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Thread: Relative v > c

  1. #91
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    Quote Originally Posted by NorthernBoy View Post
    CERN. Except that we get much, much closer to c than one part in one hundred there.

    Did you seriously not know that this is what we did there?
    I do not have a problem with that.

    That has nothing to do with what we are taking about.

    We are not talking about the absolute standard of the velocity limiting c.

    We are talking about total relative velocity.

    These are two different concepts.

  2. #92
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    Quote Originally Posted by abcdefg View Post
    I am trying to say if a particle is moving in the positive x-direction at .99c and there is spherical acceleration in all directions off that particle, then those accelerated in the direction of the positive x-axis are limited to less than .01c.
    Yes, if you believe in Ritz theory. No, if you know relativity. Speeds do not just add up in the real world, the composition for speed is more complicated.
    So, with v1=.99c and with:

    w=(v1+v2)/(1+v1*v2/c^2)

    you have to answer the new question:

    Q8: what is the correct upper limit for v2? (Hint, it is not .01c as you claimed)


    Those in the negative x-axis direction are limited to -1.99c.
    Nope.

    Q9: what is the correct upper limit for v2 in the negative x direction. Hint, it is not -1.99c

    So, you now have Q1-Q9 to answer

  3. #93
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    Quote Originally Posted by macaw View Post
    Based on what you have posted, you are. So where is the answer to Q1-Q7?
    Think this though.

    I posted the light would catch the other rocket.
    Clearly, this implies I support an absolute constant c and not one riding with the frame.

  4. #94
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    Quote Originally Posted by abcdefg View Post
    Well, first off, there is a point as they are accelerating at which from rocket to rocket, the instantaneous v of each is .5 c. Taking the position of either rocket, the other rocket disappears since length contraction becomes 0. After that point, length contraction crosses into the complex number system.

    Next, if I shot a laser from the front of the rocket to the back and out some hole toward the other rocket and I had clocks measuring the time such that I calculate the speed if the light, a speed of c to the frame would never reach the other rocket moving at relative speed v = 1.98c. It is claimed the speed of light to an inertial frame is always c.
    wow your more confused in the post then your last big post.

  5. #95
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    Quote Originally Posted by abcdefg View Post
    Think this though.

    I posted the light would catch the other rocket.
    Clearly, this implies I support an absolute constant c and not one riding with the frame.
    Just answer Q1-Q9. Ok?

  6. #96
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    Quote Originally Posted by abcdefg View Post
    This calculation does not control the speed of the rockets. These rockets are at .99c.
    Only to a third observer on the earth

    To them self their v = 0 and the other is at 0.99994949750012625624968435937579c


    Quote Originally Posted by abcdefg View Post
    For example, here lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.

    This is legitimate.
    No this is you confused and not understanding SR

  7. #97
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    Quote Originally Posted by WayneFrancis View Post

    To them self their v = 0 and the other is at 0.99994949750012625624968435937579c
    Nice.

    No this is you confused and not understanding SR
    It is precisely Ritz' theory.

  8. #98
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    Quote Originally Posted by WayneFrancis View Post
    Wrong, their relative speed to each other is not 1.98c
    Their relative speed is computed by the formula
    V1 + V2 / ( 1 + (V1V2/c2))

    so it is .99 + .99 / (1 + .992/c2)
    1.98/ (1 + 0.9801)
    or 1.98/1.9801
    or 0.99994949750012625624968435937579c

    0.99994949750012625624968435937579c != 1.98c



    So you've shown you don't know what formulas to use...congratulations.
    Thanks you.

    Now, I will see if I can get your answer to work.

    I start with ship A which is the one moving left.

    I burn to attain 0.99994949750012625624968435937579c. Now, I should be in the same frame as ship B.

    Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.99c which is close to the original launch frame or close to the earth frame. I did not make it into the frame of ship B because it is traveling at .99c relative to the original launch frame.
    So, this is not the total relative velocity.

    Your equation did not work.

  9. #99
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    Quote Originally Posted by WayneFrancis View Post
    wow your more confused in the post then your last big post.
    OK, I will redo it.

  10. #100
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    Quote Originally Posted by abcdefg View Post
    Answer to Q7
    I do not know where you got the above and so I would have a hard time proving it to be false.
    Exactly, you don't know what formula to use thus you can't prove it false. But you've been given the formula. The fact you don't understand it is another fact. You can lead a horse to water...



    Quote Originally Posted by abcdefg View Post
    Further, I do not know how you would control the speed of the rockets with each going .99c in opposite directions by imposing an equation on them.
    .99c according to who? The observer on Earth. The rest of your complaint seems to be leaving something out between your brain and what your fingers typed. You might as well complain about ice cream tasting blue.

  11. #101
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    Quote Originally Posted by macaw View Post
    Nice.



    It is precisely Ritz' theory.
    Ritz' theory of light emission has to do with light spedd adding to the source speed. We are talking about frames.

    I will let Einstein tell you about the independence between the two concepts of light and frames.

    Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate."
    http://philsci-archive.pitt.edu/arch.../01/Norton.doc

  12. #102
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    Quote Originally Posted by abcdefg View Post
    I do not have a problem with that.

    That has nothing to do with what we are taking about.
    Of course it does. We have one particle travelling at nearly c in one direction, and one travelling at c in the opposite direction, and you are asking questions about their relative speed.

    How is this not analagous to your two rockets problem?

  13. #103
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    Quote Originally Posted by abcdefg View Post
    Thanks you.



    Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.
    You can't, you are subtracting speeds again.
    Q10: What is the correct computation?

    Your equation did not work.
    The questions are accumulating, you now have Q1-Q10 to answer.

  14. #104
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    Quote Originally Posted by abcdefg View Post

    I will let Einstein tell you about the independence between the two concepts of light and frames.
    That would not be necessary, please answer Q1-Q10.

  15. #105
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    Quote Originally Posted by WayneFrancis View Post
    .99c according to who? The observer on Earth.
    He's still doing it, isn't he, refusing to add the critically important fact about which observer he's talking about.

    I do hope he stops soon.

  16. #106
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    Quote Originally Posted by abcdefg View Post
    OK, let's take that particle at 0.99999c amd accelerate it to 0.99999c just as everyone has been telling me.

    If this acceleration is in the direction of the positive x-axis, then you make sense if the first particle is moving in that direction. But, if the particle is shot in the direction of the negative x-axis, then this shot particle will return to the earth frame.

    So, velocity additions are directionally based.
    pleas read http://en.wikipedia.org/wiki/Velocity

  17. #107
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    Quote Originally Posted by macaw View Post
    Yes, if you believe in Ritz theory. No, if you know relativity. Speeds do not just add up in the real world, the composition for speed is more complicated.
    So, with v1=.99c and with:



    w=(v1+v2)/(1+v1*v2/c^2)

    you have to answer the new question:

    Q8: what is the correct upper limit for v2? (Hint, it is not .01c as you claimed)
    Answer Q8

    Now, let's say what this means.

    First there exists a rest frame and two other observers one at v1 and one at v2.

    So, this is a 3 observer model. So, right off it is wrong.
    But, under this false model as compared to the one offered on this thread is.

    w=(v1+v2)/(1+v1*v2/c^2) < c

    (v1+v2) < c( (1+v1*v2/c^2) )
    c(v1+v2) < c^2 + v1*v1
    cv1 + cv2 < c^2 + v1 * v2

    cv2 - v1 * v2 < c^2 - cv1
    v2( c - v1 ) < c( c - v1 )

    v2 < c.

  18. #108
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    Quote Originally Posted by abcdefg View Post
    Thanks you.

    Now, I will see if I can get your answer to work.

    I start with ship A which is the one moving left.

    I burn to attain 0.99994949750012625624968435937579c. Now, I should be in the same frame as ship B.
    You are not getting it. I'll explain this fully, unlike you.

    Ships A and B launch in opposite directions.
    Ships C launches with B in the same direction of B
    All ships appear to accelerate to .99c from an observer on Earth.

    After they all accelerated up to .99c, from the point of view of the Earth, The following will be true

    A observes B & C speeding away from them at 0.99994949750012625624968435937579c
    A observes Earth speeding away from them at .99c
    B & C observe A speeding away from them at 0.99994949750012625624968435937579c
    B & C observe Earth speeding away from them at .99c

    If at this point in time C starts accelerating toward A then B would observe C to have the same time and space contraction as A when it reached a relative velocity, from B, of 0.99994949750012625624968435937579c.

    This is not the same as being in the same frame becuase A & C are still seperated by a large distance and it would require more changes of velocity from one or both of the ships, A & C, before they are actually in the same frame.

    This all brings us to your last big post where we kept explaining to you that you where not taking into account all the pertinent details.

    Quote Originally Posted by abcdefg View Post
    Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.99c which is close to the original launch frame or close to the earth frame. I did not make it into the frame of ship B because it is traveling at .99c relative to the original launch frame.
    So, this is not the total relative velocity.

    Your equation did not work.
    No you have mixed frames. IE if you don't know how to apply the formulas then yes you will get garbage answers. So once again you have only shown that you are not educated in SR. No big revelation to the rest of us.

  19. #109
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    Quote Originally Posted by macaw View Post
    Yes, if you believe in Ritz theory. No, if you know relativity. Speeds do not just add up in the real world, the composition for speed is more complicated.
    So, with v1=.99c and with:

    w=(v1+v2)/(1+v1*v2/c^2)

    you have to answer the new question:

    Q8: what is the correct upper limit for v2? (Hint, it is not .01c as you claimed)





    Nope.

    Q9: what is the correct upper limit for v2 in the negative x direction. Hint, it is not -1.99c

    So, you now have Q1-Q9 to answer
    Answer Q9
    Correct limit in the negative direction -c.

    Now, what is the total relative velocity differential?

    |-c| + |c| = 2c.

    This is the most general equation to describe the total relative velocity change to a moving particle.

    In general, the following equation holds for any alteration to the speed of a relativistic particle.

    -c < v0 + v cos(Θ) < c.

    where v0 is the speed of the relativistic particle

    and Θ is the angle to the direction of the particle/object's travel.

  20. #110
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  21. #111
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    Quote Originally Posted by abcdefg View Post
    Answer Q8

    Now, let's say what this means.

    First there exists a rest frame and two other observers one at v1 and one at v2.

    So, this is a 3 observer model. So, right off it is wrong.
    But, under this false model as compared to the one offered on this thread is.

    w=(v1+v2)/(1+v1*v2/c^2) < c

    (v1+v2) < c( (1+v1*v2/c^2) )
    c(v1+v2) < c^2 + v1*v1
    cv1 + cv2 < c^2 + v1 * v2

    cv2 - v1 * v2 < c^2 - cv1
    v2( c - v1 ) < c( c - v1 )

    v2 < c.
    Good, you got the right answer. Now , let's see the answers to the other questions, Q1-Q7,Q9-10. Looks like you are contradicting yourself, you are back to the stance v>c. Why?

  22. #112
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    Quote Originally Posted by abcdefg View Post
    Answer Q9
    Correct limit in the negative direction -c.
    Q11: How did you get the answer? Show the steps, like in Q8.

    Now, what is the total relative velocity differential?

    |-c| + |c| = 2c.
    The above is closing speed, not relative speed.

    Q12: Explain the difference between closin and relative speed so we can see that you are starting to learn.

  23. #113
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    Bad answers to all of them. Especially since you are contradicting the correct answers for Q8,9 you just managed to find. You realize that you are just contradicting your own solutions, right? Try again.
    Last edited by macaw; 2009-Oct-29 at 07:55 PM.

  24. #114
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    Quote Originally Posted by abcdefg View Post
    Thanks you.

    Now, I will see if I can get your answer to work.

    I start with ship A which is the one moving left.

    I burn to attain 0.99994949750012625624968435937579c. Now, I should be in the same frame as ship B.

    Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.99c which is close to the original launch frame or close to the earth frame. I did not make it into the frame of ship B because it is traveling at .99c relative to the original launch frame.
    So, this is not the total relative velocity.

    Your equation did not work.
    I think you are missing that Velocities are relative and not additive. Ship 1 and 2 are moving each at .99c in oppostie (or some) direction relative to a rest frame. Relative to each other, they are moving at 0.99994949750012625624968435937579c. However when Ship 1 measures the distance covered between itself and Ship 2, relative to thier originating point and from its point of view, then it can calculate that to an apparent seperation velosity of 1.98c.

    The reaosn this happens is that both Ship 1 and Ship 2 when they measure from thier point of view, are in both a time dilated state and length contracted state. In this state space and time are compressed to the point that the other ship is moving at 0.99994949750012625624968435937579c relative to them and the originating point is moving at .99c relative to them.

    Another way to look at it is like this. Motion=energy=mass (simplified). As something moves faster it increases in energy. According to e=mc2 it means it is increasing in mass. As it's mass increases, it's gravitional effects on space/time are larger also. The faster something moves, the more it -warps- space time. Time moves slower for them, while at the same time, they are moving though compressed (less) space.

    If somehow you could have any object with mass at a relative velosity of >=c you have a problem. At light speed a object with mass has infinate energry, which also implies, infinate mass and infinate gravity. The moment any object with mass reached lightspeed, it would collapse into a black hole like object from it's own infinate gravity.

    You can have appearent seperation velocities that seem like they are >c, but the actual velocities of any object with mass must always be < c. Only massless particles or waves (photons) can have a v=c, and only objects with negative masses (hypothetical tachyons) can have a v>c .

  25. #115
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    Quote Originally Posted by WayneFrancis View Post
    Exactly, you don't know what formula to use thus you can't prove it false. But you've been given the formula. The fact you don't understand it is another fact. You can lead a horse to water...





    .99c according to who? The observer on Earth. The rest of your complaint seems to be leaving something out between your brain and what your fingers typed. You might as well complain about ice cream tasting blue.
    I said this already, .99c according to the acceleration equations.
    v(BT) = c tanh(a BT/c)

  26. #116
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    Quote Originally Posted by dgavin View Post
    I think you are missing that Velocities are relative and not additive. Ship 1 and 2 are moving each at .99c in oppostie (or some) direction relative to a rest frame. Relative to each other, they are moving at 0.99994949750012625624968435937579c. However when Ship 1 measures the distance covered between itself and Ship 2, relative to thier originating point and from its point of view, then it can calculate that to an apparent seperation velosity of 1.98c.
    Correct.

    The reaosn this happens is that both Ship 1 and Ship 2 when they measure from thier point of view, are in both a time dilated state and length contracted state.

    Incorrect.


    Another way to look at it is like this. Motion=energy=mass (simplified). As something moves faster it increases in energy. According to e=mc2 it means it is increasing in mass.

    Incorrect.

    As it's mass increases, it's gravitional effects on space/time are larger also.

    A classical fallacy.


    The faster something moves, the more it -warps- space time. Time moves slower for them, while at the same time, they are moving though compressed (less) space.
    Another fallacy.


    At light speed a object with mass has infinate energry, which also implies, infinate mass and infinate gravity.

    Another fallacy.

    The moment any object with mass reached lightspeed, it would collapse into a black hole like object from it's own infinate gravity.
    Yet another incorrect statement.


    You can have appearent seperation velocities that seem like they are >c,

    Actually, they are.

  27. #117
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    Quote Originally Posted by NorthernBoy View Post
    Of course it does. We have one particle travelling at nearly c in one direction, and one travelling at c in the opposite direction, and you are asking questions about their relative speed.

    How is this not analagous to your two rockets problem?

    Well, I an in one rocket and I know my speed is v(BT) = c tanh(a BT/c) according to the acceleration equations.

    This rocket is B moving to the right.

    Now, I apply a reverse burn of a and BT and then the speed is -v(BT)=-.99c from its original value. This only made it to the earth frame. This is not enough.

    Now, I have to do it again, a burn of a at BT. At this point, the rocket B is in the frame of A and the total speed change from the original B frame is -v(BT) - v(BT) = -2*v(BT) = -2*.99c = -1.98c

    The total relative speed between the two frames is 1.98c.

  28. #118
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    Quote Originally Posted by abcdefg View Post

    total speed change from the original B frame is -v(BT) - v(BT) = -2*v(BT) .
    Nope, it isn't, you are making the same mistake again. There is no surprise you are getting wrong answers. So, why don't you go back and answer Q1-Q12. Please do not point to your previously already refuted posts. Answer in line.

    I know my speed is v(BT) = c tanh(a BT/c) according to the acceleration equations.
    What you don't know is that the above has been derived using the correct speed composition formula, i.e. w=(u+v)/(1+uv/c^2) and not the incorrect one that you insist on using. I gave you the site where you learned it. The irony is that you sometimes even link in the site. See the first formula in the derivation?

  29. #119
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    Out of interest I used the (correct) formula w=(v1+v2)/(1+v1*v2/c^2) to calculate the relative closing speed of two cars heading for a head-on crash at 100 km/h each.

    Turned out they were closing at what seemed to be 200 km/h as we'd "expect". So I then multiplied their speed by 10 and re-calculated...

    Code:
    All speeds in km/h
    
                     c                
    
     1,079,252,848.800                 
    
                                                     (correct)        ("expected")      (difference)
                    v1                  v2                   w               v1+v2         (v1+v2)-w
    
               100.000             100.000             200.000             200.000             0.000 
             1,000.000           1,000.000           2,000.000           2,000.000             0.000 
            10,000.000          10,000.000          20,000.000          20,000.000             0.000 
           100,000.000         100,000.000         199,999.998         200,000.000             0.002 
         1,000,000.000       1,000,000.000       1,999,998.283       2,000,000.000             1.717 
        10,000,000.000      10,000,000.000      19,998,283.095      20,000,000.000         1,716.905 
       100,000,000.000     100,000,000.000     198,297,563.311     200,000,000.000     1,702,436.689 
     1,000,000,000.000   1,000,000,000.000   1,076,121,453.793   2,000,000,000.000   923,878,546.207
    (scroll right)


    Turns out they need to be going pretty darn fast before the relativistic effects become noticable. (Note how w tends to c).

    It's hardly surprising that people need to cling to the idea that relative speed is v1 + v2 (or v1 - v2, whatever). It works fine for the speeds we deal with in normal life. I remember myself in high school absolutely refusing to "believe" my physics teacher when he pointed this out to me. I'd have learned more, and faster, if I were more open to change of my preconceptions.
    I don't see any Ice Giants.

  30. #120
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    Quote Originally Posted by pzkpfw View Post
    Out of interest I used the (correct) formula w=(v1+v2)/(1+v1*v2/c^2) to calculate the relative closing speed of two cars heading for a head-on crash at 100 km/h each.

    I'd have learned more, and faster, if I were more open to change of my preconceptions.
    I find it ironic that abcdefg is attempting to use the formula for accelerated motion that was derived based on the relativistic speed composition in order to .... prove his incorrect use of the galilean speed composition.
    Last edited by macaw; 2009-Oct-29 at 09:14 PM.

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