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A moving object will have relativistic length contraction. About which point on the object would it contract? For instance, would the contraction be towards the point on the object that is closest to the observer?

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closed as unclear what you're asking by WillO, John Rennie, Jon Custer, GiorgioP, ZeroTheHero Apr 4 at 4:42

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  • $\begingroup$ I agree that it will be contracted in the direction parallel to the motion. My question is about the point about which the contraction takes place. $\endgroup$ – Dirac May 5 '18 at 5:15
  • $\begingroup$ What on earth does this mean? According to me, at any given moment, the front of my rocket ship is 100 meters from the back. According to you, at any given moment, the front of my rocket ship is 80 meters from the back. From those data, how do you define a "point about which the ship is contracted"? $\endgroup$ – WillO Apr 3 at 4:25
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The only well-defined point around which to define contraction is the center of the object along its direction of motion. This value is independent of the apparent length of the object. Additionally, the contraction has nothing to do with the direction that points towards the observer. The object will only be contracted along the spatial dimension in which it is traveling. So, if it is a rod traveling along its axis, it will not change in radius.

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  • $\begingroup$ Agreed that the object will contract along the spatial dimension in which it is traveling. Agreed that a rod will not change radius. However, what is the proof that it contracts about it's center? Also, how do you define center here? Is it the center of gravity? Is it the geometric center? $\endgroup$ – Dirac May 5 '18 at 5:28
  • $\begingroup$ In my question, I am referring to the point on the object along the axis of motion that is closest to the observer $\endgroup$ – Dirac May 5 '18 at 5:34
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    $\begingroup$ Contraction does not depend on a point; it occurs to the same degree regardless of the central point, and only depends on the velocity. Because the geometrical center is the only thing that is also independent of velocity and/or contraction, it represents the “center” when discussing Lorentz contractions. The center is the only geometrical invariant. Even when the length changes, the center moves along as always. Unlike general relativity, length in special relativity does NOT take mass into account. $\endgroup$ – Josh McK May 5 '18 at 5:53
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    $\begingroup$ Lorentz contractions don’t give a **** about mass. The only other thing to care about is length. No matter how many times you length contract your sphere and rod, the geometric center will be the same (different, depending on how long you make your rod). The point is that once you choose the lengths, the geometric center is constant. The center of mass is totally irrelevant. Imagine shrinking things to a point. You’ll always hit the same center. Expand them, and the geometric center will be the same. It doesn’t matter how much you send to left or right from that point. $\endgroup$ – Josh McK May 5 '18 at 6:28
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    $\begingroup$ Furthermore, length contraction is defined between two events. Your scenario does not define two clear instances of measuring lengths at specificied locations and times. $\endgroup$ – Josh McK May 5 '18 at 6:35
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The contraction is toward the front. The reason it appears to be contracted when travelling away (or you away from it) is that the light from the back has less distance to travel, so the back is seen moving before the front, & appears to partly catch up to it, making the object look shorter. L'/L= 1-v/c the same as Doppler formula for frequency. The same is true if the object is travelling cross ways. Light coming out at an angle towards the back intercepts the observer like if you throw a rock from a moving vehicle at a pole while passing it (you have to throw it partly backwards), so again, the light from the back has less distance to travel. L'/L=√(1-(v/c)²) (Lorentz Contraction). The light coming out backward also makes the object look partly turned away (the light you see is from it's rear end). If it is moving towards you (or you towards it), it would appear longer because the front would be seen moving first. L'/L=1+v/c

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