Special Relativity is wrong because time dilation is false. Time dilation cannot be true.
Special Relativity is wrong because time dilation is false. Time dilation cannot be true.
Welcome to the forum.
That's a pretty big claim, that flies in the face of "quite a lot" of science.
This is important: Have you read the rules? Have you read the advice threads? Will you be able to defend this claim?
NASA: "New Zealand is seldom photographed from orbit because it is one of the cloudier parts of planet, and because crew sleep periods often occur when the ISS passes over the area."
They have plans here for a "beautiful" test of the "one way speed of light" in 2012 http://arxiv.org/ftp/arxiv/papers/1011/1011.1318.pdf
"It only takes one experiment to dismantle a whole theory" (Hawking).
There may yet be a surprise.
Make all distance light travels in 1 second.
I make rest frame observer go to ( 1/5, 0, 0 )
I make moving observer go to ( -1/5, 0, 0 )
Set v = 3/5.
I must find t.
vt = x - x'/[1 /( √( 1 - vē/cē ) ) ] =1/5 -(-1/5)(4/5) = 9/25.
So, t = (9/25)(5/3) = 3/5.
I use Einstein
t' = ( t - vx/cē )[1 / √( 1 - vē/cē )]
t' = ( 3/5 - (3/5)(1/5) )(5/4) = 3/5.
That means t = t'.
Time dilation false.
NASA: "New Zealand is seldom photographed from orbit because it is one of the cloudier parts of planet, and because crew sleep periods often occur when the ISS passes over the area."
Yes, I already make math here.
It is lost.
I will try again.
Make all distance light travels in 1 second.
I make rest frame observer go to ( 1/5, 0, 0 )
I make moving observer go to ( -1/5, 0, 0 )
Set v = 3/5.
I must find t.
vt = x - x'/[1 /( √( 1 - vē/cē ) ) ] =1/5 -(-1/5)(4/5) = 9/25.So, t = (9/25)(5/3) = 3/5s.
I use Einstein
t' = ( t - vx/cē )[1 / √( 1 - vē/cē )]t' = ( 3/5 - (3/5)(1/5) )(5/4) = 3/5s.
No, you cannot claim that t'=t means that there is no time dilation.
First off, t'=t doesn't prove anything because it has nothing to do with time dilation.
Second off, t'=t implies
t=(t-vx/c^2)/sqrt(1-(v/c)^2)
or t=vx/c^2*gamma/(gamma-1) where gamma=1/sqrt(1-(v/c)^2)
The above doesn't mean anything, there can always be a time t that satisfies the above, so you found no "contradiction"
In reality, the time dilation gets derived totally differently:
dt'=dt/sqrt(1-(v/c)^2) (see any introductory textbook in relativity). so, dt' and dt are NOT equal (unless v=0).
A few observations:
1. If you want to disprove SR, you must first learn SR, you haven't done so.
2. Secondly, you need to learn the meaning of time dilation, you demonstrated that you don't know what it is.
3. If you think that you found an inconsistency, you didn't. You just got your basic math and physics wrong.
NASA: "New Zealand is seldom photographed from orbit because it is one of the cloudier parts of planet, and because crew sleep periods often occur when the ISS passes over the area."
You have shown a profound misunderstanding of relativity is all you have shown.
You say time dilation is false, but if I use the equation:
t'=γ(t-vx/c^{2}) With v=.6c and x=0, I get
t'=1.25t
Odd. This says that the transformed second is a quarter longer.
All you have shown is that you can set up a situation where the two clocks show the same time.
I'll assume English is not your first language.
Your thought experiment is confusing. You can't have a "rest frame" move...if it moves it isn't a "rest frame"
so lets start at the beginning
If you have something in motion relative to an observer the moving observer will experience "less time"
So if you have one observer in motion travelling .5c then the equation is
Δt' = Δt / √ 1 - v^{2}/c^{2}
so we get Δt' = Δt / √ 1 - (.5^{2}/1)
which is
Δt' = Δt / √ 1 - .25
Δt' = Δt / √ .75
Δt' = Δt / 0.86602540378443864676372317075294
so if Δt = 5 Δt' = 5.7735026918962576450914878050195
5 != 5.7735026918962576450914878050195
Δt != Δt'
The only time they would be the same is with v = 0 IE the to observers are not in relative motion to each other.
Yes, I can claim there is not time dilation.
Einstein makes time dilation as follows.
A clock move vt. He plugs into equation t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
t'=(t-vvt/c^2)*\gamma = t(1-v^2/c^2)*\gamma = t/\gamma.
He then said this clock may take any polygonal path from point A to point B.
What he did is assume the clock is moved from the origin. He then claimed he proved for any point A and B a clock could be moved and there is time dilation. I am simply pointing out his mistake.
When the origins of the 2 inertial coordinates systems are at the same place, time is marked in each coordinate system.
What I showed is a clock is moved from point A which is ((-1/5)/\gamma,0,0) ((-1/5,0,0) in the moving coordinates (to point B (1/5,0,0) and each clock elapses the same amount of time for the two clocks to be at the same place.
These are not my sayings, these are Einstein's since I use only Einstein's equations.
It is agreed you can calculate as you did. If I add 2 + 2, can I say everytime I add two number I get 4?
You must prove time dilation is true for any polygonal path like Einstein said. I simply show a path where a clock is moved from point A to point B and the stationary clock and trhe moving clock elapse the same time with no time dilation.
I can see where you miss understsand.
The time dilation that is derived must have clock that is moving only at origin of rest coordinates as its start position.
So, all you did in above is to make that true and then say true always. Einstein say clock moved from A to B and time dilation true.
I show case a clock at point A is moved to B and not time dilation.
That's a bad claim that you keep repeating. So, according to the BAUT rules, I am going to start asking some questions that I will expect you to answer.
Q1: Do you know that there are experiments that confirm relativistic time dilation?
Q2: Did you understand that time dilation is not about the coordinate "t" but about the time interval "dt"?
No, Einstein doesn't "make time dilation". Nature makes the time dilation. And it doesn't happen in the way you wrote it, it happens differently.Einstein makes time dilation as follows.
A clock move vt. He plugs into equation t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
t'=(t-vvt/c^2)*\gamma = t(1-v^2/c^2)*\gamma = t/\gamma.
Q3: Please derive the correct time dilation formula:
dt'=\gamma *dt
You can try using any textbook.
The GPS functionality proves Einstein correct and proves you wrong.He then said this clock may take any polygonal path from point A to point B.
What he did is assume the clock is moved from the origin. He then claimed he proved for any point A and B a clock could be moved and there is time dilation. I am simply pointing out his mistake.
I know what you showed, I proved it wrong.What I showed is a clock is moved from point A which is ((-1/5)/\gamma,0,0) ((-1/5,0,0) in the moving coordinates (to point B (1/5,0,0) and each clock elapses the same amount of time for the two clocks to be at the same place.
Q4: Did you understand my proof? Can you play it back to me, the way I played back your proof to you when I proved it incorrect?
Q5: You are using the Lorentz transforms, are you aware that time dilation (and length contraction) are a direct consequence of the Lorentz transforms. By using the Lorentz transforms you are automatically accepting the notion of time dilation. Do you have a problem with this?
Last edited by macaw; 2010-Nov-25 at 06:57 PM.
No, you just took a very special case
To try to translate your experiment.
Observer/clock A is located at (-1/5, 0, 0)
Observer/clock B is located at (1/5, 0. 0)
Now, observer/clock B moves with a velocity of 3/5
Then for some reason you use: x' = γ (x - v t)
but you apply this to the location of A and the location of B, which does not make sense, because you can only do that when A is in the origin, now you have to use that x is equal to 2/5, the distance between A and B when B did not have a velocity.
and you transform this to find the "time" t
Then you take the next equation: t' = γ (t - vx/c^{2})
Once more you use 1/5 instead of 2/5 for x and then you find somehow for this special case that t' = t.
So finally: you just found a special case which, when you do the transformation incorrectly, you get that t' = t, which has nothing to do with reality or with SR being incorrect.
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Many apologies. I assumed it was understood the time starts when the origins are at the same location. So, the interval for this is from the time the origins are same to the time the two clocks are at the same location.
I can show more. How much elapsed time from the origins at the same location and moving observer ( -1/5, 0, 0 ) at the same location as ( 1/5, 0, 0 ).
( -1/5, 0, 0 ) is moving so length contraction.
In time t at v = 3/5c, ( -1/5, 0, 0 ) must move a distance vt to be at ( 1/5, 0, 0 ).
vt = 1/5 - (-1/5)/(1 / √( 1 - vē/cē )) = 1/5 + 1/5(4/5) = 9/25
so (3/5)t = 9/25.
t = 3/5. This is a time interval. Then I used Einstein t' equation to get the moving clock interval.
I set x = vt just like Einstein and use Einstein equation.
Between the quantities x, t, and t', which refer to the position of the clock, we have, evidently, x=vt and
t' = t' = ( t - vx/cē )[1 / √( 1 - vē/cē )]
Therefore,
t' = t √( 1 - vē/cē ) = t - (1 - t √( 1 - vē/cē ))t
whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 - t √( 1 - vē/cē ) seconds per second.
http://www.fourmilab.ch/etexts/einst...el/specrel.pdf
I proved clock moved from point A to point B and moving clock and rest clock both elapsed 3/5s. Moving clock moved a distance vt just like Einstein said. You did not prove this wrong.
I used x' = γ (x - v t) because it you know x and x' then you solve to t. That is how long it takes for x' to move to x when time begins when origins at the same location.
x' / γ = x - vt,
t = ( x - x' / γ ) / v
Then I use Einstein to find the elapsed time to the moving clock at (-1/5, 0, 0) in the moving coordinates to go to (1/5, 0. 0).
t' = ( t - vx/cē )[1 / √( 1 - vē/cē )]
Einstein does not make only coordinates at the origin. I do not understand why you use 2/5.
Can we determine what it is that you accept, i.e. what can we use as a starting point?
1) Do you accept the validity of the two postulates of SR, e.g. speed of light in vacuum being the same in all inertial frames, etc?
2) Do you accept that the Lorentz transform follows from the two postulates of SR?
No that is incorrect. That is NOT how long it takes to come from x' to x, that is easily calculated by (x' - x) / v = 2/5 / 3/5 = 2/3 seconds
That whole interpretation does not make any sense at all.
so you forget simply gamma here?x' / γ = x - vt,
t = ( x - x' / γ ) / v
I get the same time as you without the gamma.
I used 2/5 because that is the distance between 1/5 and -1/5.Then I use Einstein to find the elapsed time to the moving clock at (-1/5, 0, 0) in the moving coordinates to go to (1/5, 0. 0).
t' = ( t - vx/cē )[1 / √( 1 - vē/cē )]
Einstein does not make only coordinates at the origin. I do not understand why you use 2/5.
Sorry to say that your whole setup does not make sense and your explanation cannot be understood.
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It is getting worse and worse.
Nope. tusenfem just showed you that this is wrong. Twice.In time t at v = 3/5c, ( -1/5, 0, 0 ) must move a distance vt to be at ( 1/5, 0, 0 ).
So, you do not understand what you are doing. Thank you.vt = 1/5 - (-1/5)/(1 / √( 1 - vē/cē )) = 1/5 + 1/5(4/5) = 9/25
so (3/5)t = 9/25.
t = 3/5. This is a time interval.
But you used it in an incorrect way. Several of us showed you several different ways that you don't know what you are doing. tusenfem has made the effort to do it correctly for you, twice.Then I used Einstein t' equation to get the moving clock interval.
Here it is again, in symbolic form only:
so:
When you try to measure time dilation , you need to do the measurements in the same place in the primed frame, so dx'=0. This means dx=vdt. (tusenfem showed you that dx=2/5 and dt=dx/v=2/3 in an earlier post). Substitute in the second expression and you get:
dt'=\gamma (dt-(v/c)^2*dt)=dt/\gamma
or:
dt=\gamma *dt'
So, dt' and dt are not equal (unless v=0 and \gamma=1).
Last edited by macaw; 2010-Nov-26 at 05:35 PM.