1 Jay |
Length contraction confusion | dinsdag 11 december 2001 7:13 |

2 Dirk Van de moortel |
Re: Length contraction confusion | dinsdag 11 december 2001 20:18 |

3 Eric Prebys | Re: Length contraction confusion | dinsdag 11 december 2001 20:21 |

4 Tom Roberts |
Re: Length contraction confusion | woensdag 12 december 2001 3:02 |

5 Nicolaas Vroom |
Re: Length contraction confusion | maandag 17 december 2001 11:31 |

6 Nicolaas Vroom |
Re: Length contraction confusion | zondag 23 december 2001 10:33 |

S.R. says that when measured from a stationary system a moving object's length is contracted (relative to its rest length).

However, imagine two masses at rest in the stationary system, one at the origin (x1=0) and the other some distance away (x2=k). At time t0 in this system, velocity v is imparted to these masses. At some later time t1, the distance between these masses is measured in the system. If t = (t1-t0), the first mass will be at x1=v*t+0 and the second mass at x2=v*t + k. This simply follows from the definition of 'velocity'.

The distance between them is then x2-x1 = v*t+k - (v*t+0) = k. In other words, the same length as the rest length.

How can this be reconciled with what S.R. says about length contraction?

"Jay"

> |
S.R. says that when measured from a stationary system a moving
object's length is contracted (relative to its rest length).
However, imagine two masses at rest in the stationary system, one at the origin (x1=0) and the other some distance away (x2=k). At time t0 in this system, velocity v is imparted to these masses. At some later time t1, the distance between these masses is measured in the system. If t = (t1-t0), the first mass will be at x1=v*t+0 and the second mass at x2=v*t + k. This simply follows from the definition of 'velocity'. |

> |
The distance between them is then x2-x1 = v*t+k - (v*t+0) = k. In other words, the same length as the rest length. How can this be reconciled with what S.R. says about length contraction? |

It does not follow from the definition of velocity but from the Galilean transformation of coordinates.

Since the two masses are living in the moving frame where they have

The coordinates in the other frame are:

Dirk Vdm

> |
S.R. says that when measured from a stationary system a moving object's length is contracted (relative to its rest length). However, imagine two masses at rest in the stationary system, one at the origin (x1=0) and the other some distance away (x2=k). At time t0 in this system, velocity v is imparted to these masses. At some later time t1, the distance between these masses is measured in the system. If t = (t1-t0), the first mass will be at x1=v*t+0 and the second mass at x2=v*t + k. This simply follows from the definition of 'velocity'. |

No, it doesn't follow from the "definition of velocity"; it follows from Galilean transformations. Lorentz transformations give a different relationship between the coordinates in the two frames.

> |
The distance between them is then x2-x1 = v*t+k - (v*t+0) = k. In
other words, the same length as the rest length.
How can this be reconciled with what S.R. says about length contraction? |

If you use the correct transformations, you'll get the right answer.

--
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Eric Prebys, Fermi National Accelerator Laboratory

Office: 630-840-8369, Email: prebys@fnal.gov

WWW: http://home.fnal.gov/~prebys

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> | S.R. says that when measured from a stationary system a moving object's length is contracted (relative to its rest length). |

Yes. Because to measure the length of a moving object one must mark both ends of the moving object _simultaneously_ and then measure the distance between the marks. In SR what one means by "simultaneous" differs for relatively-moving observers.

> |
However, imagine two masses at rest in the stationary system, one
at the origin (x1=0) and the other some distance away (x2=k). At time
t0 in this system, velocity v is imparted to these masses. At some
later time t1, the distance between these masses is measured in the
system. If t = (t1-t0),
the first mass will be at x1=v*t+0 and the second mass at x2=v*t + k.
This simply follows from the definition of 'velocity'.
The distance between them is then x2-x1 = v*t+k - (v*t+0) = k. In other words, the same length as the rest length. How can this be reconciled with what S.R. says about length contraction? |

I read your situation differently from either Dirk van de Moortel or Eric Prebys.

For descriptive purposes I assume k>0 and v>0, and v is parallel to the x axis in the +x direction. My moving observer will use an x' axis parallel to the x axis.

Yes, place those two masees there, and then start them moving with the same velocity v WRT THE "STATIONARY SYSTEM", and make them start SIMULTANEOUSLY IN THE "STATIONARY SYSTEM". Then anytime later one measures the distance between them in the stationary system and obtains k, the original distance between them. Remember that to make this measurement one marks both of them _SIMULTANEOUSLY_ in the "stationary system" and measures the distance between the marks.

The reconcilation with SR is that if you use a ruler moving along with them to measure the distance between them, you will obtain a distance larger than k. And if an observer moving with velocity v wrt the "stationary system" had been watching the whole scenario, he would simply say that originally both masses were moving in the -x' direction, and the one at x=k stopped moving before the one at x=0, so of course they are more than the distance k apart (to him).

Tom Roberts tjroberts@lucent.com

Tom Roberts

> | Jay wrote: |

> > |

> |
Yes. Because to measure the length of a moving object one must mark both ends of the moving object _simultaneously_ and then measure the distance between the marks. In SR what one means by "simultaneous" differs for relatively-moving observers. |

IMO the best way to study length contraction is to perform an experiment and to study that only in one (rest) frame.

IMO the simplest experiment is the following.
Take one rod with two mirrors A,B at both end.

A-------------O------------B

There is one Observer O in the middle.
(At rest)

O sends out a light pulse to A,B and observes that the reflection
is simultaneous.
The time is reflection time is t = OAO = OBO

Next you move the rod with a speed v to the right (you start far away from the left of A) and you perform the same experiment in such a way that the reflection time t = OAO.

(t is time at the leading edge of the pulse)

If length contraction is applicable than you will see that at the same moment when you receive the signal via mirror A you will not receive the signal via mirror B. The reflection via B should come earlier.

The bets are on if this is true.

No moving observers are involved. No moving clocks are involved.

> |
Yes, place those two masees there, and then start them moving with the same velocity v WRT THE "STATIONARY SYSTEM", and make them start SIMULTANEOUSLY IN THE "STATIONARY SYSTEM". Then anytime later one measures the distance between them in the stationary system and obtains k, the original distance between them. Remember that to make this measurement one marks both of them _SIMULTANEOUSLY_ in the "stationary system" and measures the distance between the marks. The reconcilation with SR is that if you use a ruler moving along with them to measure the distance between them, you will obtain a distance larger than k. And if an observer moving with velocity v wrt the "stationary system" had been watching the whole scenario, he would simply say that originally both masses were moving in the -x' direction, and the one at x=k stopped moving before the one at x=0, so of course they are more than the distance k apart (to him). Tom Roberts tjroberts@lucent.com |

Nicolaas Vroom

> |
Tom Roberts |

> > | Jay wrote: |

> > > |

> > |
Yes. Because to measure the length of a moving object one must mark both ends of the moving object _simultaneously_ and then measure the distance between the marks. In SR what one means by "simultaneous" differs for relatively-moving observers. |

> |
IMO the best way to study length contraction is to perform an experiment and to study that only in one (rest) frame.
IMO the simplest experiment is the following.
Take one rod with two mirrors A,B at both end. Next you move the rod with a speed v to the right (you start far away from the left of A) and you perform the same experiment in such a way that the reflection time t = OAO. (t is time at the leading edge of the pulse) If length contraction is applicable than you will see that at the same moment when you receive the signal via mirror A you will not receive the signal via mirror B. The reflection via B should come earlier. The bets are on if this is true. |

No bets, no opinions received. That means I still am not sure if you can demonstrate and understand length contraction without any moving observer. (That means length contraction is only a function of space, not of space and time or of spacetime)

> | No moving observers are involved. No moving clocks are involved. |

Created: 23 December 2001

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