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Thread: Relative v > c

  1. #61
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    Quote Originally Posted by dgavin View Post
    Because Ship 1 and Ship 2 have the same relativistic frame, thier clocks are the same.
    So two ships moving apart at close to c both have the same relativistic frame?

    This is a very ATM claim, and you should start your own thread to discuss it.

    The rest was soo poorly written as to be close to meaningless, I'm afraid.
    Last edited by NorthernBoy; 2009-Oct-29 at 04:38 PM.

  2. #62
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    Quote Originally Posted by macaw View Post
    So, we are done, you realize that all you have been posting is mistakes. You can learn how to correct them by reading (and learning) a chapter of relativistic kinematics. Start with the correct derivation of relative speed



    You don't. The laws of physics control the speed. You only need to make the effort to learn them. Much better use of your time than posting elementary mistakes that don't pass the mustard.
    In the design of LT and the kinematics of SR
    Einstein:
    Let there be given a stationary rigid rod; and let its length be l as measured by a measuring-rod which is also stationary. We now imagine the axis of the rod lying along the axis of x of the stationary system of co-ordinates, and that a uniform motion of parallel translation with velocity v along the axis of x in the direction of increasing x is then imparted to the rod. We now inquire as to the length of the moving rod, and imagine its length to be ascertained by the following two operations:--

    Furthermore,

    Einstein stated tB - tA = rAB/(c-v)

    As such, it is hypothesized that v < c or you come up with a negative time.
    Thus, all kinematical calculations are under the postulate that v < c.
    So, your equation is simply built under that condition and that is proof by vacuous implication. In oither words, you have built into your equations that v < c and thus, it is forced to be true by assumption.

    These v's are calculated according to the constant acceleration of a and BT. They sum to 1.98c.

    In addition in your kinematics, light leaves rocket one at c regardless of the motion of the rocket. It arrives at the other rocket at a distance t(c + .99c) in an effort to determine dx/dtau. Well, you lost the distance the distance the light source rocket traveled during that time with this analysis. So, you only have half the total distance.

  3. #63
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    Quote:
    Originally Posted by abcdefg

    Einstein wrote (Einstein Archive, 20-040; translation based on Stachel, 2002, p. 189):Einstein: Your attempt to replace special relativity with the assumption that the velocity of light is constant relative to the source of light was first advocated by Ritz. This assumption is compatible with Michelson’s experiment and with aberration.
    http://philsci-archive.pitt.edu/arch.../01/Norton.doc
    Quote Originally Posted by macaw View Post
    ...but incompatible with Sagnac, Fizeau and Ives-Stilwell and a bunch of other experiments. Nice try but you fail again
    I am not a Ritz person.

    I am pointing out Einstein's position. So your comment does not apply to me.

    In particular, I posted my agreemet with this statement.


    Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate and I considered what would be more probable, the principle of the constancy of c, as was demanded by Maxwell’s equations, or the constancy of c, exclusively for an observer sitting at the light source. I decided in favor of the first, since I was convinced that each light [ray] should be defined by frequency and intensity alone, quite independently of whether it comes from a moving or a resting light source."
    http://philsci-archive.pitt.edu/arch.../01/Norton.doc

    Key Concept: Special Relativity in One Sentence.
    All speeds are relative, except for the speed of light, which is absolute.
    http://www.astronomy.ohio-state.edu/...6/notes23.html


    In particular, I do not run with the argument that the speed of light is c in the frame, that is light emission theory.

  4. #64
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    Quote:
    Originally Posted by iquestor
    MDT-1 said:

    To me, this is the important point. Once the laser is fired from rocket 1, it will approach rocket 2 at the speed of light. The relative velocity of the two rockets are entirely moot, since the laserlight is not part of either system. This is a basic tenant of TOR. Therefore, Since the laster travels at c, and c > .99c, it will overtake the rocket.
    Quote Originally Posted by macaw View Post
    Yes, of course but it will take abcdefg another 1500 posts to get this concept since he already "knows" that relativity is wrong :-)
    Perhaps you should have read my opening statement.

    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.

  5. #65
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    Quote Originally Posted by NorthernBoy View Post
    Oh god.

    OK, here's how this one works. I see one rocket travelling at 0.99c to the left, and one travelling at 0.99c to the right. In my frame, the distance between them does indeed increase at pretty much 2c. No problem there, nothing in my frame is moving at greater than c, no laws being broken.

    If we instead ask what speed the observers in one of the rockets will see the other rocket moving at, then we work this out to be just a little under c. So, in each rocket's frame, nothing is moving at greater than c either.

    No paradox, no contradiction, no problem.
    Macaw already tried this.

    This is a 3rd observer.

    We are taking the position of either rocket and so this does not apply.

  6. #66
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    Quote Originally Posted by NorthernBoy View Post
    No, it is not, no matter how many times you assert otherwise.

    If the speed of one rocket viewed from earth is v, and the speed of the other is w (in opposite directions), then the speed of one rocket as seen from the other is

    (v+w)/(1-vw/c^2)

    You are incorrectly adding their speeds, so, of course, get meaningless answers.
    Again, you are introducing a 3rd observer in an attempt to understand the problem.

    There are only two observers and we can lose the earth and just have them start in a frame by themselves.

    There is no 3rd observer.

  7. #67
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    Quote Originally Posted by abcdefg View Post
    In oither words, you have built into your equations that v < c and thus, it is forced to be true by assumption.
    It is also observed to be true, which rather contradicts what you wrote up there.

    Take a particle that decays int two daughter particles, which will move at 0.99999c relative to it, and accelerate it to 0.99c, and we get the addition of velocities just as others have been telling you.

    I know that you don't like this fact, but we've been doing your rocket experiment millions of times per day in labs for years now, and always the numbers come out in line with standard SR and not in line with your bizarre distortion of it.

  8. #68
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    Quote Originally Posted by abcdefg View Post
    Again, you are introducing a 3rd observer in an attempt to understand the problem.

    There are only two observers and we can lose the earth and just have them start in a frame by themselves.

    There is no 3rd observer.
    No, I don't need a third observer. Set these rockets off as you say, and we know that each will view the speed of the other as being less than c. You are apparently claiming that each will see the other doing a speed greater than c, but you have not justified this.

    As I say above, we've done the experiment, and observed it.

  9. #69
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    Quote Originally Posted by NorthernBoy View Post
    Because, when doing so, you are applying an incorrect expression for addition of velocities. Several people have pointed it out to you.

    Instead of ignoring the point again and again, why not actually discuss the central disagreement for once.

    First, do you understand why people are criticising this step of your thought experiment?

    Secondly, can you please explain why you do not think that you should use the equation I give above for the relative velocities, but instead should use simple addition?

    This is as bad as someone claiming that adding a litre of water to a litre of ethanol gives you two litres of mixture.

    This is not a valid comparison above.

    Further, the addition of velocities does not apply because they are moving in opposite directions.

    I have already been through this with macaw.

    If you are going .4c and you shoot a particle at .8c in the same direction, the sum velocity will be < c.

    These two are moving in opposite directions at .99c. Now folks will assert this is a kind of sum velocity argument. But it is not simply because there is no addition to velocity in the same direction. Further, the addition of velocity lesson is simply to proven one object cannot exceed c which is fine and not done here.

  10. #70
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    Quote Originally Posted by NorthernBoy View Post
    No, I don't need a third observer. Set these rockets off as you say, and we know that each will view the speed of the other as being less than c. You are apparently claiming that each will see the other doing a speed greater than c, but you have not justified this.

    As I say above, we've done the experiment, and observed it.
    Where exactly do you have a reference frame moving to the left at .99c and it compares another moving to the right at .99c?

    You do not.

    I will think about doing the kinematics to this mathematically .

    I will be back.

  11. #71
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    Quote Originally Posted by abcdefg View Post
    Question 1: What is the relative speed of the rockets? 1.98c
    You see, this is your normal sneaky trick, that you arogantly assume no-one will pick you up on (despite everyone catching you in it every time).

    You, as always, leave out the question of which observer you are claiming the measurement for.

    The only observer who would measure a speed of separation of 1.98c is an observer on earth. As you yourself have repeatedly told us, your situation has no such observer, so no, the relative speed is not 1.98c according to either of your two observers.

    Why do you insist on always making this same dishonest switch? Why do you come up with a number that applies to one observer, and then either try to remove that observer, or to pretend that another one made it?

    If you are so convinced that you are right why do you have to use such a dishonest argument to try to prove your point? Do you see anyone else trying to do this? Do you see anyone else needing to underspecify the situation to leave wriggle room to avoid being pinned down?

    This is a direct question, which you are required to answer. Why do you refuse to state to which observer an observation is attributed?

  12. #72
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    Quote Originally Posted by abcdefg View Post
    Where exactly do you have a reference frame moving to the left at .99c and it compares another moving to the right at .99c?
    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?

  13. #73
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    Quote Originally Posted by NorthernBoy View Post
    (v+w)/(1-vw/c^2).
    Do you by chance have a sign wrong here? If v=w=0.99c you get a very large result.

  14. #74
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    Quote Originally Posted by abcdefg View Post
    This is not a valid comparison above.

    Further, the addition of velocities does not apply because they are moving in opposite directions.
    Now that is just silly. The formula works for particles moving in opposite directions, or in the same direction.

  15. #75
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    Quote Originally Posted by Perikles View Post
    Do you by chance have a sign wrong here? If v=w=0.99c you get a very large result.
    Yes, well caught, it needs to be a plus sign on the bottom. Original post edited to correct this.

  16. #76
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    Quote Originally Posted by NorthernBoy View Post
    It is also observed to be true, which rather contradicts what you wrote up there.

    Take a particle that decays int two daughter particles, which will move at 0.99999c relative to it, and accelerate it to 0.99c, and we get the addition of velocities just as others have been telling you.

    I know that you don't like this fact, but we've been doing your rocket experiment millions of times per day in labs for years now, and always the numbers come out in line with standard SR and not in line with your bizarre distortion of it.
    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.

  17. #77
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    Quote Originally Posted by NorthernBoy View Post
    Now that is just silly. The formula works for particles moving in opposite directions, or in the same direction.
    This means when a particle is shot down the positive x-axis at .99c, and then accelerated back the other direction, it never goes back. That is odd.

  18. #78
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    Rocket A is the one moving left and rocket B is the one moving right.

    Rocket A is a rocket in a rocket in a rocket.

    So, Rocket A opens up the back and the rocket in a rocket, C, takes off with the same a and burn time. Rocket C is now back in the same frame as when A and B started.

    So, that is currently a total speed of v between A and B, but C is still not in the frame of B. So, that is still not enough.

    Now, rocket C opens up the back and rocket D takes off at a with the same burn time. It enters the frame of B after the burn. D is now v compared to C.

    The piecewise addition is 2v as the total speed between the two or D could have never entered the frame of B.

  19. #79
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    abcdefg, without looking them up or copying them from somewhere, give definitions for the following terms, and for each, describe what that definition means with respect to the situation described in the OP.

    Be as precise as possible. Don't look it up, just repeating some quote won't give anyone insight into what the definition means to you. Don't give a lame answer like "I use them just as everyone else."

    frame of reference
    inertial
    coordinate system

    When you're done, you can go read this thread and understand why I'm asking this.
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  20. #80
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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.
    Still you do not say according to whom these numbers are.

    You need to say relative to what, or to whom, these things are meant to be.

    That being the case, the above makes no sense at all. If the particle is already going at 0.99999c, then you do not need to accelerate it again to reach this speed, it is already doing it.

    I suspect that you were trying to describe some other situation, but in your normal way, you have not explained the situation fully, and so have left ambiguity or outright error in it.

    Can you possibly try again?

    And can you please do so without changing the situation which you first proposed? Again, this is a very bad habit of yours, to start off in one situation, then to switch as soon as someone engages you in discussion about it.

  21. #81
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    Quote Originally Posted by abcdefg View Post
    This means when a particle is shot down the positive x-axis at .99c, and then accelerated back the other direction, it never goes back. That is odd.
    Nope, I can't make any sense of this sentence at all.

    abcdefg, your written English is getting worse and worse. If you are claiming that you cannot move a particle in the direction of positive x at this speed, and then slow it and move it iin the direction of egative x, then you are mistaken.

    That is such an absurd claim, though, that I cannot believe that this is what you really meant.

    I can promise you that if you are willing to stick to one thought experiment, and to stop your ridiculous habit of refusing to say which obervations are made by whom, that I will be able to help you with this.

    I can also promise you that if you insist on posting things in such mangled English, that this will never go anywhere.

    Please, read your posts back, and understand that you need to make them readable to other people, not just to you. Your writing is really really weak, and it is painful to have to extract your meaning before I can comment on it.

  22. #82
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    Quote Originally Posted by slang View Post
    abcdefg, without looking them up or copying them from somewhere, give definitions for the following terms, and for each, describe what that definition means with respect to the situation described in the OP.

    Be as precise as possible. Don't look it up, just repeating some quote won't give anyone insight into what the definition means to you. Don't give a lame answer like "I use them just as everyone else."

    frame of reference
    inertial
    coordinate system

    When you're done, you can go read this thread and understand why I'm asking this.
    I will answer off the top of my head.
    1) frame of reference - Taking the position of a particular inertial frame and performing the LT calcs of other frames relative to the chosen one.
    2) inertial - I use this to describe the members of one frame being "at rest" with one another.
    3) coordinate system - Well this could be Cartesian, Minkowsky and more general topological spaces.
    But, in SR, it would be a Euclidian Cartesian space within the inertial frame and it would be Minkowsky looking at other frames.


    How does this affect the fact one rocket is heading down the positive x-axis at .99c and the other is heading down the negative x-axis such that .99c -.99c -.99c = .99c - 2*.99c = -.99c?

    2v or not 2v, that is the question.

  23. #83
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    Quote Originally Posted by abcdefg View Post
    I am not a Ritz person.
    Based on what you have posted, you are. So where is the answer to Q1-Q7?

  24. #84
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    Quote Originally Posted by abcdefg View Post
    I will answer off the top of my head.
    1) frame of reference - Taking the position of a particular inertial frame and performing the LT calcs of other frames relative to the chosen one.
    2) inertial - I use this to describe the members of one frame being "at rest" with one another.
    3) coordinate system - Well this could be Cartesian, Minkowsky and more general topological spaces.
    But, in SR, it would be a Euclidian Cartesian space within the inertial frame and it would be Minkowsky looking at other frames.
    Oh, that's just horrible. If you are going to redefine these terms in that way, then of course no-one is going to understand what you are talking about.

    An inertial frame is simply one in which newton's laws apply, i.e. one which is not accelerating or rotating. Within that, objects can be moving in any way that they like, but an object not acted upon by a force will continue moving in a straight line at a constant speed.

    Point 1) just makes no sense at all.

  25. #85
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    Quote Originally Posted by abcdefg View Post
    I will answer off the top of my head.
    Thanks, that's the idea. I won't be the one judging the correctness, I'm not qualified for that. Some around here will be able to do so, I'm sure.

    1) frame of reference - Taking the position of a particular inertial frame and performing the LT calcs of other frames relative to the chosen one.
    Thanks. You're defining what frame is, by using the word frame. That's not very helpful. So, the next word to define, in the context of SR discussions is:

    frame

    3) coordinate system - Well this could be Cartesian, Minkowsky and more general topological spaces.
    That's not defining what it is and what it means to you, that's just listing some kinds.
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  26. #86
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    I don't know how to use the LT in this case, of course i'm not an expert on this matters, wonder if someone could do the calculation, to get the speed between both ships

    Is interesting, because IMO, there should not be communication between both rockets for the 3th observer, while there could be communication in-between the rockets for the rockets themselve, of course , is just my opinion.

  27. #87
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    Quote Originally Posted by NorthernBoy View Post
    Nope, I can't make any sense of this sentence at all.

    abcdefg, your written English is getting worse and worse. If you are claiming that you cannot move a particle in the direction of positive x at this speed, and then slow it and move it iin the direction of egative x, then you are mistaken.

    That is such an absurd claim, though, that I cannot believe that this is what you really meant.

    I can promise you that if you are willing to stick to one thought experiment, and to stop your ridiculous habit of refusing to say which obervations are made by whom, that I will be able to help you with this.

    I can also promise you that if you insist on posting things in such mangled English, that this will never go anywhere.

    Please, read your posts back, and understand that you need to make them readable to other people, not just to you. Your writing is really really weak, and it is painful to have to extract your meaning before I can comment on it.
    OK, sorry and thanks. If something is unclear, please say so and I will keep writing it differently until it makes sense to you. I do not mind.

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

    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.

  28. #88
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    Quote Originally Posted by slang View Post
    Thanks, that's the idea. I won't be the one judging the correctness, I'm not qualified for that. Some around here will be able to do so, I'm sure.



    Thanks. You're defining what frame is, by using the word frame. That's not very helpful. So, the next word to define, in the context of SR discussions is:

    frame



    That's not defining what it is and what it means to you, that's just listing some kinds.

    Frame is a collection of observers at rest with one another.

    A coordinate system is some general set of points operating under a distance formula.

  29. #89
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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?
    I keep a log of answers to your questions.

    Answer log macaw
    Q1, Q2 http://www.bautforum.com/against-mai...ml#post1610203
    Q3 http://www.bautforum.com/against-mai...ml#post1610214
    Q4, Q5 http://www.bautforum.com/against-mai...ml#post1610255
    Q6 http://www.bautforum.com/against-mai...ml#post1610255
    Q7 http://www.bautforum.com/against-mai...ml#post1610276

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    Quote Originally Posted by abcdefg View Post
    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
    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

    Quote Originally Posted by abcdefg View Post
    Question 2: If each shot a laser at each other as a signal, will they ever see the light? Yes.
    So you've shown you don't know what formulas to use...congratulations.

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