# What is wrong with deriving length contraction from the spacetime interval in this way?

My understanding of the spacetime metric is as follows: If Alice and Bob witness two light flashes $$E1$$ and $$E2$$, and Alice and Bob measures the distance between the position of the two light flashes as $$\Delta x_A$$ and $$\Delta x_B$$ respectively, and they also measure the time that elapsed between witnessing the two events happening as $$\Delta t_A$$ and $$\Delta t_B$$ respectively, then

$$(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2 - c^2(\Delta t_A)^2.$$

I have been able to derive time dilation from nothing but this equation, as follows:

$$c^2(\Delta t_B)^2 -(\Delta x_B)^2 = c^2(\Delta t_A)^2 -(\Delta x_A)^2$$ $$c^2(\Delta t_B)^2\left(1 - \frac{(\Delta x_B)^2}{c^2 (\Delta t_B)^2}\right) = c^2(\Delta t_A)^2\left(1 -\frac{(\Delta x_A)^2}{c^2\Delta t_A^2}\right)$$ $$(\Delta t_B)^2\left(1 - \frac{(\Delta x_B)^2}{c^2 (\Delta t_B)^2}\right) = (\Delta t_A)^2\left(1 -\frac{(\Delta x_A)^2}{c^2\Delta t_A^2}\right)$$ $$(\Delta t_B)^2\left(1 - \frac{v_B^2}{c^2}\right) = (\Delta t_A)^2\left(1 -\frac{v_A^2}{c^2}\right)$$

$$$$\Delta t_B\sqrt{1 - \frac{v_B^2}{c^2}} = \Delta t_A\sqrt{1 -\frac{v_A^2}{c^2}}$$$$

Letting $$\gamma_B = \frac{1}{\sqrt{1 - \frac{v_B^2}{c^2}}}$$

we therefore find

$$$$\Delta t_B = \Delta t_A \frac{\gamma_B}{\gamma_A}$$$$

In the case when the events are at the same position in Alice's reference frame, $$\gamma_A$$ becomes 1, and this equation is the familiar time dilation equation.

Here is where the real question begins: When I try to exploit this same line of reasoning for length contraction, I run into issues.

Lets assume that Alice observes these flashes at each end of a measuring stick, which is at rest with respect to Alice, both at the same time in her reference frame. Then $$\Delta t_A = 0$$, and we get

$$(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2$$

We assume now that Alice (and, hence, the measuring stick) are moving at a non-0 velocity $$v_B$$ when observed by Bob. Now we do the same sort of manipulations we did in the previous case, factoring out a $$\Delta x_B^2$$

$$(\Delta x_B)^2\left(1 - \frac{c^2}{v_B^2}\right) = (\Delta x_A)^2$$

Immediately I start to see something has gone wrong, since $$(1 - \frac{c^2}{v_B^2})$$ must be negative (by what I have heard about physics, $$v_B^2 < c^2$$), thus one of the $$\Delta x$$s must be imaginary. Indeed if we continue $$(\Delta x_B)^2 = (\Delta x_A)^2\frac{1}{(1 - \frac{c^2}{v_B^2}) }$$ $$(\Delta x_B)^2 = (\Delta x_A)^2\frac{v_B^2}{(v_B^2 -c^2) }$$ $$(\Delta x_B)^2 = (\Delta x_A)^2\frac{v_B^2}{-c^2(-\frac{v_B^2}{c^2} + 1) }$$ $$(\Delta x_B)^2 = -(\Delta x_A)^2\frac{v_B^2}{c^2}\gamma^2$$

But this implies that

$$\Delta x_B = i \Delta x_A\frac{v_B}{c}\gamma$$

That is, we have gotten an imaginary value for the observed length of the measuring stick by Bob! That doesn't make any sense!

I know there are other derivations of length contraction, but I am just not sure why this one isn't working. I have gone over the reasoning several times but I can't seem to find a flaw in it. Specifically, why should this work for time dilation but not for length contraction? Where did I go wrong?

• For lenght contraction both, dta and dtb must be zero, you can have this because the intervals are not equal, as you can easily see in a minkowski diagram. The measurements are simultaneous in both frames of reference.
– user65081
Commented Nov 11, 2021 at 1:07
• @Wolphramjonny took me a while to wrap my head around it, but I see what you are saying now. You mean that dtb in this case is the time between the two flashes, but not the time it takes for the measuring stick to move from point a to point b. I.e. this reasoning implicitly assumes that the flashes are simultaneous in Bob's frame. Is that right? Commented Nov 11, 2021 at 1:40
• the flashing will be out of synch in one of the two refernce frames. What you do to measure the lenght is to mark in your axis the locations of both the front and the end of the stick simultaneously in your reference system. nobody needs to move. so dtb is zero, forget about the flashes. dta is also zero. but you cannot equate the intervals dsa and dsb because they are not the same, the measurements of the front and back of the stick do not correspond to the same two events in the two reference systems.
– user65081
Commented Nov 11, 2021 at 1:54
• @Wolphramjonny If you put this in an answer I will accept it Commented Nov 11, 2021 at 1:57
• It is not necessary, I am glad it helped you.
– user65081
Commented Nov 11, 2021 at 1:58

What your analysis is doing is finding the spatial separation in Bob's frame of two simultaneous events in Alice's- ie of two events in Alice's frame that happen at different times in Bob's.

What you need to do instead is to compare the spatial separation of the two ends in Alice's frame with their separation at two simultaneous moments in Bob's frame. The way to do that is to consider the worldlines of the two ends of the object in Alice's frame (they will be parallel lines in the direction of her time axis) and find their spatial separation along a line of constant time in Bob's frame- that will give you the length of the object in Bob's frame.

The reason why length contraction occurs is that from Alice's perspective time Bob is noting the positions of the two ends of the object at two separate times- specifically, he's measuring the position of the leading edge of the object earlier than the position of the trailing edge, which allows the trailing edge a bit of time to move forward in the intervening period, thus giving a shortened result for the length.

You can see how it happens of you consider two people a distance apart on a platform, who are trying to measure the length of a passing train. They decide to do it by one of them noting the position of the front of the train as it passes and the other noting the position of the rear, then measuring the distance along the platform between the two positions they have noted. Clearly that only works if they note the position of the two ends at exactly the same time- if they note the positions at two different times, the train will have moved between the two measurements and they will get the wrong result. That's in effect what happens with length contraction.

If Alice and Bob witness two light flashes $$E1$$ and $$E2$$ [...]

Apparently were considering two inertial frames, with Alice being a member of one of these two, and Bob being a member of the other;

and two events (which are in turn observable, or "flashing") such that one (unnamed) member of Alice's frame and one (unnamed) member of Bob's frame were participating (in coincidence, passing each other) in one of these event, and another (though, in some cases, not necessarily distinct from the aforementioned) (unnamed) member of Alice's frame and another (though, in some cases, not necessarily distinct from the aforementioned) (unnamed) member of Bob's frame were participating (in coincidence, passing each other) in the other event; and further,

• that the distance between these two identified members of Alice's inertial frame is denoted as $$\Delta x_A$$ (and that these two members are also called "the two ends of Alice's measuring stick"),

• that the distance between these two identified members of Bob's inertial frame is denoted as $$\Delta x_B$$ (and that these two members are also called "the two ends of Bob's measuring stick"),

• that the duration of one of the end's of Alice's "measuring stick" from its instant of having taken part in one of the two mentioned events (having met and passed one of the ends of Bob's "measuring stick"), until its instant simultaneous to the instant of the other end of Alice's "measuring stick" having taken part in the other event (having met and passed the other end of Bob's "measuring stick") is denoted as $$\Delta t_A$$, and

• that the duration of one of the end's of Bob's "measuring stick" from its instant of having taken part in one of the two mentioned events (having met and passed one of the ends of Alice's "measuring stick"), until its instant simultaneous to the instant of the other end of Bob's "measuring stick" having taken part in the other event (having met and passed the other end of Alice's "measuring stick") is denoted as $$\Delta t_B$$,

then $$(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2 - c^2(\Delta t_A)^2$$

Right; where $$c$$ symbolizes signal front speed, of course. (My slightly verbose explanation of all symbols in your equation was to emphasize that, despite appearance, it is a coordinate-free statement.)

Let's assume that Alice observes these flashes at each end of a measuring stick, which is at rest with respect to Alice, both at the same time in her reference frame.

Let's (correspondigly) call the two "flash" instants or indications of the two ends of Alice's "measuring stick" simultaneous to each other (according to Einstein's definition of how this ougth to be measured).

Then $$\Delta t_A = 0$$, and we get $$(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2$$.

Correct.

We assume now that Alice (and, hence, the measuring stick) are moving at a non-0 velocity $$vB$$ when observed by Bob.

Bob and all relevant members of Bob's inertial frame measure

• that Alice and all members of Alice's inertial frame move uniformly wrt. Bob's inertial frame, and

• that they all move at equal (and, necessarily constant) non-zero speed $$vB$$.

(Also, Alice and the members of Alice's frame can determine vice versa the constant speed $$vA$$ of Bob and all members of Bob's inertial frame; and they find out that $$vA = vB$$.)

Now [...] factoring out $$(\Delta x_B)^2$$

... yes ...

[we get] $$(\Delta x_B)^2(1 - \frac{c^2}{v_B^2}) = (\Delta x_A)^2$$

No! -- $$vB$$ is certainly not defined as ratio between $$(\Delta x_B)$$ and $$(\Delta t_B)$$!

(And also, generally the values of these two differently defined quantities are not equal.)

Instead, with $$\Delta t_A = 0$$, it can be derived (by other, more elementary arguments) that

$$(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2 = (\Delta x_B)^2 \, \left( 1 - \frac{vB^2}{c^2} \right),$$

a.k.a. "length contraction".
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