Realtivity of Simultaneity Paradox - take #1749

roosh

Hey, it's me again, the boy who cried wolf only this time I swear it's a wolf

I wrote an essay on an alternative interpretation to Einstein's Special Relativity. The essay is a bit on the long side and incorporates many of the ideas that have been discussed on this forum such as, time as a dimension of the universe and the "Block Universe" structure. Included in the essay is [yet another] paradox that arises as a result of the concept of Relativity of Simultaneity - in particular as it applies to the Einstein Clock Synchronisation Convention.

As some of you may know, it's not the first time I've "discovered" a paradox inherent in relativity, I've "found" loads, only to have some helpful individuals [one in particular] point out the error in my reasoning. I had gotten to the stage where I was able to self-correct, but on this occasion I haven't been able to. Perhaps it's because I haven't been actively engaged in thinking about it for a while. It could be that it's a genuine paradox, but if I were a betting man....

So, I turn to the helpful oracle that is boards. Can you help me identify the error [if any] in my reasoning? I have the details saved as a pdf on my google drive because I'm having to do all of this on my phone and the essay contains a number of pictures.
https://drive.google.com/file/d/1OIGFUtIPPHTJyqRWeyO5zVnlJVGV79HS/view?usp=sharing

Essentially, it is Einstein's thought experiment with 2 relatively moving observers (train and platform), including the clock synchronisation set-up (light pulse mid-way between the two clocks).

Any feedback would be greatly welcomed.

If you want to see the whole essay for context, you can find it here:
https://drive.google.com/file/d/1zDNHs76Qk3av7nzhqDrxif9JG2NDk22b/view?usp=drivesdk

Bit cynical

I think perhaps the problem is in the last paragraph where you say that in the case of the domino cascade, addition of velocities does apply.

This is wrong. Even though we're not talking about light, addition of velocities does not apply to the dominoes. What happens is that the velocity is transformed according to the Lorentz transformation such that even though, from the point of view of the stationary observer, the dominoes start at different times, the velocities (from the same stationary frame) are such that the cascades meet in the middle at the same time.

The fact that they meet in the middle as viewed by the stationary observer means that the same number of dominoes will be toppled in each direction.

roosh

Bit cynical wrote: »

I think perhaps the problem is in the last paragraph where you say that in the case of the domino cascade, addition of velocities does apply.

This is wrong. Even though we're not talking about light, addition of velocities does not apply to the dominoes. What happens is that the velocity is transformed according to the Lorentz transformation such that even though, from the point of view of the stationary observer, the dominoes start at different times, the velocities (from the same stationary frame) are such that the cascades meet in the middle at the same time.

The fact that they meet in the middle as viewed by the stationary observer means that the same number of dominoes will be toppled in each direction.

Ah right, that sounds all too familiar now! I should have remembered that, sorry! I thought it was always light signals I had used before, although recall using sound signals in one but I can't remember if it was the same "problem". I just hope I didn't use dominoes in an argument bcos then my memory is on the way out

I must look up the transformation of velocities again, thanks BC

roosh

Bit cynical wrote: »

I think perhaps the problem is in the last paragraph where you say that in the case of the domino cascade, addition of velocities does apply.

This is wrong. Even though we're not talking about light, addition of velocities does not apply to the dominoes. What happens is that the velocity is transformed according to the Lorentz transformation such that even though, from the point of view of the stationary observer, the dominoes start at different times, the velocities (from the same stationary frame) are such that the cascades meet in the middle at the same time.

The fact that they meet in the middle as viewed by the stationary observer means that the same number of dominoes will be toppled in each direction.

Just thinking about the question more and I think I understand it better now. I tried simplifying the set up to just 3 dominoes - one at either clock and one in the middle.

As the dominoes start falling they are no longer "part" of the inertial frame of reference of/at rest to the train. So, as the domino from the rear of the train starts moving towards the mid-point, its velocity changes from VTrain to Vtrain+Vdomino. If we imagine a light clock with the domino then with the increase in velocity, as observed from the platform, the photon in the dominoes light clock will trace an even longer path than the light clock at rest relative to the train. The domino at the front of the train will have a new velocity of Vtrain - Vdomino. This will mean that the photon in its frame of reference will trace a shorter path and time will speed up for it, relative to the train and the other domino. This, in effect, means the observer on the platform will see the domino at the rear of the train fall at a slower velocity than that of the domino from the front.

Would that be correct?

Morbert

As the dominoes start falling they are no longer "part" of the inertial frame of reference of/at rest to the train. So, as the domino from the rear of the train starts moving towards the mid-point, its velocity changes from VTrain to Vtrain+Vdomino. If we imagine a light clock with the domino then with the increase in velocity, as observed from the platform, the photon in the dominoes light clock will trace an even longer path than the light clock at rest relative to the train. The domino at the front of the train will have a new velocity of Vtrain - Vdomino. This will mean that the photon in its frame of reference will trace a shorter path and time will speed up for it, relative to the train and the other domino. This, in effect, means the observer on the platform will see the domino at the rear of the train fall at a slower velocity than that of the domino from the front.

Would that be correct?

You have the right conclusion. The rate at which dominoes fall would be different for the left-hand and right-hand sides, and would be slower for the left-hand side.

But note that this conclusion doesn't depend on the velocity of any domino. Even if we model a falling domino as a point-like event, the conclusion would hold.

More generally: All reference frames will agree about which events are coincident. If two rows of collapsing dominoes meet at the same place and time according to one reference frame, they will do so according to all reference frames.

roosh

Morbert wrote: »

You have the right conclusion. The rate at which dominoes fall would be different for the left-hand and right-hand sides, and would be slower for the left-hand side.

But note that this conclusion doesn't depend on the velocity of any domino. Even if we model a falling domino as a point-like event, the conclusion would hold.

More generally: All reference frames will agree about which events are coincident. If two rows of collapsing dominoes meet at the same place and time according to one reference frame, they will do so according to all reference frames.

Thanks Morbert, I remember you explaining that before but I must be forgetting a part of the explanation. I can't remember why it is the case that he will measure a difference in rate of the falling dominoes.

In my mind I'm picturing the domino from the rear of the train falling first and the domino at the front of the train starting to fall after that. I was assuming that he must measure/calculate the velocity of each domino, *relative to the table/train*, as being the same, because the dominos are in motion along with the train. That is, prior to the dominoes falling, the observer on the platform would calculate/measure the velocity of the dominoes relative to the train as being 0, but when they start to fall they will have a positive velocity relative to the train, as measured from the frame of reference on the platform.

What is it that causes the velocity of the left domino to be measured as faster than the other, relative to the train, from the referenceframe of the platform?

EDIT: My sticking point might become clearer if I deconstruct the problem and explain how I am looking at things. If we ignore the issue of simultaneity for the moment and just consider things from the perspective of the observer on the platform.

They see the train passing at a suitable [for the purposes of the question] fraction of the speed of light. On the train are a set of dominoes. The dominoes are knocked from left to right only. The observer on the platform measures the velocity of the dominoes as Vtrain + VDominoes.

Now, if we consider the same scenario, but this time the dominoes are knocked from right to left. The observer on the platform will measure the velocity of the dominoes as Vtrain - VDominoes.

Here, there is no need to perform a Lorentz Transformation of velocities bcos all the measurements are done in the frame of reference of the observer on the platform.


I was expecting the velocity of the Dominoes (VDominoes) to be the same just with the signs reversed. Why is this not the case?

2nd EDIT: Is it because the dominos moving from left to right contract more while those going in the opposite direction "uncontract"?

Bit cynical

roosh wrote: »

They see the train passing at a suitable [for the purposes of the question] fraction of the speed of light. On the train are a set of dominoes. The dominoes are knocked from left to right only. The observer on the platform measures the velocity of the dominoes as Vtrain + VDominoes.

Now, if we consider the same scenario, but this time the dominoes are knocked from right to left. The observer on the platform will measure the velocity of the dominoes as Vtrain - VDominoes.

Here, there is no need to perform a Lorentz Transformation of velocities bcos all the measurements are done in the frame of reference of the observer on the platform.

I was expecting the velocity of the Dominoes (VDominoes) to be the same just with the signs reversed. Why is this not the case?

2nd EDIT: Is it because the dominos moving from left to right contract more while those going in the opposite direction "uncontract"?

There's also the fact that the forces themselves transform. So while the forces with the frame moving with the train is symmetrical, the forces as viewed from the stationary frame are not.

http://www.sciencebits.com/Transformation-Forces-Relativity

So the whole analysis can be done from the point of view of the stationary frame. We can imagine that the dominoes at each end are toppled at different times. The forces on the leading dominoes are different to those on the trailing dominoes and therefore the velocities of the toppling are different but are such that toppling meets at the middle of the moving train.

Although this would be complicated to model, it would still be plausible physics from the point of view of the stationary observer. He would simply see different forces at work in the leading and trailing dominoes.

Having done the analysis, if we then switch the view to the frame of the moving train, we see the dominoes launched at the same time with equal velocities towards the centre, meeting at the middle.

roosh

A separate question I was wondering about.

In the light clock thought experiment, if we introduce a 2nd light clock oriented at a right angle to the first i.e. one clock is oriented perpendicular to the x-axis, while the other is oriented perpendicular to the y-axis - i.e. one perpendicular to the direction of motion and one in-line with the direction of motion.

Shouldn't the observer on the platform see the "time" speed up and then slowdown in the clock that is perpendicular to the y-axis while time only slows down for the clock that is perpendicular to the direction of motion.

Also, if the clocks are set up so that 2 of the mirrors are side-by-side i.e. the top mirror of the perpendicular-to-the x-axis-clock is beside the left mirror of the perpendicular-to-the y-axis-clock, and we imagine the photon in each clock starting from the opposite mirror. Add to this that the clocks makes a click every time it reflects off a mirror.

In the rest frame of the clocks, the photons in each clock would arrive at the opposite mirror simultaneously combining to make one louder click, while according to the observer on the platform, the photons should not arrive in sync and should therefore make 2 distinct clicks.

Is that correct?