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Thread: Relative v > c

  1. #1
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    Relative v > c

    We have two identical rockets.

    One rocket is located on one side of the earth and one is located directly on the other side of the earth.

    At a predetermined time, each takes off with burns such that they each reach 0.99c, one rocket is therefore going along the position x-axis and the other is heading along the negative x-axis.

    Question 1: What is the relative speed of the rockets? 1.98c

    Question 2: If each shot a laser at each other as a signal, will they ever see the light? Yes.

  2. #2
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    What is your point?

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    Quote Originally Posted by abcdefg View Post
    Question 1: What is the relative speed of the rockets? 1.98c
    Of course not, because you can't just add the velocities like that. There is a factor in there involving a v^2/c^2 (I forget what exactly) which always limits the addition of velocities to a maximum of c. So if you are in one rocket, you see the Earth moving way at 0.99c, and the other rocket moving away in the same direction at something like 0.995c. This answers the second question as well.

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    He's putting this issue under a third observer, here we go again

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    Quote Originally Posted by Kwalish Kid View Post
    What is your point?
    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.

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    Quote Originally Posted by zerocold View Post
    He's putting this issue under a third observer, here we go again
    No I am not. I am taking the position of either of the rockets.

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    Quote Originally Posted by Perikles View Post
    Of course not, because you can't just add the velocities like that. There is a factor in there involving a v^2/c^2 (I forget what exactly) which always limits the addition of velocities to a maximum of c. So if you are in one rocket, you see the Earth moving way at 0.99c, and the other rocket moving away in the same direction at something like 0.995c. This answers the second question as well.
    This calculation does not control the speed of the rockets. These rockets are at .99c.

    For example, here lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.

    This is legitimate.

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    Quote Originally Posted by abcdefg View Post
    Taking the position of either rocket, the other rocket disappears since length contraction becomes 0.
    Bad word salad, as usual.

    Q1: Prove that "the other rocket disappears"
    Q2: prove that "length contraction becomes zero"

    After that point, length contraction crosses into the complex number system.
    Wrong again.
    Q3: Prove that "length contraction crosses into complex numbers"


    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.

    Bzzt, wrong, it reaches the other rocket. I know that this is a shock to you but physics doesn't work the way you confabulate.


    It is claimed the speed of light to an inertial frame is always c.
    ...and it is. The fact that you don't know this, is just your perenial problem.

  9. #9
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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.

    For example, here lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.
    So what? the light speed is still c in the respective experiments. That is, unless you are abcdefg and you don't understand what the text is saying.

  10. #10
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    Let me understand where is the "contradiction" ,not saying if is right or wrong, just trying to understand where is your ultimate goal.

    The relative speed between the 2 spaceships can't be higher than c

    The relative speed between a 3th guy, can be

    The light will achieve the spaceships in the frame between them

    The light will not reach the spaceships in a observer's frame

    That is?, not saying is right or wrong, just waiting for the debate...or macaw

    Edit, lol he came faster than i expected

  11. #11
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    Quote Originally Posted by abcdefg View Post
    No I am not. I am taking the position of either of the rockets.
    No, you are not. You are taking the position of a third entity, the first rocket is moving with +.99c, the seconf rocket is moving with -.99c producing what is known as a closing speed of 1.98c wrt the said third entity. But you wouldn't happen to know that and you are inadvertently think otherwise as reflected by your new "discoveries".

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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.
    .
    That is a particularly meaningless statement if you do not define your frame of reference. If you are standing on the surface of the Earth, the statement is true. The calculation does 'control' the speed of the rocket, in the sense that in the frame of reference of one rocket, it 'controls' the speed of the second rocket as measured in that reference frame. If you do not grasp this, there is not much hope.

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    Quote Originally Posted by zerocold View Post

    That is?, not saying is right or wrong, just waiting for the debate...or macaw

    Edit, lol he came faster than i expected
    Faster than the speed of light

  14. #14
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    lol, i hope the debate will be civilized

    These are cool debates, i must say

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    Quote:
    Originally Posted by abcdefg
    Taking the position of either rocket, the other rocket disappears since length contraction becomes 0.
    Quote Originally Posted by macaw View Post
    Bad word salad, as usual.

    Q1: Prove that "the other rocket disappears"
    Q2: prove that "length contraction becomes zero"
    Answer Q1, Q2:
    Let's look at length contraction from either rocket.

    Obviously, as each burns to .99c and exactly in the same way in opposite directions, there exists an instant where the relative v is c, ie one is moving in the direction of the negative x-axis and one in the positive direction.

    L' = L(sqrt( 1 - (v/c)2) = L(sqrt( 1 - (c/c)2) = L*0 = 0.

    That implies the length of the rocket becomes 0. Now, a length of zewro means the rocket disappears.

  16. #16
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    Quote Originally Posted by abcdefg View Post
    Answer Q1, Q2:
    Let's look at length contraction from either rocket.

    Obviously, as each burns to .99c and exactly in the same way in opposite directions, there exists an instant where the relative v is c, ie one is moving in the direction of the negative x-axis and one in the positive direction.

    L' = L(sqrt( 1 - (v/c)2) = L(sqrt( 1 - (c/c)2) = L*0 = 0.
    Bad answer:

    1. Massive objects cannot attain c.
    2. Length contraction is a relativistic effect, it does not happen in the proper frame of the rocket, so contrary to your new "discovery", the length of the rocket is still...L :-)


    That implies the length of the rocket becomes 0.
    Nope, bad juju. See above.


    Now, a length of zewro means the rocket disappears.
    Still bad juju. You failed Q1,2. Let's see what you do with Q3.

  17. #17
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    Quote Originally Posted by macaw View Post

    Wrong again.
    Q3: Prove that "length contraction crosses into complex numbers"
    [Edit] Answer to Q3

    Oh, once each rocket reaches .99c and rocket A decides to perform a length contraction calculation, it is

    L' = L sqrt( 1 - (v/c)2 ) = L sqrt( 1 - (1.98c/c)2 ) )
    = L sqrt( -2.9204 )

  18. #18
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    Quote Originally Posted by macaw View Post
    So what? the light speed is still c in the respective experiments. That is, unless you are abcdefg and you don't understand what the text is saying.
    I am betting I know exactly what I am saying.

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    Quote Originally Posted by macaw View Post
    No, you are not. You are taking the position of a third entity, the first rocket is moving with +.99c, the seconf rocket is moving with -.99c producing what is known as a closing speed of 1.98c wrt the said third entity. But you wouldn't happen to know that and you are inadvertently think otherwise as reflected by your new "discoveries".
    Sorry, I am not trying do "discoveries" in this ATM.

    I am not taking the position of a 3rd entity. I am taking the position of a rocket. The relative spedd is 1.98c from one rocket to another.

  20. #20
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    Quote Originally Posted by abcdefg View Post
    Oh, once each rocket reaches .99c and rocket A decides to perform a length contraction calculation, it is

    L' = L sqrt( 1 - (v/c)2 ) = L sqrt( 1 - (1.98c/c)2 ) )
    = L sqrt( -2.9204 )
    Bad juju again. The v that goes in the length contraction formula is the relative speed. You are plugging in the [b]closing[b] speed and you wonder why you are getting rubbish . Take a deep breath, read the difference between the two speeds and try again.

    Q4: What is the relative speed between the rockets?
    Q5: What is the correct length contraction?

    Please do not rush trying to answer with another set of mistakes. Your elementary mistakes are not "discoveries".

  21. #21
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    Quote Originally Posted by abcdefg View Post
    Sorry, I am not trying do "discoveries" in this ATM.

    I am not taking the position of a 3rd entity.
    Then you don't really know what you are doing <shrug>

    I am taking the position of a rocket. The relative spedd is 1.98c from one rocket to another.
    Nope, it isn't. Try again. Correctly this time. pay attention, Perikles already clued you in with the correct formula.

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    Quote Originally Posted by zerocold View Post
    Let me understand where is the "contradiction" ,not saying if is right or wrong, just trying to understand where is your ultimate goal.

    The relative speed between the 2 spaceships can't be higher than c

    The relative speed between a 3th guy, can be

    The light will achieve the spaceships in the frame between them

    The light will not reach the spaceships in a observer's frame

    That is?, not saying is right or wrong, just waiting for the debate...or macaw

    Edit, lol he came faster than i expected
    Edit, lol he came faster than i expected
    LOL he was quick was he not?

    My ultimate goal is to show relativity cannot handle this motion and to show it must be stated that show SR applies only to relative v < c.

    Next, I want to show the speed of light is an absolute constant, otherwise causality is broken.

    Light will reach the other spaceship.

  23. #23
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    Quote Originally Posted by abcdefg View Post
    Obviously, as each burns to .99c and exactly in the same way in opposite directions, there exists an instant where the relative v is c, ie one is moving in the direction of the negative x-axis and one in the positive direction.
    This does not happen - read my post above.
    Quote Originally Posted by abcdefg View Post
    I am betting I know exactly what I am saying.
    Possibly, but it is complete nonsense. I can see now why others are not posting in this thread.

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    Quote Originally Posted by macaw View Post
    Bad answer:

    1. Massive objects cannot attain c.
    2. Length contraction is a relativistic effect, it does not happen in the proper frame of the rocket, so contrary to your new "discovery", the length of the rocket is still...L :-)
    Good answer I had.
    Neither rocket attained c. I thought I stated that clearly.

  25. #25
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    You are wrong on this abc the speed between the 2 ships should be 0.99999XXXc, in their frame

  26. #26
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    Quote Originally Posted by Perikles View Post
    Of course not, because you can't just add the velocities like that. There is a factor in there involving a v^2/c^2 (I forget what exactly) which always limits the addition of velocities to a maximum of c. So if you are in one rocket, you see the Earth moving way at 0.99c, and the other rocket moving away in the same direction at something like 0.995c. This answers the second question as well.
    Yep. Hint for abcdefg:

    Rocket 1 moves with v1=+v wrt the observer
    Rocket 2, moves with v2=-v wrt the observer. This is equivalent with saying that the observer is moving with +v wrt Rocket 2.

    So, we have the situaltion:

    Observer moving with +v wrt Rocket 2
    Rocket 1 moving with +v wrt observer.

    Q6 for abcdefg: What is the speed of Rocket 1 wrt Rocket 2. Please do not answer 2v.

  27. #27
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    Quote Originally Posted by abcdefg View Post
    Good answer I had.
    Neither rocket attained c. I thought I stated that clearly.
    Then they definitely cannot attain 1.98 c, right?

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    Quote Originally Posted by abcdefg View Post
    [My ultimate goal is to show relativity cannot handle this motion and to show it must be stated that show SR applies only to relative v < c.
    You failed. Now you have to answer Q1-Q6.

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    Quote Originally Posted by zerocold View Post
    You are wrong on this abc the speed between the 2 ships should be 0.99999XXXc, in their frame
    True but he doesn't want to learn. He'd much rather post mistakes and try to pass them as "discoveries".

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    Quote Originally Posted by macaw View Post
    Bad juju again. The v that goes in the length contraction formula is the relative speed. You are plugging in the [b]closing[b] speed and you wonder why you are getting rubbish . Take a deep breath, read the difference between the two speeds and try again.

    Q4: What is the relative speed between the rockets?
    Q5: What is the correct length contraction?

    Please do not rush trying to answer with another set of mistakes. Your elementary mistakes are not "discoveries".

    Answer Q4. I answered 1.98c.
    Here, if you have a car driving one way on a highway and one the other each at .4c, the relative speed between them is .8c. You add them in this case.

    Now, if I had a car driving at .4c and tried to accelerate to c I could not do this. That would product v < c because nother travels faster than the spedd of light.

    You see, I am looking past this though. Neither are breaking this speed limit rule, so the answer is 1.98c.

    Answer to Q5: What is the correct length contraction?
    I already this answered and I am not going to change the answer.
    L' = L sqrt( -2.9204 ).

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