How Does Length Contraction Affect Markings on a Moving Conveyor Belt?

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serp777
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Homework Statement



A relativistic conveyor belt is moving at speed 0.5c
relative to frame S.Two observers standing beside the
belt, 10 ft apart as measured in S, arrange that each
will paint a mark on the belt at exactly the same instant
(as measured in S). How far apart will the marks
be as measured by observers on the belt?


Homework Equations



gamma = 1/√(1-β^2) = 1.1547
L = L0/gamma

The Attempt at a Solution


This problem seems very easy and I think I might be missing something as a result.

I simply viewed the problem as the conveyor belt being stationary (and therefore the observers) and the people marking the belt to be moving at speed 0.5c.

I then used the length contraction formula to calculate that the markings on the conveyor belt as the length between the two people marking would be length contracted.

L = 10/ 1.1547 = 8.66 feet between the two markings.



is this right or am I missing something?
 
on Phys.org
Hello, serp777.

You have to be careful when using the length contraction formula: L = L0/gamma

L0 is the "proper" length of an object. This is the length of an object as measured in the reference frame in which the object is at rest.

L is the "non-proper" length. This is the length of an object as measured in a frame in which the object is moving and in which the length is defined as the distance between the locations of the ends of the object when the ends are observed simultaneously in this frame.

Once the two marks have been painted on the belt, you can think of the section of the belt between the marks as an object.

Which reference frame measures the proper length of this object and which frame measures the non-proper length?
 
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The proper length would be measured by the observers on the moving object while the non proper length would be measured by the stationary observer? Is the numerical answer incorrect even though i manipulated the equation wrong? Thanks for your help