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Newbie question on adding velocities

  1. Mar 14, 2013 #1
    Hi all, I'm new to these boards. I have a basic conceptual knowledge of physics, but I was just watching some videos about special relativity and the following question struck me. I'm sure this is probably a pretty simple question but it's really been bothering me...

    I understand how "c" is constant and absolute and that it's improper to simply add velocities, but let's take the scenario where I'm on a spaceship traveling at the speed of light minus, say, 1 foot per second. I'm standing still (relative to the ship). I throw a baseball forward at 4 ft/s. Now to a stationary observer you can't add the velocities and say the ball is going faster than light. However in my own frame of reference, I'm at rest, so wouldn't I see the ball traveling away from me at 4 ft/s? How does one resolve this? I figure it has something to do with length contraction of the ball so it's not "really" going at 4 ft/s, but isn't the spaceship length contracted as well? Your thoughts?

    Thanks!

    Please feel free to be "mathy" in your response.
     
  2. jcsd
  3. Mar 14, 2013 #2
    To you, of course, the ball will seem to travel at 4 ft/s relative to yourself. To someone else standing nearby as you whiz past at just under c, the ball will seem to be barely moving relative to you. This is because of two factors:

    * Your ruler is length-contracted in the bystander's frame of reference, so at the point where you claim the ball has moved 4 feet relative to you, the bystander disagrees because to him your ruler is much shorter than it should be.

    * Your clock runs slow in the bystander's frame of reference, so at the point when you say one second has elapsed since you released the ball, the bystander says much more time has actually gone by.

    If the bystander uses his own ruler and his own clock, he measures a much smaller relative speed than 4 m/s.
     
  4. Mar 14, 2013 #3
    That makes a lot of sense, thanks! I assume that since my ruler is "contracted" from the bystander's point of view, then the ship itself is contracted as well (since either the ship or the ruler could be used to measure the ball's distance).
     
  5. Mar 14, 2013 #4

    HallsofIvy

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    The formula is this: if you are moving, relative to me, at speed u, and you throw a ball at speed, relative to you, v, then I will observe the ball moving at speed, relative to me, [itex](u+ v)/(1+ uv/c^2)[/itex]
    If u and v are both very small reative to c, this will be very close to u+ v. But if either both are close to c, this will be just below c.
     
  6. Mar 14, 2013 #5
    It goes without saying here that a single observer with a ruler and a clock will not be able to come to these conclusions. He will need to establish his inertial frame first which applies to the various locations that the traveler will be moving through, encompassing the various events in this scenario. Such as by this method: http://en.wikipedia.org/wiki/Einstein_synchronisation
     
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