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Two balls in space

  1. Aug 12, 2011 #1
    Some friends and I were out at a pub last night and this question came up.

    If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be travelling the speed of light.

    As the observer in the center I would be able to see light from both object but it would red shifted quite a bit. But what would an observer on each of the balls see of the opposite ball?

    Does light propagate through space at the same speed regardless of the speed of it source relative to another object?
  2. jcsd
  3. Aug 12, 2011 #2
    No, they would be travelling at a velocity:
    \frac{\frac{c}{2} + \frac{c}{2}}{1 + \frac{1}{c^{2}} \, \frac{c}{2} \, \frac{c}{2}} = \frac{c}{1 + \frac{1}{4}} = \frac{4 \, c}{5} = 0.8 \, c
    where c is the speed of light in vacuum.
  4. Aug 12, 2011 #3
    Speed of light = 299792458*m/s

    Half of the speed of light = 149896229 m/s

    I throw ball one at 149 896 229 m/s in one direction.

    I throw ball two in exactly the opposite direction at 149 896 229 m/s.

    Both balls are moving away from me at 149 896 229 m/s. An observer on ball one would see me moving away from him at 149 896 229 m/s. why would he not see ball 2 moving away at 149 896 229 m/s + 149 896 229 m/s which is my speed plus his which is the speed of light.

    Two trains moving towards each other at 50 miles per hour each have a closing speed of 100 miles per hour. When they pass reach other the effective distance between them at any given time will be the sum of there two speeds.
  5. Aug 12, 2011 #4


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    This is a valid question that everybody asks at some point. The problem is that adding the two speeds is technically wrong, it works for everyday life but in reality for moving observers instead of adding the speeds there's an equation that shows that you can never observe something moving at or faster than the speed of light. Look at what dickfore has shown you, this is the answer.

    No matter how fast you are travelling you will always measure the speed of light to be the same. For the two observes they are experiencing time dilation, because of this they measure the closing speed as less than the speed of light.
  6. Aug 12, 2011 #5
    The Law of composition of velocities (in the same direction in space) in Special Theory of Relativity is:
    v_{CA} = \frac{v_{CB} + v_{BA}}{1 + \frac{v_{CB} \, v_{BA}}{c^{2}}}
    This is quite different from what you had used based on Galilean transformation (non-relativistic kinematics).
  7. Aug 12, 2011 #6
    Okay lets try this from another angle.

    Lets add observers C & D they are exactly 10 light seconds from the guy in the middle throwing the balls. The balls are thrown at the same time with observers A & B on board. They travel for exactly 20 seconds each and are caught by observers C & D. Because C & D and the thrower are stationary relative to each other they have not experienced any time dilation. Observers on the balls A & B did relative to each other. A saw B moving away at .8c and visa versa.

    C & D observed 20 seconds of travel time and there is now a physical distance of 20 light seconds between the balls. A & B's watches are no longer in synch with C & D's and the throwers because of time dilation. However they still covered the 20 light seconds in 20 seconds to the stationary observers which means relative to themselves they were travelling at the speed of light as observed by others and measured in distance and time.

    Can you explain in laymens terms please. Remember that this was a pub discussion. I would not know where to begin with the formulas above.

    Many thanks for this guys. Very interesting way to pass time at work on a friday :)
  8. Aug 12, 2011 #7
  9. Aug 12, 2011 #8
    Is that the right link. It just leads to a page that says this.

  10. Aug 12, 2011 #9
    I don't have the slightest idea what you are trying to describe. Care to present a sketch of the situation you are describing?
  11. Aug 12, 2011 #10


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    I don't really know what you are saying here, let me try to sketch it out.


    Each full stop is half a light second. C and D are 10 light seconds away from o (thrower). A and B travel at half the speed of light. C, o and D observe A and B travelling at half the speed of light. A and B can measure their speeds by observing C, o and D. A will observe B receding from it at 80% of the speed of light and vis versa. What are you struggling with here?

    EDIT: It occurs to me that the last thing I said may be wrong, A and B may observe the distance between them increasing faster than light (I'm not sure if that's true). If so what is your point?
  12. Aug 12, 2011 #11
    Good sketch ryan. Was doing one up in paint shop which would have been embarrassing to post to the wild.

    It takes A 20 seconds to travel to C. And 20 seconds for B to travel to D. There are 20 light seconds of distance between C & D.

    So after 20 seconds both a & b have arrived at C & D respectively so the total distance between them is 20 light seconds. So A & B relative to each other have travelled apart at the speed of light as observed by C & D but only observed each other move apart at 80% c.

    Is that a bit clearer?
  13. Aug 12, 2011 #12


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    Yes C and D would measure A and B moving away from each other as C but that doesn't mean anything FTL is actually occurring. A and B would not measure their receding speed as this.
  14. Aug 12, 2011 #13
    Is that simular to saying the space between the [STRIKE]balls[/STRIKE], objects is NOT increasing by 299 792 458 m/s from the frame of either of the objects.

    But it is increasing by 299 792 458 m/s wrt the observer who is in the middle.
  15. Aug 12, 2011 #14


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    Okay, the problem is that Mustakafine has confused his high school physics with real physics. Dickfore and Ryan are both correct in that this is in the wrong forum, but let's at least settle this permanently:

    Mustakafine is trying to apply the following math:

    A central observer (O) witnesses two vehicles (A and B) travel away from his point of reference at 60 mph in opposite directions. After one hour's time, length AO is 60 miles and length OB is 60 miles. By back calculating the relative speed, he determines they were travelling 120mph away from each other, or 60mph+60mph.

    THIS IS AN APPROXIMATION!!!!!!!!!!!!!!!!! It is not how the universe actually works. If you had an accurate enough stop watch, and a perfect enough measurement of velocity, you would find that the two cars were actually travelling away from each other at 119.9999999999998mph as given by the following formula:

    v_{CA} = \frac{v_{CB} + v_{BA}}{1 + \frac{v_{CB} \, v_{BA}}{c^{2}}}

    119.9999999999998 = \frac{60 + 60}{1 + \frac{60*60} {c^{2}}}

    If you re-solve that problem with the very high speed you're talking about, you will find that they are NOT receding from each other at the speed of light.

    EDIT: Try this out in Excel (I just did) and compare the results of the correct calculation and the approximation by addition. You can see that as you approach the speed of light, your error in the calculation gets higher and higher!

    At one billionth the speed of light, Excel doesn't have enough decimal places to represent the error. Effectively zero. But at one millionth the speed of light (670 mph), the error between the correct value and the approximation of adding the two values is 1 x 10-10%.

    As you continue you will see that at a hundredth the speed of light (6.7 million mph) the error between your approximation by adding and the correct value is 0.01%. At a tenth the speed of light (67 million mph) the error is 1%. And at the speed of light (670 million mph) the difference between your approximation and the real value is 100%!

    The error in your calculation for half the speed of light is 25%. This means that your answer (the speed of light) is 25% higher than the correct answer. 1.25X = Y where X is the correct answer and Y is your answer (the speed of light) then you should get Y/1.25 = 0.8c

    Dickfore gave you the CORRECT answer originally.
    Last edited: Aug 12, 2011
  16. Aug 12, 2011 #15
    Thanks guys I get it now.

    Didnt know this was the wrong forum for this type of question first day here and until we have another pub night trying to bash Einstein after a couple beers probably the last.
  17. Aug 12, 2011 #16


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    In general, this has proved to be fruitless.

    For everyone.
  18. Aug 12, 2011 #17
    God you physics guys are a bit uptight. The responders did not have to take any to answer this but it was a genuine question.
  19. Aug 12, 2011 #18


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    I think the reason you got a less than warm welcome is the fact that these answers are readily available to anyone with the inclination to look them up.

    If you look back, both Ryan and Dickfore answered you correctly and with great compassion and willingness to explain. Dickfore gave you the formula to use, and Ryan gave you a nice explanation.

    But that wasn't enough. You pressed on with additional posts. It would probably have been less frustrating if you weren't the seven millionth person to register an account on here for the sole purpose of belligerently rejecting Einstein, or fumblingly denouncing Gauss.

    So, yes, this thread has proved frustrating for us "uptight physics guys."
  20. Aug 12, 2011 #19


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    Don't worry about it. Your question is a perfectly valid question and most of the individuals here are not like what you, unfortunately, have experienced in this thread. Just remember that many of the vector space operations you learned usually apply to [itex]\mathbb{R}^{n}[/itex] and the minkowski space of SR has some different properties from [itex]\mathbb{R}^{n}[/itex] and in GR you have to use laws that do not have a preferred coordinate system.
    Last edited: Aug 12, 2011
  21. Aug 12, 2011 #20
    A way to avoid all this 'confusion' is to use rapidities instead of speed.
    Unlike speed rapidity is additive.
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