Einstein got it all wrong?

Discussion in 'Physics & Math' started by scifes, Mar 26, 2011.

  1. Janus58 Valued Senior Member

    Messages:
    2,394
    Because the time that passes on the rocket according to the rocket will be the same as long as the conditions are that the clock on the rocket reads zero when it passes a point that is a distance 'd' from the Earth as measured by the Earth, as per your initial set up. Thsi does not however mean that the rocket measures its distance from the Earth to be 'd' at the instant that its clock reads zero.

    Imagine that there is a measuring rod extending outward from the Earth of length 'd' as measured by anyone at rest with respet to the Earth.

    The stipulation is that we start counting time on the rocket from the moment it passes the point a distance 'd' from the Earth. This is the same as saying that we start counting time from the moment the rocket is next to the end of the measuring rod.

    The fact that the rocket starts counting time when it passes the end of the rod is a fact according to either the Earth or Rocket (this is an aspect of the "one reality" you are so fond of mentioning.)

    However, according to the rocket frame (IOW, anyone at rest with respect to the rocket.), the Earth, and its measurng rod, is length contracted, and thus the end of the rod is less than 'd' from the Earth and the rocket starts counting time when it is less than 'd' from Earth by its own measurement.

    Thus the rocket will measure that 3096.63 secs passed on its clock because that's how long it took the shorter distance at 0.51c, while the Earth will measure 3600 sec on its own clock, but measure 3096.63 seconds passing on the rocket clock dur to time dilation.

    IOW, both the Earth and Rocket will agree that the rocket clock starts when it passes the end of the rod and reads 3096.63 sec when it reaches the Earth, they just won't agree as to why.

    Your analysis is flawed because you neglect to account for length contraction and the relativity of simultaneity.
     
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  3. Emil Valued Senior Member

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    Wrong.
    Right.
     
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  5. RJBeery Natural Philosopher Valued Senior Member

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    I'm not sure what you're asking. Are you wondering whether we need a "real" object to measure in order to account for length contraction? Because it appears to me that you and Janus agree that no real object is necessary. Or rather are you asking what "causes" length contraction?
     
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  7. AlphaNumeric Fully ionized Registered Senior Member

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    I take it from your silence in regards to my questions and this question of your own that you are unfamiliar with any actual special relativity.

    Even if there is something wrong with special relativity you're not going to find it by being utterly ignorant of it. The use of \(\beta = \frac{v}{c}\) is introduced in the very first few lectures or pages of any book on relativity. It is the parameter which tells you how relativistic an effect will be. If something is much much slower than the speed of light, say v=0.1c then this means that \(\beta\) is much much smaller than 1. Since things like Taylor and infinite binomial expansions are defined in terms of 'small' parameters \(\beta\) is the natural choice. In addition it is common in relativity to use natural units, where c=1, thus all v converts to \(\beta\) anyway.
     
  8. Emil Valued Senior Member

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    2,801
    For me so clearly is the principle of relativity (and I thought for others ) that I did not even comment on the idea that you can define a velocity with a single object.
    To define a speed you need two objects, which have a speed relative to each other.
    If you have a single object then you can define a velocity.
    Not for length, but for the distance between the objects which are in a speed relative to each other.
    It means that my words are very imprecisely understood (possible that my fault) and I work hard to be more precise.I am in total disagreement with Janus58.

    As an example for "distance contracts "
    If I travel to the sun at speeds V1 relative to the earth.
    On the same direction there is also an object whose speed relative to me it is V2.
    To calculate "distance contracts " I use speed V1 or V2?
     
  9. RJBeery Natural Philosopher Valued Senior Member

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    4,222
    Wait, what? How can you say that measuring SPEED requires two bodies, but measuring VELOCITY (which is really just speed with direction) doesn't??
    What does "distance contracts" mean? Do you mean, "how far away am I from the sun at a particular point in time"? If that's the case then the distance would change depending on the observer. In other words there is no absolute answer but I think it would be appropriate to use YOUR speed relative to the SUN and forget the rest.
     
  10. Emil Valued Senior Member

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    2,801
    Circular reasoning.
     
  11. Emil Valued Senior Member

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    2,801
    Oops......edit : If you have a single object then you can not define a velocity.
    I knew only "length contraction". The notion of "distance contraction" was introduced by Janus58, because of that I asked your opinion.
    But at the same time I rejected the "distance contraction".
     
  12. RJBeery Natural Philosopher Valued Senior Member

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    4,222
    Got it. They are the same thing to me because length is simply the distance between end points. If we stretch a rubber band between you and the sun, which you are racing towards at V1, do you not agree that the distance and length are equivalent?

    (admittedly, the math involved might be complicated)
     
  13. AlphaNumeric Fully ionized Registered Senior Member

    Messages:
    6,702
    No, it isn't. It comes from the fact that the Lorentz dilation factor \(\gamma(v) = (1-\frav{v^{2}}{c^{2}})^{-\frac{1}{2}}\) has a singularity at \(v=c\). This can be written as \(\gamma(\beta) = (1-b^{2})^{-\frac{1}{2}}\).

    The relativistic mass formula is \(E = \gamma mc^{2}\). If you expand this as a Taylor expansion to quadratic order you get \(E = mc^{2} + \frac{1}{2}mv^{2} + \ldots\). This demonstrates that for small \(\frac{v}{c} = \beta\) the relativistic energy is close to the sum of the rest mass energy and the kinetic energy, the low speed Newtonian approximation is recovered. Since [tez]\gamma[/tex] has a singularity at \(\beta = 1\) as the speed of an object increases the total energy increases, the higher order terms become important. These are an example of the relativistic corrections to Newtonian mechanics. Thus \(\beta\) is a natural unit to use, because at \(\beta = 0\) you have no motion at all and as \(\beta \to 1\) you have a system becoming more and more relativistic.

    It isn't circular reasoning. The fact you don't bother to look up anything due to either stupidity or dishonesty doesn't mean the information isn't there for you to find. Rejecting something you're too dishonest/lazy/stupid to even look up is a detestable attitude to have.
     
  14. Janus58 Valued Senior Member

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    2,394
    Since you are traveling at v1 with respect to the Sun and the Earth, (we will neglect the Earth's orbital motion with respect to the Sun for now.), Then you use v1 to determine what the Earth-Sun distance as measured by you compared to what it is as measured by someone at rest with respect to the Sun and the Earth because that is your relative velocity to the frame fro which you are comparing lengths. It makes no difference for you that some object moving at v2, measures it to be some other distance.
     
  15. Emil Valued Senior Member

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    2,801
    Yes, if the distance or length are on the objects.

    I understand it this way:
    Theory Special of Relativity talks about length contraction between two objects that have a speed relative to each other.
    So length contraction refers to the length on the bodies, not outside the bodies.
     
  16. Janus58 Valued Senior Member

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    2,394
    They are one in the same.

    It is very simple: You have a long rigid rod floating in space, it is 1 km long. Next to each end, unattached to or touching the rod, floats a small object.

    A spaceship flies by at .866c on a course that parallels the rod. This results in a length contraction in the rod that makes the rod only 1/2 km long according to the spaceship. Are you trying to say that the objects being unattached to the rod or each other will remain 1 km apart according to the spaceship?

    According to the rod, nothing has changed and the objects are still nest to the ends. So if you are, you are supporting a position that leads to contraction.

    For instance, if the rocket fires a laser at one of the ends of the rod as it passes, according to it, there is nothing in the way to block the laser from hitting the end of the rod. However, according to the rod, the object is still next to the end and can block the laser. Thus according to the ship, the laser hits the end of the rod and leaves a mark, but according to the rod, the laser is blocked and the no mark is left.

    What happens when we bring the rod and ship together and moving at the same speed. Doe the mark mysteriously disappear in front of the passengers of the ships eyes, or does a mark suddenly appear out of nowhere according to someone at rest with respect to the rod?

    Both notions are equally ridiculous. The only thing that makes sense is for the distance between the objects to contract along with the rod itself so that both rod and space ship agree that the objects remain next to the ends of the rods at all times.

    And once we conclude this, it is easy to see that it really doesn't matter is the rod is even there, because there is no physical connection between rod and objects. If the distance between the objects decrease when its there, its going to decrease if it is not there. It is an inescapable conclusion.
     
  17. Emil Valued Senior Member

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    2,801
    You can not use the Lorentz transformation, in the assumption for the demonstration of Lorentz transformation.
     
  18. Emil Valued Senior Member

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    2,801

    I just travel. Possible that I'll overcome the sun or maybe I'll stop soon, I still do not know.
     
  19. RJBeery Natural Philosopher Valued Senior Member

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    Then you'll get multiple answers. What exactly are you getting at? Can you provide a definition of length that differs from Janus' description that makes sense? I'd be very interested to hear it, particularly if it provided a single answer irrespective of frame of measurement.
     
  20. Emil Valued Senior Member

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    2,801
    You can have a speed relative to the road. (if the "road" is an object).
    But you can not have a velocity relative to space.
     
  21. Emil Valued Senior Member

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    2,801
    Yes, that's what I deny . There is only one Reality.
    The distance from the Sun to the Earth, is 149,597,870.7 km.
     
  22. OnlyMe Valued Senior Member

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    Length contraction applies to an object in motion, not to distances between objects. While from a moving frame of reference an object in stationary frame of reference may appear length contracted, that is a perceived condition.

    For an observer in relativistic motion, unaware of their motion and time dilation, time will be experienced as "normal" and the diastase shorter than expected, but this is a perceived condition.
     
  23. Pete It's not rocket surgery Registered Senior Member

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    10,167
    Sure, but some quantities that we relative measures rather than fundamental properties of reality.
    You seem to accept this for the measure of velocity.
    Other relative measures are position, time, direction, and distance.
    These are all measure that emerge as relationships between things, rather than fundamental properties.
     
    Last edited: Apr 6, 2011

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