Trouble with Lorentz transformations

1. Apr 8, 2010

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.

Last edited by a moderator: Apr 25, 2017
2. Apr 8, 2010

starthaus

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.

Last edited by a moderator: Apr 25, 2017
3. Apr 9, 2010

pc2-brazil

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.

4. Apr 9, 2010

starthaus

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.

Last edited: Apr 9, 2010
5. Apr 10, 2010

pc2-brazil

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.

6. Apr 10, 2010

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

7. Apr 10, 2010

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?

8. Apr 10, 2010

pc2-brazil

Thank you for the clarifications in your previous post.

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.

9. Apr 10, 2010

starthaus

No, with t'1 = t'2

10. Apr 10, 2010

starthaus

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

No, you got a different result

11. Apr 11, 2010

pc2-brazil

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').

12. Apr 12, 2010

starthaus

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.

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

13. Apr 12, 2010

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

14. Apr 12, 2010

pc2-brazil

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.

15. Apr 12, 2010

starthaus

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

16. Apr 12, 2010

starthaus

Yes.

No.

17. Apr 12, 2010

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.

18. Apr 12, 2010

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.

19. Apr 12, 2010

starthaus

Last edited: Apr 12, 2010
20. Apr 13, 2010

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-!

Last edited: Apr 13, 2010
21. Apr 13, 2010

starthaus

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.

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.

22. Apr 13, 2010

stevmg

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

23. Apr 13, 2010

starthaus

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

Read the math, it is all in the equations.

I have given the rules repeatedly in this thread.

Last edited: Apr 13, 2010
24. Apr 13, 2010

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.

25. Apr 13, 2010

starthaus

You click "New Topic"

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

They aren't. See above.

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

Not exactly, see here

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

Last edited: Apr 13, 2010
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