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Special relativity help

  1. Jan 3, 2005 #1
    Hi guys,

    I was pointed to here from another forum and was wondering if you could help me with this question. I would show you my working for it...but unfortunatly i have none as i cant even begin to start on it :( I have thought that maybe it is to do with simultaneous equations but cannot see a way of implimenting that. Every time i try and solve it i end up proving what i started with in the first place (like 1=1 or something silly :uhh: ). Anyway, ive read the sticky and i appreciate that im not going to be hand-fed the answer but any pointers would be fantastic. Thanks very much for any help! And heres the question...

    The speed of light is now 20ms^-1, and someone gets on a bike and goes at 19.999ms^-1. The person on the bike is 60 years old and you are observing the person on the bike moving past you, you are 20 years old. What is the minimum time the person on the bike would have to continue cycling for so at the end of the journey they are younger than you?
     
  2. jcsd
  3. Jan 3, 2005 #2

    jtbell

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    Staff: Mentor

    Let's call the time right now, t = 0. You want to find the time at which your age equals his age, right? So, first write down an equation that tells how old you are at some arbitrary time t. Next, write down an equation that tells how old he is at time t, taking time dilation into account. I'll let you figure out what comes next! :smile:
     
  4. Jan 3, 2005 #3

    DB

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    I think you should use the Lorentz factor for time dilation

    [tex]T_1=T_0\sqrt{1-\frac{v^2}{c^2}}[/tex]

    [tex]T_1[/tex] being the age of person travelling at "v"
    [tex]T_0[/tex] being age the observer

    Don't hold me to it but I think it's the right equation. :rolleyes:
     
  5. Jan 4, 2005 #4
    ahh thanks :) I worked it out to be 40.4 years? Doh...it all seems so obvious now. Im gonna attempt the rest of the questions i have but if i cant do them you might see me return :yuck:
     
  6. Jan 4, 2005 #5
    You might want to check how the question was asked. 40 plus years seems like a long time for someone of that age to pedel at that speed if that is the time frame your ? is asking for.
    Plus getting your age to 60.4 might be worth double checking.
     
  7. Jan 4, 2005 #6

    DB

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    I'm new to this stuff, so your as right as I am but I think you made a couple of mistakes. I see you've been working hard on it so here we go...
    I'm not sure how exact you want to be, but let's say that in order for the biker to be considered younger, he must be 1 year youger.

    I apologize I should have stated better [tex]T_1[/tex]and[tex]T_0[/tex] then you might have understood more.
    [tex]T_1[/tex]:The passage of time of the biker traveling at "v"
    [tex]T_0[/tex]:The passage of time of you, being stationary

    The biker travelling at 19.999 is traveling at 99.995% the speed of light

    [tex]T_1=T_0\sqrt{1-\frac{19.999^2}{20^2}}[/tex]

    [tex]T_1=T_0\sqrt{1-\frac{\sim399.9}{400}}[/tex]

    [tex]T_1=T_0\sqrt{1-.9999000025}}[/tex]

    [tex]T_1=T_0\sqrt{9.99975*10^-5}[/tex]

    [tex]T_1=41y*0.009999875[/tex]
    41 years = 14965 days
    41 years because that how long it would take for you to be older than the biker.

    [tex]T_1= 14965 *(0.00999875)\approx149.63 days[/tex]

    Therefore The passage of time of the biker [tex]T_1[/tex] was 149.63 days

    41 years to you, as you know, is not 41 years to the biker. So after 41 years you see the biker return and he says that he has only been gone for 149 days. So the minimum time the person on the bike would have to continue cycling for so at the end of the journey they are younger than you, according to his watch would be 149 days!

    Now I'm really not sure if I'm right at all, and there are alot of smarter people that can help you better than I can, but I hope this is the right way to solve your problem, sometimes I wish life came with an anwser book. :rofl:
     
  8. Jan 4, 2005 #7
    I don't think this is write. While the person at rest is getting older, the one on the bike is getting older too. However, they don't age at the same rate. (The biker's aging process is slower because of the time dilation)

    G
     
  9. Jan 4, 2005 #8

    DB

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    Well, I dont exactly see what your saying but I can tell you that your right by the fact that the person on the bike is older aswell, though only 149 days older, whereas the person at rest is 41 years older.
     
  10. Jan 6, 2005 #9
    Ok im still confusing myself over this question :( I thought i got it right but i just confused myself when i read back over what i've done. I'm going to try and attempt to write down what ive got so far :)

    My current age is related by:

    [tex]A = 20 + t_1[/tex]
    Where A is my age at a given time.

    The bikers age is:
    [tex]B = 60 + t_0[/tex]
    Where B is the bikers age at a given time.

    [tex]t_1[/tex] and [tex]t_0[/tex] are related by:

    [tex]t_1 = t_0\gamma[/tex](1)

    I want to know the time when A = B:

    [tex]20 + t_1 = 60 + t_0[/tex]
    Rearranged:
    [tex]t_1 = 40 + t_0[/tex](2)

    Combinging (1) and (2):

    [tex]40 + t_0 = t_0\gamma[/tex]

    [tex]\gamma[/tex] is roughly 100 (cant be bothered to do proper calcs at the minute) and so

    [tex]40 = 99t_0[/tex]

    So [tex]t_0[/tex] is about 0.4 of a year..which is how long the biker should cycle for, is that about right?
     
  11. Jan 6, 2005 #10
    Ok heres another question that is confusing me. I really dont know where to begin, infact im not even sure what half of it means (such as relative speed etc) :( I dont think we've been taught it properly yet. Anyway...

    There are 2 astronauts in space, and at the exact moment they pass each other they sync their clocks. They pass each other at a relative speed of 0.8c. Astronaut B says that when his clock reaches 10mins, he will send a flash of light towards astronaut A. I wont type out all the questions (there are lots) but if anyone could give me pointers on the first few then maybe i can help apply them to the rest of the questions. The first Q is:

    According to A, what does B's clock read when he sends the signal? What does A's clock read when B sends the signal?

    At the bottom of the question there is a hint saying that both must agree on A's clock reading when the pulse arrives, which is 30 mins. Any help appreciated!
     
  12. Jan 6, 2005 #11

    DB

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    Well, we both got the same answer. 0.4 of 365 is 146 days and I got a 149 with more precise calcs but there close enough. I would say thats the right answer but I'm not 100% sure. It does make sense according to special relativity.

    As for this question, I find it really hard to understand aswell, and to be honest that hint seems to make things worse. :yuck: But I think a start would be to take into account how far they will be from eachother at 10 mins, traveling at 0.8c, and how long it would take light to reach astronaut A remebering that astronaut a is much futher than he was when they first passed. And I think Lorentz time dilation should be used too. It's a tough problem.
     
  13. Jan 6, 2005 #12
    Hey Exelus, I'm in your class and the answer 40.4 years is correct as far I have been able to work out and I confirmed it with someone else from our class.
    I hear that Physics for Poets is very useful for the astonaut question.
    Basically the time of the other astronaut seems slower to the other astronaut. So the signal sent by the astronaut seems to be sent at something like 16 minutes to the other one. And when the light travels from one to the other they both see it coming/going at c compared to themselves no matter their speed. I just used the following 3 formulae and lots of head-banging and i got some decent answers.

    d = v * t

    L = Lo * SqrRt(1 - v^2 / c^2)

    To = T * SqrRt(1 - v^2 / c^2)

    (sorry don't know how to do any flashy formula stuff, i think the formulae were in our lecture notes though)
     
  14. Jan 7, 2005 #13
    To be exact this is 0.404 years

    To give you a connection between these two answers:

    [tex]t_1 = t_0\gamma[/tex]

    [tex]\gamma = 100[/tex] (approx.)

    [tex]t_1 = 40.4 years[/tex]

    When I read your question, it sounds like they are asking the time with respect to the person at rest. That's why the answer to this question is 40.4 years.

    Gamma
     
  15. Jan 7, 2005 #14
    Also in your second question, you have decided to not to write all the question. Are there any important info there?

    g.
     
  16. Jan 7, 2005 #15
    Hi moth! Glad to see im not the only person struggling on these questions..infact everyone ive talked to so far on the course is struggling so we are gonna meet up on monday to do em!

    Gamma, I havent written out the whole question however i have written everything you are told about the astronauts, the rest of the questions are just a continuation of the first one, such as "according to B how far apart are the ships when he sends the signal? According to A how far apart are the ships when B sends the signal?" etc.

    All the numbers and everything have been posted :(
     
  17. Jan 7, 2005 #16

    DB

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    It will take the biker 40.4 years traveling at .9c for 41 years to pass by on earth? That doesn't make sense. I still think that it would take the biker .4 of a year (146 days) for 41 years to pass on earth. (41 years because that way the 20 year old at rest would be 61)
     
  18. Jan 7, 2005 #17
    This is how I approach this problem. When ever they talk about relative velocities of two objects, it is easier to convert the problem such that one is at rest (this is my way of approach).

    Lets say astr. A and B are travelling in opposite directions with a velocities [tex]V_A[/tex] and [tex]V_B[/tex] respect. This situation is equalent to A staying at rest and B is moving with a velocity [tex]V_A + V_B[/tex] with respect to A. Now, when the signal is released, clock B reads 10 minutes. Since A knows that (they agreed to )
    clock A should read the same time 10 minutes. When the signal is realeased the distance beween A and B is [tex](V_A + V_B) 10[/tex]. Time taken for the signal to reach A therefore is, [tex]\frac {(V_A + V_B) 10}{c} = 8 min.[/tex]. So, when A got the signal it shoul read 10 + 8 = 18 min.

    Actually I don't understand the hint.

    gamma
     
  19. Jan 7, 2005 #18
    Start by simplifying the problem by attaching a rigid latticework of metre sticks and clocks centred on A. This forms the conceptual inertial frame of reference of A. When the flash of light is emitted from B, the time on B's clock as read by a nearby observer in A's frame will be 10 min. Whereas the time read by the observer's own personal wristwatch in A will be dilated to 10[itex]\gamma[/itex] min. The astronaut will also receive the photons of the flash and the photons of the clock face simultaneously. Astronaut A will thus see B's clock to read 10 min. To find the time measured on astronaut's A's clock, we note that the astronaut's clock is synchronised with the distant observer's clock. Thus astronaut A's clock reads 10[itex]\gamma[/itex] min when the flash is emitted in A's frame of reference.
     
  20. Jan 7, 2005 #19
    Actually, if an observer records A's velocity as [tex]V_A[/tex] and B's velocity as [tex]V_B[/tex], then the relative velocity is not [tex]V_A + V_B[/tex] since velocities don't add like real numbers in special relativity.
    This is not correct. The distance between A and B in A's frame of reference is [itex]vt_\mathrm{flash}[/itex] where [itex]v = 0.8c[/itex] is the relative velocity between A and B, and [itex]t_\mathrm{flash} = 10\gamma [/itex] min is the time of the flash as measured by A.
    Ignoring your previous errors, this is still not right. The question asked for A's time when the flash is emitted, not the time when the photons from the flash were received by A.
     
  21. Jan 7, 2005 #20
    Dear Jdstokes,

    Is there any other way to find relative velocities when dealing with special relativity?

    Still I believe va +vb = is the relative velocity = 0.8c = v (in your solution). This is only a stretagy for solution.

    I totally agree with you on [tex]\gamma10 min[/tex] now :blushing: . Lets see if I got it right.

    According to A, when the light flashed, clock A reads [tex]\gamma10 \min[/tex]

    According to A, when the light flashed, clock B reads 10 min.


    A receaves the signal at [tex]10 \gamma + \frac{10\gamma v}{c} = 18\gamma = 30 min.[/tex]

    where [tex]\gamma = [/tex] 1/ sqrt(1-v^2/c^2) = 1/0.6


    gamma
     
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