Trouble with Lorentz transformations

[stevmg]

Yes, I do remember the lack of simultaneity (even if it exists in one time frame) in different time frames. I will do the math and see what I come up with.

I will assume a rod of 12 ly in S' which is moving at 0.6c to the right and assume that t₁' = t₂'. I will assume t₁' = t₂' = 0. Also, x₁' = 0 and x₁ = 0. x₂' = 12.

I will assume that t₁ = 0 with t₁' (and also t₂'). The question then is what is t₂? What is x₂?

Am I on the correct path? Do have all my assumptions correct so far?

This is NOT a homework problem. I go so far back that when I went to college, there were no colleges.

 

[jtbell]

Yes, you have a valid set of assumptions. Let's see what you get as a result!

 

[stevmg]

itbell and/or starthaus -

Moving right along:

We will need \gamma for furture calculations.

\gamma = 1/\sqrt{1 - v^2/c^2} = 1/\sqrt{1 - 0.6c^2/c^2} = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.8 = 1.25

From Lorentz,
x₁ = \gamma(x₁' + vt₁')
x₂ = \gamma(x₂' + vt₂')

t₁ = \gamma(t₁' + vx₁'/c²)
t₂ = \gamma(t₂' + vx₂'/c²)

x₁' = 0, t₁' = 0
x₂' = 12, t₂' = 0
x₁ = 0, t₁ = 0

v = 0.6, \gamma = 1.25

Now to calculate x₂ and t₂:

x₂ = 1.25(12 + 0.6*0) = 15 ly
t₂ = 1.25(0 + 0.6*12) = 9 yr

Now, something is wrong.

If I use 9 as t₂ and v = 0.6, x₂ = 0.6*9 = 5.4 ly. That makes more sense. What is wrong with the equation for x₂?

I am missing something...

Just tell me, at least, does t₂ = 9? Does x₂ = 5.4 ly?

Ooooooh, I just got it. According to starthaus I want to set t₁ = t₂ = 0. When you do that, (x₂' - x₁') = \gamma(x₂ - x₁)
or 12 ly = 1.25 (x₂ - x₁)
(x₂ - x₁) = 9.6 ly. If x₁ = 0, then x₂ = 9.6
If x₁' = 0, x₂' = 12
t₁' = 1.25(0 - 0.6*0) = 0
t₂' = 1.25(0 - 0.6*9.6) = -7.2 yr

Now, does that make any damn sense? Please explain to me how that would happen?

I assume my calculations must be wrong - somewhere.

 

[stevmg]

Another question. In the last paragraph of Section XII, "Rods and Clocks in Motion" in Einstein's book Relativity there is an equation:

t = x/(\sqrt{1 - v^2/c^2})

Is that a misprint? Should it be t = t'/(\sqrt{1 - v^2/c^2})?

 

[starthaus]

If I use 9 as t₂ and v = 0.6, x₂ = 0.6*9 = 5.4 ly.

No, v is the speed between frames, you have no right to write (excuse the pun)
x₂ = 0.6*9

The correct formula is:

x₂ = \gamma(x'₂ + vt'₂)

You already knew that.

 

[jtbell]

x₂ = 1.25(12 + 0.6*0) = 15 ly
t₂ = 1.25(0 + 0.6*12) = 9 yr

Those calculations are correct. So now you know that in the unprimed frame, the left end of the rod is at x_1 = 0 at t_1 = 0, and the right end of the rod is at x_2 = 15 at t_2 = 9. You want to find the length of the rod in the unprimed frame, but you can't simply take the difference x_2 - x_1 = 15 - 0 = 15, because x_1 and x_2 are "measured" at different times, and the rod has moved in between. So you need to "correct" the position of one end of the rod to make the times match.

In the unprimed frame, the left end of the rod is at x_1 = 0 at t_1 = 0, and is moving to the right with speed 0.6. Where will it be (what is x_1) nine yr later (at t_1 = 9), and what is x_2 - x_1 at t_1 = t_2 = 9?

Or instead, you can "correct" the position of the right end of the rod. At t_2 = 9, it's at x_2 = 15 and is moving to the right with speed 0.6. Where was it nine yr earlier (at t_2 = 0), and what is x_2 - x_1 at t_1 = t_2 = 0?

 

[stevmg]

itbell and/or starthaus -

Moving right along:

We will need \gamma for furture calculations.

\gamma = 1/\sqrt{1 - v^2/c^2} = 1/\sqrt{1 - 0.6c^2/c^2} = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.8 = 1.25

From Lorentz,
x₁ = \gamma(x₁' + vt₁')
x₂ = \gamma(x₂' + vt₂')

t₁ = \gamma(t₁' + vx₁'/c²)
t₂ = \gamma(t₂' + vx₂'/c²)

x₁' = 0, t₁' = 0
x₂' = 12, t₂' = 0
x₁ = 0, t₁ = 0

v = 0.6, \gamma = 1.25

Now to calculate x₂ and t₂:

x₂ = 1.25(12 + 0.6*0) = 15 ly
t₂ = 1.25(0 + 0.6*12) = 9 yr

The correct formula is:
x2 = (x'2 + vt'2) - from above post by starthaus
to starthaus - isn't that what I wrote in this part (see the last line in the above quote?)
If so, then 15 ly is a correct calculation and so would be the 9 years for t₂ = 1.25(0 + 0.6*12) = 9 yr. However, I would have to adjust backwards the length at t = 0 (in other words -9*0.6 = -5.4. 15 - 5.4 = 9.6, so the length x₂ when t = 0 was actually 9.6 ly.

Guess what, uising the length contraction formula we get the same result. 12 * (1/\gamma) = 12/1.25 = 9.6 ly.

Now, something is wrong.

If I use 9 as t₂ and v = 0.6, x₂ = 0.6*9 = 5.4 ly. That makes more sense. What is wrong with the equation for x₂?

I am missing something...

Just tell me, at least, does t₂ = 9? Does x₂ = 5.4 ly?

Yes, starthaus, I know this is wrong and I think my calulations above go along with you but please give a thumbs up or another correction, if necessary.

Ooooooh, I just got it. According to starthaus I want to set t₁ = t₂ = 0. When you do that, (x₂' - x₁') = \gamma(x₂ - x₁)
or 12 ly = 1.25 (x₂ - x₁)
(x₂ - x₁) = 9.6 ly. If x₁ = 0, then x₂ = 9.6
If x₁' = 0, x₂' = 12
t₁' = 1.25(0 - 0.6*0) = 0
t₂' = 1.25(0 - 0.6*9.6) = -7.2 yr

Now, does that make any damn sense? Please explain to me how that would happen?

I assume my calculations must be wrong - somewhere.

Actually, using t₁ = t₂ = 0 as per starthaus gives these weird answers, but the "proper time" both ways comes out the same:

SQRT[(-7.2 - 0)² - (12 - 0)²] = \sqrt{-92.16

and

SQRT[(0 - 0)² - (9.6 - 0)² = \sqrt{-92.16

 

[starthaus]

Actually, using t₁ = t₂ = 0 as per starthaus gives these weird answers, but the "proper time" both ways comes out the same:

\sqrt{(-7.2 - 0)^2 - (12 - 0)^2} = \sqrt{-92.16

and

\sqrt{(0 - 0)^2 - (9.6 - 0)^2} = \sqrt{-92.16

it is getting worse and worse. Somehow you manage to twist and missapply everything I tried to teach you. One more time, you can EITHER:

1. Calculate x_1-x_2 for t_1=t_2

OR

2. Calculate x'_1-x'_2 for t'_1=t'_2

In EITHER case you must apply ONLY the Lorentz transforms. IF you do the things right, you will get:

1. x_1-x_2=(x'_1-x'_2)/\gamma

OR

2. x'_1-x'_2=(x_1-x_2)/\gamma

It is that simple.

 

[stevmg]

you can EITHER:

1. Calculate x_1-x_2 for t_1=t_2

OR

2. Calculate x'_1-x'_2 for t'_1=t'_2

In EITHER case you must apply ONLY the Lorentz transforms. IF you do the things right, you will get:

1. x_1-x_2=(x'_1-x'_2)/\gamma

OR

2. x'_1-x'_2=(x_1-x_2)/\gamma

It is that simple.

Starthaus - I will go back to post 1 of pc2-brazil and go through your steps again but using my numbers.

I knew that when I was getting square roots of negative numbers for "proper times", albeit that they were "equal" I knew I was in the spacelike area of the spacetime coordinate system. I guess I "proved" that one cannot be in two places at the same time no matter how fast you travel but two different points (the beginning and end of a rod) can be in two different places at the same time, which is the direction you are trying to send me.

Now, did you see my post about "Clocks and Rods in Motion" above from Einstein's Relativity, Section XII?

Any thought on that equation:

t = x/\sqrt{1 - v^2/c^2}

Is that right? Dr. Einstein did not go through the algebraic derivation of that equation based on his preceding paragraphs. Should it be t = t'/\sqrt{1 - v^2/c^2}?

 

[Cyosis]

A simple check of dimensions will give you the answer. Gamma is a dimensionless quantity so on the left we have the dimensions of time and on the right side dimensions of length. This is impossible. Whenever in doubt, check the dimensions if they are not the same on both sides it is wrong.

 

[stevmg]

Cyosis -

It would appear that my edition of Relativity has a misprint as this has the form of
t = \gammax.

Mine is the Tess edition (15th edition) of Relativity, The Special And General Theory
ISBN: 1-57912-515-8. I will have to go to the library and look up a totally different edition of the same works and see if this "misprint" exists there, too.

Steve G
Melbourne, FL

 

[starthaus]

Now, did you see my post about "Clocks and Rods in Motion" above from Einstein's Relativity, Section XII?

Any thought on that equation:

t = x/\sqrt{1 - v^2/c^2}

Is that right? Dr. Einstein did not go through the algebraic derivation of that equation based on his preceding paragraphs. Should it be t = t'/\sqrt{1 - v^2/c^2}?

yes, it is a typo

 

[stevmg]

To starthaus:

Using the algebraic derivation below by pc2-brasil, the answer becomes thus:

x_2' = 12, x_1' = 0, v = 0.6c
x_2 - x_1 = \(1/\gamma)*(x_2' - x_1')
\gamma = 1/\sqrt{1 - v^2/c^2} = 1/\sqrt{1 - v^2/c^2}
= 1/\sqrt{1 - (0.6c)^2/c^2} = 1.25
(x_2 - x_1) = (x_2' - x_1')/\gamma = (12 - 0)/1.25 = 9.6 ly

Thank you, I understand it now.
So, in order to derive the length contraction result using the transformation with t' (x = γ(x' + vt')), I have to find the relation between t'₁ and t'₂, knowing that t₁ = t₂:
t_1=\gamma(t'_1+vx'_1/c^2)
t_2=\gamma(t'_2+vx'_2/c^2)
t_1=t_2
\gamma(t'_1+vx'_1/c^2)=\gamma(t'_2+vx'_2/c^2)
t'_1+vx'_1/c^2=t'_2+vx'_2/c^2
t'_2-t'_1=-\frac{v(x'_2-x'_1)}{c^2}
Now, I can make x₂ - x₁:
x_2-x_1=\gamma(x'_2+vt'_2-x'_1-vt'_1)
x_2-x_1=\gamma[x'_2-x'_1+v(t'_2-t'_1)]
Substituting the expression found for (t'₂-t'₁)
x_2-x_1=\gamma\left[x'_2-x'_1-\frac{v^2(x'_2-x'_1)}{c^2}\right]
x_2-x_1=(x'_2-x'_1)\gamma(1-\frac{v^2}{c^2})
x_2-x_1=(x'_2-x'_1)\gamma\times\gamma^{-2}
x_2-x_1=\frac{x'_2-x'_1}{\gamma}
which is the length contraction result.
This same result can be obtained in a faster way by using x'₁ = γ(x₁ - vt₁) and x'₂ = γ(x₂ - vt₂), and calculating x'₂ - x'₁ with t₁ = t₂.

I can do it in steps and I did just that in my post (#37) and got the right answer (9.6 ly) but the intermediate steps looked weird but were mathematically correct. You thought I was screwing up but I wasn't. I was going up the hill a different way but I still got to the top. If you don't believe me, query itbell as he or she told me to do it this way.

Also, my "proper time" calculations based on the "weird" results where we reached the same square root (\sqrt{-92.16} also confirms the calculations. x₂ - x₁ = 9.6 ly
and that is exactly what your method did but not the same way.

Your method is a lot easier and pc2-brazil's algebra derives the general approach to this problem.

pc2-brazil thanked you in advance. I have to thank you after-the-fact.

By the way, my user name (stevmg) isn't too secret coded. I've been around a long time.

Steve G
Melbourne, FL

 

[stevmg]

Cyosis:

It's a misprint.

I found the same book online in .pdf format and the equation
t = \gammax should be t = \gamma*1 or

t = 1/\sqrt{1 - v^2/c^2}

He's actually comparing what t is for t' = 0 and t' = 1

Makes sense as one second in "moving" time is greater than 1 second in "stationary" time.

I know, there is NO preferred frame of reference, but this compares one that is moving (O') with one that is "stationary" (O)

 

[Cyosis]

I know it is a misprint I told you in post 40 and gave you a simple procedure for checking such things. If t is a time that new equation doesn't make any sense dimensionally either unless 1 has the unit of time somehow.

 

[stevmg]

Actually, Cyosis, you told me how to check for the error. In other words you implied it was in error but didn't say it was. Now I couldn't conceive of Dr. Einstein publishing a book with such an error. So, I wanted somebody else out there who actually saw it printed correctly to tell me so as I had only one book at my disposal. The .pdf "book" on line confirmed that it was a typo.

When I was in charge of teaching residents in Pediatrics, one of my staff had the habit of telling the residents "You might want to do something (whatever the something was." What he meant was "Do this something." Not "You might want to..." I had to correct him many times and tell him, "iIf you want to tell someone to do something, don't suggest it, tell them." That's where I am coming from. You have to tell me something in an idiot-proof fashion or I will find the idiot's way to screw it up. Years of experience in this regard.

Thanks for all your help so far and that goes to
starthaus
itbell
Fredrik
DaleSpam
JesseM
and all the others

 

[Cyosis]

Perhaps I didn't spell it out. The main point of my post was to provide you with a method you can use to check an equation. I can't stress enough how helpful dimensional analysis is. It won't prove that an equation is correct, but if the dimensions on the left don't equal the dimensions on the right you know for sure the equation is wrong. Anyhow I don't want to go off topic too much, but it's important to learn to review your or someone else's results and ask yourself if they make sense.

 

[stevmg]

Hey, Cyosis -

No problem. I get it. When I took Physics and Chemistry in high school and college, that was one way we would keep tabs on any gross errors in equations.

BUT...

When it came to me to even question Dr. Einstein (or a book he authored) I was a little hesitant to assume anything and sought to get additional information. As it turns out, my questioning was justified and starthaus and you confirmed this. Good to see, though, that it was a misprint and not Dr. Einstein.

What I said about the residents was true - I always gave them "no s--t" instructions whenever I wanted them to do something. I never implied anything. In medicine, if something can be done wrong, it will be.

Makes you really confident of health care, right?

 

[yuiop]

What I said about the residents was true - I always gave them "no s--t" instructions whenever I wanted them to do something. I never implied anything. In medicine, if something can be done wrong, it will be.

Always safest to be explicit. Ask the blacksmith who took a red hot horse shoe from the fire and said to his apprentice, "When I nod my head, hit it hard with the hammer."

 

[stevmg]

Oh my God!

language is not precise. No way it can be made to be. Sometimes over explicitness can be too cumbersome but common sense still prevails.

Sometimes there just is no cure for stupid and I "ain't" so smart.

language Myths by Laurie Bauer & Peter Trudgill.

 

[starthaus]

The correct formula is:
x2 = (x'2 + vt'2) - from above post by starthaus
to starthaus - isn't that what I wrote in this part (see the last line in the above quote?)
If so, then 15 ly is a correct calculation and so would be the 9 years for t₂ = 1.25(0 + 0.6*12) = 9 yr. However, I would have to adjust backwards the length at t = 0 (in other words -9*0.6 = -5.4. 15 - 5.4 = 9.6, so the length x₂ when t = 0 was actually 9.6 ly.

Guess what, uising the length contraction formula we get the same result. 12 * (1/\gamma) = 12/1.25 = 9.6 ly.

I can't follow all your numerical calculations, as a rule, you should be able to express your ideas in symbolic form. I repeat, you have no right to write what amounts to x_2=v*t since v is the speed between the frames.

Actually, using t₁ = t₂ = 0 as per starthaus gives these weird answers, but the "proper time" both ways comes out the same:

SQRT[(-7.2 - 0)² - (12 - 0)²] = \sqrt{-92.16

and

SQRT[(0 - 0)² - (9.6 - 0)² = \sqrt{-92.16

I never wrote such stuff, nor did I direct you to. Negative proper time is a clear indication of another error in your calculations.

 

[stevmg]

Let's put this to bed:

To wit, assume that there are two reference frames O (the "stationary" one) and O' (the "moving" one.) The moving frame (O') is going at 0.6c to the right.
The length of the rod is (in this example) 12 ly in the O' frame and is stationary with respect to the O' frame. This is a specific case of the x₂' - x₁' which pc2-brazil discusses. v = 0.6c and is the speed between the frames with O stationary and O' moving at the 0.6c to the right.
Thus, gamma is 1.25.
\gamma = 1/\sqrt{1 - v^2/c^2}
\gamma = 1/\sqrt{1 - (0.6c)^2/c^2}
\gamma = 1/\sqrt{1 - 0.36
\gamma = 1/\sqrt{0.64 = 1/0.8 = 1.25
By convention, we will assume that all measurements and times start from x = 0, x' = 0, t = 0 and t' will calculate to zero which we will see below.
By the specific example I gave, we were given a specific length in O' which I chose to be 12 ly. This would represent x₂' - x₁' as stated a few paragraphs above.
x₁ = 0, t₁ = 0
x₁' = \gamma*(x₁ - v*t₁) = 1.25(0 - 0.6*0) = 0 If I am not mistaken then this is the correct Lorentz transformation regarding distance.
t₁' = 1.25(t₁ - v*x₁) = 1.25(0 - 0.6*0) = 0 Likewise, this should be the correct Lorentz transformation with regards to time.
By the conditions imposed in the discourse between starthaus and pc2-brazil, t₁ = t₂. This was so because the L = (x₂ - x₁) was being measured or calculated in the stationary frame given the measure of L' = (x₂' - x₁') which is the length in the moving frame (or O').
t₁ = t₂ = 0. This is my adaptation of the requirement that t₁ = t₂.
Because of the relativity of simultaneity one cannot assume that t₁' = 0, so this must be calculated, if you need it. The question did not ask for t₂' but merely for the value of x₂ - x₁. We will use the given that t₁ = t₂ and with that we are to calculate x₂ - x₁. This means (and pc2-brazil states so) we are given x₂' - x₁' and which, in my particular case here, is 12 ly as stated above. Because of the values of x₁ = 0, t₁ = 0 and, x₁ = 0 and t₁' = 0 which are stated above, all we need to find is x₂ which will then be the value of x₂ - x₁. We can calculate t₂' which will give the value of t₂' - t₁'. Again, this was not asked for but doing this reveals some interesting intermediate results which only makes sense in the time-space Minkowski system.

We develop two equations in two unknowns which will give us answers for x₂
(which is the answer we are after) and t₂' which we can accompish at the same time.

Equation 1:
x₂ = \gamma*[(x₂' - x₁') + vt₂'] Now, I am positive that this is the correct form of the Lorentz transformation when going from O' back to O.
x₂ = 1.25*[12 ly + (0.6)t₂']
Equation 2:
t₂' = \gamma*[t₂ -vx₂] = 1.25*[0 - (0.6)x₂]
t₂' = -1.25*0.6x₂
t₂' = -0.75x₂
substituting equation 2 into equation 1:
x₂ = 1.25[12 ly + (0.6)(-0.75x₂]
1.5625x₂ = 15 ly
We divide both sides by 1.5625 and we get:
x₂ = 9.6 ly
This also represent (x₂ - x₁) = L from pc2-brazil. This is because x₁ = 0, so the value of x₂ is the L.
This is the right answer and does equal L'/\gamma or 12/1.25 which is the original pc2-brazil notation.

If we back substitute x₂ = 9.6 ly into t₂' = -0.75x₂ we get t₂' = -7.2 yr.

I also derived using the above equations but substituting for x₂ with t₂' into equation 1 and proceeded to solve the new equation for t₂'. This likewise came out with the same answer: -7.2 yr.

pc2-brazil's algebraic derivation "buried" the negative value of t₂' and it was not evident when you just use the differences of the x's and t's. When we approached it in two steps, we found it out again, if we looked for it. Now we are dealing with coordinates in space-time and because of the placing the origin of O' on top of the origin of O, we have a negative number for t₂' but it is all relative, hence, no harm no foul.

Since this not a trip but a rod, we can have two different but simultaneous events, and they are simultaneous because they are being done at the same time in frame O. If one were himself or herself in O' we would "see" the measurement at the front of of the rod being done 7.2 years (-7.2, 12, 0, 0) ahead of the measurement at the back end of the rod (0, 0, 0, 0).

In most texts I have seen, they gloss over all this middle men part of the calculations and just state (using the terminology of pc2-brazil and starthaus) L = L'/\gamma which yields the same results and does not require delving into Minkowski space.

 

[starthaus]

In most texts I have seen, they gloss over all this middle men part of the calculations and just state (using the terminology of pc2-brazil and starthaus) L = L'/\gamma which yields the same results and does not require delving into Minkowski space.

The standard texts are very simple, they start with the Lorentz transform:

x'=\gamma(x-vt)
t'=\gamma(t-vx/c^2

then, they calculate:dx'=\gamma(dx-vdt)
dt'=\gamma(dt-vdx/c^2

In order to calculate the length L'=dx' as a function of the length L=dx
the texts make use of the fact that dt'=0 (you must mark both ends of L' simultaneously in frame F'). This gives:

dt=vdx/c^2

Substitute in the first equation and you will get:

L'=L/\gamma

None of your the impossible to follow manipulations. For example:

Equation 2:
t₂' = \gamma*[t₂ -vx₂] = 1.25*[0 - (0.6)x₂]
t₂' = -1.25*0.6x₂
t₂' = -0.75x₂

is not usable since t'_1=0 is DIFFERENT from your t'_2 above. The theory of measurement REQUIRES that you mark both ends of the rod simultaneously, i.e. t'_1=t'_2. So, the fact that you obtained the correct result for L' is purely accidental.

 

[stevmg]

Posts 1 - 5 of this topic says the opposite of what you just stated above in post 53. In those earlier posts you stated that t₁' does not equal t₂' and you wanted t₁ = t₂ because we were measuring the result in the unprimed frame.

The answer of 9.6 ly is not accidental. It follows from the already proven L = L'/\gamma.

Remember, as per pc2-brazil, we were given (x₂' - x₁'. That's what pc2-brazil stated and you went along with. He wanted to use t₁' = t₂' but you changed that to t₁ = t₂ and stated that t₁' could not equal t₂' because of the relativity of simultaneity. Just reread those posts. I didn't write them, you and pc2-brazil did.

 

[starthaus]

Posts 1 - 5 of this topic says the opposite of what you just stated above in post 53. In those earlier posts you stated that t₁' does not equal t₂' and you wanted t₁ = t₂ because we were measuring the result in the unprimed frame.

If you want to measure x'_2-x'_1 you need to make t'_2=t'_1.
If you want to measure x_2-x_1 you need to make t_2=t_1.
I think I already told you this a few times.

Remember, as per pc2-brazil, we were given (x₂' - x₁'. That's what pc2-brazil stated and you went along with. He wanted to use t₁' = t₂' but you changed that to t₁ = t₂ and stated that t₁' could not equal t₂' because of the relativity of simultaneity. Just reread those posts. I didn't write them, you and pc2-brazil did.

At this rate , you'll never learn anything.

 

[stevmg]

Well, actually I have learned a lot.

Too bad that we are not in a classroom with a blackboard, where the give and take is quicker and sometimes the light bulb goes off sooner. Right now, you're right. I am stuck in neutral. But I can tell you this:

If you have a rod of length L' as measured in O' and with O' moving at v relative to O, by length contraction when you transform this to what an observer in O (unprimed frame) sees, he/she will see a rod of length L = L'/\gamma.

JesseM and DaleSpam went over that with me (as well as with many others) in the Einstein train thread. That isn't hard to understand. Dr. Einstein demonstrated that in a simplistic way (he was great at breaking things down to simple elements) in Section XII. "Rods and Clocks in motion" in Relativity.

If you want to have one more pass before you say the hell with me, consider below

"If you want to measure x₂' - x₁' you need to make t₁' = t₂'
If you want to measure x₂ - x₁ you need to make t₁ = t₂"
This is quoting what you have stated above several different times in the thread.

It appears that in the case that I presented, we are to apply the 12 ly (x₂' - x₁') to O' (because that is where it is "measured") and use t₁' = t₂' which would be zero. We would then transform this to O via Einstein-Lorentz and would wind up with an x₁ and x₂ which would be the length (L) in O. Are we on the same sheet of music? I know it seems redundant and beyond elemental but just answer the question. In other words, we "measured" L' and from that will calculate L.

 

[starthaus]

It appears that in the case that I presented, we are to apply the 12 ly (x₂' - x₁') to O' (because that is where it is "measured") and use t₁' = t₂' which would be zero. We would then transform this to O via Einstein-Lorentz and would wind up with an x₁ and x₂ which would be the length (L) in O. Are we on the same sheet of music?

Nope. This is wrong. Remember the rule, in order to get a valid answer in O you need to measure x_2-x_1 for t_2=t_1. When you "Lorentz transform" t'_2 and t'_1 into O, you will NOT get t_2=t_1, so you will not get a valid measurement. Why is this so, Steve?

 

[stevmg]

That is because of the relativity of simultaneity, i.e., when t₂'=t₁' in O', when these are transformed, the "corresponding t's" are not equal any longer. They are only equal in the original frame (O') because they were "set" that way by virtue of a measurement of the front and back of the rod being done together. The measurement of the rod in O' means that all elements from front to back are measured at the same time.

There's the "jump" where I am hung up. I comprehend the measurement in O' as pc2-brazil presented (x₂' - x₁' = L')

But I want to stop here and have you "OK" what I have said so far. If so, I will then proceed.

 

[starthaus]

That is because of the relativity of simultaneity, i.e., when t₂'=t₁' in O', when these are transformed, the "corresponding t's" are not equal any longer. They are only equal in the original frame (O') because they were "set" that way by virtue of a measurement of the front and back of the rod being done together. The measurement of the rod in O' means that all elements from front to back are measured at the same time.

Yes, this is correct. Now you understand why I objected to your method?

 

[stevmg]

I will understand it more when I successfully complete the problem.

As I read pc2-brazil's opening statement (post 1, this thread) it appears that he knows L' (or x₂' - x₁') and, by consequence, t₁' = t₂' because he measured the position of the front of the rod (x₂') at the same time (t₂' = t₁') he measured the position of the back of the rod (x₁').

That is the core of my confusion: when you used the word "measure" it would seem that you mean "calculate." We are given x₁', x₂', t₁' = t₂' so we would not need to measure x₂' - x₁' therefore we would want to calculate x₂ - x₁ or L.

We would not get a "good" calculation of x₂ - x₁ unless t₁ = t₂. If I calculate t₁ and t₂ from the data given on x₁', x₂', t₁' and t₂', I will NOT get t₁ = t₂ because of this relativity of simultaneity.

At this point I will stop again, to wait for the "OK" on what I have just stated. So, let's make sure I have this correct so far.