Why is length contracted, and not ELONGATED?

[dreamLord]

Silly question, I assume, but I can't quite figure it out. I've just started reading up on Special Theory of Relativity, and only have knowledge about Lorentz Transformations.

Anyway, supposing a rod is at rest in the frame S', which is moving with velocity v with respect to frame S. Then the length of the rod in frame S' is the true length L(0), and is given by x2' - x1', where x2' and x1' are the end points of the rod in frame S'.

Now we have to find the length of the rod in frame S, which is given by x2 - x1, where x2 and x1 are the end points of the rod in frame S.
x2 is related to x2' by x2 = γ(x2' - vt)
and x1 is related to x1' by x1 = γ(x1' - vt)

Then length in frame S = x2 - x1 = γ(x2' - x1')
Or length L = γL(0)

Hence length to an observer in frame S is elongated.

May I know where I went wrong?

 

[ghwellsjr]

If you would have also calculated the times of the two events in frame S you would see that they are at different times meaning that the length that you are calculating for the rod is while it is moving. What you need to do is find two events that are at the same time in frame S (but different times in frame S') that correspond to the end points of the rod. Do you think you can figure that out?

 

[dreamLord]

No sorry, I don't get it. I thought I had taken care of that part by putting t1 = t2 = t in the Lorentz transformation for x2 and x1?

EDIT : Are you saying that the measurement of x1 and x2 is not simultaneous, while measurement of x1' and x2' is ?

What is wrong with this method? All the texts that I have consulted find out x2' and x1' in terms of x2 and x1, and then subtract THOSE ; but to find the length of the rod in frame S, shouldn't we find out x2 - x1 ?

 

[ghwellsjr]

You said the rod is at rest in frame S', correct? If that is true, then it doesn't matter what time coordinates you use to specify the ends of the rod in frame S'--you'll still get the same length, which is called the Proper Length, not the true length, of the rod. So now if you change the time coordinate for one end of the rod in frame S' and transform the events for both ends of the rod, you'll get any length you want but for different time coordinates for the two ends. What you need to do is pick a pair of time coordinates for the two ends of the rod in frame S' such that after you transform the events into frame S, the events are at the same time and that will give you the correct contracted length for the moving rod.

It is probably easiest to select one end of the rod to be at the origin of frame S' when you do this.

 

[ghwellsjr]

No sorry, I don't get it. I thought I had taken care of that part by putting t1 = t2 = t in the Lorentz transformation for x2 and x1?

You're transforming from frame S' to frame S which means you have set t1' = t2' = t.

EDIT : Are you saying that the measurement of x1 and x2 is not simultaneous, while measurement of x1' and x2' is ?

The way you did it, x1 and x2 are not simultaneous (but they should have been), while x1' and x2' were simultaneous (but they should not have been).

What is wrong with this method? All the texts that I have consulted find out x2' and x1' in terms of x2 and x1, and then subtract THOSE ; but to find the length of the rod in frame S, shouldn't we find out x2 - x1 ?

In whichever frame the rod is not moving, you can allow the events to be non-simultaneous so that the events for the frame in which the rod is moving will be simultaneous.

 

[dreamLord]

I think I understand where I went wrong. While transforming x2' and x1' to x2 and x1, I neglected the fact that the measurements of x2 and x1 would not end up being simultaneous, while x2' and x1' would be. If I had transformed x2 and x1 to x2' and x1' instead, I would have gotten simultaneous measurements for x2 and x1, while x2' and x1' would be not simultaneous - which doesn't matter because the rod is at rest in frame S'.

Is that correct?

 

[ghwellsjr]

Well, technically that is correct, but that only works if you already know the contracted length of the rod in frame S. If the idea is that you're supposed to figure it out from its rest length in frame S' and its speed in frame S, then you have to transform from S' to S and not the other way around.

Of course, since you probably already know the contracted length, you can use it to determine two events in S' that are the correct length and at different times such that when you transform them back to S, they will be simultaneous. But it seems like you ought to figure it out from S' without knowing the contracted length in S.

 

[dreamLord]

I'm afraid I don't quite understand how I would find the contracted length (assuming I don't already know it) from the rest length and the speed in frame S? If I transform from S' to S, then I end up with length elongation and not contraction...

 

[ghwellsjr]

Well one way is to do it by trial and error. But as I said before, you ought to set one end of the rod at the origin of both frames to minimize the computation. Then start with the length of the rod as the x' coordinate for the other event and the t' coordinate can be zero. Transform just the t' coordinate to t and see if it is zero which it obviously won't be. Then increase it and repeat. See if t gets closer to zero. If it does, then repeat until you get zero. If it doesn't, then you have to decrease t' until t comes out zero. When you finally get a value of t' that produces t=0, then you can use that value of t' to calculate x which will be the correct contracted length of the rod.

Once you do this, you will quickly see what value of t' to use to get t=0, especially if the rest length of the rod is 1.

 

[dreamLord]

Okay, I understood your method. But how then is the standard derivation given in most texts correct? I still don't quite understand why to find the length of the rod in frame S we do not find out x2 and x1, but instead find out x2' and x1'.

If we consider time dilation in the same scenario for two events, then we will find out t2 and t1 which are in the frame S, and then subtract them to get the time interval for frame S. Should we not operate on coordinates of the S frame for length contraction as well?

 

[ghwellsjr]

Let me work it out for you and see if this make sense:

One end of the rod is at the origin of the rest frame. The other end of the rod is at x' so the rest length (or the Proper Length) of the rod is x'.

Assuming that we are using units such the c=1, the time coordinate in the moving frame is t = γ(t'-vx'). We want to make this evaluate to zero. Obviously, we want to set t'=vx'.

Now we calculate the x coordinate using x = γ(x'-vt') where we substitute vx' for t' and we get x = γ(x'-vvx'). We do a little manipulation to get x = γx'(1-v²). We remember that γ = 1/√(1-v²) and so γ² = 1/(1-v²) or (1-v²) = 1/γ². Substituting this into our previous equation we get x = γx'/γ² = x'/γ, showing that the moving rod is contracted by the reciprocal of the Lorentz factor.

 

[dreamLord]

Yeah I understood your method earlier itself. What I'm having trouble understanding is the proof/derivation given in most texts (which is the same as this : http://en.wikipedia.org/wiki/Length_contraction#Derivation )

What I don't get is why we are transforming x1 and x2 to x1' and x2'. Did my time dilation example/question make sense? In that to find the interval we find out the difference between t2 and t1 in the SAME frame as that of which we need the interval ; why shouldn't we find out the difference between x2 and x1 in the SAME frame as that of which we need the length?

 

[ghwellsjr]

Okay, I understood your method. But how then is the standard derivation given in most texts correct? I still don't quite understand why to find the length of the rod in frame S we do not find out x2 and x1, but instead find out x2' and x1'.

I don't know what the standard derivation in most texts looks like. Can you link to an online reference?

We start with the length of the rod in its rest frame S'. In the previous post, I put one end of the rod at the origin instead of calling it x1. I put the other end at x' instead of calling it x2'. This is just to reduce the computation.

If we consider time dilation in the same scenario for two events, then we will find out t2 and t1 which are in the frame S, and then subtract them to get the time interval for frame S. Should we not operate on coordinates of the S frame for length contraction as well?

We want to make sure that t2 is equal to t1 so they subtract to zero. We make this happen by selecting values of t1' and t2' such that when we transform the two events from frame S' to frame S, they come out equal. The time dilation will calculate automatically if we use the Lorentz transformation correctly.

As you can see in my previous post, the computation is complicated enough when I set one end of the rod at the origin. Trying to do this more generally with the two ends of the rod located anywhere will be even more complex and I leave that exercise to you. If you understand what I did in my previous post, you should be able to extrapolate that to the more general case.

 

[ghwellsjr]

Yeah I understood your method earlier itself. What I'm having trouble understanding is the proof/derivation given in most texts (which is the same as this : http://en.wikipedia.org/wiki/Length_contraction#Derivation )

What I don't get is why we are transforming x1 and x2 to x1' and x2'.

This is the first time I have ever looked at that derivation and it is quite interesting. It doesn't really transform x1 and x2 to x1' and x2', it merely is manipulating the equations to derive the end result. It might have been a little more helpful if they had included a few more intermediate steps but if you mentally do the substitutions, you can see how the math works out.

Did my time dilation example/question make sense? In that to find the interval we find out the difference between t2 and t1 in the SAME frame as that of which we need the interval ; why shouldn't we find out the difference between x2 and x1 in the SAME frame as that of which we need the length?

I like my derivation (which isn't original to me) better than the wikipedia one because it addresses the issue that in the rod's rest frame, we are free to select any time coordinates for the end point events of the rod which allows us to pick some that result in the event in the moving frame being at the same time. So in my derivation, we are forcing the time difference between the two events to be zero in the same frame in which we are calculating the position difference which is the contracted length.

Again, I want to make it clear that the wikipedia derivation is not actually transforming from one frame to another, it is mathematically manipulating the equations in a clever (and not so obvious) way to produce the desired result.

 

[dreamLord]

To be honest, I like your way more too. It's much more logical.

The wikipaedia one (and the one I've found in Kleppner, Berkeley and a few national texts) is, like you said, mathematical manipulation which doesn't make sense to me in a logical manner. Still, if you can manage to extract a logical reason from that method, please do so, because I will have to be writing this method in my examinations.

However, how can you say that they are not transforming from one frame to another?

 

[MikeLizzi]

To be honest, I like your way more too. It's much more logical.

The wikipaedia one (and the one I've found in Kleppner, Berkeley and a few national texts) is, like you said, mathematical manipulation which doesn't make sense to me in a logical manner. Still, if you can manage to extract a logical reason from that method, please do so, because I will have to be writing this method in my examinations.

However, how can you say that they are not transforming from one frame to another?

May I suggest the following analysis from my web site:

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

 

[Fredrik]

Silly question, I assume, but I can't quite figure it out. I've just started reading up on Special Theory of Relativity, and only have knowledge about Lorentz Transformations.

Anyway, supposing a rod is at rest in the frame S', which is moving with velocity v with respect to frame S. Then the length of the rod in frame S' is the true length L(0), and is given by x2' - x1', where x2' and x1' are the end points of the rod in frame S'.

Now we have to find the length of the rod in frame S, which is given by x2 - x1, where x2 and x1 are the end points of the rod in frame S.
x2 is related to x2' by x2 = γ(x2' - vt)
and x1 is related to x1' by x1 = γ(x1' - vt)

Then length in frame S = x2 - x1 = γ(x2' - x1')
Or length L = γL(0)

Hence length to an observer in frame S is elongated.

May I know where I went wrong?

The fact that you're denoting the S' coordinate of the right endpoint of the rod by x2' suggests that you didn't realize that you need to consider three events, not two.

I will call the end of the rod that has the smaller position coordinate in both coordinate systems the left end of the rod. Let A be any event on the world line of the left end of the rod. Let B be the event on the world line of the right end of the rod that's simultaneous in S with A. Let C be the event on the world line of the right end of the rod that's simultaneous in S' with A. (I recommend that you draw a spacetime diagram for the coordinate system S. Draw the world lines of both ends of the rod, and a simultaneity line of S' through A).

Let L₀ be the rest length of the rod. Since the rod is at rest in S', L₀ is equal to the coordinate length in S'. So L₀=x'_C-x'_A. We are looking for the relationship between L₀ and the coordinate length of the rod in S, which will be denoted by L. By definition of L, we have L=x_B-x_A.

For any event E, I will denote the coordinates of E in the coordinate systems S and S' by
$$S(E)=\begin{pmatrix}t_E\\ x_E\end{pmatrix}$$ and $$S'(E)=\begin{pmatrix}t'_E\\ x'_E\end{pmatrix}$$ respectively. When I don't feel like LaTeXing matrices, I'll write S(E)=(t_E,x_E) and S'(E)=(t'_E,x'_E) instead. I will use units such that c=1. In these units, the relationship between S'(E) and S(E) is given by
$$\begin{pmatrix}t'_E\\ x'_E\end{pmatrix}=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t_E\\ x_E\end{pmatrix}.$$ This is just the Lorentz transformation in matrix form.

Now suppose that S was chosen so that the world line of the left end of the rod goes through the event at the origin. Then we can choose A to be that event. Then we have S(A)=(0,0), S'(A)=(0,0), S(B)=(0,L) and S'(C)=(0,L₀). The coordinates in S of points on the world line of the right end of the rod are given by the map
$$t\mapsto\begin{pmatrix}t\\ vt+x_B\end{pmatrix}.$$
If we denote that map by R, we have
$$S(C)=R(t_C)=\begin{pmatrix}t_C\\ vt_C+x_B\end{pmatrix}=\begin{pmatrix}t_C\\ vt_C+L\end{pmatrix}.$$ So $x_C=vt_C+L$. We use this result in the last step of the following calculation.
$$\begin{pmatrix}0\\ L_0\end{pmatrix}=\begin{pmatrix}t'_C\\ x'_C\end{pmatrix} =\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t_C\\ x_C\end{pmatrix} =\gamma\begin{pmatrix}t_C -vx_C\\ -vt_C+x_C\end{pmatrix} =\gamma\begin{pmatrix}t_C -vx_C\\ L\end{pmatrix}.$$
So $L_0=\gamma L$, or equivalently, $$L=\frac{L_0}{\gamma}.$$

 

[Chestermiller]

Silly question, I assume, but I can't quite figure it out. I've just started reading up on Special Theory of Relativity, and only have knowledge about Lorentz Transformations.

Anyway, supposing a rod is at rest in the frame S', which is moving with velocity v with respect to frame S. Then the length of the rod in frame S' is the true length L(0), and is given by x2' - x1', where x2' and x1' are the end points of the rod in frame S'.

Now we have to find the length of the rod in frame S, which is given by x2 - x1, where x2 and x1 are the end points of the rod in frame S.
x2 is related to x2' by x2 = γ(x2' - vt)
and x1 is related to x1' by x1 = γ(x1' - vt)

Then length in frame S = x2 - x1 = γ(x2' - x1')
Or length L = γL(0)

Hence length to an observer in frame S is elongated.

May I know where I went wrong?

The easiest way to do this is to write:

x2' = γ (x2 -v t2)

x1' = γ (x1 -v t1)

so

x2' - x1' = γ (x2 - x1) - γv (t2 - t1)

But when reckoning the length of the moving rod from the S frame, you need to measure the locations of both ends of the rod at the same time, so

t2 = t1,

and hence, x2 - x1 = (x2' -x1') / γ

When reckoned from the S frame, the length of the rod is shorter than in its own rest frame.

 

[ghwellsjr]

To be honest, I like your way more too. It's much more logical.

The wikipaedia one (and the one I've found in Kleppner, Berkeley and a few national texts) is, like you said, mathematical manipulation which doesn't make sense to me in a logical manner. Still, if you can manage to extract a logical reason from that method, please do so, because I will have to be writing this method in my examinations.

However, how can you say that they are not transforming from one frame to another?

If you just write out the equations for the Lorentz transformation process, are you transforming from one frame to another? Or are you merely establishing a mathematical relationship? In any case, let me restate the derivation from wikipedia with some extra steps and using gamma:

The rod is at rest in frame S' and moving at v in frame S. There are a lot of equations that we can write down but we arbitrarily pick two that relate the end points of the rod in S' to their events in S. Note that we are not concerned with the events in S' because we are ignoring the corresponding time coordinates in frame S'. This is because we realize that we can specify any time coordinates for the events in frame S' and it won't change the calculation of the length of the rod since it is at rest. So we specify the two ends of the rods in frame S' as x1' and x2' and we write:

x1' = γ(x1-vt1) and x2' = γ(x2-vt2)

Now we want t1 to be equal to t2 so we replace both of them with simply t.

x1' = γ(x1-vt) and x2' = γ(x2-vt)

Now we note that since L(0) is equal to x2' - x1' we can write:

L(0) = x2' - x1' = γ(x2 -vt) - γ(x1-vt)
L(0) = γx2 - γvt - γx1 + γvt
L(0) = γx2 - γx1 + γvt - γvt
L(0) = γx2 - γx1
L(0) = γ(x2-x1)

But since we also know that L (the length in frame S) is equal to x2-x1 we can make that substitution and get:

L(0) = γL

And rearranging we get the desired final (and correct) result:

L = L(0)/γ

Note that this derivation is very similar to yours in post #1 where you made the mistake of implicitly setting t = t1' = t2'. Instead, I explicitly set t = t1 = t2. So in effect, you were doing the calculation for the situation where the rod is at rest in frame S and you were calculating its length in frame S'. Remember, the equations are establishing a relationship between the coordinates of events in two different frames. You can use the Lorentz transformation equations to solve for the coordinates in either direction but it is more complicated going in the opposite direction and you have to remember that the sense of v is in the opposite direction.

 

[harrylin]

I don't know what the standard derivation in most texts looks like. Can you link to an online reference? [..].

I also learned to obtain it directly from the Lorentz transformations; maybe because I was taught that way, it looks simpler to me than your method.

 

[ghwellsjr]

I also learned to obtain it directly from the Lorentz transformations; maybe because I was taught that way, it looks simpler to me than your method.

Now that I have learned how the derivation works, especially with the added steps and using gamma, I agree. It turns out to be very simple.

 

[bobc2]

Silly question, I assume, but I can't quite figure it out. I've just started reading up on Special Theory of Relativity, and only have knowledge about Lorentz Transformations.

Anyway, supposing a rod is at rest in the frame S', which is moving with velocity v with respect to frame S. Then the length of the rod in frame S' is the true length L(0), and is given by x2' - x1', where x2' and x1' are the end points of the rod in frame S'.

Now we have to find the length of the rod in frame S, which is given by x2 - x1, where x2 and x1 are the end points of the rod in frame S.
x2 is related to x2' by x2 = γ(x2' - vt)
and x1 is related to x1' by x1 = γ(x1' - vt)

Then length in frame S = x2 - x1 = γ(x2' - x1')
Or length L = γL(0)

Hence length to an observer in frame S is elongated.

May I know where I went wrong?

The math has been handled very well here, so I can't add anything more to that. However, dreamLord, it might be good to keep track of the physical picture implied by the Lorentz transformaions as I'm trying to illustrate with the sketch below. We have red and blue meter sticks moving in opposite directions along the X1 axis of the black rest frame. But, the larger implication of the Lorentz transforms in the context of Minkowski 4-dimensional space is that the red and blue meter sticks are 4-dimensional objects. So,we actually have here the 3-D cross-section views of the 4-D universe at one instant of time for each observer. And the reason for the perceived length contraction is immediately obvious.

 

[ghwellsjr]

And the reason for the perceived length contraction is immediately obvious.

It's not immediately obvious to me even after many minutes of trying to figure out what you are trying to show.

You mentioned a black frame and I think you are showing axes for red and blue frames, so are we supposed to think about three frames at the same time?

You said you have a red and a blue meter stick but I see what appears to be two red and two blue meter sticks. That confuses me.

Plus you show all four meter sticks aligned with lines of simultaneity. The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?

Since you said that both meter sticks are moving in opposite directions in the black frame and since everything looks symmetrical to me, I'm assuming that they are moving at the same speed, in which case they should be length contracted by the same amount and I would think they should be aligned parallel to the black x-axis but I don't see that.

So it's immediately obvious to me that a sketch like this needs some background training in order for it to communicate to other people.

 

[ghwellsjr]

Here's another way for an observer to measure the length of a moving rod, or for a moving observer to measure the length of a stationary rod. Let's say the rod is 1 unit in length (using units where c=1) stretching from coordinates [t,x] of [0,0] to [0,1]. The observer is moving in the x direction at 0.8c. In the rest frame of the rod, he will be at the origin at time 0. Since v = d/t and t = d/v, the time in the rod's rest frame that he will reach the other end of the rod will be t = d/v = 1/0.8 = 1.25. The event of him arriving at the far end of the rod in the rod's rest frame is [1.25,1]. Transforming this event into the observer's rest frame we get [0.75,0]. Now he can calculate how long the rod is using d = vt = (0.8)(0.75) = 0.6. This is the same answer we would get using the reciprocal of the Lorentz factor, 1/γ = √(1-v²) = √(1-0.8²) = √(1-0.64) = √(0.36) = 0.6.

Isn't that fun?

 

[bobc2]

I guess I did a pretty poor job of laying out the sketch without a useful explanation. You and the others have done such a good job of explaining the contraction with the Lorentz transformations, I thought I might be able to compliment that with the space-time sketch. It's an old habit started in my first graduate course in special relativity when our prof required us to do so many of the old paradox problems for homework. He required us to first do it with just the math and then repeat it using space-time diagrams. I got in the habit of thinking of the universe as 4-dimensional, although my prof never used the term, "block universe" and never discussed the reality of the 4-dimensional objects.

Does the little tutorial on 4-dimensional space and space-time diagrams in this thread help any?

https://www.physicsforums.com/showthread.php?p=4139454#post4139454

If not I can add in some basic analytic geometry on the front end.

It's not immediately obvious to me even after many minutes of trying to figure out what you are trying to show.

I'm trying to show two 4-dimensional meter sticks. Each one is slanted in the opposite direction relative to the black coordinate system (the vertical black line is the black X4 axis and the horizontal black line is the black X1 axis).

You mentioned a black frame and I think you are showing axes for red and blue frames, so are we supposed to think about three frames at the same time?

Yes. I wanted to show red and blue moving in opposite directions with respect to the black rest frame.

You said you have a red and a blue meter stick but I see what appears to be two red and two blue meter sticks. That confuses me.

The two slanting blue lines represent the front and rear ends of the blue meter stick, displayed as a 4-dimensional object (X2 and X3 coordinates are supressed for ease of viewing). Likewise the two red slanted lines represent front and rear ends of the 4-dimensional red meter stick.

Plus you show all four meter sticks aligned with lines of simultaneity. The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?

If you follow the blue line of simultaneity, you will see the front and rear points on the blue meter stick that are simultaneous. And continuing along that blue line of simultaneity you will see the front and rear points of the red stick that are simultaneous in blues simultaneous space (an instantaneous 3-dimensional cross-section vew, cut across the 4-dimensional universe). So, just working within blue's inertial frame you can directly compare the lengths of the blue and red sticks as they appear in blue's 3-dimensional world. Then you can do the same thing for the red world, using the red simultaneous space.

Since you said that both meter sticks are moving in opposite directions in the black frame and since everything looks symmetrical to me, I'm assuming that they are moving at the same speed,...

That is correct, they are moving along the black X1 axis in opposite directions at the same speed. That's why the 4-dimensional stick objects are slanted in opposite directions with the same slant angle relative to the black X4 axis. Note that as you move along the blue 4-dimensional stick in the blue X4 direction, you are increasing the distance along the black X1 axis. Or, another way of saying it, if you are at rest in the black rest frame, that means you are moving straight forward into time, i.e., moving along the black X4 axis at the speed of light--and as you move along black X4, the blue meter stick increases its distance away from you (black) along the black X1 axis. Also if you are red guy, moving along the red X4 at light speed, you also see the blue meter stick increasing its distance away from you along the red X1 axis.

... in which case they should be length contracted by the same amount...

As mentioned above, the blue guy "sees" the red guy's stick contracted as compared to his own (just comparing lengths within the blue simultaneous space) and the red guy "sees" the blue guy's stick contracted as compared to his own (comparing lengths just within the red simultaneous space). And the reason I've used the symmetric Loedel space-time diagram is that you are able to compare the lengths directly, since the line lengths in the sketch have exactly the same actual distance calibrations (you can't compare distances directly between black and the moving frames without resorting to the hyperbolic calibration curves).

...and I would think they should be aligned parallel to the black x-axis but I don't see that.

Take another look at the diagram. If the blue and red 4-dimensional sticks were aligned parallel to the black X4 axis, that would mean that the blue and red sticks were at rest in the black rest system. For example, follow the blue stick along the 4th dimension, assuming the blue stick is parallel to the black X4 axis: In that case, as time passes (as you move along blue's stick along the direction of the X4 axis) you would see that the blue stick front and rear positions along black's X1 axis have not changed.

So it's immediately obvious to me that a sketch like this needs some background training in order for it to communicate to other people.

Apparently that is the case. I'll have to give more thought as to how to do a better job of presenting the space-time diagrams. Any suggestions?

 

[bobc2]

Post Script. I probably should just give it up as a fruitless effort. I had forgotten the lengthy question and answer sessions it took to get this concept across to some of my students.

 

[Fredrik]

I think there's just too much information in that diagram. (I didn't try to determine if that information is correct or not). I would prefer one with just the necessary information drawn in. Something like this:

If you disregard my poor drawing skills (the slope of the x' axis isn't exactly right), the only problem with this diagram is that a person who doesn't understand that the scales on the axes are fixed by the invariant hyperbolas might interpret it as saying that $L_0=x'_C-x'_A>x_B-x_A=L$. I would recommend section 1.7 in Schutz to anyone who's struggling with that. In particular fig. 1.11 at the bottom of page 17.

To see that the inequality actually goes the other way, what you have to do is to draw a curve of the form $-t^2+x^2=\text{constant}$ with the constant chosen so that the curve goes through C. The point where this curve intersects the x-axis will have coordinates (0,L₀). This point will be to the left of B on the x axis. So $L_0<L$.

 

[bobc2]

Here's the picture with the hyperbolic calibration curves. Of course these are the time-like curves. I could generate the space-like curves as well but just happened to have these handy, so am specifically illustrating the time dilation rather than the length contraction (a little off point, but I think you get the point about the need for hyperbolic calibration curves).

 

[bobc2]

Here are the calibration curves for time-like and space-like. Red and blue inertial frames are included. I still think it's better to avoid the calibration curves by just showing two observers moving in opposite directions. Both time dilation and length contraction are directly viewed. I insisted my students understand these diagrams--it just took a little more time for some of them.

 

[PAllen]

Are you perhaps confusing the Enzyte Theory of Relativity with the Einstein Theory of Relativity?

---

Sorry mentors, tried and tried to resist, couldn't any longer (stop it).

Do what you may.

 

[ghwellsjr]

The fact that you're denoting the S' coordinate of the right endpoint of the rod by x2' suggests that you didn't realize that you need to consider three events, not two.

The above comment makes me ask you the same question I asked bobc2 (which he didn't answer):

The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?

There are only two events under consideration, not three or more. These are the end points of the moving stick in frame S and they must be simultaneous in that frame. The distance between the two events is L. In the rod's rest frame S', those two events are not simultaneous but since the rod is at rest, the difference in the x-coordinates of the two events is equal to the Proper Length of the rod, L₀.

I think there's just too much information in that diagram. (I didn't try to determine if that information is correct or not). I would prefer one with just the necessary information drawn in. Something like this:

I agree that there's too much information in bobc2's drawing but there's too much in yours too, as far as the number of events goes. The two events we care about are labeled A and B. C should not be shown as an event. Erase that dot and the C. If you try to show that there are two simultaneous events in frame S', then you have missed the whole point of the derivation. Events A and B are not simultaneous in frame S' and that's the way we want it to be.

If you disregard my poor drawing skills (the slope of the x' axis isn't exactly right), the only problem with this diagram is that a person who doesn't understand that the scales on the axes are fixed by the invariant hyperbolas might interpret it as saying that $L_0=x'_C-x'_A>x_B-x_A=L$.

But that is (almost) the correct interpretation if you recognize that x'_C is equal to x'_B so you should have said $L_0=x'_B-x'_A>x_B-x_A=L$. However, you really need to show the grid lines for the two frames, otherwise, it's impossible to tell what the value of x'_B is or that it is larger than x_B. The whole purpose of a Minkowski diagram is to show the two sets of coordinates for the two frames so that you can see that they are different for the two frames and it's those numbers that show that L is less than L₀, not the fact that two line segments are actually drawn to two different lengths.

I would recommend section 1.7 in Schutz to anyone who's struggling with that. In particular fig. 1.11 at the bottom of page 17.

To see that the inequality actually goes the other way, what you have to do is to draw a curve of the form $-t^2+x^2=\text{constant}$ with the constant chosen so that the curve goes through C. The point where this curve intersects the x-axis will have coordinates (0,L₀). This point will be to the left of B on the x axis. So $L_0<L$.

You already stated the correct relationship between L and L₀ at the end of post #17, why'd you change your mind? This last paragraph is all wrong. We don't need hyperbolic calibration curves, we just need marked grid lines. Why don't you or bobc2 draw a correctly annotated Minkowski diagram that illustrates the derivation of the length contraction that is the subject of this thread?

 

[Fredrik]

I made a mistake in my comments about my spacetime diagram. I will have more to say about that when I've figured out how to draw the diagram in a math program.

Edit: I'm trying to relearn how to do this in Mathematica. That will definitely take a while, and I will have to do something else in about ten minutes. So I will have to return to this later. I have posted a reply to ghwellsjr below, but the diagram and my new comments about it will have to wait.

 

[bobc2]

This sketch describes the situation originally set up in the post. Here we can consider red to be the rest system, S, and blue to be the S' system moving with respect to S at relativistic speed. There is no need for calibration curves for this kind of symmetric diagram (Loedel space-time diagram). It is clear that red will "see" a contracted rod, length L, as compared to the length, L', as "seen" in the blue frame S'.

 

[Fredrik]

You already stated the correct relationship between L and L₀ at the end of post #17, why'd you change your mind? This last paragraph is all wrong.

I didn't change my mind, but I agree that the last paragraph is all wrong. I must have made some elementary mistake when I wrote that post. I certainly didn't realize that what I was saying was the opposite of what I had said before.

Why don't you or bobc2 draw a correctly annotated Minkowski diagram that illustrates the derivation of the length contraction that is the subject of this thread?

I can't speak for him, but I drew a primitive diagram in Paint because I use math software so seldom that I forget everything I know about them between each time. Of course, now that I see that I have made a blunder, I will have to relearn some of it to draw an exact diagram that includes the relevant hyperbola. I will have to leave my computer for at least an hour 20 minutes from now, so I probably won't be able to do that for a few hours.

I would say that the subject of this thread is length contraction, not some specific calculation. I didn't even look at the derivation in post #1. I just saw that the result was wrong, and that the notation looked weird to me. Then I posted a correct derivation, and later the spacetime diagram that I had described in words. The diagram is correct, but I made some comments about it that are incorrect.

I agree that there's too much information in bobc2's drawing but there's too much in yours too, as far as the number of events goes. The two events we care about are labeled A and B. C should not be shown as an event. Erase that dot and the C. If you try to show that there are two simultaneous events in frame S', then you have missed the whole point of the derivation. Events A and B are not simultaneous in frame S' and that's the way we want it to be.

We definitely need three events. The motion of the (component parts of) the rod is a congruence of curves in spacetime. I only drew two of them (the world lines of the endpoints). The two observers (or rather, the coordinate systems associated with their motions) are slicing spacetime in different ways, into hypersurfaces that they think of as "space at a specific moment in time". What an observer thinks of as "the rod right now" is the union of the congruence and his simultaneity line through the origin. To the observer who's comoving with the rod (and using coordinate system S'), "the rod right now" is the line AC. To the observer who sees the rod moving at velocity v (and is using coordinate system S), "the rod right now" is the line AB. This disagreement is the reason why there is such a thing as length contraction.

The above comment makes me ask you the same question I asked bobc2 (which he didn't answer): There are only two events under consideration, not three or more. These are the end points of the moving stick in frame S and they must be simultaneous in that frame. The distance between the two events is L. In the rod's rest frame S', those two events are not simultaneous but since the rod is at rest, the difference in the x-coordinates of the two events is equal to the Proper Length of the rod, L₀.

"Proper length" is assigned only to spacelike curves. You don't seem to have thought about what curve we're talking about here. The "proper length of the rod" is the proper length of the curve AC. This is the significance of the event C.

However, you really need to show the grid lines for the two frames, otherwise, it's impossible to tell what the value of x'_B is or that it is larger than x_B. The whole purpose of a Minkowski diagram is to show the two sets of coordinates for the two frames so that you can see that they are different for the two frames and it's those numbers that show that L is less than L₀, not the fact that two line segments are actually drawn to two different lengths.

The only thing we need to add is an invariant hyperbola through C, to see where it intersects the x axis.

 

[bobc2]

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).

 

[PAllen]

We've seen plenty of confusion here between terms proper distance, proper length, and rest length. I believe ghwellsjr is following one common usage where proper length (not distance, which applies between a pair of events along a specified simultaneity line) of an effectively rigid body is the same as rest length = length in a frame in which it is at rest. In such a frame, there is no requirement to measure length of a rigid object between simultaneous events, because nothing changes or moves in this frame. You can note the coordinate position of one end of the rigid body at time t1, and the position of the other end at time t2, and subtract, and still get proper length = rest length, because neither end is moving in this frame. With this understanding, only two events are needed (that are simultaneous for the frame in which the rod is moving). This same observation is also the basis of the 'cute' derivation in the wikipedia entry attributed to Born.

 

[ghwellsjr]

This sketch describes the situation originally set up in the post.

I don't think so. My comments about Fredrik's diagram applies to this new one also. The only difference is he moved your pairs of events closer so that X₁ and X'₁ were at the same time and appear at the same place on the diagram. But you are also missing the point of this thread--there are only two events. They are simultaneous in the frame in which the rod is moving and they are not simultaneous in the frame in which the rod is at rest.

Here we can consider red to be the rest system, S, and blue to be the S' system moving with respect to S at relativistic speed.

Well now this is a confusing statement. First off, the systems are moving with respect to each other. We should never think of one of them at rest and the other one moving. What dreamLord proposed in his OP is that the rod is at rest in S' and moving in S. That seems at odds with your attaching "rest" to S and "moving" to S'.

There is no need for calibration curves for this kind of symmetric diagram (Loedel space-time diagram).

I agree and that's how I ended my previous post, "we don't need hyperbolic calibration curves, we just need marked grid lines". But you left out the marked grid lines. That's what is going to tell the story on a Minkowski diagram. Why didn't you put them in like I requested?

It is clear that red will "see" a contracted rod, length L, as compared to the length, L', as "seen" in the blue frame S'.

That's not at all clear and I don't know what you mean by "see" or why you put them in quotes. You have arbitrarily decided to assign the primed coordinates and L₀ to the longer rod just so that it will appear longer on the diagram than the other rod. But your mistake is that you don't realize that I could just as legitimately assign the primed coordinates and L₀ to the rod that is drawn shorter on the diagram. You need to get rid of either one of those rods. You need to draw just one rod and one pair of events in your diagram, it doesn't matter which one, but you can't label the events exclusively to one frame. The same events belong to both frames but with different coordinates for each frame. That's where you need to show the grid lines for each frame so that we can see that those two events are simultaneous in the frame in which the rod is moving and not simultaneous in the frame in which the rod is at rest.

 

[ghwellsjr]

We've seen plenty of confusion here between terms proper distance, proper length, and rest length. I believe ghwellsjr is following one common usage where proper length (not distance, which applies between a pair of events along a specified simultaneity line) of an effectively rigid body is the same as rest length = length in a frame in which it is at rest. In such a frame, there is no requirement to measure length of a rigid object between simultaneous events, because nothing changes or moves in this frame. You can note the coordinate position of one end of the rigid body at time t1, and the position of the other end at time t2, and subtract, and still get proper length = rest length, because neither end is moving in this frame. With this understanding, only two events are needed (that are simultaneous for the frame in which the rod is moving). This same observation is also the basis of the 'cute' derivation in the wikipedia entry attributed to Born.

Exactly.

And the only reason I brought up Proper Length in post #4 is because dreamLord incorrectly used the term "true length" in his OP. The "true" length of the rod is L₀ in frame S' where it is at rest and L (which is contracted) in frame S where it is moving. Both are "true" lengths as defined in their respective frames. But I was trying to dissuade him from using the term "true length".

 

[ghwellsjr]

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).

You see how you have labeled the event at the origin with X'₁ and with X₁? Now you should add two more labels to your other two events so they both say X'₂ and X₂. All events on a Minkowski diagram have two labels, one for each coordinate system. Then you need to delete one of those events and the rod that goes with it, it doesn't matter which one and put in those grid lines and you will see that in whichever frame the events defining the end points of the rod are simultaneous, the distance between the rods is shorter than in the other frame in which the events are not simultaneous but in which the rod is at rest.

The issue is not whether one of the events is at the origin or not, it doesn't matter. You could have done the same thing with your other diagram--one rod, one pair of events.

That is, if you want to illustrate the problem that was presented by dreamLord in this thread. If you are showing a different problem then you need to say so.

 

[Fredrik]

I made a couple of mistakes in this thread. First, there was this one:

The fact that you're denoting the S' coordinate of the right endpoint of the rod by x2' suggests that you didn't realize that you need to consider three events, not two.

As PAllen explained in post #36, we can use the fact that (in my notation) $x'_B=x'_C$ to calculate $L_0$ as $x'_B-x'_A$ instead of as $x'_C-x'_A$. So you can certainly do a calculation that ends with the correct result $L=L_0/\gamma>L_0$ without ever mentioning the event C explicitly.

I don't mind using this trick to make the calculation slightly shorter, but I think it would be a bad idea to use it to hide what I think is the most important thing about this problem: The observers are referring to different slices of the congruence when they talk about "the rod right now".

This is clear in the spacetime diagram I posted. Unfortunately I made a major blunder when I wrote down my comments to the diagram:

If you disregard my poor drawing skills (the slope of the x' axis isn't exactly right), the only problem with this diagram is that a person who doesn't understand that the scales on the axes are fixed by the invariant hyperbolas might interpret it as saying that $L_0=x'_C-x'_A>x_B-x_A=L$. I would recommend section 1.7 in Schutz to anyone who's struggling with that. In particular fig. 1.11 at the bottom of page 17.

To see that the inequality actually goes the other way, what you have to do is to draw a curve of the form $-t^2+x^2=\text{constant}$ with the constant chosen so that the curve goes through C. The point where this curve intersects the x-axis will have coordinates (0,L₀). This point will be to the left of B on the x axis. So $L_0<L$.

The diagram is not misleading, at least not in the way I said it would be. That first inequality (the one I said is wrong) is correct, as can be seen from the formula $L=L_0/\gamma$ that I derived in post #17.

I didn't notice that the result I obtained here is different from the one I obtained before. I didn't even look at what I had written long enough to realize that, because I thought I remembered that a naive interpretation of the diagram would give you the wrong idea about what length is greater. But I remembered it wrong. A naive interpretation of the spacetime diagram for time dilation will give you the wrong idea, but this one won't.

This can be seen by actually drawing the invariant hyperbola through C. With v=0.6 and L=1, the diagram with the hyperbola looks like this:

As you can see, the hyperbola intersects the x-axis to the right of the point (0,L), not to the left as I said in the quote above. Since that intersection point is known to have coordinates (0,L₀), this means that $L_0>L$, which is consistent with the formula $L=L_0/\gamma<L_0$.

 

[bobc2]

You see how you have labeled the event at the origin with X'₁ and with X₁? Now you should add two more labels to your other two events so they both say X'₂ and X₂. All events on a Minkowski diagram have two labels, one for each coordinate system.

Certainly not. I did not make up the labels for the rod. In the original post it was clear that he wanted two ends of the rod to be labeled so as to have two events that marked the right and left ends, X1' and X2', as rod ends in a simultaneous 3-D space (identified in my sketch as "Simultaneous Blue") corresponding to the S' frame moving relative to an S rest frame. Labels X1 and X2 were explicitly identified as end labels for a simultaneous space in the S rest frame.

Then you need to delete one of those events and the rod that goes with it...

No. I'm trying to emphasize the physical clarity of the problem by utilizing a picture of the 4-dimensional rod. He is specifically talking about a physical rod at rest in the S' frame that is in motion relative to a rest frame, S. So the rod stays.

it doesn't matter which one and put in those grid lines and you will see that in whichever frame the events defining the end points of the rod are simultaneous...

What you are suggesting only works if your definition of simultaneous space refers to a hyperspace with some kind of hypertime. You should spell it out if you intend something unconventional. We are referring here to a 3-dimensional simultaneity (X2 and X3 are supressed in our diagrams).

I'll put in the grid lines for you later. However, I've shown in the sketch the plane of simultaneity for blue and for red. Only lines parallel to the "Simultaneous Blue" can represent a simultaneous space for blue. And only lines parallel to the "Simultaneous Red" can represent simultaneity for red, unless you reject the Einstein special relativity theory.

...the distance between the rods is shorter than in the other frame in which the events are not simultaneous but in which the rod is at rest.

It is seen directly in the sketch that the cross-section view of the 4-dimensional rod, i.e., the length of the rod, is shorter in the rest S frame (red frame) than it is in the moving blue frame (the blue frame is the one in which the rod is at rest).

The issue is not whether one of the events is at the origin or not, it doesn't matter. You could have done the same thing with your other diagram--one rod, one pair of events.

Of course. That's why I did it that way originally--and because in the original setup he used the Lorentz transformation with the offset displacement rather than a simple boost. I did it this last way because I was getting the impression it might be more clear to you (just using a boost).

That is, if you want to illustrate the problem that was presented by dreamLord in this thread. If you are showing a different problem then you need to say so.

I thought I was giving a physical picture consistent with resolving the problem he was having the way he confused the use of the gamma expression. My sketch goes to his original question, "...where I went wrong?" His analysis was indicating that the rod length in the simultaneous red space was longer than the rod length in the simultaneous blue space--which was backwards. It was hoped that the sketch would at least give him some physical insight and guidance.

 

[Fredrik]

This sketch describes the situation originally set up in the post. Here we can consider red to be the rest system, S, and blue to be the S' system moving with respect to S at relativistic speed. There is no need for calibration curves for this kind of symmetric diagram (Loedel space-time diagram). It is clear that red will "see" a contracted rod, length L, as compared to the length, L', as "seen" in the blue frame S'.

The statement that S is the "rest system" is misleading, since none of them is at rest in an absolute sense (S' has velocity v in S, and S has velocity -v in S'), and (as your diagram shows) the rod is at rest in S', not S.

Also, the labels $x_1,x_2,x'_1,x'_2$ are a bit odd since each event is assigned coordinates by both coordinate systems. You could e.g. call the events that have S coordinates $x_1$ and $x_2$ 1 and 2 respectively, and the events that have S' coordinates $x'_1$ and $x'_2$ 3 and 4 respectively. Then you can define $L_0=x'_4-x'_3$ and $L=x_2-x_1$.

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).

I like this better just because it looks less cluttered, but I would still put coordinate-independent labels on the events, since the diagram doesn't favor one coordinate system over the other.

You see how you have labeled the event at the origin with X'₁ and with X₁? Now you should add two more labels to your other two events so they both say X'₂ and X₂. All events on a Minkowski diagram have two labels, one for each coordinate system.

This makes sense, but I would recommend coordinate-independent labels (1,2,3 or A,B,C) instead of one label for each coordinate system ($x_1,x'_1,x_2,x'_2,x_3,x'_3$).

Then you need to delete one of those events and the rod that goes with it, it doesn't matter which one and put in those grid lines and you will see that in whichever frame the events defining the end points of the rod are simultaneous, the distance between the rods is shorter than in the other frame in which the events are not simultaneous but in which the rod is at rest.

He certainly doesn't need to delete an event, and there's just one rod in the diagram (represented by the entire region between the world lines of the end points). What you are referring to are two spacelike curves whose proper lengths are equal to the coordinate lengths of the rod in the two coordinate systems.

You have arbitrarily decided to assign the primed coordinates and L₀ to the longer rod just so that it will appear longer on the diagram than the other rod. But your mistake is that you don't realize that I could just as legitimately assign the primed coordinates and L₀ to the rod that is drawn shorter on the diagram.

This is wrong. The diagram shows that the rod is at rest in S'. (The world lines of both endpoints of the rod are parallel to the t' axis). The line segment labeled L₀ (in the diagram involving four events) is drawn along a simultaneity line of S' and it starts and ends on the world lines of the endpoints of the rod, so it's clear that that line segment, not the other one, represents the rest length of the rod.

That is, if you want to illustrate the problem that was presented by dreamLord in this thread. If you are showing a different problem then you need to say so.

The OP asked about length contraction. bobc2's diagram illustrates length contraction. Hence it's not a different problem.

 

[greswd]

How did this thread manage to get so long?

It's simple. Both ends of the rod can be represented by two parallel lines in either frame.

Find the positions of both ends of the rod in the "moving frame" at ONE point in time in the moving frame.

This means we need to consider:

t1-vx1/c^2 = t2 -vx2/c^2


Simplify to show that the length is contracted by a factor of gamma.

 

[ghwellsjr]

This thread got to be so long because it wasn't until post #15 that dreamLord finally disclosed why he rejected all my other attempts to explain length contraction:

To be honest, I like your way more too. It's much more logical.

The wikipaedia one (and the one I've found in Kleppner, Berkeley and a few national texts) is, like you said, mathematical manipulation which doesn't make sense to me in a logical manner. Still, if you can manage to extract a logical reason from that method, please do so, because I will have to be writing this method in my examinations.

However, how can you say that they are not transforming from one frame to another?

Once he stated why he needed help with that one particular derivation, I gave him an explanation in post #19 and he hasn't posted since, so I can only assume that he was satisfied.

But then bobc2 posted what he called an "immediately obvious" sketch explaining length contraction and in my response to him that it was not "immediately obvious" to me, I asked him:

Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?

But rather than answer that question, he continued to explain his sketch, all the while I'm trying to relate it to the OP's concern. Now it is becoming obvious that all these added posts are not attempting to explain the OP's concern but to show other methods for illustrating length contraction, which is OK, I just wish when I ask if they are other methods, people will simply answer yes.

 

[bobc2]

This thread got to be so long because it wasn't until post #15 that dreamLord finally disclosed why he rejected all my other attempts to explain length contraction:

Once he stated why he needed help with that one particular derivation, I gave him an explanation in post #19 and he hasn't posted since, so I can only assume that he was satisfied.

But then bobc2 posted what he called an "immediately obvious" sketch explaining length contraction and in my response to him that it was not "immediately obvious" to me, I asked him:

But rather than answer that question, he continued to explain his sketch, all the while I'm trying to relate it to the OP's concern. Now it is becoming obvious that all these added posts are not attempting to explain the OP's concern but to show other methods for illustrating length contraction, which is OK, I just wish when I ask if they are other methods, people will simply answer yes.

You're right, ghwellsjr. Actually, I wasn't trying to show a different derivation of length contraction (what I presented was not a derivation). I thought I was supplementing your excellent mathematical presentation to show in picture form the results of your analysis. You have always presented some of the most useful explanations of Lorentz transformations when people show up on the forum seeking help, and you certainly did a good job on this thread.

I think I assumed too much in thinking it would be immediately obvious--I was obviously wrong about that and for a while didn't realize the additional confusion I was bringing to the thread. But, I think you've got the situation back in proper perspective. Thanks.

And thanks to Fredrik for pointing out details in my sketches that could have been represented in a more correct and consistent fashion.

 

[ghwellsjr]

How did this thread manage to get so long?

It's simple. Both ends of the rod can be represented by two parallel lines in either frame.

Find the positions of both ends of the rod in the "moving frame" at ONE point in time in the moving frame.

This means we need to consider:

t1-vx1/c^2 = t2 -vx2/c^2

Simplify to show that the length is contracted by a factor of gamma.

Another reason this thread is so long is because after the OP's question is sufficiently answered, someone else comes along claiming to have a "simple" answer but doesn't provide enough explanation to support that claim. I have tried to understand your post and I'm afraid I need help. Here's as far as I have gotten:

You start with an equality, the two sides of your equation came from the Lorentz Transformation equation for time:

t' = γ(t-vx/c^2)

So for the two events under consideration I'm assuming you started with:

t1' = γ(t1-vx1/c^2) and t2' = γ(t2-vx2/c^2)

Then you decided* that t1' equals t2', correct? So then:

γ(t1-vx1/c^2) = γ(t2-vx2/c^2)

You divided out gamma to get your starting equation:

t1-vx1/c^2 = t2-vx2/c^2

Now you say "simplify". Alright, I think you mean to calculate x1-x2 which is the length of the rod moving in the frame S. Here's where I go with that:

t1-t2 = vx1/c^2-vx2/c^2 = (v/c^2)(x1-x2)

(c^2/v)(t1-t2) = x1-x2

Now I would have thought that the next step would be to set t1 equal to t2 but that only means the equality holds true when x1 also equals x2 so that the length of the rod moving in frame S is zero which isn't where we want to go.

So I'm stumped. I need help. I can't figure out your "simple" explanation nor your request to "simplify".

* NOTE: You implicitly set t1' equal to t2'. This is exactly what the OP did in his original post and which I pointed out in post #5 and which leads to the conclusion that t1 cannot also be equal to t2 which is what we need to calculate the length of the rod moving in frame S and to compare it to the length of the rod stationary in frame S'.