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Moving and not moving people using Lorentz transformation

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Ok I have a moving person (primed) going 50 m/s in the positive x direction, and I have someone stationary (unprimed) observing them.
    At t = 0, the moving person is at x(0) = 100m
    Write an equation for the object’s position as a function of time x(t)
    seen by the observer in the unprimed (stationary) reference frame.

    And it says the only unknown in my expression should be the variable t.

    So to be clear, the stationary reference frame is unprimed. The moving reference frame is primed.

    2. Relevant equations

    [tex]x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
    [tex]x=\frac{x'-vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    3. The attempt at a solution
    First of all, the wording of this confuses me. From what I've been reading in the book, the first equation is for the stationary (unprimed) observer. And the primed (moving) reference frame is the second equation.
    But the question asks me to write an equation for x(t) for the unprimed observer. Shouldn't that be x'(t)?

    Also, since the primed reference frame is moving with the person who is moving in the primed reference frame, does that mean their position is always 0? From their perspective, it's the unprimed observer who we consider stationary who is moving.

    So I get...
    [tex]0=\frac{100-50t}{\sqrt{1-\frac{50^2}{(3x10^8)^2}}}[/tex]

    So I replaced all the variables I knew, which was x, given in the original problem as 100. And v, given as 50. And the assumed x'(t) of 0. Is that right? I think I'm way off and not even close to understanding what I'm being asked to do here.

    Thanks
     
  2. jcsd
  3. Sep 7, 2013 #2

    rude man

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    The last equation is incorrect. Let v << c and that should be obvious. Since x' = 0 then the person who is stationary in the primed system is moving in which direction as seen from the person in the unprimed system? Remember, v is the speed of the primed system as seen from the unprimed system so if v > 0 then the primed system is moving in which direction as seen by the unprimed system?

    No. That's why it's x(t) instead of x'(t). x(t) is measured in the unprimed ("stationary"} system. x'(t)= 0 always.
    Yes. The person is stationary in the primed system. In other words, that person is the primed system. There is no motion in the primed system so x' = 0 always.

    Fix you equation for x(t) and realize v is a positive number. v is defined as the primed system moving wrt the unprimed system.

    Finally, you need to write t' as a function of t (think time dilation). You want x(t), not x(t'). At t = 0 the primed person is 100m from the unprimed observer, at which time t' = 0 also. Then as t increases, how does t behave compared to t'?
     
  4. Sep 7, 2013 #3

    vela

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    A slight correction: it's actually x'(t') = constant. The person moving in the unprimed frame is stationary in the primed frame, but not necessarily at the origin.

    The equations you wrote down for the Lorentz transformations work for two frames where the origin of the primed and unprimed frames coincide. Following this convention, you should be able to see that x'(t') isn't equal to 0.
     
  5. Sep 7, 2013 #4

    rude man

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    Big 10-4.
     
  6. Sep 7, 2013 #5
    Thanks for the response.
    You're right, it's supposed to be [itex]x=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]
    In my book it has two formulas. One of them is [itex]x'=γ(x-vt)[/itex]. [itex]γ=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]. The other is [itex]x=γ'(x'+vt')[/itex]. For the first one, it says "We require a linear transformation so that each event in system K corresponds to one, and only one, event in system K'." And that's the equation they give. The second equation they give says "We can use similar arguments from the standpoint of an observer stationed in system K' (moving reference frame) to obtain an equation similar to the previous one." So according to this book, [itex]x=γ'(x'+vt')[/itex] is from the standpoint of the observer in the moving reference frame.
    So as far as the question on my homework goes, it says "equation for the objects position as a function of time x(t) seen by the observer in the unprimed reference frame." So that corresponds to the first equation of the two I just typed in this paragraph, correct?
    This is a bit confusing to me. So t' changes as t changes, but it changes slower. Not sure how I'd write that out in equation form. Do I use the same formulas, but solve for t'?
    The person moving in the unprimed frame should be stationary in the unprimed frame, since he's moving with the frame, right?
    Wouldn't x'(t') equal zero since no matter how much time elapses in the primed frame, the person never moves, because the primed frame is moving with him?

    Thanks for the responses guys.
     
  7. Sep 7, 2013 #6

    vanhees71

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    Either I don't understand this question in the original posting or it's a trick question.

    If I understand the somwhat unclearly formulated problem correctly, it's already everything said, how the primed person is moving wrt. to the unprimed. There is nothing to calculate, just write down the formula for uniform motion ;-).
     
  8. Sep 7, 2013 #7

    vela

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    No, stationary doesn't imply x'=0. A person sitting at x'=0 for all t' is indeed stationary in the primed frame, but just because someone's stationary doesn't mean the person is located at the origin.
     
  9. Sep 7, 2013 #8

    vela

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    It's kind of trick question in a sense. The idea is to get students to recognize that there's really no relativity needed to answer the question.
     
  10. Sep 7, 2013 #9

    rude man

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    Isn't he located at the origin because he is the origin? We don't have motion within the primed system.

    x = (x' + vt' )γ + 100
    but x' = 0
    so x = vt'γ + 100
    but t = t'γ
    therefore x = vt + 100.
     
    Last edited: Sep 7, 2013
  11. Sep 7, 2013 #10

    vela

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    You're changing the equations. You can't have both ##x = \gamma(x'+vt')## and ##x = \gamma(x'+vt') + 100## be true at the same time.
     
  12. Sep 7, 2013 #11

    rude man

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    I previously acknowledged that I omitted the 100m.
     
  13. Sep 8, 2013 #12
    Ok this makes no sense to me.
    Here's the equation:
    [tex]x=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    I'm supposed to write out an expression for x(t)... but there's no t in that equation. It's t'.
    Am I supposed to be creating this expression:
    [tex]x(t)=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    Can I legally just turn that x into x(t)? Even though there's no t in the expression?
     
  14. Sep 8, 2013 #13

    vela

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    Reread posts 6 and 8.
     
  15. Sep 8, 2013 #14
    Oh yeah, duh, t and t' are equal.

    If they weren't equal, creating such an equation would not be legal?
     
  16. Sep 8, 2013 #15

    rude man

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    Yes, but you need to add 100 m to x(t) since the fellow at x' = 0 starts at 100 m from the fellow at x = 0.

    You obviously must replace t' with t though. What is the relationship between t' and t? Remember I suggested 'time dilation'?
     
  17. Sep 8, 2013 #16

    rude man

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    Uh-uh.
     
  18. Sep 8, 2013 #17
    I was trying to figure out where to put that 100 m. I don't replace x(t) with it? I just add it to the equation? The problem says I should have only 1 variable in the expression, and that's t. So if x(t) isn't 100 m, what is it? Does the x(t) count as a variable that should be represented by a constant?
    The relationship between t' and t is:
    [tex]t'=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
    And
    [tex]t=\frac{t'-\frac{v}{c^2}x'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
    So when v << c, t = t'
     
    Last edited: Sep 8, 2013
  19. Sep 8, 2013 #18

    vela

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    You've missed the point again. Look, the problem says at t=0, the person is at x(0)=100 m. The person is moving at 50 m/s. Write down x(t). The stuff about the second frame has absolutely nothing to do with this problem.
     
  20. Sep 8, 2013 #19

    rude man

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    The second equation is appropriate but incorrect, so fix it. (Actually, here it doesn't matter, but fix it anyway).

    x' = 0, remember? The guy in the frame x' IS the x' frame, and he is always at x' = 0.
    So, use that info to get t(t') and from there x(t). Don't forget the 100 m.
     
  21. Sep 8, 2013 #20
    I have to write down x(t) and replace all the variables except t with values. If x(t) is a variable, I don't know how to replace that with a value while keeping t as a variable.
    I keep forgetting about the +. It's correct in my notes.
    Is t(t') just the equation I typed, but replacing t with t(t')? I think a big part of my problem is creating these variable as a function of some other variable expressions is something I've never done in any class I've ever taken, so I'm not sure on the rules.
     
    Last edited: Sep 8, 2013
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