# Why can't Lorentz boosts be used for contraction?

Let's say you have a rod that is 10 meters long. Observer O sees the ends of the rod at (t=0, x=0), and (t=0, x=10). Observer O' moves at speed v = 0.8c relative to O. What is the length of the rod in O's perspective?

Using the length contraction formula L' = γL, we find that O' sees the rod as 6 meters long. Geometrically, the ends of the rod are at (t'=0, x'=0) and (t'=0, x'=6).

However, things seem to be different if you use the Lorentz boosts, t' = γ(t-vx/c^2) and x' = γ(x-vt). The ends of the rod are at (t'=0, x'=0) and (t'=13.3/c, x'=3.3).

Even without the foreknowledge that the length contraction formula is the right way to answer this question, it seems like the Lorentz boosts gives the wrong answer, as it mixes length and time for the second end of the rod. However, I don't see why you *can't* apply it here. All the Lorentz boosts tells you is the coordinates of the ends of the rod. But it appears that in this scenario Lorentz boosts gives you the "incorrect" coordinate of the second end of the rod (while the contraction formula + the geometry gives you the "correct" coordinate). Why?

## Answers and Replies

robphy
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You need to boost the worldlines of the ends of the rod.
Then, find the intersections of those boosted worldlines with the observer's space-axis (where t'=0).

PeterDonis
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Let's say you have a rod that is 10 meters long.

You left out a key item: which observer is at rest relative to the rod? I assume you intended it to be observer O, but you should not leave these things to be assumed.

Observer O sees the ends of the rod at (t=0, x=0), and (t=0, x=10).

More precisely, observer O sees the ends of the rod at the instant t = 0 of his time at the events you give.

Using the length contraction formula L' = γL, we find that O' sees the rod as 6 meters long. Geometrically, the ends of the rod are at (t'=0, x'=0) and (t'=0, x'=6).

More precisely, observer O' sees the ends of the rod at the instant t' = 0 of his time at the events you give.

things seem to be different if you use the Lorentz boosts

That's because you are leaving out relativity of simultaneity. The pair of events that observer O uses to determine the length of the rod is different from the pair of events that observer O' uses to determine the length of the rod, because they are using different definitions of simultaneity--different sets of events happen at the same time according to each of them.

So it is true that, if you take the pair of events that observer O uses to determine the length of the rod, and Lorentz transform them into the rest frame of observer O', you do not get a pair of events with a spatial separation of 6 units. But you also do not get a pair of events that happen at the same time according to observer O'. This is obvious since the t' coordinates of the two events are not the same. So these two events do not determine the "length" of the rod according to observer O'; they don't happen at the same time according to him. You have to find a different pair of events that do happen at the same time according to observer O', in order to show what length he determines the rod to have.

Yes, by 10 meters I meant in O's frame. t is in the frame of O, t' is in the frame of O'.

I'm confused about when to apply Lorentz boosts now. Let's say a ball is attached at the second end of the rod. This case suggests that in the frame of O', the ball is now located (t'=0, x'=6). But isn't the ball at some instant an 'event'? And to transform an event from one reference frame to another you use Lorentz boosts, which tells you that the ball is actually located at (t'=13.3/c, x'=3.3).

Orodruin
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But isn't the ball at some instant an 'event'?
No, the ball has a world line. It passes through a large number of different events. An event is a given place and a given time. The ball exists for an extended time period and so corresponds to several events, its world line.

For a given time in a particular frame, you can pick the event which is specified by that time and the position of the ball at that time. Picking a different frame to specify "the time", events that were simultaneous with the previous convention will no longer be that.

Ibix
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I'm confused about when to apply Lorentz boosts now.
I think you are doing it right, you just aren't quite getting what it's telling you.

The point is that in the O' frame, the rod is moving at 0.8c. Since, in this frame, you aren't measuring the position of the ends simultaneously, you aren't measuring length. Instead, you are mixing length and speed. Think of trying to measure the length of a beetle sitting on a ruler. It's easy if the beetle is still, but if it keeps walking along you need to record its head and tail position simultaneously or else you can get absurd answers.

The beetle is just sloppy experimental procedure. Here you've had the same mistake happen as a result of the relativity of simultaneity. Your simultaneous measurement of position in frame O is not simultaneous in O'. Unfortunately, it was O' where you needed to take care. The solution is to find a pair of events that are simultaneous in O' and work with those.

Yes, by 10 meters I meant in O's frame. t is in the frame of O, t' is in the frame of O'.

I'm confused about when to apply Lorentz boosts now. Let's say a ball is attached at the second end of the rod. This case suggests that in the frame of O', the ball is now located (t'=0, x'=6). But isn't the ball at some instant an 'event'? And to transform an event from one reference frame to another you use Lorentz boosts, which tells you that the ball is actually located at (t'=13.3/c, x'=3.3).

Ball reaches the end of the rod at t'=0s, x'=6m seen from O' perspective certainly is an event but that is already how O' sees it. If you were to do the lorentz transformations to see where this _instance_ of the ball is located from the perspective of O, you would certainly NOT get t=13.3/c and x=3.3 (not x' and t').

Since the rod is not moving relative to O, the end of the rod at 10 meters will ALWAYS be at 10 meters. So, logically, when doing the Lorentz transformations, you should end up with x=10m seen from the perspective of O, for ANY ball that O' measures to be on the worldline of the rod's end.

x = γ(x'-v*t') = 1.66... * 6 m = 10 m as expected.

As for your initial calculations, how did you calculate x' = γ(x-v*t) and end up with 3.3? I am no expert so maybe i am wrong, but as far as i understand it

x' is the same independent on if O' would be moving towards the edge of the rod at 10m or away from it. In the case of O' moving away, your value t' = 13.3/c is correct. If O' was moving towards the edge of the rod at 10m, the value shoud be negative at t' = -13.3/c (actually -13.3m/c, again if i am not mistaken - which is a very small value in seconds, hence it would have been much better to consider the rod to measure 10 lightsecond, in which case -13.3ls/c would have been exactly -13.3s and the calculation would have been easier).

x' SHOULD be x' = γ(x-v*t) = 1.666.. * 10m = 16.666...m in both cases, never 3.3m.

So the events e1(x/t) and e2(x2/t2) seen from the perspective of O as e1(0s/0s), e2(10m/0s) are seen from the perspective of O' as e1'(0m,0s), e2'(16.666...m, 13,3m/c) in the case of O' moving away of the other edge of the rod(the edge which is not local to himself) and e1'(0m,0s), e2'(16,666...m, -13.3m/c) if he is moving towards the edge that isn't local to himself.

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Yes, it should be t' = -13.3/c and x' = 16.6.

I'm still struggling to understand this. Do you only "find where the points are simultaneous" in the frame of O' when they are simultaneous in the frame of O? And all other times just use the Lorentz boost as normal?

robphy
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Think geometrically [Draw a Spacetime Diagram].
The Lorentz transformation relates the two observers' coordinate-assignments of an event.

The problem of length contraction is finding the observer-dependent spatial-separation between two parallel worldlines.
You must use the Lorentz transformation to boost the worldline of each end of the rod.
Find two events on the worldline of the left end and apply the transformation. Connect the boosted events to get the boosted left-end worldline.
Do the same with the right-end worldline.

Now consider the spatial-axis of your observer making the measurement (that is, the line of simultaneity according to that observer).
That axis intersects each worldline, at events (call them) P and Q. The length measured by that observer ( the apparent length of the rod) is the spatial-separation between P and Q.

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Yes, it should be t' = -13.3/c and x' = 16.6.

I'm still struggling to understand this. Do you only "find where the points are simultaneous" in the frame of O' when they are simultaneous in the frame of O? And all other times just use the Lorentz boost as normal?

Well, i was about to actually answer your actual question, hence i was about to give you two events e1(x,t) and e2(x2,t2), which both are located on the two worldline edges of your object, which are NOT simultaneous events, hence t != t2 from the perspective of O but ARE simultaneous events from the perspective of O', hence e1'(x',t') , e2(x2',t2') with t'=t2'.
But then i ran into issues with there actually being inverse Lorentz transformations which to me make absolutely no sense, so i got into trouble with the positive and negative signs.

You might want to try e1(0m, 0s) and e2(10m, 8m/c)

use the Lorentz transformation on those two and see what you get. Obvislou e1 translates into e1'(0m,0s) so you only have to care for the latter.

If i didn't miscalculate it, you should get e2'(6m, 0s), hence, you would then have two events which are simultaneous and which you can use to calculate the length.

It helps to know, that any two events which are space separated and happen simultaneous from the perspective of an observer A, will never happen simultaneous from the perspective of an observer A' which is moving relative to A at any given velocity.
Hence, only two *space/time separated* events which are NOT simultaneous from the perspective of A, could ever end up being simultaneous from the perspective of A' moving relative to A.
And like it or not, that is how we usually measure "stuff". We measure their endpoints or instances of their endpoints which "happen" at the same time/simultaneous.

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Nugatory
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Even without the foreknowledge that the length contraction formula is the right way to answer this question, it seems like the Lorentz boosts gives the wrong answer, as it mixes length and time for the second end of the rod. However, I don't see why you *can't* apply it here. All the Lorentz boosts tells you is the coordinates of the ends of the rod. But it appears that in this scenario Lorentz boosts gives you the "incorrect" coordinate of the second end of the rod (while the contraction formula + the geometry gives you the "correct" coordinate). Why?

The length of the rod is the difference between the coordinates of the end points AT THE SAME TIME. In the frame in which the rod is at rest, I find that at time t=0 one end is at x=0 and the other end is at x=10; the length is 10 meters. However, when I transform the the two events (x=0,t=0) and (x=10,t=0) into the primed frame, I find that those two event do not happen at the same time in that frame (their t' coordinates are different) so it is a mistake to use the difference between the x' coordinates of those two events to calculate the length in the primed frame. Instead I have to find where the endpoints of the rod lie AT THE SAME TIME in the primed frame, use that to calculate the length in the primed frame.

This is much easier to see if you draw a spacetime diagram showing the x and t axes, the x' and t' axes, and the worldlines of the two ends of the rod - so much easier that it would be a good exercise for you to try it for yourself.

Mister T
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Let's say a ball is attached at the second end of the rod. This case suggests that in the frame of O', the ball is now located (t'=0, x'=6). But isn't the ball at some instant an 'event'?

The ball being located at x'=6 at time t'=0 is indeed an event. The value of x' is the location, the value of t' is the clock-reading.

And to transform an event from one reference frame to another you use Lorentz boosts, which tells you that the ball is actually located at [...]

It will give you the location x and clock-reading t.

I think what's confusing you is your use of the word "when". A phrase like "when the ball is located at x'=6 the clock reads t'=0" has meaning only if the frame of reference is known. In this case it's implied that the frame of reference is the primed frame, so we're good. But if a clock that's at rest in the primed frame reads t'=0, that is in and of itself an event. If that clock is not located at x'=6 then observers in motion relative to it will not agree that the ball's location is x'=6 "when" that clock reads t'=0.