# I Definition and measurement of proper length

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1. Oct 25, 2016

### Frank Castle

As I understand it, the proper length, $L$ of an object is equal to the length of the space-like interval between the two space-time points labelling its endpoints, i.e. (in terms of the corresponding differentials) $$dL=\sqrt{ds^{2}}$$ (using the "mostly plus" signature).
Furthermore, this is only equal to the objects rest length, $L_{0}$ when the two space-time points are simultaneous, i.e in the inertial frame for which $dt=0$ and as such (in terms of the corresponding differentials) $$dL_{0}=\sqrt{dx^{2}+dy^{2}+dz^{2}}$$

I'm not sure I quite understand this last point, however. What is the intuition for why the two events have to be simultaneous?

For example, suppose one wishes to measure the length of a rod and one is in an inertial frame that is at rest with respect to the rod. Utilising the definition of proper length, $dL_{0}=\sqrt{ds^{2}}$, why is it that the two events (labelling the two endpoints of the rod) have to be simultaneous, since if one is at rest with respect to the rod, then surely if one measures the endpoints of the rod at two different times to determine the distance this won't affect the result as the endpoints are at rest with respect to the observer?!

Is the point that the rest length of the rod is simply determined by Pythagoras's theorem: $$dL_{0}=\sqrt{dx^{2}+dy^{2}+dz^{2}}$$ and so the proper length, $L$ is equal to the rest length, $L_{0}$ only in the case where the two events labelling the endpoints of the rod are simultaneous, i.e. when $dt=0$, such that $$dL=\sqrt{ds^{2}}=\sqrt{dx^{2}+dy^{2}+dz^{2}}=dL_{0}\;?$$ (Similar to how the proper time of an object is defined as the length of the time-like interval between two space-time points along the objects worldline, $d\tau=\frac{1}{c}\sqrt{-ds^{2}}$, and this coincides with coordinate time, $t$ in the case where one is in the rest frame of the object, i.e. (in terms of the corresponding differentials) when $dx=dy=dz=0$, such that $d\tau=\frac{1}{c}\sqrt{-ds^{2}}=dt$?!)

2. Oct 25, 2016

### PeroK

It's nothing to do with relativity.

If you wanted to measure the length of a moving train, let's say passing through a long station. You are at one end of the platform and your friend is at the other end of the platform. Your friend is there to observe the front of the train and you observe the rear. He dutifully notes the front of the train passing the far end of the platform and you observe the rear of the train passing the near end of the platform. And, so, you conclude that the train is the exactly length of the platform. And this is true of any moving train. Because of where you are standing, you always get a length of precisely the distance between you. And, if one of you moves further away from the other then all your measurements of length for moving trains will increase to the new distance between you.

You only get the correct length of the train when you observe the position of the rear of the train and he observes the position of the front of the train at the same time.

In an extreme case, if the end of the train leaves London at 12:00 and the front arrives in Edinburgh at 16:30 and you don't care about the time difference in these observations, then you'd conclude that the train was 400 miles long!

3. Oct 25, 2016

### Frank Castle

What has confused me is an example where one is at rest in a laboratory an one wishes to determine the "proper" height of the laboratory. In this example, the author states that an observer will measure the height at a given instant in time, $dt=0$, further more, since the laboratory is assumed to have a uniform shape (i.e. its width and depth remain constant throughout), i.e. $dx=dy=0$, it follows that $$\text{height}=\int_{\text{bottom}}^{\text{top}}\sqrt{ds^{2}}=\int_{0}^{h}dz=h$$ where the $z$-axis is taken to be the vertical axis, with $z=0$ the coordinate for bottom of the laboratory and $z=h$ the coordinate of its top.

Are they simply noting that the proper length of an object is always equal to its rest length, which is by definition the simultaneous length between its two endpoints, and as such, in the rest frame of the object $l=\int\sqrt{dx^{2}+dy^{2}+dz^{2}}$, however, in an inertial frame moving at some velocity $v$ with respect to the object, the proper length will be given by $l=\int\sqrt{-c^{2}dt'^{2}+dx'^{2}+dy'^{2}+dz'^{2}}$?!

4. Oct 25, 2016

### PeroK

The proper length is the rest length, by definition. That is to say, the length as measured in a reference frame where the object is at rest.

Note that the good thing about measuring something at rest is that you can measure where the two ends are at different times, as the object is not going anywhere! This is what we normally do if we want to measure the length of a train. We do this when it is not moving. In this case, you drop the $\Delta t$ and:

$L_0 =\sqrt{\Delta x^{2}+\Delta y^{2}+ \Delta z^{2}}$

If an object is moving in your frame of reference then:

a) It is harder to measure its length (in your reference frame) as you need to synchronise two measurements.

b) You get a shorter measure of length than its proper length.

So, you must have $\Delta t' = 0$ in this case. Otherwise, you are measuring a combination of its length and how far it has moved in $\Delta t'$.

In this case:

$L'=\sqrt{\Delta x'^{2}+\Delta y'^{2}+ \Delta z'^{2}}$ (note that you must have $\Delta t' = 0$ in this case)

And, of course, $L_0 = \gamma L'$, where $\gamma$ is the gamma factor for the relative velocity between reference frames.

Last edited: Oct 27, 2016
5. Oct 25, 2016

### Frank Castle

Ah ok. So in the example I've gave is it simply that one is in the rest frame of the laboratory and so its (vertical) length is simply given by the proper length $L_{0}=\sqrt{\Delta x^{2}+\Delta y^{2}+ \Delta z^{2}}$, regardless of whether the coordinates at the top of the lab, and those of the bottom of the lab are measured simultaneously or not?!

In general, I have seen proper length defined as the length between two space-like separated events such that $$dL_{0}=\sqrt{ds^{2}}$$ so I was just trying to see how this tallies up with the definition above. Is it simply that, in general, one will be moving a some velocity relative to the object that one wishes to measure and so the measurement of the two space-time points must be made simultaneously, otherwise the measured length would not equal the actual (proper) length of the object?!

6. Oct 25, 2016

### Staff: Mentor

It's generally best to use the term "interval" when you're talking about two arbitrary events, and reserve "proper length" for the exact sense that @PeroK is using it. However, the English language is inherently imprecise, so you should not be too concerned when you find even reputable sources failing to respect this nicety - the math is unambiguous, and it's what you should be paying attention to.

7. Oct 25, 2016

### Frank Castle

Ok, so is it simply that the proper length equals the arc length of a space-like interval in the inertial frame in which the two space-time events separated by the space-like interval are simultaneous?

8. Oct 25, 2016

### PeroK

Yes, exactly.

That's the next step. By a number of means, you can show that the interval $ds^2$ is frame invariant. So, for a spacelike interval, the length of the spacetime interval is equal to the proper length of the object - which is the length of the spacetime interval in the particular reference frame where the object is at rest.

Note: I see that you have this in post #3. Sorry, I missed that and was still working from your OP. So, in any reference frame:

$L_0 = \sqrt{-c^2 \Delta t'^2 + \Delta x'^{2}+\Delta y'^{2}+ \Delta z'^{2}}$

Edit: see further explanation below.

Last edited: Oct 25, 2016
9. Oct 25, 2016

### Frank Castle

Using this more general definition though, doesn't the length of the spacetime interval only equal the proper length if the two events are simultaneous?! If two events are space-like separated, then one can always find a frame in which they are simultaneous - is this the inertial frame that we identify with the rest frame of the object whose length that we wish to measure, since then: $dL_{0}=\sqrt{dx^{2}+dy^{2}+dz^{2}}=\sqrt{ds^{2}}$?!

10. Oct 25, 2016

### PeroK

To summarise and be more precise.

1) You can measure the proper length of an object in its rest frame by the simple:

$L_0 = \sqrt{\Delta x^{2}+\Delta y^{2}+ \Delta z^{2}}$

2) If you measure the length of an object that is moving relative to you, you get a length of:

$L = \sqrt{\Delta x'^{2}+\Delta y'^{2}+ \Delta z'^{2}}$

And you must, of course, have $\Delta t' = 0$ between the measurements of the two ends.

3) The length of a spacetime interval is defined as:

$ds^2 = -c^2 \Delta t^2 + \Delta x^{2}+\Delta y^{2}+ \Delta z^{2}$

This is frame invariant, so:

$ds^2 = ds'^2$ across all reference frames

4) If you measure the proper length of an object in its rest frame, then the proper length is only equal to the length of the spacetime interval if the measurements are simutaneous - or, at least, if you take the same $t$ coordinate:

$L_0 = \sqrt{\Delta x^{2}+\Delta y^{2}+ \Delta z^{2}} = \sqrt{-c^2 \Delta t^2 + \Delta x^{2}+\Delta y^{2}+ \Delta z^{2}}$

Only when you take $\Delta t = 0$

Now, if you transform these particular coordinates to another reference frame, then:

$L_0 = \sqrt{-c^2 \Delta t'^2 + \Delta x'^{2}+\Delta y'^{2}+ \Delta z'^{2}}$

But, this isn't really a measurement of length of the object in the primed frame, as in the primed frame you will have a non-zero time difference.

11. Oct 25, 2016

### Frank Castle

Is the point here that the spatial coordinates in the unprimed frame get transformed into a mixture of temporal and spatial coordinates in the primed frame, hence in the primed frame the space-time length $\sqrt{-c^2 \Delta t'^2 + \Delta x'^{2}+\Delta y'^{2}+ \Delta z'^{2}}$ is equal in value to $L_{0}$, but doesn't correspond to a measurement of the actual physical length of the object. If one wishes to measure the actual length of the object in the primed frame one would have to measure its endpoints at the same time (i.e. simultaneously), such that $L=\sqrt{\Delta x'^{2}+\Delta y'^{2}+ \Delta z'^{2}}$.

12. Oct 25, 2016

### PeroK

I think that's exactly the point.

13. Oct 25, 2016

### Frank Castle

Ok cool. I think I understand now. Thanks for your help!

14. Oct 25, 2016

### robphy

For simplicity, consider an extended object travelling inertially.

The proper length (or rest length) is the "distance between the parallel worldlines of its endpoints", geometrically speaking.
(It's not enough to consider two arbitrary spacelike-related events on those worldlines.)

In Euclidean geometry, given a point on one line,
we compute this distance by constructing the segment perpendicular to that line through that point,
then determining the length to the other parallel line.

We do the same in Minkowski spacetime geometry,
where we use a hyperbola (or a "causal diamond") to tell us how to define "perpendicular".
Physically, the events on this perpendicular segment are simultaneous according to that extended object.

The same idea works for a Galilean spacetime diagram,
using the Galilean spacetime definition of "perpendicular".

15. Oct 27, 2016

### vanhees71

16. Oct 27, 2016

### Frank Castle

So is the point that we want our notion of the proper length of an object to coincide with its rest length, and since this quantity should be Lorentz invariant we need to construct it out of Lorentz invariant measure of distance, i.e. the metric $\Delta s^{2}$. However, this is a measure of distance in space-time, whereas the rest length of the object is a spatial distance. As such, in order for the two notions to coincide, if we wish to determine the proper length of an object we must measure the endpoints of the object simultaneously (regardless of whether we are at rest with respect to the object or not), such that $$L=\sqrt{\Delta s^{2}}=\sqrt{\Delta x^{2}+\Delta y^{2}+\Delta z^{2}}$$ Or am I just overthinking this? Is it just that we wish to measure a spatial length, and this is achieved by simply using Pythagoras's theorem. In the rest frame one can measure the endpoints without doing so simultaneously since the object will always be at rest with respect to the observer. However, in another (inertial) frame, moving at some velocity relative to the object, it is important that one makes the measurements of the endpoints simultaneously, since otherwise one will not obtain a true measurement of the length of the object as there would be an additional contribution to the length measurement arising from the distance that the moving observer travels in the non-zero time interval?!

17. Oct 27, 2016

### vanhees71

Defining something in a special frame of reference, well-defined by the physical situation is an invariant definition. Here we consider "proper length" of some body like a rigid rod. Then there's a reference frame, where the rod is at rest as a whole, and you define it's proper length as measured in this reference frame, $L_0$. You can express this definition of proper length in any (local) inertial reference frame in a covariant way, given by Eq. (1.2.21).

18. Oct 27, 2016

### Frank Castle

I'm just wondering why one has to read of the coordinates of the endpoints simultaneously (as written at the bottom of page 9 of the notes, above Eq. (1.2.17))?

19. Oct 27, 2016

### vanhees71

It's the definition of a length measurement: The observer reads off the values of the spatial coordinates of the endpoints of the rod, simultaneously to measure its length (by definition). Also note that you never see the length contraction of a quickly moving rod. You rather see it rotated since you don't see the ends of the rod simultaneously (except in special cases) due to the finiteness of the speed of light.

20. Oct 27, 2016

### Frank Castle

But if an observer is in the rest frame of the rod then surely it doesn't matter if you measure the endpoints at two different points in time, since its length isn't going to change?!