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I was studying how the Lorentz transformations are derived, and found this table in Special Relativity for The Enthusiastic Beginner by David Morin. The primed frame $S'$ is taken to have a velocity solely along the positive x-direction of the unprimed $S$ frame.

The table that specifies four conditions for deriving the Lorentz transformations. Taken from "Special Relativity for the Enthusiastic Beginner", by David Morin

I justified the four conditions to myself as follows:

  1. Time dilation can be stated as $\Delta t_{other frame} = \gamma \Delta t_{proper}$. The proper time is measured on a stationary clock in the concerned frame, so the condition $\Delta x' = 0$ and the result $\Delta t = \gamma \Delta t'$ made sense. This was because $\Delta x' = 0$ implied that the clock was stationary in the concerned frame, and $\Delta t'$ was the proper time.

The statements 3 and 4 could also be reconciled with a qualitative understanding of the rear clock ahead effect and that of relative velocity (thought I am a bit clunky with putting that into words here. If needed I will elaborate.) The conditions for length contraction and the results were confusing, however.

The way I understand it:

For length measurements in a particular frame, the measurements should be simultaneous in that particular frame.

So, with the condition $\Delta t' = 0$, the implication was that the length measurement was being taken in the primed frame. The basic idea of length contraction is $\Delta l_{other frame} = {\Delta l_{proper}}/{\gamma}$. So with the measurement being made in the primed frame, proper length would be the length measured in the primed frame, i.e. $\Delta x'$. Since the length observation in any other frame will be contracted, I arrived at $$\Delta x = \Delta x' / \gamma$$

This contradicts what the table says: $$\Delta x' = \Delta x / \gamma$$

So I was wondering what the fallacy in my logic was. I would really appreciate if someone worked through my argument and explained its mistake.

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  • $\begingroup$ Note that "proper length" means "rest length", which is the length of an object measured in the frame in which it is at rest (say S). The rest length is always the largest length of the object, while in all other frames (say S") the object is in motion and simultaneous measurements of the endpoints of this moving object provide lengths which are shorter with respect to the rest length, thus the name "length contraction". $\endgroup$ – Batiatus Aug 18 '20 at 16:10
  • $\begingroup$ @Batiatus I understand what length contraction is. I was confused as to the logic for the conditions specified in the table: specifically, why $\Delta x' = \Delta x/\gamma$ (and not the other way around) when the measurement is made for $\Delta t'=0$ $\endgroup$ – Myungjin Hyun Aug 25 '20 at 9:46
  • $\begingroup$ My answers to What is wrong in this false derivation of length “dilation”? and Why is this assumption made in deriving time dilation? might be helpful. $\endgroup$ – Philip May 28 at 14:56
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You interpretations of conditions 1,3 and 4 are spot on. However, in concrete terms condition 2 means that the length contraction formula for a ruler in S' being viewed from S only holds true if the ruler is stationary in S' (ie the positions of both its ends are pined down at the same time t').

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I think the confusion with the Length Contraction item in the table (in section 2.1.1) lies in a subtle switch of situation between his "introduction of Length Contraction in section 1.3.3" and his "derivation of the Lorentz Transformation in section 2.1.1". (I think spacetime diagrams and additional notation would have helped distinguish the situations. Symbolic equations are occasionally not enough.)


When Morin introduces Length Contraction in section 1.3.3,
person-B (GROUND) is measuring the apparent-length of train-A (MOVING-TRAIN) that has rest length $L_A$. His analysis determines that person-B measures $$L_{B}=\frac{L_{A}}{\gamma}.\qquad(1.19)$$ At the expense of extra notation, this might be better expressed (using "wrt" = "with respect to") as $$L_{A,wrtB}=\frac{L_{A,wrtA}}{\gamma}.$$ Then, Morin summarizes the result as $$L_{observed}=\frac{L_{proper}}{\gamma}.\qquad(1.20)$$

On a spacetime diagram drawn by person-B (GROUND),
we draw the blue parallel timelines (worldlines) of the front and back of train-A. Note

  • OQ (the proper length of train-A) and
  • OM (person-B's measured-length of train-A).

robphy-Morin-lengthContraction-1

Note that MQ (along the timeline of the front of train-A) is orthogonal to OQ (along the spaceline $t'=0$). So this is a Minkowski-right triangle $OQM$ with right-angle at $Q$, where $\theta$ is the equal to the rapidity ($v_{B,wrt A}=\tanh\theta$ and $\gamma=\cosh\theta$).
In this triangle, $OM$ is the hypotenuse (since it is opposite the right-angle) and $OQ$ is the adjacent-side.

So, $\displaystyle\gamma=\cosh\theta=\frac{ADJ}{HYP}=\frac{OQ}{OM}=\frac{L_{A,wrtA}}{L_{B,wrtA}}$ which can be written as

\begin{align} (HYP)&=\frac{(ADJ)}{\cosh\theta}\\ OM&=\frac{OQ}{\cosh\theta}=\frac{OQ}{\gamma}\qquad(1.19, 1.20) \end{align}


Now, when deriving the Lorentz transformation in 2.1.1,

Consider a reference frame $S'$ moving relative to another frame $S$, as shown in Fig. 2.1. Let the constant relative speed of the frames be $v$
....
Our goal is to look at two events and relate the $\Delta x$ and $\Delta t$ in $S$ to the $\Delta x'$ and $\Delta t'$ in $S'$.

In this section, Morin sets up the form of the Lorentz transformation with $\Delta x =A \Delta x' +B\Delta t'$.
Morin wants to use "Length Contraction" to obtain the coefficient-$A=\gamma$ by setting up $\Delta x= \gamma \Delta x'$.

Morin-table-2.1.1

From the table, choose two events with $\Delta t'=0$ (so we don't need to worry about the so-far unknown coefficient-$B$): events $O$ and $Q$, which are simultaneous according to person-A (MOVING-TRAIN) $$\Delta t'_{OQ}=(t'_Q-t'_O)=0.$$

Now, here is the important part.

For this derivation, Morin wants (according to his goal) $$\Delta x_{OQ}=(x_Q-x_O)\qquad\mbox{ in terms of $\Delta x'_{OQ}=(x'_Q-x'_O)$ } $$ but this does not directly involve $\Delta x_{OM}=(x_M-x_O)$ from 1.3.3!
Instead, it involves a different pair ($O$ and $N$) of simultaneous events according to person-B (GROUND), where $$ \Delta x_{OQ}=\Delta x_{ON}.$$ That is to say, use the event $N$ so that $x_Q = x_N$.
(What is special about event $M$ is that $x’_Q=x’_M$, which doesn’t help us with $x_Q$ like $N$ does.)

robphy-Morin-lengthContraction-2

On a spacetime diagram drawn by person-B (GROUND),
consider a differently-sized train at rest according to person-B (GROUND, WITH REST-TRAIN) that person-A (MOVING-TRAIN) measures using events $O$ and $Q$.
Draw red parallels to B's timeline through event-O and through event-Q.
Note:

  • ON (the proper-length of this differently-sized train-B),
    where $N$ is the event on the front parallel that person-B says simultaneous with $O$.
  • OQ (person-A's measured-length of this differently-sized train-B).

Note that $ONQ$ is a Minkowski-right triangle, with right-angle at $N$ (so $OQ$ is the hypotenuse) and the same $\theta$ as before. Thus (following the ideas of above), \begin{align} (HYP)&=\frac{(ADJ)}{\cosh\theta}\\ OQ&=\frac{ON}{\cosh\theta}=\frac{ON}{\gamma} \end{align}

Compare the role of $OQ$ here and in (1.19) above.


Expressing these results in the $\Delta x$-notations (needed for 2.1.1)

from 1.3.3, \begin{align} (HYP)&=\frac{(ADJ)}{\cosh\theta}\\ OM&=\frac{OQ}{\cosh\theta}=\frac{OQ}{\gamma} \qquad(1.19,1.20)\\ \Delta x_{OM}&=\frac{\Delta x'_{OQ}}{\cosh\theta}=\frac{\Delta x'_{OQ}}{\gamma} \qquad(1.19,1.20 \mbox{ where $OQ$ is the proper-length of train-A}) \end{align}

from 2.1.1, \begin{align} (HYP)&=\frac{(ADJ)}{\cosh\theta}\\ OQ&=\frac{ON}{\cosh\theta}=\frac{ON}{\gamma}\\ \Delta x'_{OQ}&=\frac{\Delta x_{ON}}{\cosh\theta}=\frac{\Delta x_{ON}}{\gamma}=\frac{\Delta x_{OQ}}{\gamma} \qquad(1.20 \mbox{ where $OQ$ is the observed-length of train-B}) \end{align} so that $\Delta x_{OQ} = \gamma \Delta x'_{OQ}$, implying that coefficient-$A$ is equal to $\gamma$.

Again, I think spacetime diagrams and additional notation would help distinguish the situations.


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  • $\begingroup$ @Glowingbluejuicebox Does this answer your bountied-question? $\endgroup$ – robphy May 27 at 20:44

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