Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz Transformations Help Mathematica

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

    [tex]t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    and i rearrange to get:

    [tex]v=\sqrt{c^2(1-\frac{t_0^2}{t^2})}[/tex]

    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.

    thanx
    Woody
     
  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

    Pete
     
  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.

    Pete
     
  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.

    Thanx
     
  11. Aug 2, 2004 #10
    pmbphy

    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

    Notice:

    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

    Pete
     
  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

    Pete
     
    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.

    Pete
     
  18. Aug 2, 2004 #17

    DW

    User Avatar

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

    DW

    User Avatar

    Actually,
    (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.

    Pete
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Lorentz Transformations Help Mathematica
  1. Help With Mathematica (Replies: 2)

  2. Mathematica help (Replies: 1)

  3. Help in Mathematica (Replies: 15)

  4. Help with Mathematica? (Replies: 1)

Loading...