# I Doubt regarding Lorentz contraction

1. Oct 13, 2018

### Kaguro

Hello all.

I am having some small trouble with applying the lorentz transformations to calculate lorentz contraction. Here's what I did:

Let O be the rest system and O' be the system moving with velocity v w.r.t O along x axis. Consider a rod lying in the O' system with ends x1' and x2'.
Length of rod in O' system is:

L' = x2' - x1'
measured at the same instant t'.

In the O system,
x2 = $\gamma$(x2'+vt')
x1 = $\gamma$(x1'+vt')

So,
L = x2-x1 = $\gamma$(x2-x1) = $\gamma$(x2' - x1')
so,
L = $\gamma$(L')

But... that's not quite right... L should always be smaller than L'...
Where did I go wrong?

2. Oct 13, 2018

### m4r35n357

Have you looked here?

Also the third term in your penultimate equation looks like it should not be there. I think you are processing the right way, but misinterpreting which frame to use where. Anyhow, the answer is in the link.

3. Oct 13, 2018

### Kaguro

Yes, I have seen the "correct" way to derive. But Can you tell me what's wrong with my approach?

4. Oct 13, 2018

### PeroK

If you calculate the times, in the O frame, for these events, you will see that they are not the same. You have one end of a moving rod at location x1 at some time, and the other end at location x2 at a different time. The distance between x1 and x2 is not, therefore, the length of the rod!

5. Oct 13, 2018

### Ibix

Your rod appears to be moving in O. In which frame is measuring at the same time important?

6. Oct 13, 2018

### Kaguro

At the time when one end of the rod is at x1, where is x2 at that time??

7. Oct 13, 2018

### PeroK

That's what you have to work out.

Note: when you apply the Lorentz Transformation, you need the times of the (transformed) events in frame O to be the same. That means you cannot simply transform simultaneous events from frame O'.

8. Oct 13, 2018

### Staff: Mentor

One end of the rod is at position $x_1$ at time $t_1$. The other end of the rod is at position $x_2$ at time $t_2$. How far and in which direction does that end move during the time interval $t_2 - t_1$, traveling at speed $v$? You need to be careful with + and - signs here. It might help to draw a diagram.

9. Oct 13, 2018

### Kaguro

What I have understood is:
Length is the difference in space coordinates of two events whose time coordinates are same.
I choose rod to be at rest in O'. So, these events x1' and x2' should be occurring at the same time t1' = t2'.

Now, I need to transform these events separately in O. And the time coordinates of these will NOT be same.

So, I write:
x1 = $\gamma$(x1' + vt1') at time t1.

Now I need x2 at time t1, not time t2... ( or maybe x1 and x2 both at t2)
But when I use the time transformation equation t = $\gamma$(t' + vx'/c^2) , this happens:

t1 = $\gamma$(t1' + v*t1'/c^2)
I need x3(t=t1), so that L = x3 - x1

x3 = $\gamma$(x2' + v*t1')

So,
L = $\gamma$(x2' -x1' + v$\gamma$(t1' + vt1'/c^2 ) - vt1'))

Well... Now what?

10. Oct 13, 2018

### PeroK

You've lost me at this point.

More simply, you need $t_2 = t_1$. Where does that lead?

11. Oct 13, 2018

### Arkalius

Let's try it with some numbers. You have a rod that's 5 units long and is stationary in our current frame of reference. So, we have our left end of the rod at 0 and the right end at 5. We'll pick 0 for our time coordinate to make things easy. So we have two events we'll transform, (0, 0) and (0, 5). Let's transform into a frame moving +0.6c, so our Lorentz factor is 1.25.

The origin is fixed in a transformation so $x'_1 = 0, t'_1 = 0$ still. Let's transform the other one.
$x'_2 = 1.25 ( 5 - 0.6 \cdot 0 ) = 6.25$
$t'_2 = 1.25 ( 0 - 0.6 \cdot 5 ) = -3.75$

So, our other coordinate has transformed to (-3.75, 6.25). But, we need to find it's x coordinate when the time coordinate is 0. Since this worldline is moving 0.6c in the -x direction, we can simply calculate $x'_3 = 6.25 + (0 - -3.75) \cdot -0.6 = 4$, so the point on the worldline at t=0 in this frame is at x=4. So, the width of our rod is 4 in this frame, which is what we expect with $\frac 5 \gamma = \frac 5 {1.25} = 4$

12. Oct 13, 2018

### stevendaryl

Staff Emeritus
To apply the Lorentz transformations, it's important to clearly define what are the relevant events. An event is a specific place at a specific time, and a coordinate system assigns 4 numbers, x, y, z, and t, to each event.

For the events involved in measuring the length of an object, let's ask how you would measure the length of a moving train. You need to have someone observing one end of the train, to report something like "The left end of the train was at $x=x_1$ at time $t=t_1$". You also need someone else to report something like "The right end of the train was at $x=x_2$ at time $t=t_1$. The two x-measurements must be at the same time. Then you can subtract the x-coordinates to get the length: $L = x_2 - x_1$.

Now, in relativity, the notion of "at the same time" is frame-dependent. So if you're measuring the length of a train moving along the x-axis in two different frames, you need at least three different events:
• $e_1$: The event where the left end of the train is measured by someone in frame $F_1$ and someone in $F_2$
• $e_2$: The event where the right end of the train is measured by someone at rest in frame $F_1$. To compute a length in frame $F_1$, you need for $e_1$ and $e_2$ to have the same time coordinate, according to $F_1$
• $e_3$: The event where the right end of the train is measured by someone in frame $F_2$. $e_3$ and $e_1$ must have the same time coordinate, according to $F_2$.
So considering only the x-coordinate, we have the following values:
1. $x_1, t_1$, the $F_1$ coordinates of $e_1$
2. $x_1', t_1'$, the $F_2$ coordinates of $e_1$
3. $x_2, t_2$, the $F_1$ coordinates of $e_2$
4. $x_2', t_3'$, the $F_2$ coordinates of $e_2$
5. $x_3, t_3$, the $F_1$ coordinates of $e_3$
6. $x_3', t_3'$, the $F_2$ coordinates of $e_3$
The length of the train in $F_1$ is $x_2 - x_1$
The length of the train in $F_2$ is $x_3' - x_1'$

The Lorentz transformations relate $x_j, t_j$ to $x_j', t_j'$. The fact that $e_1$ and $e_2$ are simultaneous, in frame $F_1$ relates $t_1$ and $t_2$. Similarly, $t_1'$ is related to $t_3'$.

A few more relationships: If the train is at rest in frame $F_2$, then that tells you something about the relationship between $x_2'$ and $x_3'$. And if you let $L'$ be the length of the train in $F_2$, then you should have all you need to figure out $x_2 - x_1$.

13. Oct 13, 2018

### Mister T

And you are interested in the length measured in $O$.

You are choosing events that occur at the same time in $O'$, don't you agree that's not correct, based on your definition of length?

14. Oct 13, 2018

### Staff: Mentor

I would recommend to not just transform events, but transform the worldlines. One end of the rod has a worldline x=0, and the other end has the worldline x=L. If you transform them you will get two worldlines in the other frame. What are those lines?

15. Oct 17, 2018

### Kaguro

After doing lots of reading for a few days, I seem to have understood the main problem.

Actually, m4r35n357 was right...
The explanation given in wikipedia is the easiest to understand, if you consider it patiently.

After transformation of the endpoint-events, they end up having different times.This was not a problem during transformation to the proper frame, as rod was in rest. So, if I measure one end now and other 10 hours later, still its fine, because rod is at rest in O' frame.

But in the other case, i.e. transforming into the O frame, rod is no longer at rest. So we use the time transformation equation to get times of transformed events. During this time interval, O' has moved some distance. This is to be subtracted from the apparent length! And you end up with:

That's really simple now, I guess...

Thanks everyone, for helping me out!

16. Oct 17, 2018

### Demystifier

Here you assume $t'_1=t'_2$, but that's wrong. In the O system you must take $t_1=t_2$.