# B Finding the length of a metre stick from a moving frame of reference

#### etotheipi

Suppose we have a stationary metre stick with one end positioned at the origin of a stationary frame of reference, pointing along the positive x axis.
The world lines of both ends of the metre stick are consequently vertical. Now consider another primed frame moving at velocity v relative to the stationary frame, calibrated such that both frames measure a time of 0 when the origins coincide.

The lines corresponding to x' = 0 and t' = 0 are drawn on the spacetime diagram below (from the Theoretical Minimum lecture series)

He says that the length of the metre stick in the moving frame of reference is represented by the distance from the origin to the intersection of the lines x = 1 and t' = 0 (between the two dots). I can't understand why this is the case.

It makes some sense to imagine that the length of the metre stick in the primed frame will be obtained from setting t' = 0 and measuring the x' coordinate at the end of the metre stick, but what has intersection with the x = 1 line got to do with this - that is, what is stopping us from labelling the position of the end of the metre stick anywhere else along the t' = 0 line?

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#### PeroK

Homework Helper
Gold Member
2018 Award
Suppose we have a stationary metre stick with one end positioned at the origin of a stationary frame of reference, pointing along the positive x axis.
The world lines of both ends of the metre stick are consequently vertical. Now consider another primed frame moving at velocity v relative to the stationary frame, calibrated such that both frames measure a time of 0 when the origins coincide.

The lines corresponding to x' = 0 and t' = 0 are drawn on the spacetime diagram below (from the Theoretical Minimum lecture series)

He says that the length of the metre stick in the moving frame of reference is represented by the distance from the origin to the intersection of the lines x = 1 and t' = 0 (between the two dots). I can't understand why this is the case.

It makes some sense to imagine that the length of the metre stick in the primed frame will be obtained from setting t' = 0 and measuring the x' coordinate at the end of the metre stick, but what has intersection with the x = 1 line got to do with this - that is, what is stopping us from labelling the position of the end of the metre stick anywhere else along the t' = 0 line?

View attachment 247571
The vertical line at $x=1$ represents the worldline of one end of the metre stick. The dot represents the event corresponding to where that end is at time $t' =0$. The origin represents the event corresponding to the other end at time $t' =0$.

The length of the metre stick in the primed frame is, by definition, the distance beween simultaneous measurements of the position of each end: in this case at $t' =0$.

The end of the metre stick is not at any other point on the line $t' =0$, so you cannot use any other point in a measurement of the length of the stick.

#### David Lewis

Suppose we have a stationary metre stick with one end positioned at the origin of a stationary frame of reference, pointing along the positive x axis.
Am I correct in assuming the x-axis is the horizontal line labeled t = 0?

#### etotheipi

The vertical line at $x=1$ represents the worldline of one end of the metre stick. The dot represents the event corresponding to where that end is at time $t' =0$. The origin represents the event corresponding to the other end at time $t' =0$.

The length of the metre stick in the primed frame is, by definition, the distance beween simultaneous measurements of the position of each end: in this case at $t' =0$.

The end of the metre stick is not at any other point on the line $t' =0$, so you cannot use any other point in a measurement of the length of the stick.

I'm still slightly confused about the whole procedure. As far as my understanding goes, the line x' = 0 represents all the (x,t) points in our reference frame which 'Lorentz transform' (so to speak) to x' = 0, and similarly for the line t' = 0. So where the line t' = 0 intersects the line x = 1, for some value of t in our reference frame, the corresponding coordinates in the primed frame are (x', 0)? Is this the right way to interpret the primed lines in the diagram?

Sorry if I'm making absolutely no sense, it seems as though this should be really straightforward but for some reason it won't click.

#### etotheipi

Am I correct in assuming the x-axis is the horizontal line labeled t = 0?
I think so

#### vela

Staff Emeritus
Homework Helper
I'm still slightly confused about the whole procedure. As far as my understanding goes, the line x' = 0 represents all the (x,t) points in our reference frame which 'Lorentz transform' (so to speak) to x' = 0, and similarly for the line t' = 0. So where the line t' = 0 intersects the line x = 1, for some value of t in our reference frame, the corresponding coordinates in the primed frame are (x', 0)? Is this the right way to interpret the primed lines in the diagram?
That's right. So what's still confusing you? Is it the notion that the observer at rest in the unprimed frame and the observer at rest in the primed frame don't agree on what "at the same time" means?

#### etotheipi

That's right. So what's still confusing you? Is it the notion that the observer at rest in the unprimed frame and the observer at rest in the primed frame don't agree on what "at the same time" means?

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#### jbriggs444

Homework Helper
For some reason PF is not letting me type a response, so sorry about the image!

I don't understand how the metre stick is represented by the line between the two dots in the moving frame.
It has been going wonky on me lately also. It is intermittent.

The meter stick at a point in time is a row of events. For instance, the meter stick at t=0 is a row of points where the left end was at t=0, where the 1 cm mark was at t=0, where the 2 cm mark was at t=0 and so on. This is a row of events in four dimensional space-time.

The meter stick at t'=0 is a different row of events. The left hand end event is the same. But it also contains the event where the 1 cm mark was at t'=0, where the 2 cm mark was at t'=0, etc. This is a different row of events.

Both rows are straight lines in four dimensional space. Both rows start at the same event. The two rows end at two different events. We can eliminate two out of the four coordinates and project the two lines in 4-space onto a single two dimensional sheet of paper. [It's Minkowski (hyperbolic) geometry rather than Euclidean, so there is still weirdness to deal with. Rotations are hyperbolic rather than spherical]

#### PeterDonis

Mentor
It has been going wonky on me lately also. It is intermittent.
If you're using Firefox, a recent update might have made third-party content blocking more aggressive, and since a lot of the supporting media for this site comes from a different domain (bernhardtmedia.com IIRC), to Firefox it looks like third-party content. I was able to fix the wonky behavior by disabling content blocking for physicsforums.com. You should be able to do that by clicking on the icons to the left of the site address in the address bar, and then clicking "Turn off Blocking for this site".

#### etotheipi

Ahh, I think I mostly understand now.

The world lines of both ends of the metre stick show the path of the metre stick relative to both different coordinate systems. That is, at t = 0 for the stationary frame, the two ends are simply 1 unit apart on the x axis. Conversely, when t' = 0 for the moving frame of reference, the length of the metre stick is given by the distance between the two dots where the world lines intersect the x' axis.

I think my confusion was due to somehow thinking of the world lines as only part of the stationary reference frame, whilst in reality they correspond to paths through spacetime, each point of which is measured slightly differently by the two different coordinate systems.

"Finding the length of a metre stick from a moving frame of reference"

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