Trouble with Lorentz transformations

[pc2-brazil]

Good evening,

As an effort for trying to understand Lorentz transformations, I'm trying to use them to derive the "length contraction" result.
Consider two reference frames, O (non-primed) and O' (primed), moving with respect to each other with a velocity v. Consider them to be under http://en.wikipedia.org/wiki/Lorent...rmation_for_frames_in_standard_configuration".
Choosing the non-primed reference frame, the Lorentz transformations for position (x-axis) and time (t-axis) will be:
x = γ(x' + vt')
t = γ(t' + vx'/c²)
The inverse transformations will be:
x' = γ(x - vt)
t' = γ(t - vx/c²)
Where γ is the Lorentz factor and c is the speed of light in vacuum.

Now, I will try to derive the length contraction result.
Suppose I have a thin rod moving along with the primed reference frame. One end of the object is at x'₁ x'₂. Then, the length of the object in the primed reference frame is L' = x'₂ - x'₁.
To find the corresponding coordinates of the ends of the rod in the non-primed frame (x₁ and x₂), I will use the inverse transformation for position:
x₁ = γ(x'₁ + vt')
x₂ = γ(x'₂ + vt')
The length of the rod as measured by the non-primed frame will be:
x₂ - x₁ = γ(x'₂ + vt') - γ(x'₁ + vt')
x₂ - x₁ = γ(x'₂ - x'₁)
L = γL'
This is wrong. I should have obtained L' = γL, because the non-primed frame sees the rod shorter than the primed frame does. In other words, the length measured by the non-primed frame should be shorter than the proper length.
What am I thinking wrong?

Thank you in advance.

 

[starthaus]

Good evening,

As an effort for trying to understand Lorentz transformations, I'm trying to use them to derive the "length contraction" result.
Consider two reference frames, O (non-primed) and O' (primed), moving with respect to each other with a velocity v. Consider them to be under http://en.wikipedia.org/wiki/Lorent...rmation_for_frames_in_standard_configuration".
Choosing the non-primed reference frame, the Lorentz transformations for position (x-axis) and time (t-axis) will be:
x = γ(x' + vt')
t = γ(t' + vx'/c²)
The inverse transformations will be:
x' = γ(x - vt)
t' = γ(t - vx/c²)
Where γ is the Lorentz factor and c is the speed of light in vacuum.

Now, I will try to derive the length contraction result.
Suppose I have a thin rod moving along with the primed reference frame. One end of the object is at x'₁ x'₂. Then, the length of the object in the primed reference frame is L' = x'₂ - x'₁.
To find the corresponding coordinates of the ends of the rod in the non-primed frame (x₁ and x₂), I will use the inverse transformation for position:
x₁ = γ(x'₁ + vt')
x₂ = γ(x'₂ + vt')
The length of the rod as measured by the non-primed frame will be:
x₂ - x₁ = γ(x'₂ + vt') - γ(x'₁ + vt')
x₂ - x₁ = γ(x'₂ - x'₁)
L = γL'
This is wrong. I should have obtained L' = γL, because the non-primed frame sees the rod shorter than the primed frame does. In other words, the length measured by the non-primed frame should be shorter than the proper length.
What am I thinking wrong?

Thank you in advance.

You need to calculate x₂ - x₁ for the condition: t₂ = t₁. You are inadvertently calculating it for t'₂ = t'₁, this is why you get the error.

 

[pc2-brazil]

You need to calculate x₂ - x₁ for the condition: t₂ = t₁. You are inadvertently calculating it for t'₂ = t'₁, this is why you get the error.

But why shouldn't t'₂ equal t'₁?
I thought this was implied, because, for the primed referential, the times of measurement of the ends of the rod were the same.

Thank you in advance.

 

[starthaus]

But why shouldn't t'₂ equal t'₁?
I thought this was implied, because, for the primed referential, the times of measurement of the ends of the rod were the same.

Thank you in advance.

Because of relativity of simultaneity. You are measuring in the unprimed frame, you need to mark the endpoints simultaneously in the unprimed frame. This means t₂ equals t₁ thus guaranteeing that t'₂ is NOT equal to t'₁.

 

[pc2-brazil]

Because of relativity of simultaneity. You are measuring in the unprimed frame, you need to mark the endpoints simultaneously in the unprimed frame. This means t₂ equals t₁ thus guaranteeing that t'₂ is NOT equal to t'₁.

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.

 

[stevmg]

The following two equations are the easiest way to remember Einstein's depiction of the Lorentz transformation (one-dimension). This\\ey appear on page 34 of his book "Relativity."

x' = gamma(x - vt)
t' = gamma(t - vx/c^2)
Thus x'_2 = gamma(x_2 - vt) and x'_1 = gamma(x_1 - vt)
We don't need the the equations for t_1, t'_1, t_2 and t'_2 from here on.
x'_2 - x'_1 = gamma(x_2 - x_1) or (x'_2 - x'_1)/gamma = x_2 - x_1

Remember, gamma = 1/[SQRT(1 - v^2/c^2)]. Gamma is always >=1, so 1/gamma is always <=1.

This makes sense. You give us x'_2 and x'_1 so x_2 - x_1 is always <= x'_2 - x'_1
L' = (x'_2 - x'_1) and L = (x_2 - x_1)
Thus L' = gamma*L which is the way it should be...

The same as starthaus above...

Steve G

 

[stevmg]

H-E-L-P!

How do I partially quote a post?

How do I refer to a prior post on the same thread or different thread by hyperlink?

 

[pc2-brazil]

Thank you for the clarifications in your previous post.

H-E-L-P!

How do I partially quote a post?

How do I refer to a prior post on the same thread or different thread by hyperlink?

First, click the "Quote" button below the post you want to quote. Then, you enter a page similar to the "New reply" page in which the whole post you quoted appears written between two "QUOTE" tags. To partially quote, just erase the part of the quote that you don't want, keeping only the relevant part.
To refer to a prior post on the same thread, first you need to get its address. To do this, click on the small number that appears in the top-right side of the post. Then, copy the address in the address bar of your browser. For example, for the post marked with "#5" in this thread:
https://www.physicsforums.com/showpost.php?p=2665478&postcount=5
This hyperlink can be used to refer to this post.

 

[starthaus]

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.

No, with t'₁ = t'₂

 

[starthaus]

Thus L' = gamma*L which is the way it should be...

No, that's clearly wrong. The correct answer is L'=L/gamma

The same as starthaus above...

Steve G

No, you got a different result

 

[pc2-brazil]

No, with t'₁ = t'₂

I'm measuring from the unprimed frame, so I need to mark the endpoints simultaneously in the unprimed frame. Thus, t₂ equals t₁.
With t₁ = t₂:
x'₂ - x'₁ = γ(x₂ - vt₂) - γ(x₁ - vt₁)
x'₂ - x'₁ = γ(x₂ - x₁)
L' = γL
This is the result I was expecting. The observer in the unprimed frame sees the rod shorter than its proper length (L').

 

[starthaus]

I'm measuring from the unprimed frame, so I need to mark the endpoints simultaneously in the unprimed frame.

The math that you wrote below contradicts your above statement.

Thus, t₂ equals t₁.
With t₁ = t₂:
x'₂ - x'₁ = γ(x₂ - vt₂) - γ(x₁ - vt₁)

When you write x'₂ - x'₁ it means that you are measuring in the primed frame.

x'₂ - x'₁ = γ(x₂ - x₁)
L' = γL

One more time, L' = γL is wrong. If you did things correctly you should have gotten L' = L/γ

 

[stevmg]

Then pc2-brazil had it right the first time...

L = $$\gamma$$L'

Starthaus, you are right... due to time dilation (with concomitant length contraction), L' will always be less than L. I got inverted. Damn, am I getting sloppy!

Steve G

 

[pc2-brazil]

The math that you wrote below contradicts your above statement.
When you write x'₂ - x'₁ it means that you are measuring in the primed frame.

I still think that that is not necessarily true, since I'm analyzing the same situation as before, in which the unprimed frame measures the length of an object that is moving with the primed frame. Thus, t₁ = t₂.
So, I can choose to write x₂ - x₁ and, by Lorentz transformations, obtain an expression containing x'₂ - x'₁, or I can write x'₂ - x'₁ and obtain an expression containing x₂ - x₁. But, in both cases, I have to use t₁ = t₂ and obtain L' = γL, since the situation I want to analyze is the same, and not a different one where I'm measuring from the primed frame.

 

[starthaus]

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.

Yes, this is now fully correct. Post #14 is not.

 

[starthaus]

I still think that that is not necessarily true, since I'm analyzing the same situation as before, in which the unprimed frame measures the length of an object that is moving with the primed frame. Thus, t₁ = t₂.
So, I can choose to write x₂ - x₁ and, by Lorentz transformations, obtain an expression containing x'₂ - x'₁, or I can write x'₂ - x'₁ and obtain an expression containing x₂ - x₁.

Yes.

But, in both cases, I have to use t₁ = t₂ and obtain L' = γL,

No.

 

[stevmg]

DQ = Dumb question. Use the above

If I have a stick 1 meter long and it slides on a table frictionless.

If it moves at 0.6c, will it fall through a hole in the table 0.9 meter?

What is x_1, what is x_2, what is x'_1, what is x'_2 (therefore what is L and what is L'?)

My gut feeling is that L' = 1 m (moving at 0.6c). L calculates to 0.8 m by length contraction and it will fall through the 0.9 m hole. Now, using what you discussed above can you show us by use of the Lorentz transformations rather than just the length contraction formula.

In the text Special Relativity by AP Frenchf M.I.T., on page 97, he goes over this exact problem presented by pc2-brazil. He assumes the measurement of x'_1 and x'_2 are done simultaneously in the S' frame of reference and he derives L' = x'_2 - x'_1 = L/gamma = (x_2 - x_1)/gamma. Now it would appear that by length contraction that the moving frame (x'_2 - x'_1) should be greater than the "static" frame (x_2 - x_1) so that it would "contract" to x_2 - x_1 but that is the unit length in S' is less than the unit length in S and thus the length L in S is greater than the length L'. S is the "static" F.O.R. (the x_1 and x_2) and S' is the "moving" F.O.R. (the x'_1 and x'_2.)

I know, there is no such thing as a preferred or static F.O.R., but you know what I mean here.

 

[stevmg]

You have to keep oriented as to who is what here. A rod which is in motion relative to a reference frame is shorter than it would be in its own frame. Thus, the 1 meter rod is actually 0.8 m in the reference frame while it is 1 m in its own (moviing) frame. It will fit through the 0.9 m gap in the table as described.

In this sense, the x'_2 - x'_1 is greater than the x_2 - x_1 which sort of contradicts what you guys have shown. What you have shown is that the moving frame measurement is smaller than that of the reference frame. It is true that the moving frame measurement transforms to something shorter in the reference frame even though it measures longer in its own frame.

Gotta be careful here and not blindly follow equations without keeping track of your orientation.

 

[starthaus]

In this sense, the x'_2 - x'_1 is greater than the x_2 - x_1

What is x'_2-x'_1 in your mind?
What is x_2-x_1?

which sort of contradicts what you guys have shown.

If that is so, you are contradicting the whole world of mainstream physics.

What you have shown is that the moving frame measurement is smaller than that of the reference frame.

This is what mainstream physicists agree on. Do you think it is wrong? Why?

Gotta be careful here and not blindly follow equations without keeping track of your orientation.

Yep, you have to pay attention.

 

[stevmg]

Starthaus:

Here simple example which goes along with what you say:

Take the Earth and a star some 7.2 lt-yr away.
Assume there is a reference frame S' that moves at 0.6c to the right with respect to the "stationary" F.O.R. S. Assume there is a rocket ship that travels in S at 0.8c to the right for 9 hours.

As per convention let us assign x_1 = 0 and t_1 = 0 and x'_1 = 0. Now, we state that t_2 = 9. Thus, after 9 hours (t_2) in the S F.O.R. the coordinate of that rocket ship is x_2 = 9*0.8 = 7.2 lt-yr (which coincides with the position of that star.) Thus x_2 - x_1 = 7.2 - 0 = 7.2.

Because v (the velocity of S' is 0.6c, gamma = 1/SQRT[1 - 0.6^2] = 1/0.8 = 1.25

Using Lorentz for distance, x'_2 = gamma[(x_2 - vt_2)) = 0.8[7.2 - 9*0.6) = 2.25
Thus x'_2 - x'_1 = 2.25 - 0 = 2.25. This is less than the 7.2 lt-yr difference cited above for x_2 - x_1. I've set t_1 = t'_1 = 0 and t_1 = t'_1 (because we used the same instantaneous measurement of the x_2 - x_1 as you cited in one of your explanations.) I made no such assumption for t'_2.
Clearly, t'_2 = gamma[t_2 - vx_2/c^2] = 1.25 (9 - 0.6*7.2) = 5.85 years. This not = to t'_1 of 0.

5.85 - 0 = 5.85 or t'_2 - t'_1 which is not 9 or t_2 - t_1.

But, there is another question associated with this:

Assume a rod is of sufficient length that at 0.8c moving to the right, it measures 7.2 lt-yr. in the S F.O.R. This rod would have to be 12 lt-yr in length in the S' F.O.R. for it to do this.

It appears that this rod is longer in S' (12) or x'_2 - x'_1 than in S (7.2) or x_2 - x_1.

What gives?

H-E-L-P-!

 

[starthaus]

Starthaus:

Here simple example which goes along with what you say:

Take the Earth and a star some 7.2 lt-yr away.
Assume there is a reference frame S' that moves at 0.6c to the right with respect to the "stationary" F.O.R. S. Assume there is a rocket ship that travels in S at 0.8c to the right for 9 hours.

As per convention let us assign x_1 = 0 and t_1 = 0 and x'_1 = 0. Now, we state that t_2 = 9. Thus, after 9 hours (t_2) in the S F.O.R. the coordinate of that rocket ship is x_2 = 9*0.8 = 7.2 lt-yr (which coincides with the position of that star.) Thus x_2 - x_1 = 7.2 - 0 = 7.2.

Because v (the velocity of S' is 0.6c, gamma = 1/SQRT[1 - 0.6^2] = 1/0.8 = 1.25

Using Lorentz for distance, x'_2 = gamma[(x_2 - vt_2)) = 0.8[7.2 - 9*0.6) = 2.25
Thus x'_2 - x'_1 = 2.25 - 0 = 2.25. This is less than the 7.2 lt-yr difference cited above for x_2 - x_1.

Good, so ths time you got it right, $$x'_2-x'_1 < x_2-x_1$$
So, you understood my correction and made the appropiate calculations this time.

I've set t_1 = t'_1 = 0 and t_1 = t'_1 (because we used the same instantaneous measurement of the x_2 - x_1 as you cited in one of your explanations.) I made no such assumption for t'_2.
Clearly, t'_2 = gamma[t_2 - vx_2/c^2] = 1.25 (9 - 0.6*7.2) = 5.85 years. This not = to t'_1 of 0.

Now you are doing something different, you are measuring a trip, not a ruler. Since your trip requires elapsed time, you can't mark its ends simultaneously either in S , nor in S'.
Do you understand the difference from the example in the OP? It is a totally different scenario.

But, there is another question associated with this:

Assume a rod is of sufficient length that at 0.8c moving to the right, it measures 7.2 lt-yr. in the S F.O.R. This rod would have to be 12 lt-yr in length in the S' F.O.R. for it to do this.

v=0.8 means gamma=0.6

So, the proper length of the rod would have to be 7.2/0.6=12

It appears that this rod is longer in S' (12) or x'_2 - x'_1 than in S (7.2) or x_2 - x_1.

What gives?

Nothing exotic, the proper length of the rod had to be 12ly in S' in order for it to be measured as 0.6*12=7.2ly in S.

 

[stevmg]

Now you are doing something different, you are measuring a trip, not a ruler. Since your trip requires elapsed time, you can't mark its ends simultaneously either in S , nor in S'.
Do you understand the difference from the example in the OP? It is a totally different scenario.

v=0.8 means gamma=0.6

So, the proper length of the rod would have to be 7.2/0.6=12

Nothing exotic, the proper length of the rod had to be 12ly in S' in order for it to be measured as 0.6*12=7.2ly in S.

Starthaus - this part above I don't get. I know how to do it but I don't get it.
What is the difference between a trip and a ruler?

Why is the length in S' longer than in S?

Again, I know how to do it but don't have a feel for it. I don't know when to "go for the gusto" and use what rule.

Steve G

 

[starthaus]

Starthaus - this part above I don't get. I know how to do it but I don't get it.
What is the difference between a trip and a ruler?

All points on along a ruler share the same time. No two points along a trajectory (trip) share the same time. Big difference.

Why is the length in S' longer than in S?

Read the math, it is all in the equations.

Again, I know how to do it but don't have a feel for it. I don't know when to "go for the gusto" and use what rule.

Steve G

I have given the rules repeatedly in this thread.

 

[stevmg]

How do I start a new thread?

My question would be:

How do you mathematically equate a Riemann sum as area under the curve to an anti-derivative? How do you prove that, theoreticlly, theone is equalent to the other?

Assuming the function is continuous between points a and b, there is always a Riemann sum and thus the function is integrable.

An anti-derivative is an algebraic manipulation which converts a new algebraic function to the function at hand such that the function at hand is the derivative of the new function. This may not always be possible to obtain such as the anti-derivative to y = e^(-x^2) but is integrable because it is continuous through the domain of x.

 

[starthaus]

How do I start a new thread?

You click "New Topic"

My question would be:

How do you mathematically equate a Riemann sum as area under the curve to an anti-derivative?

You don't. The Riemann sum is a number and the anti-derivative is a symbolic function.

How do you prove that, theoreticlly, theone is equalent to the other?

They aren't. See above.

An anti-derivative is an algebraic manipulation which converts a new algebraic function to the function at hand such that the function at hand is the derivative of the new function.

This isn't right, you may want to get a calculus refresher.

This may not always be possible to obtain such as the anti-derivative to y = e^(-x^2)

Not exactly, see here

but is integrable because it is continuous through the domain of x.

Yes, so? What does all this have to do with this thread?

 

[stevmg]

Nothing -

That was supposed to be on the "new" thread I asked about how to start.

Where is the "new topic" button? I can't find it.

I can finish my questions there.

 

[stevmg]

Stathaus -

Go here:

https://www.physicsforums.com/showthread.php?p=2671272#post2671272

Steve G

 

[stevmg]

Starthaus -

In the problem of the rod of 12 ly length is S' while moving at 0.6c to the right measures 7.2 ly in S. Now this is a rod and the time is the same at all points on the rod whether you are looking in S' or S. True? If so, that's why we use the length contraction formula

L_S = L_S' X (gamma)^-1

True?

I know this seems elemental but I am just trying to confirm one thing at a time here.

 

[jason12345]

I'm measuring from the unprimed frame, so I need to mark the endpoints simultaneously in the unprimed frame. Thus, t₂ equals t₁.
With t₁ = t₂:
x'₂ - x'₁ = γ(x₂ - vt₂) - γ(x₁ - vt₁)
x'₂ - x'₁ = γ(x₂ - x₁)
L' = γL
This is the result I was expecting. The observer in the unprimed frame sees the rod shorter than its proper length (L').

Correct. As you've shown, you have to be careful which frame you choose to mark the end points simultaneously in, because this will give rise to the end points being marked at different times in the other frame. Since the end points are fixed for all t in the stationary frame of the rod, it makes sense to measure them simultaneously in the other frame.

 

[starthaus]

Starthaus -

In the problem of the rod of 12 ly length is S' while moving at 0.6c to the right measures 7.2 ly in S. Now this is a rod and the time is the same at all points on the rod

Yes

whether you are looking in S' or S.

False. This can happen only in one frame at a time. Remember relativity of simultaneity?

 

[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.