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Lorentz Transformations Help Mathematica

  1. Aug 2, 2004 #1
    Ok I need some help. I have.


    and i rearrange to get:


    Ok I am setting C=1
    tnot = 518400
    t = 4.7304 *10^17

    as you can imagine my answer i know will be .999999999999 with alot of 9'sV.

    The problem is I can't get anything anywhere to solve this exactly. Everything rounds to 1, even mathematica, somebody help me with this because after i solved this i wanted to graph this function and examine it closer but if i can't even get an exact answer i don't think it's gonna be easy to graph.
    Last edited: Aug 2, 2004
  2. jcsd
  3. Aug 2, 2004 #2
    and this does have to do with SR not just math so mentors please don't warn me. i thought if i posted it in math no one would know what lorentz was. so sorry if this is bad post.
  4. Aug 2, 2004 #3
    and if you don't understand why i need exact solution it's cause an answer of .9999c might correspond to 200 years where .999999999999999999 to 2 billon so i need exacts. sorry to bother yall.

  5. Aug 2, 2004 #4
    That relation can be re-written using the following approximation

    [tex](1 + d )^n \approx 1+d[/tex]

    The ratio v/c becomes

    v/c = 0.9999999999999999999999994

    That's 29 0's followed by a 4

  6. Aug 2, 2004 #5
    sweetness, but how did you do that, can you show me a derivation if its not so long you don't want to.?
  7. Aug 2, 2004 #6
    what is the d for??? and where did the v and t's go?
  8. Aug 2, 2004 #7
    The first thing I did was to factor out the c and divide through by c since it sems that you're looking for the ratio v/c right? d is any number such that |d| << 1 and n is any number You are now left with a relation of the form

    (1 - r^2)^(1/2)

    where r = t0/t. So if I want to use that relation I'd have to set

    d = -r^2

    n = 1/2

    As we say in the world of math and physics plug and chug, which means to place in the actual values and calculate.

  9. Aug 2, 2004 #8
    so you just said 1 + (-r^2)^1/2 and said r = 518400. 4.730 *10^17
  10. Aug 2, 2004 #9
    Pete you know way more than me so I am not saying this is bad solution, but that still doesn't help me because I get 1- tiny tiny number and stinking mathmatica after .0000001 goes to 0 that was why i posted this thread, do you see what I am saying, btw, how are you getting the answer like you calculator or what, or maybe I am misunderstanding you still.

  11. Aug 2, 2004 #10

    your message box is full lol it wont let me text ya

    ok my calc and mathmatica are rounding it's drivin me batty how are you inputing it into. I know i keep asking you the same question, but are you putting it in as d or actual numbers. cause i get sqrt. of 1 - something tiny so it just cause to one takes root and answer is one but we both know that its like .99999999999999999999
  12. Aug 2, 2004 #11
    The approximation is

    (1-r^2)^2 ~ 1 - (1/2)r^2

    (1/2)r^2 ~= 6x10-29


    1- (1/2)r^2 = 1 - 6x10-1 = 0.4
    1- (1/2)r^2 = 1 - 6x10-2 = 0.94
    1- (1/2)r^2 = 1 - 6x10-3 = 0.994
    1- (1/2)r^2 = 1 - 6x10-4 = 0.9994
    1- (1/2)r^2 = 1 - 6x10-5 = 0.99994
    1- (1/2)r^2 = 1 - 6x10-29 = 0.99999999999999999999999999994

    I think I missed one of the 9's at first. There should be 28 9s followed by a 4

  13. Aug 2, 2004 #12
    I just keep on don't I,

    where does this come from. not that it's wrong
  14. Aug 2, 2004 #13
    Its rooted in calculus and power series expansion. It's a truncated Maclurin series expansion of a function. Any function which has all of its derivatives well defined has a series expansion. So all one does is tak f(x) = (1 + x)^n and expand it in powers of x. When |x| << 1 only the first power of x need be retained. Then, as in this problem, you can let x = -(t0/t)2

    Last edited: Aug 2, 2004
  15. Aug 2, 2004 #14
    that is what i was thinkin but not sure. thanx

    i'm gonna work with this but i still think my calc is not gonna handle it, butyou've done more than your job.

    Thanx alot
  16. Aug 2, 2004 #15
    do you square the t's afterwards or before.
  17. Aug 2, 2004 #16
    Iit doesn' matter.

  18. Aug 2, 2004 #17


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    [tex](1 + d )^n \approx 1+nd[/tex] and only for |d|<<1.
  19. Aug 2, 2004 #18


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    (1-r^2)^(1/2) ~ 1 - (1/2)r^2
  20. Aug 2, 2004 #19
    I keep getting x to be 6.00488*10^-25
    and my answer to have 24 0's where's the discrpancy, doing it just like you.

    i'm off by factor of 4 o's but have the same numbers other than o's
    Last edited: Aug 2, 2004
  21. Aug 2, 2004 #20
    What did you use as x? I used x = (1/2)(t0/t)2.

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