Terrel rotation of a falling stick

Question

The first option is obviously incorrect. This is because relativistic effects will cause some apparent rotation (an optical illusion).

Now when I try to visualize the problem, it seems to me that c) should be the correct option since the left end of the stick is closer to the eye as compared to the right end. To balance that the stick might appear to be rotated anticlockwise. This was my reasoning.

The answer turns out to be b).

Now, how to approach such problems where it seems that mere visualization is needed but which apparently doesn't work (at least for me)? What is the better way of looking at this problem and solving it?

Answer 1

More often than not, in textbooks the wording "looks to an observer $X$" is intended to mean "is in observer $X$'s frame". Thus I don't think the question is about Terrell rotation, but rather about simple Lorentz transformations.

I'll proceed assuming this.

Let's consider a thin horizontal stick that's falling down with velocity $v_{\text s}$. The worldline of its points will be described as (I'm using the units where $c=1$):

$$p_{\text s}=\pmatrix{t\\ x\\ y}= \pmatrix{ t\\ x\\ v_{\text s}t}.$$

The Lorentz transformation along the $x$ axis into a frame moving with velocity $v_x$ with respect to the original frame is

$$L=\pmatrix{ \gamma & -\gamma v_x & 0\\ -\gamma v_x & \gamma & 0\\ 0 & 0 & 1},$$

where

$$\gamma=\frac1{\sqrt{1-v_x^2}}.$$

Applying it to $p_{\text s}$, we get

$$p_{\text s}'=\pmatrix{t'\\ x'\\ y'}=Lp_{\text s}=\pmatrix{ \gamma(t-v_x x)\\ \gamma(x-v_x t)\\ v_{\text s} t }.$$

To completely switch to the point of view of the moving observer, we should express the worldline of the stick in terms of the observer's time $t'$, rather than the original frame's time $t$:

$$p_{\text s}'=\pmatrix{ t'\\ \frac x\gamma-v_x t'\\ v_{\text s}(x v_x+\frac{t'}\gamma) }.$$

Here we can see that $x'$ still grows with $x$ growing, which means there's no unexpected inversion. Now the last part is the $y'(t')$. Inserting $t'=0$ with $x=0$ and then with $x=1$ into the third component of $p_{\text s}'$ and taking into account that $v_x>0$ and $v_{\text s}<0$, we find that the left-hand end of the stick should be higher than the right-hand one. This makes the answer unambiguously (b).

Terrell rotation

You may still be curious how to work out the actual Terrell rotation, even though it wasn't the question posed by the textbook. For this we have to do the following.

• Pick a distance $d'$ of the observer from the plane of the stick's motion
• Choose a point $x'$ along the stick that we want to watch
• Note that the signal from the point on the stick $(x',y',0)$ to the observer at $(0,0,d')$ must travel for time $\sqrt{x'^2+y'^2+d'^2}$, so the time of emission of a signal that we receive at time $t'_{\text{reception}}$ is

$$t'_{\text{emission}}=t'_{\text{reception}}-\sqrt{x'^2+y'^2+d'^2}.$$

• Compute the expression for $y'(x',t')$
• Use this expression to formulate the equation like $$y'=y'(t'_{\text{signal}})$$ to solve for $y'$. The resulting equation should look something like

$$y'=v_{\text s}\gamma\left(v_x x'+t'_{\text{reception}}-\sqrt{x'^2+y'^2+d'^2}\right).$$

Now, as you can see, the equation is quadratic. This means the solution won't be a straight line. It'll be a quadratic curve. Even with $v_x=0$ you'll see a curve instead of a straight line. Moreover, there are two solutions, and you must choose the appropriate one.

The result for $d'=5$, $v_x=0.3$, $v_{\text s}=-0.5$, $t'_{\text{reception}}\in[-2.5,10]$ would be

Answer 2

The keyword :

Since observer is moving rightward, the information (in this case light from the rod) from the right side will reach the observer earlier than the information from the left side (of the rod).

Derivation :

1.Picture (a) is what actually happens (and the observer is at the same reference as the rod).

2. Consider the 4th frame in (a). The right & the left side of the rod is falling at the same time.

3. Observer is moving rightward. So his/her distance to the right side is closer that the left side. Coz the light path is shorter. The fx : [Information from the right side reach the observer 1st, and left side reach later.]

4. Consider the 3rd frame in (b). The (information of) [A] rod right side is dropped on the floor reach 1st, and (information of) [B] the rod left side is dropped on the floor reach the observer later.

5. Consider the 4th frame in (b). The [A] information leaves the observer, and [B] information [B] just started reached the observer.

6. Consider the 5th frame in (b). The [B] information just reached the observer.

That is my take. Feel free to comment/add should this case has alternative perspective/derivations.

p/s: (the text below is my personal ref, not a paper/textbook. Not directly related, I just felt its worth mentioning)

At 9:12 of the research video is (imho) probably the closest physical observation mention to relativistic effect mentioned in this question (Just that the observer is at the same reference @ the object). At 5:21 is the raw picture (from the observer). Since the question mentioned qualitatively.. I think (b) is an acceptable answer (I like the rod to be a bit curved). ( :

Answer 3

A solution

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The observer moves rightwards wrt. ground. In the options are shown observer's observations, with its time increasing rightwards too.

Assuming no gravity and periodicity of snapshots, options $a$ and $d$ are ruled out based on effects of simultaneity and either $b$ or $c$ showing said effects. Moreover $b$ and $c$ differ in only which end falls first.

Even without doing any math, the right end of the rod must fall first. To see this, initially position the moving observer at rod's midpoint$^1$. As time progresses, light from the right end would reach the observer before that from the left end. Since the observer isn't aware of its own motion and speed of light is frameless, it concludes that the right end started falling sooner

Mathematically, let the moving observer's frame be $(x',t')$ and the gorund's frame be $(x,t)$. The events $$ \begin{align} A&=(x_a,t_a)=\text{left end falls} \\ B&=(x_b,t_b)=\text{right end falls} \end{align} $$ with the constraint $$ t_a=t_b=t_0\tag{0} $$

would become for the observer $A'=(x'_a,t'_a)$ and $B'=(x'_b,t'_b)$ where $$ t'_i=\gamma(t_i-\beta x_i) \tag{1} $$

with $c=1,\beta=v_{observer},\gamma=(1-\beta^2)^{-1/2}$. Using eqns. $0$ and $1$

$$ \begin{align} t'_b-t'_a&=\gamma(0-\beta (x_b-x_a))\\ &=-\gamma\beta L_0\\ &<0\hspace{4cm}(\beta>0,L_0>0) \end{align} $$ therefore $t'_a>t'_b$ and the left end is observed falling later, vindicating option $b$.

Also

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Terrel rotation of a falling stick

IMHO the question isn't about Terrel rotation. That requires evaluating actual views visible to observer, requires knowledge of the observer's position, wouldn't have kept the rod straight in the shown options, and is a more meta application of STR, that seems a bit advanced for the question.

The first option is obviously incorrect.

Good for you. To me there's nothing obvious about its incorrectness. If anything its obviously correct. All my non-relativistic life has drilled the notion down into my bones of falling rigid rods not behaving like reflecting light rays, just because they move.

Now when I try to visualize the problem, it seems to me that c) should be the correct option since the left end of the stick is closer to the eye as compared to the right end.

Doesn't that depend on where the moving observer initially was? As discussed earlier, to use that argument for the current case, you must observe from a special position - the midpoint.

Now, how to approach such problems where it seems that mere visualization is needed but which apparently doesn't work (at least for me)?

Usually one can trust and straightforwardly use Lorentz transform eqns. to give and get the correct answer. Developing visual intuition for relativistic scenarios isn't easy. Even Minkowski diagrams have a element of math to them. To see the clearest non-mathematical arguing of expected relativistic consequences in simple cases, I recommend a reread of Einstein's insightful arguments in his $1905$ STR paper.

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$^1$ for this argument, the moving observer must be so positioned that if it weren't moving, the falling of the rod's ends would appear simultaneous.