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Speed of light bothering me

  1. Jun 7, 2015 #1
    Dear PF Forum,
    There's one thing still bothering me.
    The speed of light.
    Supposed two observer, A and B
    A is from the west, B from the east.
    Separated by 100 lys.
    B sends a signal, say, B1 to A. So B1 will be received by A in 100 years, right?
    Now, supposed B travels at 0.8c and at the distance 90lys from A B sends a signal, B2.
    B2 will be received by A in 90 years, right?
    Not 50 years. I supposed we can't divide 90ly by (c+0.8c), because light travels at (of cource) the speed of light.

    And what if 1 second before B sends B2, A suddenly travels at 0.8c?
    When will B2 be received by A?
    A. 450 years?
    Because Va.t + 90ly = c.t
    0.8.t+90 = t
    t = 450
    B. 90 years?
    Because A is in the same frame reference of the signal? Or not? Because A travels 1 seconds before the signal was fired.
    C. Other?

    What if A travels 0.8c the instant B sends B2
    When will B2 be received by A?
    A. 450 years?
    B. 90 years?
    C. Other?


    What if A travels 0.8c 1 second after B sends B2
    A. 450 years?
    B. 90 years?
    C. Other?

    What if A travels 0.5c 1 second before B sends B2
    A. 270 years? Provided the calculation above.
    B. 90 years?
    C. Other?

    What if A travels 0.5c the instant B sends B2?
    What if A travels 0.5c 1 second after B sends B2?

    Thanks for any effort to explain to me
     
  2. jcsd
  3. Jun 7, 2015 #2

    PeroK

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  4. Jun 7, 2015 #3
    Thanks PeroK for your clarification

    Every frame here is in A's frame.

    The speed of light.
    Supposed two observer, A and B
    A is from the west, B from the east.
    Separated by 100 lys. [In which reference frame?]
    B sends a signal, say, B1 to A. So B1 will be received by A in 100 years, right?
    Now, supposed B travels at 0.8c and at the distance 90lys from A B [In which reference frame?]sends a signal, B2.
    B2 will be received by A in 90 years, right?[In which reference frame?]
    Not 50 years[In which reference frame?]. I supposed we can't divide 90ly by (c+0.8c), because light travels at (of cource) the speed of light.

    And what if 1 second before B sends B2, A suddenly travels at 0.8c[In which reference frame?]?
    When will B2 be received by A?[In which reference frame?]
    A. 450 years?[In which reference frame?]
    Because Va.t + 90ly = c.t
    0.8.t+90 = t
    t = 450
    B. 90 years?[In which reference frame?]
    Because A is in the same frame reference of the signal? Or not? Because A travels 1 seconds [In which reference frame?]before the signal was fired.
    C. Other?

    What if A travels 0.8c [In which reference frame?]the instant B[In which reference frame?] sends B2
    When will B2 be received by A[In which reference frame?]?
    A. 450 years?[In which reference frame?]
    B. 90 years?[In which reference frame?]
    C. Other?[In which reference frame?]


    What if A travels 0.8c[In which reference frame?] 1 second after B sends B2[In which reference frame?]
    A. 450 years?[In which reference frame?]
    B. 90 years?[In which reference frame?]
    C. Other?[In which reference frame?]

    What if A travels 0.5c 1 second [In which reference frame?]before B sends B2
    A. 270 years? Provided the calculation above.[In which reference frame?]
    B. 90 years?[In which reference frame?]
    C. Other?[In which reference frame?]

    What if A travels 0.5c the instant B sends B2?[In which reference frame?]
    What if A travels 0.5c 1 second after B sends B2?[In which reference frame?]

    Thanks for any effort to explain to me
     
  5. Jun 7, 2015 #4

    PeroK

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    In A's frame the calculation is simple: t = d/c is the time for a light signal to travel a distance d. Regardless of the speed of the light source.

    But, A is always stationary in its own reference frame. A cannot travel at 0.8c in its own reference frame. You need to rethink your questions based on that.
     
  6. Jun 7, 2015 #5
    Wait...
    Did you say, any object CAN'T TRAVEL?
    So EVERY CALCULATION IN RELATIVIY MUST BE DONE IN THE OBJECT REST FRAME?
    EVERYTHING ELSE ARE IN MOTION ACCORDING TO THE OBSERVER?
    The observer must think, or destined at rest?
    Is that so?
     
  7. Jun 7, 2015 #6

    PeroK

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    There's no way to measure absolute motion. No observer can say "I'm travel absolutely at 100m/s or 0.8c or whatever". All an observer can say is that they are moving with respect to something else.

    A can certainly know that initially they were at rest wrt B, then they accelerated and ended up travelling at 0.8c wrt to B. This implies two reference frames: the frame in which A was inititally at rest wrt to B (but now A is travelling at 0.8c) and a new frame in which A is at rest (and B is moving at 0.8c towards A). Measurements of distance and elapsed time will differ in these two frames, so you need to decide which frame of reference you are using.
     
  8. Jun 7, 2015 #7
    I miss something where. I often see the shorthand "wrt" in the answers of the previous thread.
    What is that? With respect to?
     
  9. Jun 7, 2015 #8

    PeroK

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    Yes!
     
  10. Jun 7, 2015 #9
    Okay...
    Let's say there's a signal coming 90 lys to A and C.
    While they are still at rest.
    Both synchronize their watches and reset their watches to zero
    Then suddenly A travels 0.8 c in the opposite direction of the signal. C keeps staying.
    The signal will eventually catches A, right.
    What will A watch show when A is receiving the signal?
    What will C watch show when C is receiving the signal?
     
  11. Jun 7, 2015 #10

    PeroK

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    The situation for C is simple: the combined (separation) velocity of A and the light signal is 1.8c. So, it takes 50 years by C's watch for the light to reach A. It's a common misconception that combined (separation) velocities must stay below c, but that is false. C can happily see two objects move with a combined velocity of 1.8c towards each other. And, of course, another 40 years for the light to reach C.

    But, A would observe something different. The best way to look at this is to use time dilation (or the Lorentz transformation) to map the time A receives the signal in C's frame (d = 40 lys and t = 50 ys) to A's frame (d' = 0 lys and t' = 30 ys).

    You can also use Lorentz to calculate the time and place (according to A) when C receives the signal.

    Also, a good exercise is to use Lorentz to calculate where and when the light signal is emitted in A's frame (assuming A is moving at 0.8c towards the signal). It's d = 90 lys and t = 0 in C's frame.
     
  12. Jun 7, 2015 #11
    Ahh, I must have typed incorrectly. A travels 0.8 at the SAME direction of the signal.
    I'm sorry PeroK for your inconvenience.
    And thanks for answering me.
     
  13. Jun 7, 2015 #12

    PeroK

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    The numbers are different, but the principle is the same. Solve in C's frame, then transform to A's frame using Lorentz.
     
  14. Jun 7, 2015 #13
    Sorry to ask the wrong question.
    Let's say there's a signal coming 90 lys to A and C.
    While they are still at rest.
    Both synchronize their watches and reset their watches to zero
    Then suddenly A travels 0.8 c in the SAME direction of the signal. C keeps staying.
    The signal will eventually catches A, right.
    What will A watch show when A is receiving the signal?
    What will C watch show when C is receiving the signal?
     
  15. Jun 7, 2015 #14

    PeroK

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    You do all the calculations in C's frame and then use Lorentz to transform to A's frame. If you didn't understand what I did above, then what's the point of me giving you numerical answers for a new problem if you won't understand those either?
     
  16. Jun 7, 2015 #15
    No, I didn't :smile:
    I just understand ##\gamma## in Lorentz Transformation. I'm reading Lorentz Transformation with matrix, now.
    Still thanks for the answer.
     
  17. Jun 7, 2015 #16
    It's just that we (the layman) often hears twins paradox, time machine, etc...
    At first I think understanding twins paradox is not that hard. But it's very difficult (for me).
    Not given up yet.
     
  18. Jun 7, 2015 #17
    Why 30 years?
    Is it because ##\gamma = \frac{1}{\sqrt{1-0.8^2}} = \frac{5}{3}##
    And ##\frac{50}{\frac{5}{3}}## = 30?
     
  19. Jun 7, 2015 #18

    PeroK

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    Yes, that's time dilation, which is a special case of the Lorentz Transformation where you are following A. I.e. in C's reference frame, A's watch is moving, hence time dilated. So, when the light reaches A (in the first problem that's simply after 50 years), A's watch will read (3/5)*50 = 30 yrs.

    But, the general Lorentz Transform is:

    ##t' = \gamma (t - \frac{Vx}{c^2})##

    ##x' = \gamma (x - Vt)##

    So, for events that take place at points other than where A is, you don't have simply time dilation. It's more complicated.

    It depends what you are trying to learn with this. To understand Special Relativity you do have to study it thoroughly.
     
  20. Jun 7, 2015 #19
    And as I read from this
    http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
    There are still y' and z' that you don't write. But thank you very much. Yours is more to the point! Easier to understand.
     
  21. Jun 8, 2015 #20
    No, you can calculate anything with a single reference system in which some things are moving, and often (but not always) that is also the easiest. In fact that is what you do when you talk about an object that moves at 0.8c.

    If you use a standard reference system for the physics, then this implies that you assume for objects that are not moving relatively to this system, that they are physically in rest (the objects cannot be thought to be in rest and travel at 0.8c). Consequently, from that perspective you assume that:
    - the speed of light relative to them is c in all directions
    - they are not length contracted
    - if they are clocks then they run at normal speed.
     
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