1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Length Contraction of observers approaching an object

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Two momentarily coincident observers approach a small and distant object. One measures the object to be twice as large as the other's measurement. Find their relative velocity.

    2. Relevant equations

    Lorentz Transforms

    [itex] L/L_0 = \sqrt(1 - \beta^2) [/itex]


    3. The attempt at a solution

    I tried using the length contraction formula, but it seems way too easy... I thought length contraction is only perceived in the direction of motion. So if we are approaching the object, how can we see any length contraction between the two reference frames?

    It just seems too easy for an assignment question, and my intuition is usually right. If it is, can you explain to me the flaw in my reasoning?

    Thanks for your time!
     
  2. jcsd
  3. Jul 19, 2012 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What do you use this to find?
     
  4. Jul 19, 2012 #3
    The assumption I made, which I find very, very sketchy, is that [itex]L/L_0 = 1/2[/itex] (since length contracts in the moving frame), which I then am able to use to solve for [itex]\beta[/itex]

    Relativity just isn't my thing... my instructor glazed over it very quickly without any details, but I need a lot of details to get a firm grasp of the topic - hence my difficulty.
     
  5. Jul 19, 2012 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is not what "twice as large" in this question means.

    There are two observers and one object. The way I interpret the question, the length contraction relation has to be used twice, once between each observer and the object. Then, meaning has to be attached to "twice as large"; then ...
     
  6. Jul 19, 2012 #5
    oh ok I see what you mean. Thanks...

    Am I still correct in assuming we only need to deal with length here?

    Something that still feels unclear to me is the notion of length contraction in direction of motion. If the spaceship is moving toward the object, how would they perceive length contraction?
     
  7. Jul 19, 2012 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think you will find that after using length contraction, you will have to use one more concept. This is the "then ..." part of my previous post.
    Yes, you have a point, but suppose the object is partially transparent :biggrin: There are subtleties involved in giving operational definitions of "observe length contraction".
     
  8. Jul 19, 2012 #7
    I am currently staring at [itex] 4(1- \beta_b^2) = (1 - \beta_a^2) [/itex], which I got by dividing the Lorentz factor of each spaceship and setting them equal to 2 (the ratio of the measured lengths)

    If this is correct, how can I get relative velocity from this... I suppose this is what you refer to as the additional concept?
     
  9. Jul 19, 2012 #8

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This looks good.
    Yes. If [itex]\beta_a[/itex] and If [itex]\beta_b[/itex] are the two relative velocities between the observers and the object, how do you "add velocities" to find the relative velocity between the two observers?
     
  10. Jul 19, 2012 #9
    George, just wanted to this opportunity to thank you for the help - you probably get that a lot but nonetheless, felt I should say it.

    On the last topic, I already tried solving it that way, (clearly the speed of the faster ship subtract the speed of the slower ship is the relative speed, so let u = V_b - V_a where V_b is the faster ships speed), but I'm hitting a wall - is it possible to solve for u algebraically? If so, I will try again.
     
  11. Jul 19, 2012 #10

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thanks, but this might be premature. Maybe I'm leading you down the wrong path.
    Have you taken relativistic addition of velocities in your class?
     
  12. Jul 19, 2012 #11
    We've only gotten up to the Lorentz Transforms for velocities, but I'd imagine this could be extended.

    However, I'm letting V_a and V_b represent the speed of the spaceships relative to a common fixed reference frame, i.e. the frame where the length is "proper". Shouldn't then we just use the same vector addition/subtraction of velocities?
     
  13. Jul 19, 2012 #12

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You haven't seen anything like
    [tex]\beta = \frac{\beta_b - \beta_a}{1 - \beta_b \beta_a}?[/tex]
     
  14. Jul 19, 2012 #13
    I have. This is going to be ugly, isn't it?
     
  15. Jul 19, 2012 #14

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Very. You can use [itex]4(1- \beta_b^2) = (1 - \beta_a^2)[/itex] to express [itex]\beta_b[/itex] in terms of [itex]\beta_a[/itex], but there still will be a different final answer for each [itex]\beta_a[/itex].

    This makes sense physically. No matter what [itex]\beta_a[/itex] is, you can always find a relative velocity between a and b such the "twice as large condition" satisfied.
     
  16. Jul 19, 2012 #15

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    The question seems pretty ambiguous to me. If this is exactly how it is worded on your assignment, you may want to ask your professor for clarification.

    Perhaps you are supposed to assume that the 2 observers measure not the length of the object along their approach vector, but rather the solid angle that the object appears to take up. If light is reflected off the object, toward the observers at the moment they are coincident, each observer will measure a different solid angle since one will be closer to the object than the other when the light arrives at their detector.
     
  17. Jul 19, 2012 #16

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Actually, now that you suggest this, I think it it might be something similar, but a little different. Suppose that the object is continuously emitting light.

    "Two momentarily coincident observers" could mean that they each measure an image of the object at the same time, i.e., at the coincidence event.

    "small and distant object" probably means that some type of small angle approximation can be used.

    Different measured solid angles, could mean something different like different "headlight effects".
     
  18. Jul 19, 2012 #17
    Think I got it; algebra wasn't too bad. Much appreciated!
     
  19. Jul 19, 2012 #18

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    I agree, this seems like the most reasonable interpretation.
     
  20. Jul 20, 2012 #19

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Now, I think this is a question about aberration (sometimes called the "headlight effect"), not about Lorentz contraction. Have you taken aberration?
    I think that this is a good idea:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Length Contraction of observers approaching an object
  1. Length contraction. (Replies: 6)

  2. Length contraction (Replies: 4)

  3. Length Contraction (Replies: 3)

  4. Length contraction? (Replies: 11)

Loading...