Help! Solving Einstein's Length Contraction Problem

[makphi]

TL;DR

Does a metre rod at rest relative to K appear longer or shorter in respect to K'?

I have to admit that my "best" math days are long gone. That said, I wonder if anyone can help me? I'm stuck in part 1 of Einstein's book on 'Relativity, The Special & General Theory: The behaviour of measuring rods & clocks in motion', specifically on the second equation : √(1-v^2/c^2) used to describe the length of a metre rod at rest on the x-axis in respect to K as viewed from K'. By substituting the value of x with 0 & 1 respectively and getting the difference
, I calculated the relative length of the metre rod as viewed from K' to be the reciprocal of the above equation i.e.: 1/(√(1-v^2/c^2)) meaning the metre rod would appear longer instead of shorter as viewed from K'. Where am I going wrong?

 

[A.T.]

√(1-v^2/c^2) used to describe the length of a metre rod at rest on the x-axis in respect to K as viewed from K'.
...
I calculated the relative length of the metre rod as viewed from K' to be the reciprocal of the above equation

If the equation is for the length in K', why the reciprocal?

 

[PeroK]

Summary: Does a metre rod at rest relative to K appear longer or shorter in respect to K'?

I have to admit that my "best" math days are long gone. That said, I wonder if anyone can help me? I'm stuck in part 1 of Einstein's book on 'Relativity, The Special & General Theory: The behaviour of measuring rods & clocks in motion', specifically on the second equation : √(1-v^2/c^2) used to describe the length of a metre rod at rest on the x-axis in respect to K as viewed from K'. By substituting the value of x with 0 & 1 respectively and getting the difference
, I calculated the relative length of the metre rod as viewed from K' to be the reciprocal of the above equation i.e.: 1/(√(1-v^2/c^2)) meaning the metre rod would appear longer instead of shorter as viewed from K'. Where am I going wrong?

Let me guess. You applied a Lorentz Transformation to the two ends of the rod in its rest frame and took the difference in x' coordinates in the frame where it is moving?

 

[Herman Trivilino]

I calculated the relative length of the metre rod as viewed from K' to be the reciprocal of the above equation

Taking the reciprocal is not the correct way to determine the length as measured by K'. The reason is that when the measurement is taken in K the position of each end of the rod is noted. But K' will not agree that the positions of the ends were noted at the same time. Events that are simultaneous in K are not, in general, simultaneous in K'.

 

[makphi]

Let me guess. You applied a Lorentz Transformation to the two ends of the rod in its rest frame and took the difference in x' coordinates in the frame where it is moving?

Exactly! Is that wrong?

 

[Pencilvester]

Exactly! Is that wrong?

Yes, see @Mister T ’s post, post #4.

 

[makphi]

Taking the reciprocal is not the correct way to determine the length as measured by K'. The reason is that when the measurement is taken in K the position of each end of the rod is noted. But K' will not agree that the positions of the ends were noted at the same time. Events that are simultaneous in K are not, in general, simultaneous in K'.

I just substituted the values of x=0, x=1 in the first equation (see attached) & got the difference (X1'-X0') for the length of the metre rod in K'. This turned out to be the reciprocal i.e 1/(√(1-(v^2)/(c^2))). I suspect my logic is wrong but I don't know how.
Can you show me how to arrive at √(1-(v^2)/(c^2)) using the first equation of the Lorentz transformation?

 

[makphi]

Yes, see @Mister T ’s post, post #4.

I don't get it. So how do we arrive at this conclusion using the Lorentz transformation equations:
"...
If, on the contrary, we had considered a metre-rod at rest in the x-axis with respect to K,
then we should have found that the length of the rod as judged from K1 would have been
√(1-(v^2)/(c^2)) ; ..."

 

[Ibix]

Exactly! Is that wrong?

The problem is that you substituted the x positions and used the same times in the rod's rest frame. But these measurements did not happen at the same time in the frame where the rod is moving. So you measure the length plus $v\Delta t$, which does indeed turn out to be $\gamma l$.

You need to make sure your measurements are made at the same time in the frame where the rod is moving, which means involving the time transforms.

 

[MikeLizzi]

makphi, everyone is trying to point out to you the same problem. When I first learned SR I went round and round that same problem too. Here is how I finally resolved it for myself.

http://www.relativitysimulation.com/Documents/DeterminingTheLengthOfMovingObject.html

 

[Pencilvester]

I don't get it. So how do we arrive at this conclusion using the Lorentz transformation equations:
"...
If, on the contrary, we had considered a metre-rod at rest in the x-axis with respect to K,
then we should have found that the length of the rod as judged from K1 would have been
√(1-(v^2)/(c^2)) ; ..."

You know that one end of the rod moves along the line $x’ = -vt’$. You need to find $x’$ of the other end at some constant $t’$. $t’ =0$ would be the most convenient. If you set the part of the Lorentz transformation that yields $t’$ equal to zero, you can solve an equation for $x$ as a function of $t$ (or vice versa) that defines a line in frame K such that $t’ =0$. Find the intersection of that line with the line that defines the other end of the stick in frame K, use that $x$ and $t$ in the Lorentz transformation to get $x’$, subtract the $x’$ coordinate of the other end (0 in this case), and voila, you have the length in frame K’. If you use general symbols instead of specific numbers, you can work out that the length will always be shortened by a factor of $1/\gamma$.

 

[Nugatory]

Exactly! Is that wrong?

Yes, because you have not allowed for the relativity of simultaneity. The length of the rod is the distance between where the ends are at the same time. In the frame in which the rod is at rest one end might be at the origin $(x=0,t=0)$ at time zero and the other end at the event $(x=L,t=0)$; we correctly calculate that the length of the rod is $L$. However, when we Lorentz-transform these two events to find their coordinates in the primed frame, we get $(x'=0,t'=0)$ and $(x'=\gamma L,t'=-\gamma vL)$; because the time coordinates are different these are not the positions of the end of the rod in the primed frame at the same time so $\gamma L$, the difference between the $x'$ coordinates is not the length of the rod in the primed frame.
You may find tht if you plot the paths of the two endpoints of the rod on a Minkowski spacetime diagram you'll get a better intuitive feel for what's going on, and a Google search for "relativity length contraction derivation" will find a derivation that steps through the algebra needed to find the event that corresponds to the position of the end of the rod at time zero in the primed frame, which will turn out to be $(x'=L/\gamma,t'=0)$.

And you are not the first person to have discovered this relativity of simultaneity pitfall the hard way... There's a reason why @PeroK was able to guess with high confidence what you did, and why @Orodruin's signature is what it is...

 

[Herman Trivilino]

I just substituted the values of x=0, x=1 in the first equation (see attached) & got the difference (X1'-X0') for the length of the metre rod in K'.

The only mistake you're making there is to assume that $x'_1-x'_0$ is the length of the rod. It's not. It's the difference in the positions of the ends of the rod taken at two different times.

By the way, it would make your posts more readable if you use LaTeX to write your mathematical expressions.