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 = \gammaL'

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&#039;_1+vx&#039;_1/c^2)
t_2=\gamma(t&#039;_2+vx&#039;_2/c^2)
t_1=t_2
\gamma(t&#039;_1+vx&#039;_1/c^2)=\gamma(t&#039;_2+vx&#039;_2/c^2)
t&#039;_1+vx&#039;_1/c^2=t&#039;_2+vx&#039;_2/c^2
t&#039;_2-t&#039;_1=-\frac{v(x&#039;_2-x&#039;_1)}{c^2}
Now, I can make x₂ - x₁:
x_2-x_1=\gamma(x&#039;_2+vt&#039;_2-x&#039;_1-vt&#039;_1)
x_2-x_1=\gamma[x&#039;_2-x&#039;_1+v(t&#039;_2-t&#039;_1)]
Substituting the expression found for (t'₂-t'₁)
x_2-x_1=\gamma\left[x&#039;_2-x&#039;_1-\frac{v^2(x&#039;_2-x&#039;_1)}{c^2}\right]
x_2-x_1=(x&#039;_2-x&#039;_1)\gamma(1-\frac{v^2}{c^2})
x_2-x_1=(x&#039;_2-x&#039;_1)\gamma\times\gamma^{-2}
x_2-x_1=\frac{x&#039;_2-x&#039;_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&#039;_2-x&#039;_1 &lt; 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.

 

[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&#039;_1-x&#039;_2 for t&#039;_1=t&#039;_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&#039;_1-x&#039;_2)/\gamma

OR

2. x&#039;_1-x&#039;_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&#039;_1-x&#039;_2 for t&#039;_1=t&#039;_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&#039;_1-x&#039;_2)/\gamma

OR

2. x&#039;_1-x&#039;_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&#039; = 12, x_1&#039; = 0, v = 0.6c
x_2 - x_1 = \(1/\gamma)*(x_2&#039; - x_1&#039;)
\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&#039; - x_1&#039;)/\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&#039;_1+vx&#039;_1/c^2)
t_2=\gamma(t&#039;_2+vx&#039;_2/c^2)
t_1=t_2
\gamma(t&#039;_1+vx&#039;_1/c^2)=\gamma(t&#039;_2+vx&#039;_2/c^2)
t&#039;_1+vx&#039;_1/c^2=t&#039;_2+vx&#039;_2/c^2
t&#039;_2-t&#039;_1=-\frac{v(x&#039;_2-x&#039;_1)}{c^2}
Now, I can make x₂ - x₁:
x_2-x_1=\gamma(x&#039;_2+vt&#039;_2-x&#039;_1-vt&#039;_1)
x_2-x_1=\gamma[x&#039;_2-x&#039;_1+v(t&#039;_2-t&#039;_1)]
Substituting the expression found for (t'₂-t'₁)
x_2-x_1=\gamma\left[x&#039;_2-x&#039;_1-\frac{v^2(x&#039;_2-x&#039;_1)}{c^2}\right]
x_2-x_1=(x&#039;_2-x&#039;_1)\gamma(1-\frac{v^2}{c^2})
x_2-x_1=(x&#039;_2-x&#039;_1)\gamma\times\gamma^{-2}
x_2-x_1=\frac{x&#039;_2-x&#039;_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!

Langauge 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.

Langauge Myths by Laurie Bauer & Peter Trudgill.