Lorentz Contraction and Chalk Guns

[montreal1775]

Homework Statement

This is actually an exam problem, not a homework problem. The solution have been posted, but I don't understand the solution. Here is the question, as posted:

A very long rod travels parallel and very close to a flat surface in our laboratory. The
surface contains chalk guns that are spaced apart every meter. All the chalk guns fire
simultaneously in the lab frame at t = 0 and leave marks on the rod at the point just
above them at the time they fired.

The chalk marks on the rod are measured by an observer on the rod and are
found to be 2 meters apart. How fast is the rod moving with respect to the lab?

Homework Equations

Lorentz Transformations

The Attempt at a Solution

When I took the exam, I arrived at the same answer as in the solutions. I was skeptical though, because I thought the chalk guns should appear closer together in the frame of the rod, not spaced at 2 meters. Anyway, the solutions say v=sqrt(3)*c/2. After the exam though, I tried using Mathematica to solve for v explicitly, and it only returned a complex solution. If I take the length of the chalk guns to be 1/2 meters apart in the frame of the rod, then I get the solution posted. Am I just confused, or should the distance between the chalk guns in the rod's frame be spaced closed together? Isn't that why it's called a length contraction, not dilation?

 

[Delphi51]

A moving object appears to be contracted. The people on the moving object are not aware of this. So it makes sense that the observer on the floor measures 1 m apart while the observer on the rod measures 2 m - which looks like 1 m to the stationary observer.

Another way to look at it is that the stationary observer thinks that guy on the rod has a shrunken meter stick. The contracted meter stick fits twice in the 1 m distance between marks.

 

[montreal1775]

But isn't the 1 meter distance the proper length? Since all the guns fire simultaneously in the lab frame, isn't the measured 1 meter the proper length? If so, I don't see how the distance could be measured to be greater than 1 meter in the rod frame. Even if this isn't the case, why do I get v=abs(c)*sqrt(3)*i for the velocity when I use a computer? If I consider the distance measured in the rod frame to be 1/2 meters between chalk guns, then I get the answer posted, but the question says the observer in the rod's frame measures 2 meters.

 

[Doc Al]

But isn't the 1 meter distance the proper length? Since all the guns fire simultaneously in the lab frame, isn't the measured 1 meter the proper length?

1 meter is the proper length between chalk guns.

If so, I don't see how the distance could be measured to be greater than 1 meter in the rod frame.

If the rod frame measured the distance between chalk guns, they would get 1/2 meter. But instead they measured the distance between two events: Chalk gun A firing and chalk gun B firing. In the rod frame, those events happen at different times, thus they do not represent a measurement of the distance between the two chalk guns. (The relativity of simultaneity will get you every time.)

Even if this isn't the case, why do I get v=abs(c)*sqrt(3)*i for the velocity when I use a computer?

Who knows what you did there. (Maybe that's punishment for using a computer, when a piece of paper is all you need. Just kidding! )

If I consider the distance measured in the rod frame to be 1/2 meters between chalk guns, then I get the answer posted, but the question says the observer in the rod's frame measures 2 meters.

The easiest way to solve this is from the lab frame. Since the chalk marks are made simultaneously, they do constitute a length measurement of the moving rod. The proper distance between two points in the rod frame is 2 meters, which the lab measures as 1 meter. (As Delphi51 pointed out.) Thus γ = 2.

 

[montreal1775]

Thanks for the reply. Sorry, I should have mentioned the next part of the question was to calculate the time between adjacent guns firing in the rod's frame. With v=sqrt(3)*c/2, I found the time difference to be -1/sqrt(3) *10^-9 s, which matches the solutions.

I'm having trouble finding the space-time interval to be invariant in both frames. Based on the numbers in the solution, I find (c*delta(t))^2-(delta(x))^2 to be -1 m^2 in the rod frame, and in the lab frame, it should be 1 m^2. Shouldn't they be identical? Am I just making a math error?

Thanks for the help!

 

[Doc Al]

I'm having trouble finding the space-time interval to be invariant in both frames. Based on the numbers in the solution, I find (c*delta(t))^2-(delta(x))^2 to be -1 m^2 in the rod frame, and in the lab frame, it should be 1 m^2. Shouldn't they be identical? Am I just making a math error?

Yes, they should be identical. Show how you did the calculation in the lab frame.

 

[montreal1775]

Oh, right. Stupid mistake. Thanks a lot!