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  1. #1 lorentz transformation 
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    Can someone please explain to me the equation of the lorentz teransformation.
    From what I have gathered, it's the equation of how to compare a certain position from two different objects at different velocities (if that is incorrect then please correct me.
    The equation which im having trouble with is the two:

    x'=x-vt/sqroot(1-vsq/csq)

    *sq=squared

    and the similar one for time (anyone who knows the first will know the second)

    And the general equation to be honest, its all fairly daunting me.

    To re-iterate:

    Can someone explain me the Lorentz Transformation, and the equations that go with it, in a way that isn't out of my league.

    Help would be appreciated, as then I can understand Einsteins book on Relativity more.


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  3. #2  
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    First, someone needs to know your level of mathematical understanding.

    Second, someone needs to know how well you can "see" in your minds eye those mathematical equations and associated "relevantness" to space-time (perhaps a harder thing to pass-off lightly).

    THEN, they will know how to answer your question.


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  4. #3  
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    Level of understanding - Post GCSE, with some tiddles of extra information.
    Can I understand maths equations in my head and see them? Well, it entirely depends on the context. I mean, depending on how complex it is will be the factor of if I can understand it or not.

    I may be out of my league here,

    It sucks being young and mathematically challenged!
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  5. #4  
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    Wow,

    so you are trying to "be someone"?
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  6. #5  
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    Quote Originally Posted by streamSystems
    Wow,

    so you are trying to "be someone"?
    Be someone? I just have a deep interest in science, and am reading Einsteins relativity to expand my interest. I understand it so far, but I just like to know how the mathematics works, as would anyone really. As fundamentally, it is all down to the maths.
    Plus the equation LOOKS on a base level, understandable. Well at least to me, no complicated calculus in this one. But then as usual I'm probably way out of my league.
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  7. #6  
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    NO.......you're doing just fine.

    I mean, you have extrapolated so much from me, already.....?
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  8. #7  
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    ten_ben Look here. http://www.thescienceforum.com/viewt...?p=59623#59623

    PS, by edit. If you have questions, post them in this thread. D'ya know how to do that by copy/paste on this site?
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    I apologise.

    I am a bad person to just be someone to jump in and assume things.

    It though did take a little questioning, right.........for you to find how well to be "Bambi on stilts"?

    No, Ok..

    I can answer your question.

    BUT..........Lorenz transformations are a thing of great "who really wants to be known for saying such a thing.............".

    ........otherwise, with your experience, why ask the question?

    Why not show more interest in those pioneers who made Lorenz feel outdated............like CERN hopefuls..........ideally?

    ........at least "capitalise" on your dilemma of science.,,,......ask questions that are relevant to your contemporary research phenomenaaaaaAAAAAAAA ventures...........(ps, I'm a watchin).
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  10. #9  
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    Quote Originally Posted by Guitarist
    ten_ben Look here. http://www.thescienceforum.com/viewt...?p=59623#59623

    PS, by edit. If you have questions, post them in this thread. D'ya know how to do that by copy/paste on this site?
    Thanks, i'll definately have a good look at that!
    Plus i'm sure I can do the copy and pasting in this site!
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  11. #10  
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    You're easy-2-pleeeeeease...........?

    Some people in this forum seek "fame" value............others just make those people's lives a HELLLLLLLLL.

    You need an answer?

    Consult the "science God"?

    Righhhhhhhhht?
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  12. #11  
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    Seek medical help, immediately.
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    Reading your thing, all goes swimmingly until I get to the words " transformation is orthogonal." and something about rotations. I just don't see where this rotation is, as surely if its just going at a different velocity, there would be no rotation?
    or have I just confused myself with a simple idea?
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    Hi ten_ben. Don't worry at all about fancy words like "orthogonal" for now Also don't worry about the thetas, sines and cosines. Just go ahead and draw the graphs I suggested (remember that time is the y axis, space - or distance, they are very roughly the same thing - is the x axis). See what you need to do to bring B to rest on its own coordinates (aka in its own reference frame)

    Does that work for you?
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  15. #14  
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    Jeez you lot, you've totally not answered the question!

    There can be a number of Lorentz tranasformations/manefestations a few examples are these:

    Time dilation
    Lorentz Contraction
    Relative Mass

    Each one of these has an equation, but in each of these will be a funny little y, the y is called gamma and is reffered to as the Lorentz factor, and here it is:

    1/ /---- 1-(v^2/c^2)

    ( /---- is the square root)
    (^2 means squared)

    This equation is this: The lorentz factor needed for lorentz transformations equations is....

    1 divided by the squareroot of 1 minus (the velocity (speed) of an object squared divided by the speed of light squared).

    Usually at speeds less than light, the value for the Lorentz factor is 1, at higher speeds close to light it is below that number, for instance if an object travelled at 0.75c, then in the Lorentz contraction (length contraction) the size of the object to an outside observor would be 3/4 less than what it normally would. The closer you get to light speed the shorter and shorter it would get, the same goes for time dilation, relative mass, etc.

    Hope I've been of some help, contact me if you need anymore explanation, or simply just look up Special relativty on the Net somewhere.
    "If you wish to make an apple pie from scratch, you must first invent the universe". - Carl Sagan
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    PS, the equation you have looks a bit dodgy and crappy anyway, if it is a Lorentz contraction it should look simply like this:

    x = x0 / y

    Where x is the length we percieve
    Where x0 is the length of the obejct being measured in it's own reference frame.
    Where y is the Lorentz factor.
    "If you wish to make an apple pie from scratch, you must first invent the universe". - Carl Sagan
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  17. #16  
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    Quote Originally Posted by svwillmer
    There can be a number of Lorentz transformations
    No, tough guy, there's just the one. You got it wrong, by the way. Try again, using mathematics, for preference.
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  18. #17  
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    I didn't mean THE transformation, i was reffering to the transformations of an object due to the Lorentz factor-sorry if i was a bit unclear.
    "If you wish to make an apple pie from scratch, you must first invent the universe". - Carl Sagan
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    I'm still doing my head in over this. Which is rather irritating as I feel that when I do understand it i'll wonder how I ever didn't.
    Now, is there anybody here who I can add on msn who knows this formula well, as if I can speak to someone about it, i'll probably find it far easier to get this in my head.
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  20. #19  
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    Nah, ask your questions here, in public, then everyone can learn from the replies. As it happens, I do know the formula well, but I'm off on holiday tomorrow lunch.

    What are your specific questions (like, I mean, don't just say "please explain it")
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  21. #20  
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    Quote Originally Posted by Guitarist
    Nah, ask your questions here, in public, then everyone can learn from the replies. As it happens, I do know the formula well, but I'm off on holiday tomorrow lunch.

    What are your specific questions (like, I mean, don't just say "please explain it")
    I just don't understand how the formula works. I know what its meant to be saying, but I don't see how the formula shows it.

    The fact that the Lorentz transformation is talking about viewing objects from different velocities and places = fine

    The fact that it can be represented by X'= X-VT /sqroot (1-Vsq/Csq) =not fine.

    My problem is , if I recall what streamsystems said before, I just can't visualize how the formula works. To me its just a gaggle of letters, of which I cant get to relate to one another.
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  22. #21  
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    Quote Originally Posted by ten_ben
    The fact that it can be represented by X'= X-VT /sqroot (1-Vsq/Csq) =not fine.
    Not fine with me, eiher. I dunno where you got that equation from, but, at first glance, it looks wrong to me

    if I recall what streamsystems said before, I just can't visualize how the formula works. To me its just a gaggle of letters, of which I cant get to relate to one another.
    First, at the risk of being banned: pay no attention to Streamsystems; (s)he is a nutter. Second, if you think that equations are "just a gaggle of letters", you are missing something. Maybe you feel terms (symbols) weren't defined here?

    Dunno, gotta go for now.
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  23. #22  
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by ten_ben
    The fact that it can be represented by X'= X-VT /sqroot (1-Vsq/Csq) =not fine.
    Not fine with me, eiher. I dunno where you got that equation from, but, at first glance, it looks wrong to me

    if I recall what streamsystems said before, I just can't visualize how the formula works. To me its just a gaggle of letters, of which I cant get to relate to one another.
    First, at the risk of being banned: pay no attention to Streamsystems; (s)he is a nutter. Second, if you think that equations are "just a gaggle of letters", you are missing something. Maybe you feel terms (symbols) weren't defined here?

    Dunno, gotta go for now.
    That equation s right im certain. As I copied it from Einsteins book on relativity, though my use of symbols may not be that clear.
    As for the gaggle of letters, i'm still not sure weather the v (well, in the book it isn't a v, more a weird v like shape), is the velocity or the distance. Either way, it still just confuses me, I just need an explanation of the equation.
    I must be making a mountain out of a molehill by now!
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  24. #23  
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    The Lorentz transformation between the positions and times (x, y, z, t) as measured by an observer "standing still," and the corresponding coordinates and time (x¹, y¹, z¹, t¹) measured inside a "moving" space ship, moving with velocity u are

    x¹ = x - ut/sqrt1 - u²/c²

    y¹ = y,

    z¹ = z,

    t¹ = t - ux/c²/sqrt1 - u²/c²

    There equations relate measurements in two systems, one of which in this instance is rotated relative to the other:

    x¹ = x cos Θ + y sine Θ,

    y¹ = y cos Θ - x sin Θ,

    z¹ = z.
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  25. #24  
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    ten_ben How are you doing with this? I can help with the arithmetic, and maybe, just maybe, with the physics. If you want us to show you, just shout.

    It really isn't, like, advanced math, or anything.

    P.S. Guaged is right, of course, but maybe not that helpful to you? Dunno, do say.
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  26. #25  
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    Quote Originally Posted by Guitarist
    ten_ben How are you doing with this? I can help with the arithmetic, and maybe, just maybe, with the physics. If you want us to show you, just shout.

    It really isn't, like, advanced math, or anything.

    P.S. Guaged is right, of course, but maybe not that helpful to you? Dunno, do say.
    I gave up as I found I was getting nowhere. I understand the arithmetic, ad as you sad I can see the equation isn't that confusing (its nice to have equations without "e" in them!).
    With Gauged, when he made his post, he was just telling me what I already knew.
    I understand that apparently "x¹ = x - ut/sqrt1 - u²/c² ", i mean Ive seen the equation a bundle of times. But alas, I do not understand WHY it is that.
    I mean, when Mr Lorentz sat down one day to figure out the equation, he didn't just go "Ah! x¹ = x - ut/sqrt1 - u²/c² !, how easy!". He spent time figuring it out, and eventually came up with that.
    I just don't see how he came up with it, I know it is easy, but with a no so great knowledge in maths, and books that people my age don't read, I just need some explanation to understand it.
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  27. #26  
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    IIRC (and I might not) it'd derived from the consequences of considering the speed of light constant in all reference frames. The classic example is to have one spaceship pass another at some fraction of the speed of light. During that passing one ship fires a laser at a mirror and both ships measure the time and distance it traveled. They'll get different results but both conclude that the laser was travelling at light speed. (The differences in time and distance should be a factor of 1/(1-sqrt(v^2/c^2)).) Of course, the last time I tried to do the math for this, I got it wrong, so maybe someone else can do it this time.
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  28. #27  
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    Ya, well, MagiMaster gives the sort of illustration that all books use, as he says, but which I personally find unhelpful (these illustrations appeal to our intuition, only to have some fat lady sit on them. I prefer the more abstract approach, but maybe that's just me).

    OK, first some preliminaries. It appears that, to a physicist, time is nothing more or less than that which is measured by clocks. Likewise, distance is merely what can be measured by rulers. Apparently this is not debatable, so there.

    Second, with regard to time: if my clock ticks at half the rate of your clock, then my "hour" seems to you to be twice as long as yours. So, slow clocks imply dilated time. Right? In fact time dilation is the reciprocal of clock rate.

    Let's try this first. Suppose you are on a stationary train and bouncing a tennis ball off the floor. You can time how long it takes for the ball to return to your hand, down and up..... OK? I, on the platform, will agree on that time.

    Now suppose your train is moving. If the speed of the train is constant (and the track smooth) you will find the time taken for the straight down-up ball to return to your hand to be the same as when you were stationary. But I on the platform as you whiz by, see that the balls trajectory is not up and down, but rather it is diagonal, a longer distance, so I calculate it will take longer to make the round trip. You, on the train disagree, so I can only conclude that either the ball is being accelerated by the train (true in the case of a tennis ball) or that your your clock is running slower that mine (true in the case of light, which can't be accelerated)

    So we have just 3 numbers in there: the speed of the train, the speed of my clock and the speed of your clock as calculated by me, and these can be connected by a right-angled triangle. Train speed = v is one side, what I believe your clock rate to be is at right angles to that, call it t' and what I believe to be my clock rate, say t, connecting the "free ends" diagonally.

    Let's play with that a while and see what we come up with. Remember, first we must equalize all units before we can calculate anything. (actually, that's not true in spacetime, but let's stay simple).

    Let vt be the distance the train travels in time = t (remember that as far as you are concerned, the train isn't moving; your tennis ball "experiment" proves that). Fix a constant of speed c, light speed if you want, Then ct = distance traveled in time = t and ct' = distance traveled in time = t'.

    Hello Pythagoras: (ct)<sup>2</sup> = (ct')<sup>2</sup> + (vt)<sup>2</sup> or

    (ct')<sup>2</sup> = (ct)<sup>2</sup> - (vt)<sup>2</sup>

    Now divide through by (ct)<sup>2</sup>, giving

    (ct')<sup>2</sup>/(ct)<sup>2</sup> = 1 - (vt)<sup>2</sup>/(ct)<sup>2</sup>.

    Notice that the squared c's cancel on the LHS, as do the squared t's in the 2nd term on the RHS, giving

    t'<sup>2</sup>/t<sup>2</sup> = 1 - v<sup>2</sup>/c<sup>2</sup> so

    t' = t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>1/2</sup>. Remembering that time dilation is the reciprocal of measured time, one finds that, (with an abuse of notation, where I'm now letting t' refer to time dilation, rather than measured time)

    t' = t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>.

    Bingo. Length contraction works out similarly, without the last step.
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  29. #28  
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    Nice post Guitarist
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  30. #29  
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    Thanks, Neutrino, but I wish I could persuade you guys to think in terms of coordinate transformations. For then it is easy to see, for example, that the so-called "twin paradox" isn't a paradox at all, neither has it anything to do with acceleration, still less with General Relativity.

    Ah well.
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  31. #30  
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    Guitarist.

    Coordinatinate transformations is a term employed by mathematics.

    What we are talking about is how coordinates "transform" through space and time?

    Is that correct?

    Otherwise you are a pure mathematician.

    But "coordinate transformation": can you elaborate//////what is your context of "coordintate transformation" expertise?
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  32. #31  
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    Listen to me moron. This thread was started by some guy with a genuine desire to learn, unlike you. We all here are trying to help him, so do us all a favour - eff off outta this thread, if you have nothing useful to contribute.
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  33. #32  
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    Quote Originally Posted by Guitarist
    Ya, well, MagiMaster gives the sort of illustration that all books use, as he says, but which I personally find unhelpful (these illustrations appeal to our intuition, only to have some fat lady sit on them. I prefer the more abstract approach, but maybe that's just me).

    OK, first some preliminaries. It appears that, to a physicist, time is nothing more or less than that which is measured by clocks. Likewise, distance is merely what can be measured by rulers. Apparently this is not debatable, so there.

    Second, with regard to time: if my clock ticks at half the sate of your clock, then my "hour" seems to you to be twice as long as yours. So, slow clocks imply dilated time. Right? In fact time dilation is the reciprocal of clock rate.

    Let's try this first. Suppose you are on a stationary train and bouncing a tennis ball off the floor. You can time how long it takes for the ball to return to your hand, down and up..... OK? I, on the platform, will agree on that time.

    Now suppose your train is moving. If the speed of the train is constant (and the track smooth) you will find the time taken for the straight down-up ball to return to your hand to be the same as when you were stationary. But I on the platform as you whiz by, see that the balls trajectory is not up and down, but rather it is diagonal, a longer distance, so I calculate it will take longer to make the round trip. You, on the train disagree, so I can only conclude that either the ball is being accelerated by the train (true in the case of a tennis ball) or that your your clock is running slower that mine (true in the case of light, which can't be accelerated)

    So we have just 3 numbers in there: the speed of the train, the speed of my clock and the speed of your clock as calculated by me, and these can be connected by a right-angled triangle. Train speed = v is one side, what I believe your clock rate to be is at right angles to that, call it t' and what I believe to be my clock rate, say t, connecting the "free ends" diagonally.

    Let's play with that a while and see what we come up with. Remember, first we must equalize all units before we can calculate anything. (actually, that's not true in spacetime, but let's stay simple).

    Let vt be the distance the train travels in time = t (remember that as far as you are concerned, the train isn't moving; your tennis ball "experiment" proves that). Fix a constant of speed c, light speed if you want, Then ct = distance traveled in time = t and ct' = distance traveled in time = t'.

    Hello Pythagoras: (ct)<sup>2</sup> = (ct')<sup>2</sup> + (vt)<sup>2</sup> or

    (ct')<sup>2</sup> = (ct)<sup>2</sup> - (vt)<sup>2</sup>

    Now divide through by (ct)<sup>2</sup>, giving

    (ct')<sup>2</sup>/(ct)<sup>2</sup> = 1 - (vt)<sup>2</sup>/(ct)<sup>2</sup>.

    Notice that the squared c's cancel on the LHS, as do the squared t's in the 2nd term on the RHS, giving

    t'<sup>2</sup>/t<sup>2</sup> = 1 - v<sup>2</sup>/c<sup>2</sup> so

    t' = t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>1/2</sup>. Remembering that time dilation is the reciprocal of measured time, one finds that, (with an abuse of notation, where I'm now letting t' refer to time dilation, rather than measured time)

    t' = t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>.

    Bingo. Length contraction works out similarly, without the last step.
    Thank you so much, that's absolutely fantastic! I can finally compute this transformation in my mind, it all makes sense now. I really really appreciate it!
    Apologies to all for my long time to grasp the concept, but now I am there.
    Thanks to all!
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  34. #33  
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    Well,if you enjoyed that, try this.

    Consider the rectangular coordinate set {x,t} (in fact, draw it; t is the vertical axis, x the horizontal). Let t = time, x = distance in the x direction. A body A at rest in {x,t} traces out a trajectory parallel to t. The trajectory of a body B in uniform motion in {x,t} goes off at some angle in the positive x and t directions.

    Now B's motion is said to be inertial if it is possible to manipulate the coordinates {x,t} in such a way as to bring B to rest i.e. make B's trajectory parallel to the manipulated t, call it t'. You easily see that all that's required is to "tilt" {x,t} to obtain {x',t'}. This tilt is called a rotation, and this process of coordinate manipulation is called an orthogonal coordinate transformation (orthogonal simply means that lengths and angles are conserved).

    Note that now A is moving in the {x'.t'} coordinates; in other words A and B can't agree on who is moving, and therefore can't agree on whose length is contracted and whose time is dilated.

    Now consider the body B' which starts at some x<sub>1</sub> on x and returns after time = t to x<sub>1</sub>. First note this.

    From a purely kinematic point of view, all return journeys of equal speed and duration are exactly equivalent, so let's take the easy option: B' goes off straight, turns round and retraces his outward path. Draw that with respect to the coordinates {x,t}. What do you see? (remember you can't go back in time!)

    I get a dog-leg. You'll find there is absolutely nothing, no coordinate transformation whatever, that will bring B' to rest i.e. make his trajectory parallel to t. Therefore B', unlike A, has no right to consider himself at rest in any coordinate set. Therefore A and B' both agree that only B' is moving and only B' is subject to time dilation.

    Suppose A and B' are twins. By the above, if A and B' are to meet at some time = t<sub>2</sub> both A and B' agree that Bprime's clock must have been ticking slower than A's, so he will have aged less than A. This is the so-called Twin Paradox, but it isn't a paradox!
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  35. #34  
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    To throw a spanner in the works here - everyone has tactfully assumed that length contraction perpendicular to the direction of motion is zero without actually justifying that claim. Why is that zero?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  36. #35  
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    *Sigh*

    Consider the body with dimensions L<sub>x</sub> x L<sub>y</sub> x L<sub>z</sub> in its own rest frame.

    The velocity v of this body with respect to this frame is the vector sum of its velocities in the x, y and z directions, as measured from the rest frame:
    v = v<sub>x</sub> + v<sub>y</sub> + v<sub>z</sub>

    The Fitzgerald-Lorentz contractions are

    L<sub>x</sub> = L'<sub>x</sub>(1 - v<sub>x</sub><sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>
    L<sub>y</sub> = L'<sub>y</sub>(1 - v<sub>y</sub><sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>

    etc.

    One easily sees that, when y is perpendicular to the direction of motion, v<sub>y</sub> = 0, L<sub>y</sub> = L'<sub>y</sub>(1 - 0)<sup>-1/2</sup> and therefore L'<sub>y</sub> = L<sub>y</sub>.
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  37. #36  
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    From Encarta:

    "Each observer uses a system of coordinates as the frame of reference for measurements, and these coordinates can be transformed one into the other by a mathematical manipulation. The equations for this transformation, known as the Lorentz transformation equations, were adopted by Einstein, but he gave them an entirely new interpretation. The speed of light is invariant in any such transformation.
    Microsoft ® Encarta ® Encyclopedia 2005 © 1993-2004 Microsoft Corporation. All rights reserved."


    Can we all just PAUSE for a moment and LOOK, and yes, SERIOUSLY LOOK, at what EINSTEIN was looking for: he was looking for DA MATH of DA OBSERVOR relevant to DA SPACE-TIME.

    Why sip at breast milk: does anyone see a baby sip? It graps that nipple and sucks HARD.

    Go for a REAL theory of the OBSERVOR and space-time.

    You get lost in DA WOODS of mathematics, like yuo are now with this thread, no one will get you out more than I can.

    What DA EINSTEIN was SAYIN was that DA SPEED OF DA LIGHT is CONSTANT from all REFERENCES of DA OBSERVOR.

    Let's not forget that.

    DA OBSERVOR could be ANYWHERES.

    If DA OBSERVOR could be DA ANYWHERES, WHY NOT, people, "why not" do a theory of space-time and perception regarding DA OBSERVOR of DA SPACE-TIME?

    WHAT CAN DA MATHS DO FOR THE NEED TO FIND A SYSTEM OBSERVOR REFERENCE?

    I KNOW: a human being that is able to do a theory of space-time (DA SPACE-TIME WE PERCEIVE AS SPACE-TIME) and perception.


    Still, your mathematics excites me (as you can tell).

    Please continue.



    I tell you what: you can try to push sxxx up a hill with your own bare hands, or you can get animals to crap on ground higher than you (my point being, "why make science so hard when so much else as an interpretation of space-time works?)
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  38. #37  
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    Quote Originally Posted by streamSystems

    Still, your mathematics excites me (as you can tell).

    Please continue.
    Thank you for the compliment. Let me repay in kind :Fuck off, you irritating moron.

    I am serious.
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  39. #38  
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    NO.

    OOOooooo....nooooooo.

    You are so "ill funded" as a fraternity, it spooks me to think numbers can ultimately explain my future wthout at least adressing the very insecurity of my ability to be aware.
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  40. #39  
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    Quote Originally Posted by Guitarist
    *Sigh*
    Lol, sadly it ain't so simple

    The Fitzgerald-Lorentz contractions are

    L<sub>x</sub> = L'<sub>x</sub>(1 - v<sub>x</sub><sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>
    L<sub>y</sub> = L'<sub>y</sub>(1 - v<sub>y</sub><sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>

    etc.
    The fact that you have no mixed products in these equations comes from assuming that the contractions can only occur in the direction of motion and not the direction perpendicular to it so you can't use these equations to justify that claim (classic circular reasoning).
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  41. #40  
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    Quote Originally Posted by river_rat
    ......assuming that the contractions can only occur in the direction of motion and not the direction perpendicular to it so you can't use these equations to justify that claim (classic circular reasoning).
    Umm. Did you read this bit?
    Quote Originally Posted by Guitarist
    One easily sees that, when y is perpendicular to the direction of motion, v<sub>y</sub> = 0, L<sub>y</sub>= L'<sub>y</sub>(1 - 0)<sup>-1/2</sup> and therefore L'<sub>y</sub> = L<sub>y</sub>.
    .

    What's circular? The only assumptions in there are that any 3-vector can be decomposed into it's 1-vector components, and that any 3-vector may be zero-valued on any one of these components. You don't doubt this, I'm sure.

    All the rest follows naturally, with no prior assumption of conclusion.

    Which of these two assumptions anticipates the result I was attempting to prove?
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  42. #41  
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by river_rat
    ......assuming that the contractions can only occur in the direction of motion and not the direction perpendicular to it so you can't use these equations to justify that claim (classic circular reasoning).
    Umm. Did you read this bit?
    Quote Originally Posted by Guitarist
    One easily sees that, when y is perpendicular to the direction of motion, v<sub>y</sub> = 0, L<sub>y</sub>= L'<sub>y</sub>(1 - 0)<sup>-1/2</sup> and therefore L'<sub>y</sub> = L<sub>y</sub>.
    .

    What's circular? The only assumptions in there are that any 3-vector can be decomposed into it's 1-vector components, and that any 3-vector may be zero-valued on any one of these components. You don't doubt this, I'm sure.

    All the rest follows naturally, with no prior assumption of conclusion.

    Which of these two assumptions anticipates the result I was attempting to prove?

    a 3-d CIRCULAR ENTITY IS A SPHERE.

    Doing the trig on that aint so easy (in regard to space-time)......3-d space past to future inclusive.
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  43. #42  
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    The equation (and derivation) of the contraction of L<sub>y</sub> = L'<sub>y</sub>(1 - v<sub>y</sub><sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup> in your case presupposes that the contraction is only in the direction of motion. For starters, to do the geometry behind that equation you assume that the problem is one dimensional and that the contractions are only in the direction of motion. Just think of the triangles.

    The results are right but the justification is not. I don't want to give the solution straight away as its a good exercise in SR thought experiments to prove the result.

    PS Steamsystems, your last post proves to me that you have no idea what you are talking about here. Thanks for clearing that up.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    No, thank YOU for clearing that up (where YOU stand).
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  45. #44  
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    I'll respond to river_rat in a bit, but as long as we're talking Lorentz transformations, I thought this might be some fun.

    E = mc<sup>2</sup> is perhaps the most famous equality in science; it's derivation must surely be a candidate for the most audacious. Here's Einstein (hugely paraphrased):

    Consider a material body B with energy content E<sub>initial</sub>. Let B emit a "plane wave of light" (what's that, by the way?) for some fixed period of time = t. One easily sees that the energy content of B is reduced by E<sub>initial</sub> - E<sub>final</sub>, which depends only on t.

    Let E<sub>initial</sub> - E<sub>final</sub> = L (i.e.light energy).

    Now, says Einstein, consider the situation from the perspective some observer moving uniformly at velocity = v with respect to B. Then, evidently, by Lorentz time dilation, L' depends only on t', which is t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>.

    The difference between L and L' is simply L' - L = L[(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup> - 1]. By expanding (1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup> as a Taylor series, and dropping terms of order higher than 2 in v/c, he finds that

    L(1 + v<sup>2</sup>/2c<sup>2</sup> - 1) = Lv<sup>2</sup>/2c<sup>2</sup> = 1/2(L/c<sup>2</sup>)v<sup>2</sup>.

    With a flourishing hand-wave Einstein now says something like this: the above is an equation for the differential energy of a body in relative motion; but so is E = 1/2mv<sup>2</sup>, the equation for kinetic energy - these can only differ by an irrelevant additive constant, so set

    L/c<sup>2</sup> = m and so L= mc<sup>2</sup>.

    But, says he, L is simply a "quantity" of energy, light is this case, that now depends only on m and c<sup>2</sup> so......

    E = mc<sup>2</sup>.

    Audacious, or what? To see the man in action, look here http://dbserv.ihep.su/~elan/src/einstein05c/eng.pdf

    It's a sobering thought, one that I shall probably need later tonight, and one that nut-cases in all forums all over the world will be heartened by, that with a little high-school math and a lot of intuition, one the most profound equalities in science pops out.
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  46. #45  
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    Quote Originally Posted by river_rat
    The equation (and derivation) of the contraction of L<sub>y</sub> = L'<sub>y</sub>(1 - v<sub>y</sub><sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup> in your case presupposes that the contraction is only in the direction of motion.
    While I still don't quite see your point, let's try this, although as I have trouble visualizing physical problems, it doesn't look quite right to me:

    Let a body in uniform motion have velocity v. Now v is a 3-vector and can be decomposed as v = v<sub>x</sub> + v<sub>y</sub> + v<sub>z</sub> on the coordinate set {x,y,z}.

    Assuming the following equality (from Einstein):
    v = (v<sub>x</sub> + v<sub>y</sub> + v<sub>z</sub>)/[1 + (v<sub>x</sub>v<sub>y</sub>v<sub>z</sub>)/c<sup>2</sup>]
    I find this is
    v{1/(v<sub>x</sub> + v<sub>y</sub> + v<sub>z</sub>)/[1 + (v<sub>x</sub>v<sub>y</sub>v<sub>z</sub>)/c<sup>2</sup>]}
    = v - {(v<sub>x</sub>v<sub>y</sub>v<sub>z</sub>)/[1 + (v<sub>x</sub>v<sub>y</sub>v<sub>z</sub>)/c<sup>2</sup>]}.

    Now if any or all of the v<sub>i</sub> = 0, then

    v - 0 = v self evidently. Suppose v<sub>z</sub> = 0, i.e. z is perpendicular to the direction of motion, Then

    v = v<sub>x</sub> + v<sub>y</sub> + 0 (datum) = v<sub>x</sub> + v<sub>y</sub> - 0 (from the above).

    But, by Lorentz transformation, and assuming c as a constant, length contraction L' depends only on v, so, if v<sub>z</sub> = 0, it makes no contribution to L'.

    Bah, it's rubbish! It's the right approach, I feel sure, but poorly argued.

    Gwaan, show us!
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  47. #46  
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    Hey Guitarist, lets suppose that length contracts in the direction perpendicular to motion (the other case is handled with a similar argument).

    Now lets try a little challenge, you are going to make a silver paint mark on a metal pole and plant that pole in the ground next to a railway. Lets suppose that the paint mark is 1 meter above the ground. Now I am going to climb aboard a train and drive towards the pole at a speed of 0.99c and try put a black paint mark the same distance from the ground (i.e. I measure one meter vertically from the ground and hold my paintbrush out the window at that height.

    Now in my frame of reference, the pole contracts as I see it moving towards me at 0.99c and thus my paint mark will be above yours. However, according to you my vertical distances have contracted so my paint mark will be below yours. So we have the contradiction that my paint mark is both above and below yours on the pole, and thus perpendicular directions cannot contract.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  48. #47  
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    Tell you what, this physics is hard work; so far, I've piloted a spaceship, fired lasers, bounced tennis balls, painted and planted posts and all with a clock in one hand and a ruler in the other.

    I'm exhausted!

    Anyway, I'm not sure I can get the math to work through, especially as I now see my last attempt was gibberish. I can work back from Einsteins' velocity addition formula for two successive linear transformations along the same axis, but I cannot convince myself it's a valid approach.
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  49. #48  
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    really....it's just parabolic functions instead of sin x, it's sinh x, and instead of cos x, it's cosh x. And always remember! Sinh^2x - Cosh^2 x = 1;~[/quote]
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  50. #49  
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    I think you mean hyperbolic functions Chainrule
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    OOP! thank you for pointing that out! you are absolutely right! Excuse me~
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