less. It only shows your stupidity and not even knowing the meaning of the word paradox.


It's not just a matter of labeling stuff. There is no such paradox, that's all. You've taken the wrong route to that paradox.


Oh, I get it alright. You don't even know highschool physics.


Ooops, trying to fool around with words. Highshcool kid knows the difference between instantenuous velocity and average velocity, but MacM doesn't. Well, still want to say you know highschool physics? Hahahaha.


No error what so ever. Just a unique fact if you accept Relativity. To bad you don't understand it well enough to appreciate it. Bumping your gums doesn't change the physical reality as calculated by Relativity. ds/dt can be counter intuitive (backwards) and FTL. Sorry you lose. Live with it.


Wrong again. ds/dt will give you the real velocity. Those V1 and V2 that you assumed are ds/dt, but of course not MacM's velocity. Your MacM's velocity just doesn't fit into any known physics, a dream world phsyics may be.

If you still don't know what I mean, try to add this MacM's velocity to V1 (the real velocity). Do they add up? Okay, they don't, MacM. I am giving you the answer just in case you'll fooling around with your wrong interpretation again, hehehe.


Only because you are to stupid to understand the enaing of the word paradox. In the beginning you tried to argue it didn't occur at all now you want to jpretend you knew it all along and that it really isn't a paradox. Pathetic loser.

As I said, anybody who knows a little SR know such thing. You just like to fush around with your wrong interpretation and, what, even suggesting FTL. Hahahaha, feeling high with proud that you know SR a lot, MacM?

Paul T,

Save your English course for kid at school, MacM. Talk about physics or math here, since this is "Physics & Math".

It is really amazing that you seem to think physicists can't read or don't need to know how to read and understand the meaning of words. It really is giving you a black eye.

Silly! I did, hahahaha.

Silly indeed. So why did you. Oh, I know, just like adding Alpha C., to the other discussion. It changes the subject since you can't respond to the current one.

Again, I said...save the crap for school kid. On second thought, you better keep a distance from kid, hehe. You are a dengeruos fellow, who would destroy kid's future with all your BS preaching, hahaha.

HaHa indeed. From the composition of your posts I can see why you disapprove of understanding readig, writting and other necessary skills to actually be a good scientist.

Let's see where the fool is hidding. The name is MacM. Where did you learn that ds/dt = 0 means aproaching '0'??? What's approaching zero? It's not ds/dt which is velocity. What's a silly thought you have there, MacM.

Do you really think by continueing to repeat this out of context bullsh_t is meaningful.

OK let me repeat. The speed of light is NOT 3E6 m/s Paul T. It is 3E8 m/s. HeHeHe and you claim to know Relativity and don't even know the speed of light? :bugeye:

BTW: In case you missed it. I won the discussion or is that what this BS is all about - sour grapes.

Hm, improvement. At least you are trying to give us some impression on your math ability. But, let see....it's wrong MacM. In math, ds/dt is calculus notation. You've gotten it all mix up with limit concept.

Funny. I guess I better throw away my "Schaum's - College Physics, 7/ed"

Also, instantaneous velocity is not Vo + at, hahaha. You are a real dengeruous fellow for kid MacM.

The formula for Instantenous Velocity was given above.

Vi = lim ds/dt
...dt->0

Wrong again. Instantenous velocity is not the same as average velocity. MacM, I suggest you to pay a visit to your nearby high school and talk to some of kids there to learn something about the difference of the two. But, please...don't try to teach them physics, you'll ruin their future.

Again I guess I had best throw away my College Physics books. Page 25:

"INSTANTANEOUS VELOCITY is the average velocity evaluated for a time interval that approaches zero".

What is even funnier is that I wouldn't need a college book to tell me that it was. It is just common sense. Something you clearly are lacking.

Hahaha. The truth is, you had it all screwed-up, MacM. Instantenuous velocity is actually average velocity? Hahahaha, this is an obvious indication that you need some basic physics guidance from highschool kid. Take my advise, don't be that stuborn.

Take MY advice and at least make your allegations halfway valid. Let me suggest you go read a book or two or three. Re-read my quote from the college physics text above.

Hahaha. You know what you problem is? You have serious problem with highschool physics. What's happen, MacM. Did you skip school a lot when you were young? How did you manage to get all the "high physics" knowledge without going through the basic? Ooops, you'd found a short cut, apparently.

You are simply pathetic. Do you not realize your BS is making you look pretty damn childish and stupid.? It is. Note: The above quote is from "College Physics" assh_le. Go learn some.

Yep, pretty damn simple truth...and all wrong. Go and read some calculus and understand what ds means. It's infinitesimal, MacM. The differential distance of 1 billions Ly is not infinitesimal, hahahaha. You are a funny guy, MacM.

As I said your BS is so transparent. It has no relation to anything I have said or believe. Billions of Lyr is the change in distance with change in velocity PER Relativity. Or do you still want to deny that also?

Who said nothing funny here, you are funny. You are having problem with the meaning of infinitesimal s and t. I am not talking about the english meaning of the word, the math meaning; afterall we are in this "Physics & Math" forum.

What a lot of crap. Surely you understand others reading this sees you are talking outright horsesh_t .

Just pointing out to you. Good, if you know the mistake.

It's not just a matter of labeling stuff. There is no such paradox, that's all. You've taken the wrong route to that paradox.

You are not only blind but dumb. To say "The faster I run away the closer I get", is clearly a "Paradoxial" statement. The only reason it is paradoxial and not an out right false statement is that "Paradox" means it may appear absurd but is true. Do you now want to claim it is not true, that Relativity is in error? If not then you best accept the paradox.

Ooops, trying to fool around with words. Highshcool kid knows the difference between instantenuous velocity and average velocity, but MacM doesn't. Well, still want to say you know highschool physics? Hahahaha.

Perhaps you actully think your posts has some meaning, bearing or affect on others. You are right. They are intelligent enough to see your foolishness.

Wrong again. ds/dt will give you the real velocity. Those V1 and V2 that you assumed are ds/dt, but of course not MacM's velocity. Your MacM's velocity just doesn't fit into any known physics, a dream world phsyics may be.

So stating that a rocket is gong 0.6c is V1 and a rocket is going .08c is V2, is dream world and not physics. HaHaHaHa. What a load of crap.

If you still don't know what I mean, try to add this MacM's velocity to V1 (the real velocity). Do they add up? Okay, they don't, MacM. I am giving you the answer just in case you'll fooling around with your wrong interpretation again, hehehe.

Just what in the hell are you babbling about - adding up? Nonsense. V1 and V2 were GIVEN velocities of the problem.

As I said, anybody who knows a little SR know such thing. You just like to fush around with your wrong interpretation and, what, even suggesting FTL. Hahahaha, feeling high with proud that you know SR a lot, MacM?

Feeling proud that you lost the discussion are you. You lost jerk. Go back and learn something, then come back and talk to me.

The formula for Instantenous Velocity was given above. Vi = lim ds/dt
dt-->0 There is no lim ds/dt.

You either have:
ds/dt
or
delta-s/delta-t with delta-t going to zero

There is no lim ds/dt.

You either have:
ds/dt
or
delta-s/delta-t with delta-t going to zero

Damn and I swore that is what "Schaum's College Physics, 7/ed, page 26 showed.

(Trying now to see if attchements have been fixed yet)

Nope not yet. But anyone wanting to know the truth just get a copy of the above College Text and it was printed just as I have shown.

I'll give you the benefit of the doubt until you see the re-spaced version. This forum doesn't hold columns. I had to add '....' in front of dt-->0 to make it locate underneath as now shown.

I'm looking at a Schaum book right now... they don't use that terminology because it is wrong.

You take a limit of you deltas to get your derivatives. Deltas and ds are two different things.

Since Schaums is not online (that I know of), here is a reference for you:
http://archives.math.utk.edu/visual.calculus/2/definition.12/

I'm looking at a Schaum book right now... they don't use that terminology because it is wrong.

Bullsh_t. I'm also looking at it right now also. Publication is 1978. Old but not that old. In any case you are grossly in error to attack me when I post formulas from a college physics book.

Go attack them if you don't agree with their publication.

You take a limit of you deltas to get your derivatives. Deltas and ds are two different things.

Perhaps so but that is the way it is published. In any case we all know and understand the reduction of dt to infitesimal to establish instantaneous velocity. We also understand that that velocity is in reality an average over the limited dt that is used to calculate it.

Since Schaums is not online (that I know of), here is a reference for you:
http://archives.math.utk.edu/visual.calculus/2/definition.12/

I'm looking at a Schaum book right now... they don't use that terminology because it is wrong.

Bullsh_t. I'm also looking at it right now also. Publication is 1978. Old but not that old. In any case you are grossly in error to attack me when I post formulas from a college physics book.

Go attack them if you don't agree with their publication.

You take a limit of you deltas to get your derivatives. Deltas and ds are two different things.

Perhaps so but that is the way it is published. In any case we all know and understand the reduction of dt to infitesimal to establish instantaneous velocity. We also understand that that velocity is in reality an average over the limited dt that is used to calculate it.

Since Schaums is not online (that I know of), here is a reference for you:
http://archives.math.utk.edu/visual.calculus/2/definition.12/

I took a look at your link.

dy......................delta y
---..=.....lim.......--------
dx....delta x-->0...delta x

if x is made t and y is made s then ds/dt = V; It translates to the same thing.

................ds
V'=...lim...----
....dt-->0...dt

denotes the derivative of the function f.

In any case your attack on me for posting that formula is improper. If you want to claim a modernization or change in current usage, I have no problem with that, but your post is simular to the attacks when one refers to what once was called relavistic mass. It is no longer used (in general) but is still contained in virtually every physics book. :bugeye:

MacM,

Bullsh_t. I'm also looking at it right now also. Publication is 1978. Old but not that old. In any case you are grossly in error to attack me when I post formulas from a college physics book.

Go attack them if you don't agree with their publication.


Frankly, I have never seen any physics book that uses notation for instantenuos velocity as given by you below:
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;ds
v=&nbsp;&nbsp;lim&nbsp;&nbsp;&nbsp;-------
&nbsp;&nbsp;&nbsp;&nbsp;dt-->0&nbsp;&nbsp;&nbsp;dt

You said that your Schaum's series College Physics book uses such notation and I would say it is very odd. Of all that I have seen, they all use the following notation:

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<font face=symbol>D</font>s
v=&nbsp;&nbsp;lim&nbsp;&nbsp;&nbsp;-------
&nbsp;&nbsp;&nbsp;&nbsp;<font face=symbol>D</font>t-->0&nbsp;&nbsp;&nbsp;<font face=symbol>D</font>t

Sometimes, the author also gives differential calculus notation that is v = ds/dt, but never...never in the form like the one given by you.

Obviously, I am now very interested to see your book. Could you please get that particular page scanned and posted? I want to see such notation being use in the book for the first time. I believe many others in this forum may want to see that too.

O, just for your information. I managed to peep in my local university book store the book that you referred to. It was the ninth edition Schaum's Series College Physics. Certainly not the same edition as the one you have. And, you know what, the instantenuos velocity is defined in that book as what I always find in any other standard physics book. So, I must say, yours must be a very rare publication. You should keep it properly as I believe it may become a collectiable item, you know, like those wrongly printed notes or coins. :D

Paul T,

I'll stand corrected. My book also shows the "triangular" delta s, etc.

However, since I don't do symbols in HTML and 'd' is often used to represent delta I posted it in its easier form. My error since it infers a dfferent mathematical process when combined with the 'lim'.

Once again, d is NOT delta. You can now kindly apologize for once again saying arguing something as if you understand it.

MacM,



I'll stand corrected. My book also shows the "triangular" delta s, etc.

However, since I don't do symbols in HTML and 'd' is often used to represent delta I posted it in its easier form. My error since it infers a dfferent mathematical process when combined with the 'lim'.

Let me show a little flash back.

Earlier, I said: "In math, ds/dt is calculus notation. You've gotten it all mix up with limit concept."

And you replied: Funny. I guess I better throw away my "Schaum's - College Physics, 7/ed", which means you didn't understand my comment. Basically, you had no idea that you mixed two different concepts into one.

Then, Persol said:

There is no lim ds/dt.

You either have:
ds/dt
or
delta-s/delta-t with delta-t going to zero


for which you replied: Damn and I swore that is what "Schaum's College Physics, 7/ed, page 26 showed.

So, again you missed it.

My earlier comment requires your understanding that limit and differential calculus, by convention, use different notation to describe instantenuous velocity. Since, you missed it, I knew you have no idea that limit and differential calculus are difference. This, I think is quite understandable.

Persol's comment was straight to the point, pointing out to you that <font face=symbol>D</font>s and ds are not exchangeable, because they have different meaning. Your reply to Persol clearly shows your unwillingness to try to understand what he try to tell you and I found that this is generally your habit (not trying hard enough to understand other people's posts).

Once again, d is NOT delta. You can now kindly apologize for once again saying arguing something as if you understand it.

Up your twits. Virtually everybody uses dt , etc for delta t. You are strictly argueing a mathematical formality. I choose to ignore your BS non-point.

Virtually everybody uses dt , etc for delta t.NOBODY who understands what a differential is does.

MacM,



Let me show a little flash back.

Earlier, I said: "In math, ds/dt is calculus notation. You've gotten it all mix up with limit concept."

And you replied: Funny. I guess I better throw away my "Schaum's - College Physics, 7/ed", which means you didn't understand my comment. Basically, you had no idea that you mixed two different concepts into one.

Same up yours to you. I simply thought you were commentig on the formula not the usage of symbols. When you jposted the HTML version (which you didn't before, you only stated the formula was in error). I realized what your complaint was. So once again up yours.

Then, Persol said:


for which you replied: Damn and I swore that is what "Schaum's College Physics, 7/ed, page 26 showed.

So, again you missed it.

And again it has to do with symbology rather than a simple dt or ds which generally means delta t or delta s which the triangle also means. The formula was fine except for those that want to pick BS and create an issue where there really is none.

My earlier comment requires your understanding that limit and differential calculus, by convention, use different notation to describe instantenuous velocity. Since, you missed it, I knew you have no idea that limit and differential calculus are difference. This, I think is quite understandable.

What you think really is meaningless.

Persol's comment was straight to the point, pointing out to you that <font face=symbol>D</font>s and ds are not exchangeable, because they have different meaning. Your reply to Persol clearly shows your unwillingness to try to understand what he try to tell you and I found that this is generally your habit (not trying hard enough to understand other people's posts).

Ditto. At least I don't alter their post or rewrite what they said and then attempt to argue my version of what they said. Which you and jPersol both like to do.

Now are you prepared to be a man yet and admit your errors? I admitted mine (as minor as they are). Can you find the courage yet to admit that Relativity produces Paradoxes? I thought not. :bugeye:

Same up yours to you. I simply thought you were commentig on the formula not the usage of symbols. When you jposted the HTML version (which you didn't before, you only stated the formula was in error). I realized what your complaint was.No you didn't... you continued to make it, and just now said it was a 'technicality'. Taking the limit of ds/dt is COMPLETELY different than taking the limit of delta-s/delta-t.

No you didn't... you continued to make it, and just now said it was a 'technicality'. Taking the limit of ds/dt is COMPLETELY different than taking the limit of delta-s/delta-t.


Ho-Hum, Yawn. Ignore, ignore, ignore. Non-issue. We all (including me) have the formula for instantaneous velocity. Since I don't type HTML I used 'd' for delta. Ho-Hum. What a crock of crap.

It isn't a non issue. What you originally posted is actually an equation for acceleration... just that nobody actually types that anymore.I don't type HTML I used 'd' for delta.To quote you, "bullshit". If that's all it was you would have know what I meant when I typed it and wouldn't have gone and flown off to left field.

NOBODY who understands what a differential is does.

Not when actually performing the calculation but in typing or writing in general they damn sure do. If you were doing a formal paper then perhaps (no infact) you would use proper conotation. In this forum? I've seen many short cuts but it depends on who uses them doesn't it.

I've shown the formula was extracted from my College Physics book. It indeed contains the triangular symbol. What about not typing HTML symbols do you not understand or that using 'd' as the delta means.

YOur comments would be acceptable if they were merely cautionary or clarification. They are not they are bulls_t attacks which have no merit what-so-ever..

Not when actually performing the calculation but in typing or writing in general they damn sure do.No, they don't. Sorry to disappoint. They have two completely different meaning and you need to be clear which one you are talking about. The point is that you really don't know. If you didn't then you would have found no problem with the comment There is no lim ds/dt.

You either have:
ds/dt
or
delta-s/delta-t with delta-t going to zero

YOur comments would be acceptable if they were merely cautionary or clarification. They are not they are bulls_t attacks which have no merit what-so-ever..You dumbass. It WAS cautionary. ALL that I said was the above quote. You are the idiot that decided to argue it.

MacM,

Same up yours to you. I simply thought you were commentig on the formula not the usage of symbols. When you jposted the HTML version (which you didn't before, you only stated the formula was in error). I realized what your complaint was. So once again up yours.


It's more than just a mere wrong symbol. Since, you picked 'd' to represent '<font face=symbol>D</font>', we know that you had no idea what that 'd' represent in diferential calculus, hence you mixing up two different math concepts. If you know enough basic calculus, firstly you would not pick 'd' and secondly you would have explained to us that 'd' represents '<font face=symbol>D</font>' the first instance your equation was commented.

And, now I understand why my argument using ds/dt in many of my earlier posts like hitting the wall, since you have no clue what ds/dt really mean.


And again it has to do with symbology rather than a simple dt or ds which generally means delta t or delta s which the triangle also means. The formula was fine except for those that want to pick BS and create an issue where there really is none.


No. I have never seen <font face=symbol>D</font>s being used sparingly to represent ds, simply because they have difference meaning.


Now are you prepared to be a man yet and admit your errors? I admitted mine (as minor as they are). Can you find the courage yet to admit that Relativity produces Paradoxes? I thought not. :bugeye:

First of all, your 'd' and <font face=symbol>D</font> error is not a minor error, as it concerns conceptual issue.

If your "PARADOX" has any physical meaning at all, I would not have started this thread. This thread was started simply to dismiss you "paradox" claim and my standing on that remain the same. You arrived to such "paradox" based on your everyday experience, like when you accelerate away from point A, your distance to A increases with time. This everyday experience just doesn't apply for relativistic case. It's more or less comparable to your other everyday experience like Galiliean's addition of velocities (w = v +u), which is not applicable for light (say u=c), for example. Do you see this inapplicability of (w = v +u) for light as a paradox? It is not a paradox. It is simply at odd to our everyday experience, but that's the way nature works...and hence no paradox (same thing for your "paradox").

Paul T,

Lets see if this works?

'<font face=symbol>D</font>' , Damn it does. I copied it from my Outlook version of your post. Thanks. For others that might be interested it is a "<" then "font face=symbol" and then ">D" and then "< / fonts >". Leave out the spaces in the last group.

MacM,

It's more than just a mere wrong symbol. Since, you picked 'd' to represent '<font face=symbol>D</font>', we know that you had no idea what that 'd' represent in differential calculus,

And you have no idea what the hell you are talking about.

hence you mixing up two different math concepts.

About as meaningful complaint as 'x' or '*' for times in standard math., especially when I explained why I posed 'd' and not the HTML symbol. Your BS of continueing on is very obvious. You are desperate to save face since you first said I was full of sh_t about the contrction issue. Then when you actually calculated it and saw I was right, then it becomes 'hallucination" and not real. What a joke you are. Grow up.

[qote]If you know enough basic calculus, firstly you would not pick 'd' and secondly you would have explained to us that 'd' represents '<font face=symbol>D</font>' the first instance your equation was commented.

Perhaps, you are right. I should have noted my change in the posted formula but I did not assume anybody with common sense would make an issue about it. Since we now know you don't have common sense, I'll be sure to qualify future posts just for you.

And, now I understand why my argument using ds/dt in many of my earlier posts like hitting the wall, since you have no clue what ds/dt really mean.

In your dreams assho_e

No. I have never seen <font face=symbol>D</font>s being used sparingly to represent ds, simply because they have difference meaning.

You are a pathetic prude

First of all, your 'd' and <font face=symbol>D</font> error is not a minor error, as it concerns conceptual issue.

Not minor if used improperly. This was a casual discussion and not a formal paper with actual calculations. I suspect 90% + of persons reading this understood my post to mean what I intended.

If your "PARADOX" has any physical meaning at all, I would not have started this thread. This thread was started simply to dismiss you "paradox" claim and my standing on that rema