Clocks Within Each Ship in Bell Spaceship Paradox

[A.T.]

Do you have an opinion regarding why there was so much confusion regarding Bell's Paradox, even among experts?

I only know that laymen often misunderstand length contraction as comparison between two time points (before and after acceleration), while in fact it compares two different frames. So they fail to see that keeping constant distance in the initial rest frame, implies stretching in the current rest frame.

To me the most intuitive explanation is replacing the string with a chain. The rigid chain links will get shorter in the initial frame, so they cannot span the same distance anymore.

In the case of a more solid string, the links are individual atoms which contract, or their EM fields which connect them to each other.

 

[Mentz114]

I only know that laymen often misunderstand length contraction as comparison between two time points (before and after acceleration), while in fact it compares two different frames. So they fail to see that keeping constant distance in the initial rest frame, implies stretching in the current rest frame.

In the case of a more solid string, the links are individual atoms which contract, or their EM fields which connect them to each other.

I don't think that is possible. Any apparent contraction in length is offset by a change in the internal potentials caused by time dilation. The atoms would remain in unstretched equilibrium.

Can you show equations ?

 

[A.T.]

I don't think that is possible. Any apparent contraction in length is offset by a change in the internal potentials caused by time dilation. The atoms would remain in unstretched equilibrium.

If the distances between the atoms do not change, but forces between them increase, then their potentials must be have been contracted. How else would you explain the increasing stresses, based solely on the initial rest frame?

 

[Mentz114]

If the distances between the atoms do not change, but forces between them increase, then their potentials must be have been contracted. How else would you explain the increasing stresses, based solely on the initial rest frame?

Nothing I've written disproves what you suggest so that remains open and I'm not going to attempt to pursuade you in someone else's topic.

My position is that there is nothing to explain so it's irrelevant anyway. No disrespect intended.

 

[stevendaryl]

OK. My point is that frames are irrelevant in the sense that we need only find a frame independent analysis. The problem can be expressed in a frame independent way if can find a Lorentz scalar that tells us if the ships are getting further apart. There is a tensor, $\theta_{ab}=\partial_a u_b$ which has a contraction $\theta={\theta^a}_a$ which tells us if the rockets move apart or get closer. For the case in question it is $\gamma^3 \partial_t\beta=\gamma^3\ a$. The ships are moving apart.

I'm sorry, did you already define $\theta$? The way you've defined it seems to treat $u_b$ as a vector field, but how is that defined?

I tried to come up with a covariant formulation of the problem, and it seemed sort of complicated. Here are my thoughts on this:

Suppose you just want to characterize how stretched a string is, in a covariant way. Here's my approach: We label a point along the string connecting the rockets with a number $\lambda$, which is the distance along the string from one end to that point. Then we could define $\mathcal{P}(\lambda, \tau)$ to be the event that the point labeled $\lambda$ passes through at proper time $\tau$.

In terms of $\mathcal{P}(\lambda, \tau)$ we can define two partial derivatives:

$D^\mu = \frac{\partial}{\partial \lambda} \mathcal{P}$
$U^\mu = \frac{\partial}{\partial \tau} \mathcal{P}$

$U^\mu$ is just the 4-velocity of the point on the string labeled $\lambda$

So if we start at $\lambda, \tau$ and consider a nearby piece of the string at $\lambda + \delta \lambda, \tau + \delta \tau$, the separation will be, to first order:

$S^\mu = D^\mu \delta \lambda + U^\mu \delta \tau$

Now, here's the tricky part, it seems to me. If we want to know the spatial separation between two nearby pieces of string, that means that we have to choose $\delta \tau$ so that $S^\mu$ is purely spatial, in the local comoving rest frame of the string. That means that

$U^\mu S_\mu = 0$

because the 4-velocity is purely temporal, in that comoving rest frame. So that implies that:

$U^\mu D_\mu \delta \lambda + U^\mu U_\mu \delta \tau = 0$

Since $U^\mu U_\mu = 1$ for any object (in units with c=1), it follows that:

$\delta \tau = - \delta \lambda \ U^\mu D_\mu$

This is the value of $\delta \tau$ that makes $\mathcal{P}(\lambda, \tau)$ and $\mathcal{P}(\lambda + \delta \lambda, \tau + \delta \tau)$ simultaneous, in the comoving rest frame of the string at label $\lambda$. So for this value of $\delta \tau$, the separation between the nearby pieces of string is given by:

$S^\mu = \delta \lambda (D^\mu - (U \cdot D) U^\mu)$

If there were no stretching or compression, then the separation between the two points would be of magnitude $\delta \lambda$. So a measure of the stretching or compression is the factor:

$Q = - \frac{S_\mu S^\mu}{\delta \lambda^2}$ (the minus sign is because the square of a spatial vector is negative, in my convention)

If there is no stretching or compression, then $Q = 1$. If $Q > 1$, that means the string is being stretched. If $Q < 1$, that means the string is being compressed.

So writing it out,

$Q = - (D^\mu D_\mu - (D^\mu U_\mu)^2)$ (where I again used $U^\mu U_\mu = 1$)

Just as a check, if every part of the string is at rest in Rindler coordinates, then that means

$x(\lambda, \tau) = \lambda cosh(\frac{\tau}{\lambda})$
$t(\lambda, \tau) = \lambda sinh(\frac{\tau}{\lambda})$

(The usual Rindler coordinates use $X, T$ instead of $\lambda, \tau$. The relationship is just $\lambda = X$, $\tau = \lambda gT$)

then $Q = 1$ (I'm skipping the proof). So if the string is at rest in Rindler coordinates, then it is unstretched.

On the other hand, if every part of the string accelerates together at the rate, then:

$x(\lambda, \tau) = \lambda + f(\tau)$
$t(\lambda, \tau) = g(\tau)$ (independent of $\lambda$.

Then

$Q = 1 + U^2$

where $U = \frac{dx}{d\tau}$

So $Q > 1$. So if the points on the string are undergoing simultaneous acceleration, then the string stretches.

 

[Mentz114]

I'm sorry, did you already define θ\theta? The way you've defined it seems to treat ubu_b as a vector field, but how is that defined?

I tried to come up with a covariant formulation of the problem, and it seemed sort of complicated. Here are my thoughts on this:

Yes, $u$ is a congruence (time-like vector field). I skipped a lot of detail because I wanted to keep it simple. It is fully covariant. It is done in the global frame basis where partial differentiation is covariant. I can repeat it in the local frame and (of course) get the same result.

It'll take time to understand your workings above.

 

[PeterDonis]

did you already define $\theta$? The way you've defined it seems to treat $u_b$ as a vector field, but how is that defined?

$\theta$ is just the expansion scalar; it's part of the kinematic decomposition of the timelike congruence $u_b$. See here:

http://en.wikipedia.org/wiki/Congru...atical_decomposition_of_a_timelike_congruence

What you are doing in the rest of your post is basically the same thing, just with different notation. The key is that the kinematic decomposition is frame-independent; you can compute it in any frame you like, but the scalars derived from it, such as the expansion scalar, are invariant. So the expansion scalar of the Bell congruence being positive is a frame-invariant way of saying that the string stretches.

 

[harrylin]

[..] The fact that people are using comoving frames to decide questions of physics and not getting clear answers makes me think those methods are very subtle and difficult to use. So, complicated and not needed for understanding. [..].

That was exactly Bell's point! He simply stayed in S to obtain the answer and while that's perhaps a bit subtle, his method is not difficult or complicated. And it looks much simpler to me than your method.

 

[Mentz114]

That was exactly Bell's point! He simply stayed in S to obtain the answer and while that's perhaps a bit subtle, his method is not difficult or complicated. And it looks much simpler to me than your method.

I'm not sure what you're calling 'you're method'. If you mean the expansion scalar, then I cannot take credit for that. It's a standard kinematic decomposition found in textbooks.

As for the 'explanations' why an apparently unstretched string snaps they all seem to be flawed because there is no way to define (unambiguously) a time-independent length of an object if the clocks along its length cannot be synchronised. So any object in this condition cannot, by definition, have a constant length because the length is different at every measurement ( I think stevendaryls calculatation reinforces this).

Using some bogus definition of distance it is easy to conclude that there is something to explain - but there isn't.

 

[A.T.]

Using some bogus definition of distance it is easy to conclude that there is something to explain - but there isn't.

What do you mean by "bogus "? In the initial rest frame you have a ruler at rest and synchronized clocks placed along it. You can easily measure where the ends of the string are at any time point, and thus what the length of the string is, according to that frame.

 

[harrylin]

[..] As for the 'explanations' why an apparently unstretched string snaps they all seem to be flawed [...].

Well of course, we all know the string will increasingly come under strain. But when Bell first presented this example, many people at CERN gave the wrong answer ("There emerged a clear consensus that the thread would not break!"). That was almost certainly due to wrong and perhaps overly complicated reasoning. In contrast, Bell's physical analysis is as simple and easy as it can get, and it gives the right answer. Therefore he included it in his paper "How to teach relativity".

 

[Mentz114]

What do you mean by "bogus "? In the initial rest frame you have a ruler at rest and synchronized clocks placed along it. You can easily measure where the ends of the string are at any time point, and thus what the length of the string is, according to that frame.

But the clocks on the ends of the ruler are not synchronised with the string clocks. So measuring them simultaneously ( in S) is not simultaneous in the ship frame.

I think StevenDaryls approach overcomes this, at the cost of producing a length measure which cannot remain the same. He works out a frame independent length which is based on proper time.

 

[A.T.]

So measuring them simultaneously ( in S) is not simultaneous in the ship frame.

Yeah, but it gives you a perfectly valid length measurement in S. I don't understand what is "bogus" about it.

 

[harrylin]

But the clocks on the ends of the ruler are not synchronised with the string clocks. So measuring them simultaneously ( in S) is not simultaneous in the ship frame. [..]

And there is absolutely no need to look at what is simultaneous in S'.

Bell's approach is quite similar to Einstein's approach when he made the first clocks prediction: first he worked out, based on the Lorentz transformations, what according to S the rate is of a clock in motion. That yielded a new law of physics about moving clocks. Based on physical reasoning he then predicted the retardation of a clock that moves at constant speed in a circle, without needing any more Lorentz transformations.
(once more: §4 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ )

Imagine that we next slightly complicate Einstein's example, in the spirit of the title of this thread, by having two identical clocks that are located at a distance from each other depart simultaneously, with identical acceleration profiles.
Following Einstein and Bell, it "is at once apparent" that they will remain in synch with each other according to clocks at rest in S; and that they will delay, compared to clocks in S, by the same amount.
In contrast, if one feels the need to continuously transform to instantaneous rest frames S' and S''(etc) of the moving clocks and next compare the moving clocks with each other from consecutive S' and S'', a complicated calculation follows from which one learns nothing new and that yields the same result - if no error is made.

 

[Mentz114]

Well of course, we all know the string will increasingly come under strain. But when Bell first presented this example, many people at CERN gave the wrong answer ("There emerged a clear consensus that the thread would not break!"). That was almost certainly due to wrong and perhaps overly complicated reasoning. In contrast, Bell's physical analysis is as simple and easy as it can get, and it gives the right answer. Therefore he included it in his paper "How to teach relativity".

And there is absolutely no need to look at what is simultaneous in S'.
...

Thanks. So the ship clocks are out of synch by the same amount with the ruler clocks. I just want to summarise the gist of the 'explanation'.

In the stationary frame basis every ship has the same velocity and acceleration, and the gap between them remains the same. But the string gets tighter and breaks.
So a force field of the EM type is proposed that has this effect on matter. Now the force field must increase so it cannot depend on the acceleration which is constant. It must therefore depend only on the velocity of the ships as measured in the ground frame. Now suppose the acceleration is zero - the same conditions hold as previously.

Why does the string not snap ?

(This is a bit obvious so I must have overlooked something ).

[edit]
The 'effect' obviously can depend on the magnitude of the acceleration. Getting late here. Apologies.

 

[A.T.]

In the stationary frame basis every ship has the same velocity and acceleration, and the gap between them remains the same. But the string gets tighter and breaks.

Yes, since the length didn't change, the only explanation why the stress increases is that the stress-free-length, that the string tries to reach has changed.

So a force field of the EM type is proposed that has this effect on matter.

The interactions between atoms are EM.

Now the force field must increase...

Not just increase everywhere, but contract along the direction of motion of its source. This can lead to increased forces along that direction, even if the interacting atoms keep a constant distance.

It must therefore depend only on the velocity

Yes, length contraction depends on velocity.

 

[harrylin]

Thanks. So the ship clocks are out of synch by the same amount with the ruler clocks. I just want to summarise the gist of the 'explanation'.

In the stationary frame basis every ship has the same velocity and acceleration, and the gap between them remains the same. But the string gets tighter and breaks.
So a force field of the EM type is proposed that has this effect on matter. [..]

That's perhaps a bit putting the cart before the horse. If atomic bonds are electromagnetic, then EM laws should make predictions that agree with the relativity principle. And apparently this is the case with Maxwell's laws (it's not simple though).

From the Lorentz transformations, Einstein (re-)derived the following laws of physics:
- a clock that is brought in motion will tick slower than in rest by the factor gamma (ceteris paribus)
- a ruler that is brought in motion will contract by the factor gamma (ceteris paribus)

Those laws should not be confounded with the Lorentz transformations. But they look very similar, and "length contraction" and "time dilation" similarly can have two meanings. It is increasingly my opinion that lack of distinction between the Lorentz transformations and the new laws of nature that followed from them is the main cause of why "paradoxes" with clocks and rulers continue to bug people.

To state clearly what is often brushed over:

1. SR's Lorentz transformations. Positions and distances as well as times and time periods as measured in different inertial frames are mapped to each other by means of the Lorentz transformations. This is valid between two inertial reference systems (with "Einstein synchronization").

2. Some of SR's laws of nature. The frequencies and lengths of objects (clocks and rulers) depend on their speed. This is valid for objects in any state of motion in a single inertial reference system (by a single "observer"). Of course, the effects of other physical influences must also be taken in account.

BTW the use of the "observer" short-hand only helps to forget that it does not mean objects or people. Even "frame" can be misleading in a similar way.

When mixing up these very different but similar looking notions, one could for example wrongly think that:

- the Lorentz transformations may be used for the accumulated time of an inertial clock according to an "observer" in any state of motion - the modern version of the "twin paradox".
- a non-connected system of accelerating rockets will contract as a whole - the Bell spaceship paradox.
- an accelerating system with clocks will naturally maintain Einstein synchronization (posts 1 and 8).

PS. question to 1977ub: was something like that also the reason for your confusion, or was there a different cause?

 

[bligh]

I like the thought exper of two clocks, one stays, the other acclerated to the front. stays one year and then acclerated to the rear.
But, this non stationary clock would have had it's time "dilated" during two accelerations and would therefore show less time elapse, not more as you
suggested.
Correct?

 

[bligh]

although time dilatation in accelerated clocks is well accepted, no one has shown length contraction in real life.
Correct?

 

[Ibix]

although time dilatation in accelerated clocks is well accepted, no one has shown length contraction in real life.
Correct?

Depends what you mean. There is no way to explain the cosmic ray muons without length contraction, and the explanation for the motion of a charge near a current carrying wire needs it, too.

However, no one has ever shut a 1m pole into a 0.5m barn, if that's what you mean.

 

[Stalin Beltran]

Yes the initial conditions are as described in Wikipedia.

PS. if you find that the string doesn't break, then it may be a good idea to present your calculation here for debugging. Comparing the situations at t₀=0 and t₁ >>0 suffices for the analysis.

Of course. I have it published it in my blog (It doesn't mean it is correct, it is just my analysis).
Bell’s spaceship thread does not breaks
The analysis basically goes like this:

Step 0: We have 2 spaceships, A and B, in the same frame of reference that a spaceship S. The rocket motors are about to start, but not yet. A, B and S are "simultaneous", so if motors start, they will start simultaneous to S too.

Step 1: A and B starts motors with exactly the same acceleration for a $\Delta t$ as short as we want. (A, B and S share the same frame of reference when motors start). This results in a final velocity $V_1$ for A and B because both the acceleration and the $\Delta t$'s are equal. At the end of this $\Delta t$ both A and B still share the same frame of reference because they both have the same velocity $V_1$. However S remains at rest, so S is now in another frame of reference. The relativity of simultaneity states that, at the end of this step, both A and B are not longer "simultaneous" to S, so saying that the accelerations are equal in the initial rest frame S is not possible at the end of this step. A and B are in the same frame of reference, so if they accelerate, they both will accelerate at the same time and in the same proportion.

This could be extended to N steps, and the results are the same. A and B being in the same frame of reference means they are at rest relative to each other. The distance between them will stay the same, so the string will not break.

 

[Nugatory]

Of course. I have it published it in my blog (It doesn't mean it is correct, it is just my analysis).
Bell’s spaceship thread does not breaks
The analysis basically goes like this:

Step 0: We have 2 spaceships, A and B, in the same frame of reference that a spaceship S. The rocket motors are about to start, but not yet. A, B and S are "simultaneous", so if motors start, they will start simultaneous to S too.

Step 1: A and B starts motors with exactly the same acceleration for a $\Delta t$ as short as we want. (A, B and S share the same frame of reference when motors start). This results in a final velocity $V_1$ for A and B because both the acceleration and the $\Delta t$'s are equal. At the end of this $\Delta t$ both A and B still share the same frame of reference because they both have the same velocity $V_1$. However S remains at rest, so S is now in another frame of reference. The relativity of simultaneity states that, at the end of this step, both A and B are not longer "simultaneous" to S, so saying that the accelerations are equal in the initial rest frame S is not possible at the end of this step. A and B are in the same frame of reference, so if they accelerate, they both will accelerate at the same time and in the same proportion.

This could be extended to N steps, and the results are the same. A and B being in the same frame of reference means they are at rest relative to each other. The distance between them will stay the same, so the string will not break.

There is no such thing as "being in a frame of reference", so any analysis that casually tosses that phrase around is suspect. The reference frame is simply a convention used for assigning x and t coordinates to events, so any reference frame can always be applied to everything in a given problem.

There is one reference frame in which the two spaceships are at rest relative to one another (meaning that $x_1(t)-x_2(t)$ is a constant where $x_1(2)$ and $x_2(t)$ are the x-coordinates of the leading and trailing spaceships at time $t$ using the $x$ and $t$ coordinates assigned by that frame). That is the frame in which the ground-based observer is at rest ($x_g(t)$ is constant for all $t$ where $x_g(t)$ is the x-coordinate of the ground-based observer's location at time t). In that frame the string is never at rest, so it length-contracts and breaks. In any other frame, the distance between the two ships $x_1(t)-x_2(t)$ (using the $x$ and $t$ coordinates of that frame) increases with time so the string breaks.

 

[harrylin]

I like the thought exper of two clocks, one stays, the other acclerated to the front. stays one year and then acclerated to the rear.
But, this non stationary clock would have had it's time "dilated" during two accelerations and would therefore show less time elapse, not more as you
suggested.
Correct?

Perhaps you are referring to post #13 by stevendaryl. He did not discuss a rocket that is in rest for one year, but a rocket that is accelerating for one year!

The two-way clock transport inside the rocket results in a very slight retardation of the transported clock on the rear clock. If that experiment was done in rest, the one year in the front would be totally irrelevant. And it is easy to understand that also in an accelerating rocket whatever the effect of this transport may be (that requires deeper analysis), it does not depend on the time that the clock is staying in the front of the rocket. Now consider what happens during the year of acceleration. Due to length contraction of the rocket, the front clock will at any instant be going at a very slightly lesser speed than the rear clock, so that during that time the rear clock delays on the front clock. The longer the transported clock stays in front, the greater the delay of the rear clock will be.

 

[harrylin]

[..] A and B are in the same frame of reference, so if they accelerate, they both will accelerate at the same time and in the same proportion. [..] A and B being in the same frame of reference means they are at rest relative to each other.

Coincidentally, I had just before your post identified the interpretation error behind those phrases!
See my post #87.

Proportional mapping between inertial reference frames was assumed for deriving the Lorentz transformations, and inertial A and B in rest relative to each other can according to SR be used to set up an inertial frame for describing natural phenomena. It is the misapplication of such assumptions to mapping of an inertial frame with an accelerating frame that leads to paradoxes of this kind.

 

[Stalin Beltran]

In any other frame, the distance between the two ships x1(t)−x2(t)x_1(t)-x_2(t) (using the xx and tt coordinates of that frame) increases with time so the string breaks.

I can see the frame where the spaceships are at rest. I can see the frame where the string contract. But I just don't see the frame where the distances between spaceships increases. Could you elaborate this please?

 

[Dale]

although time dilatation in accelerated clocks is well accepted, no one has shown length contraction in real life.
Correct?

Lorentz seemed to think that Michelson and Morely showed length contraction in real life. Personally, I think there is a lot of evidence showing length contraction in real life, from MM, to muons, to particle bunches in accelerators, to magnetism.

 

[1977ub]

PS. question to 1977ub: was something like that also the reason for your confusion, or was there a different cause?

Not exactly. But the confusion re laws-of-physics vs lorentz contraction is interesting part of this.

 

[Stalin Beltran]

Lorentz seemed to think that Michelson and Morely showed length contraction in real life. Personally, I think there is a lot of evidence showing length contraction in real life, from MM, to muons, to particle bunches in accelerators, to magnetism.

As far as I understand, length contraction is the way an observer traveling with speed V sees the world in the direction of movement. For example, a muon doesn't notice its clock had slow down. Instead it see its distance to Earth has contracted, and that is the way it reaches the Earth's surface.

Am I right?

 

[Nugatory]

I can see the frame where the spaceships are at rest. I can see the frame where the string contract. But I just don't see the frame where the distances between spaceships increases. Could you elaborate this please?

There is no frame in which the spaceships are both at rest (except the ground frame before and up to the moment that they light off their engines). They remain at rest relative to another (that is, moving in the same direction at the same speed) in the ground frame but of course they are not at rest in that frame once the engines have fired to start them moving. In all other frames, the ships are moving relative to one another.

 

[Dale]

As far as I understand, length contraction is the way an observer traveling with speed V sees the world in the direction of movement. For example, a muon doesn't notice its clock had slow down. Instead it see its distance to Earth has contracted, and that is the way it reaches the Earth's surface.

Am I right?

Yes. Although I am not a fan of the "observer" phrasing. I prefer to simply speak of reference frames and not insist on attaching them to observers.

 

[Stalin Beltran]

To complete the description about length contraction, we must say the length of an object is not contracted (as we usually tend to think) if you take a photograph of it. As stated in wikipedia:

"It was shown by several authors such as Roger Penrose and James Terrell that moving objects generally do not appear length contracted on a photograph. For instance, for a small angular diameter, a moving sphere remains circular and is rotated. This kind of visual rotation effect is called Penrose-Terrell rotation."

 

[Dale]

Really? That is the big "proof" you think is missing? What then, photos of different size objects and different shape objects?

If you are determined to reject SR regardless of the evidence then you can always find some experiment that was not performed and claim incomplete evidence. However, whatever missing experiments you choose to complain about, you are still left with the challenge of explaining all of the previous experiments that have been performed:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[1977ub]

To complete the description about length contraction, we must say the length of an object is not contracted (as we usually tend to think) if you take a photograph of it. As stated in wikipedia:

"It was shown by several authors such as Roger Penrose and James Terrell that moving objects generally do not appear length contracted on a photograph. For instance, for a small angular diameter, a moving sphere remains circular and is rotated. This kind of visual rotation effect is called Penrose-Terrell rotation."

Length contraction and time dilation are not directly perceived but are reconstructed ala forensic analysis and require that you are satisfied that there are clocks remote from the observer which are synchronized with his clock. A photograph does not presume to take measurements of different parts of an object simultaneously in one's own frame - and so doesn't presume to make any statement about actual lengths - contracted or uncontracted.

 

[Stalin Beltran]

Really? That is the big "proof" you think is missing? What then, photos of different size objects and different shape objects?

If you are determined to reject SR regardless of the evidence then you can always find some experiment that was not performed and claim incomplete evidence. However, whatever missing experiments you choose to complain about, you are still left with the challenge of explaining all of the previous experiments that have been performed:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Proof against Bell Paradox? Not at all. If I try to reject Bell Paradox saying that length contraction is not real that will be circular reasoning. The last post was to complete the idea of length contraction asked by bligh in post 89.

What my analysis reveals about Bell Paradox is that A and B could not accelerate simultaneously to frame S. Please note that if we found (like Bell did) that A and B are accelerating simultaneously in the S frame, then Bell is right and the string breaks.

 

[Stalin Beltran]

Length contraction and time dilation are not directly perceived but are reconstructed ala forensic analysis and require that you are satisfied that there are clocks remote from the observer which are synchronized with his clock. A photograph does not presume to take measurements of different parts of an object simultaneously in one's own frame - and so doesn't presume to make any statement about actual lengths - contracted or uncontracted.

To be honest I have not checked the Roger Penrose paper (I am looking for other facts of relativity), but It sounds good. While studying the relativity of simultaneity I found that length contraction could be an "artifact" of the relativity of simultaneity. That is, the object could actually no contract, but look contracted when we measure it.