Dear Ziaedin,

ZS: To make the problem easier suppose the OMG particle is approaching another particle. The observer in the frame of OMG particle can assume its own frame is standing still and the other particle is approaching it. In that case it takes one year for these two particles to meet.

Slow down, what event are you timing that from? We know when they meet but you have no starting point to define a duration.

ZS: But the same observer in the frame of OMG particle can also assume the OMG particle is speeding towards the other particle and thus the distance between them vanishes to zero which consequently reduces the time of travel to zero. The same two arguments can be repeated in the frame of other particle.

Yes, or in any other frame you choose, but if you want to say "it takes one year for these two particles to meet" you need an event at each end of that period, not just the meeting.

What you are saying is a completely standard aspect ant not a problem in any way, I would recommend you look up something called the "Parable of the Surveyors" to understand this better. I think it was originated by John heeler but you'll find several versions of it on the web.

Ziedin,

In which reference frame do you want the answer?

(Earth-frame or particle frame)

I think that you've described both outcomes quite ably - and so I am confused as to why you are asking.

Dear James

From any frame, say, the destination frame.

Please note that I asked this question in reference to the conclusion proposed in the linked paper/experiment at the end of my question. It is accepted by relativistic physicists as a proof for time dilation and length contraction. I wanted to highlight the consequences and contradictions between the logic, which was followed in the paper to prove special relativity, and our experimental data for light travel.

Ziedin,

In *any* frame? Then the answer might be any value!

If in the destination frame (in which the particle has a very high speed) then "almost one year"

Sorry James, I meant from either side (any of the two).

What is the answer from the frame of the particle?

Ziaedin,

The particle experiences a vanishingly brief duration between departure and arrival.

Thanks James

The problem is clear now. We have two answers as you confirmed, zero and one year. Both answers are given by the observer in its own frame. Let us rename the particle and the destination as Alice and Bob. According to special relativity both of them in their own frames should record the same time which is not what you confirmed.

The answer can be either of the above depending on how they interpret the scenario. If Alice assumes she is in fact stationary and Bob is moving towards her then the expected time of arrival is one year. However, if she thinks that she is moving with the same speed towards Bob then according to the theory of length contraction the whole distance between her and Bob reduces to zero and she gets to Bob in no time.

As this issue is a little bit elusive, one can switch the argument as one wishes to give either of the two results without being exposed. This fallacy found in the experiment I referred to at the end of my question, playing with both time dilation and length contraction at will, and the paper is not the only one.

Both times are correct but in the frame of the particle, the distance is reduced by length contraction so we see the particle approaching at nearly c, the particle sees us approaching at the same speed. The whole thing is just a geometric projection.

Dear George

I think I have given a response to your comment in my last answer. An observer in a frame is given the liberty by special relativity to assume he is moving and give zero as his travel time by reference to length contraction. But he can give one year if he assumes being stationary and the second frame is approaching him. In OMG case this is clear but in papers such as Frisch and Smith this fallacy can be hidden.

Dear Ziaedin,

ZS: An observer in a frame is given the liberty by special relativity to assume he is moving

No, that is not correct, motion is relative to the frame but "the observer" defines the frame so it would mean "the observer" was moving relative to "the observer" which is obviously a contradiction.

What SR says is that in your example we can define a frame K using the Earth to define the origin of the coordinates and we can define a different frame K' using the particle to define the origin of the coordinates. In K, the Earth is not moving while the particle moves at high speed for a year and travels a distance of nearly a light year. In K', the particle is not moving while the Earth moves at high speed and if you watched it for a year, it would travel a distance of nearly a light year. If you assume the particle was created one light year distant from Earth in frame K however, it would only travel a short distance in a short time in frame K' between creation and detection. The reality of the existence of the particle is the same, only the coordinates allocated vary with the choice of frame.

Dear George

I deliberately did not mention the earth in the question to make the problem easier to see. Please remember that according to special relativity there is no universal frame of reference. But we still have the incorrect habit to assume the earth as the universal fixed frame when we are considering the movement of particles in special relativity.

To make the problem easier suppose the OMG particle is approaching another particle. The observer in the frame of OMG particle can assume its own frame is standing still and the other particle is approaching it. In that case it takes one year for these two particles to meet. But the same observer in the frame of OMG particle can also assume the OMG particle is speeding towards the other particle and thus the distance between them vanishes to zero which consequently reduces the time of travel to zero. The same two arguments can be repeated in the frame of other particle.

My main point for asking this question is to show that this trick is inadvertently used at will in any analysis related to special relativity, especially in the paper I referred to at the end of my question.

Dear Ziaedin,

ZS: To make the problem easier suppose the OMG particle is approaching another particle. The observer in the frame of OMG particle can assume its own frame is standing still and the other particle is approaching it. In that case it takes one year for these two particles to meet.

Slow down, what event are you timing that from? We know when they meet but you have no starting point to define a duration.

ZS: But the same observer in the frame of OMG particle can also assume the OMG particle is speeding towards the other particle and thus the distance between them vanishes to zero which consequently reduces the time of travel to zero. The same two arguments can be repeated in the frame of other particle.

Yes, or in any other frame you choose, but if you want to say "it takes one year for these two particles to meet" you need an event at each end of that period, not just the meeting.

What you are saying is a completely standard aspect ant not a problem in any way, I would recommend you look up something called the "Parable of the Surveyors" to understand this better. I think it was originated by John heeler but you'll find several versions of it on the web.

Dear George

The event is still the same but the earth was swapped with a second particle, as some have the habit to assume the earth is a fixed frame and I also wanted to stay well within special relativity. Simply, the observer in the OMG particle, which is away from a second particle, one light year away, and is moving towards it, can measure zero, one year and any time between zero and one year, depending on his different assumptions and arguments using time dilation and length contraction.

In fact, special relativity has given an observer the liberty to paint space-time as he wishes. The reason is that there is no fixed universal frame. Now, what are our options, as the observer also does not know the speed of his own inertial frame?

In reference to the question, in the first case the particle in the other frame is an OMG particle, In the second case the particle in the observer frame is an OMG particle and in the last case one can say there is no OMG particle at all as both particles have certain speeds less than the speed recorded for the OMG particle.

This elasticity, given to the observer by special relativity, is the trick I am talking about. Time and space or space-time is not any more what we observe. It is what we can manipulate at will.

Dear Ziaedin,

ZS: ... does not know the speed of his own inertial frame

The words "his own inertial frame" means "the frame in which he is at rest" so the speed is zero by definition. You (1) is the only correct answer, there is no option regarding that.

ZS: It is what we can manipulate at will.

We can choose at will a definition of an inertial frame and then assign coordinates to events using that choice, but the events themselves are fixed by what happened in the experiment and for any frame, there is only one set of coordinates that applies to an event. A good example of that is the invariant interval between events, it always has the same value regardless of the choice of frame.

Again I will just say that this is all very basic relativity that you will learn in the first chapter of any textbook on the subject.

Dear George

Are you sure you are not talking about Newtonian inertial observer?

In relativistic theory “The worldline of any inertial observer is a straight line of Minkowski Spacetime” This is quoted from

Special Relativity in General Frames: From Particles to Astrophysics by Éric_Gourgoulhon, Springer-Verlag Berlin Heidelberg 2013. I have also added its mathematical definition in the attached file from the same source.

Hi Ziaedin,

ZS: Are you sure you are not talking about Newtonian inertial observer? In relativistic theory “The worldline of any inertial observer is a straight line of Minkowski Spacetime”

They are the same thing, the worldline is straight if there is no external (unbalanced) force acting on the object, essentially that is Newton's First Law. The worldlines of the OMG particle and another similar particle would be two straight lines with the hyperbolic angle between them given by the rapidity of their relative motion.

https://en.wikipedia.org/wiki/Rapidity

Dear George

But you tried to refute the second and third options used/available to an observer by saying that: The words "his own inertial frame" means "the frame in which he is at rest" so the speed is zero by definition.

We now agree that by definition the speed of the inertial frame of the observer is not zero. It is some unknown constant speed.

Now going back to the question, the third option in my previous answer is in fact the only option available to an observer which is allowed by special relativity (STR) as the other two cases are in contradiction to STR. But the second layer of trick is that physicist mainly use the first two options.

To summarize, the issue is:

• STR only acknowledges case/option three
• No one knows the exact speed of its own frame or in fact any other frame. Thus case three is hardly considered/used by physicists. (first trick)
• Physicists mainly use cases one and two in stark contradiction to STR (second trick)
• Physicists switch between cases one and two at will (third trick)

Hi Ziaedin,

ZS: But you tried to refute the second and third options used/available to an observer by saying that: The words "his own inertial frame" means "the frame in which he is at rest" so the speed is zero by definition.

I did and what I said is correct. In an earlier reply you said: "I deliberately did not mention the earth in the question to make the problem easier to see." and I am doing the same, I'm talking about two particles approaching as you did when you said: "To make the problem easier suppose the OMG particle is approaching another particle.".

Suppose there is a frame defined which allocates x and t coordinates to events relative to some (possibly hypothetical) particle or object moving inertially. We can make that more intuitive by calling the reference body "an observer". Particle A is moving left to right and particle B is moving right to left, both along the X axis of the frame. If neither particle is subject to any force, their worldlines (t plotted against x) will be straight. The angle between their worldlines is a measure of their speed and if expressed as rapidity is is a hyperbolic angle.

By definition then, speed is dx/dt and since all x values are measured in the frame, i.e. "relative to the observer", the observer's speed must be zero. That's why we sometimes call it "the rest frame of the observer".

ZS: We now agree that by definition the speed of the inertial frame of the observer is not zero.

No we don't.

ZS: It is some unknown constant speed.

Ask yourself "an unknown speed relative to what?" and you may realise the problem you face.

ZS: STR only acknowledges case/option three

No, it is case 1 by definition.

Dear George

Any definition should be based on facts especially a scientific one. We cannot define a moving frame a non-moving one just because the observer is within the frame. Please do not tell this to an officer if he stops you for cruising well above the speed limit.

I quoted the true definition. It is not mine; it is from a textbook. In fact, one of the correct statement of STR is that; there is no known fixed frame.

Please also note that your first answer in this forum is in contradiction with your later one.

GD “Both times are correct but in the frame of the particle, the distance is reduced by length contraction so we see the particle approaching at nearly c, the particle sees us approaching at the same speed.”

If any inertial observer assumes his own frame has zero speed, then the observer in the frame of the particle should never see any length contraction of the distance.

Dear Ziaedin,

ZS: We cannot define a moving frame a non-moving one just because the observer is within the frame. ... In fact, one of the correct statement of STR is that; there is no known fixed frame.

Since there is no absolute frame, "the observer" is all that can be used. As I say, "observer" is essentially a synonym for "the frame". As I said last time:

GD: Ask yourself "an unknown speed relative to what?"

ZS: Please do not tell this to an officer if he stops you for cruising well above the speed limit.

The speed is registered by the radar gun he is holding so the gun is "the observer". The numbers on its readout are only valid in the frame defined by using the gun as the origin of the coordinate system.

ZS: Please also note that your first answer in this forum is in contradiction with your later one.

None of my statements are contradictory, you need to think about what it means for that to be true, your grasp of SR at the moment is badly flawed.

ZS: If any inertial observer assumes his own frame has zero speed, then the observer in the frame of the particle should never see any length contraction of the distance.

The term "length contraction" means that objects moving relative to the observer's frame are measured to be shorter than they would be in the object's own frame.

Dear George

I gave the definition from a text book. It is not mine. You need either to show me another accepted definition or to prove the definition in the text book is not correct.

If we do not know the speed of a frame it does not mean we should assume it is not moving. The relativistic physicists since Galileo have tried hard to show us that if we feel a frame is not moving it does not mean its speed is zero. It can be moving at any constant speed. Thus, assuming zero speed for an inertial frame by a co-moving observer is pre-Galilean physics.

Dear George

If the observer in the frame of OMG particle assumes that its speed is zero then which one of the two observers can measure the time of travel as zero? According to your assumption only one year is the correct answer, measured by any of the two observers, one in the frame of OMG particle and the other in the frame of the second particle.

Dear Ziaedin,

ZS: I gave the definition from a text book. It is not mine. You need either to show me another accepted definition or to prove the definition in the text book is not correct.

No, the definition you gave is correct and so is what I said. The error is that you think they are not compatible but they are, there is nothing in anything I have said that conflicts with the definition you gave.

ZS: If we do not know the speed of a frame it does not mean we should assume it is not moving.

There is no such thing as an absolute frame so the words "it is not moving" have no meaning unless you say "it is not moving relative to ***", that is the basic understanding of all relativity, whether Galilean or Lorentzian. When we talk about "the observer", it is just a phrase which means "the origin of the frame" so it has speed zero by definition.

Dear George

If the definition is correct, then it does not say anything about the speed of an inertial observer to be zero. Only acceleration and rotation must be zero.

Dear Ziaedin,

ZS: .. the speed of an inertial observer ..

Speed relative to what?

Dear George

The question applies to your assumption as well. Zero speed in relation to what?

If you assume absolute speed of zero then this is denied in relativity.

If it is relative speed of zero then even two accelerating objects can have zero speed relative to each other. That is why in the definition of inertial observer all necessary conditions are specified. They are zero rotation and zero acceleration, nothing else.

What we can say is that the speed is constant and in relativistic physics it can be anything between zero and c.

Here is the definition of inertial observer in Einstein on line site.

"An inertial reference frame is a reference frame in which the first law of classical mechanics holds: A body on which no external forces act either remains at rest or moves with constant speed along a straight path. An inertial observer is an observer that is at rest with respect to an inertial reference frame. In the context of relativity, an inertial reference frame is one that drifts in gravity-free space without undergoing rotation or being accelerated."

Dear Ziaedin,

ZS: The question applies to your assumption as well. Zero speed in relation to what?

In relation to the frame, also known as the observer, of course. Hence, since "speed" is defined "relative to the observer", the speed of the observer relative to the observer must be zero by definition.

ZS: even two accelerating objects can have zero speed relative to each other.

They could but that would not be an "inertial frame", it is an accelerated frame.

ZS: That is why in the definition of inertial observer all necessary conditions are specified. They are zero rotation and zero acceleration, nothing else.

Yes, exactly.

EoL: "An inertial reference frame is a reference frame in which the first law of classical mechanics holds: A body on which no external forces act either remains at rest or moves with constant speed along a straight path. An inertial observer is an observer that is at rest with respect to an inertial reference frame. In the context of relativity, an inertial reference frame is one that drifts in gravity-free space without undergoing rotation or being accelerated."

Again, perfectly correct, and the worldline of any body moving inertially would then be a straight line in that frame which is the point we discussed earlier.

What I don't see is why you thought something I said earlier contradicted any of that, it seems to me that we are in complete agreement.

Dear George

When we say the observer in his own inertial frame (or simply the observer frame), the observer in the frame of OMG particle or the observer with the OMG particle, it means the speed of the observer in relation to the inertial frame or the OMG particle is zero. Here we just give a smart-eye to the inertial frame.

Now please let me go back to what I think is problematic which can be highlighted with the OMG question. I will try to rewrite what I have said with some more clarifications.

According to special relativity there is no fixed universal frame. However, relativistic physicists have given themselves the liberty to paint space-time as they wish.

The relativistic-physicist-observer with the OMG particle (particle A) - the observer hereafter-, which is one light year away from a second particle (particle B), and is moving towards it, can measure zero, one year and any time between zero and one year, depending on his different assumptions and arguments using time dilation and length contraction.

The three assumptions/options, assuming only two inertial frames exist, are,

In reference to the question, in the first case particle B is an OMG particle. In the second case the particle in the observer frame (particle A) is an OMG particle and in the last case one can say there is no OMG particle at all as both particles have certain speeds less than the speed recorded for the OMG particle.

This elasticity is the trick I am talking about. Time and space or space-time is not any more what we observe. It is what we can manipulate at will.

Time given by the observer in each assumption/option is as follows:

The third answer is in fact close to reality and special relativity. But a relativistic physicist cannot say I do not know, they try to use one of the first two cases at will ignoring the fact that those two cases are not supported by special relativity.

Dear Ziaedin,

ZS: Now please let me go back to what I think is problematic which can be highlighted with the OMG question.

I think that's a good idea, we seem to have been sidetracked a bit.

ZS: I will try to rewrite what I have said with some more clarifications.

That will help greatly, I think the problem here is that you are rushing to conclusions without carefully working through the problem. Take a bit more time and see if you can fill in the gaps. This version is better but still jumps to conclusions that aren't justified by your explanation.

ZS: According to special relativity there is no fixed universal frame.

That is correct, hence you always have to specify what frame you are using.

ZS: However, relativistic physicists have given themselves the liberty to paint space-time as they wish.

There is no absolute frame as you say, all frames are therefore equally valid, but when you choose one frame to use, you have to make clear which one it is.

ZS: The relativistic-physicist-observer with the OMG particle (particle A) - the observer hereafter-, which is one light year away from a second particle (particle B), and is moving towards it, can measure zero, one year and any time between zero and one year, depending on his different assumptions and arguments using time dilation and length contraction.

No. You said "The relativistic-physicist-observer with the OMG particle (particle A) - the observer hereafter-, which is one light year away from a second particle (particle B)". That means that at some moment on a clock synchronised with that of 'the observer', particle B is 'one light year away' as measured by rulers which are at rest relative to 'the observer'. Particle B is moving at close to c relative to 'the observer' therefore it will take just fractionally more than one year as measured by 'the observer' for B to reach 'the observer'.

ZS: The three assumptions/options, assuming only two inertial frames exist, are, ...

So far, you have only defined one frame, that of 'the observer' which is a frame in which particle A is at rest, so the rest of your text is premature. There is no concept of whether that frame is "moving" because there is no absolute frame and the observer frame cannot be moving relative to itself, obviously.

Now, try continuing your explanation from this point. I think what you want to do next is define a second frame which is based on particle B, but you cannot skip that step, see whether you can use that to explain why you think there is a problem bearing in mind what I have said about this first frame. I would just warn you that everything I've said about this frame will also apply to the new one.

The purpose of the Lorentz Transforms is to tell you what values in the second frame will be if you are given coordinates in the first frame, and vice versa, but there is no flexibility in that at all, you choose a frame and the maths tells you the answers.

Dear George

There are only two particles, one light year apart, and I have assigned a frame to each. An observer can also be assigned to each frame, like a pilot sitting in a spaceship. It is as simple as that. I even made it easier by removing the observer in the frame of particle B. Please tell me what is unclear here. If it is clear please read the rest of my previous answer.

Dear George

Regarding the Lorentz Transforms, as you might remember in another question in RG, I have shown in an article that Lorentz Transforms is based on a flawed analysis of M&M experiment. After more than one year I have not been disproved.

https://www.researchgate.net/post/Does_Michelson_and_Morley_experiment_support_length_contraction

In fact, in yet another forum I think it was agreed on the correctness of my finding. Simply put, relativistic physics is a theory with a false foundation.

https://www.researchgate.net/post/Why_is_special_relativity_still_denied_today

Dear Ziaedin,

ZS: There are only two particles, one light year apart, and I have assigned a frame to each. An observer can also be assigned to each frame, like a pilot sitting in a spaceship.

That's fine, so using your analogy, the pilot is of course at rest in the spaceship.

ZS: It is as simple as that. I even made it easier by removing the observer in the frame of particle B. Please tell me what is unclear here.

That's fine so far and it automatically gives you what option

ZS: If it is clear please read the rest of my previous answer.

In your previous message you said:

ZS: The three assumptions/options, assuming only two inertial frames exist, are, 1. The easiest and most used option is to assume the observer frame is not moving and the other inertial frame is moving in relation to it.

Each observer is at rest in his frame, just as each pilot is at rest in his own ship.

In the frame of particle A, particle A is at rest and particle B is approaching. From a distance of one light year as measured by A, particle B will take one year to arrive.

In the frame of particle B, particle B is at rest and particle A is approaching. From a distance of one light year as measured by B, particle A will take one year to arrive.

Now, what is the problem?

Dear George

Please read beyond assumptions/options 1.

The problem is that the first two options are firmly refuted by special relativity. At best, they are two possibilities among countless other possibilities. We are not allowed to choose only two possibilities from countless possibilities with no justification at all.

This question also highlights the point that special relativity takes away quantitative details from scientific studies as I pointed out in my article “Special Relativity the revival of metaphysics” attached to the question below.

https://www.researchgate.net/post/Can_special_relativity_be_categorized_as_metaphysics

Dear Ziaedin,

ZS: Please read beyond assumptions/options 1.

There is no need, option 1 is the correct and only valid answer in special relativity. What you need to do is start from that point and see why you think there is a problem. As I asked, "what is the problem"?

Dear George

One of the main points of special relativity is to defy the existence of any stationary frame. Why the frame of the observer is deemed stationary here?

Dear Ziaedin,

ZS: Why the frame of the observer is deemed stationary here?

Because "observer" is just another name for "frame".

The conclusion from SR is that you can call any frame "stationary", it's just a name that has no physical significance.

Dear George

GD - The conclusion from SR is that you can call any frame "stationary"

Can you please provide a reference?

On the contrary SR says the absolute speed of no frame is known. Zero or otherwise. Why are we scientifically allowed to assume the speed of any frame to be zero and not 10 or 100 m/s or even c?

The point I wanted to raise in this forum is that relativistic physicists subjectively assume their own frame is stationary but whenever suits them they flip the assumption and assume they move at certain speed and the second frame is stationary. The fact is that neither is correct.

Dear Ziaedin,

ZS: Can you please provide a reference?

Not easily, it's just a meaningless name. You will often see frames called K and K' and again those are just names. You could call them "Jim" and Sheila", it wouldn't make any difference.

ZS: On the contrary SR says the absolute speed of no frame is known

It doesn't say it is not known, it says there is no such thing, the concept is superfluous.

ZS: Why are we scientifically allowed to assume the speed of any frame to be zero and not 10 or 100 m/s or even c?

The velocity of an object in a frame is defined as the rate of change of location of the object relative to the origin of the frame. It follows from that definition that the velocity of the frame itself in that frame has to be exactly zero because you are measuring it relative to itself, no other value is permitted because it wouldn't be self-consistent. The same is of course true of every frame.

Dear George

• Science is against any wrong assumption and an assumption without reference to any experiment and logic. Why every resident/local observer should assume its own frame is stationary which goes against reality and even special relativity. What is the difference between this assumption and geocentric idea?
• If every local frame is assumed stationary, then why the same observer can also assume his frame is moving?

Dear Ziaedin,

ZS: If every local frame is assumed stationary, then why the same observer can also assume his frame is moving?

He can't assume that his frame is moving, relative to what would it be moving? There is no such thing as "absolute motion".

When we say "stationary", it means not moving relative to the observer, and "the observer" is just another name for the frame. We've been over this many times now Ziaedin, I don't see any way to make that any clearer for you.

Dear George

You are right, we already very well cleared the relation between an observer and his own frame. We already agreed that the observer is not moving relative to his own frame. When I say "observer's frame is moving" it means both observer and its frame are moving relative to a second frame/observer. I have not said anything which disregard this assumption. I also tried to simplify this by introducing Bob and Alice instead of the two particles to make sure I definitely assume that observer and its frame are the same. Hope this is clear now.

Returning to the question: What I have claimed is that relativistic physicists usually go for the first option below but they allow themselves switch to the second option. The fact is that both are not correct, according to reality and even special relativity. I provided an example to show the use of these two assumptions by relativistic physicists are common and well accepted.

Alice can also claim both the cases above.

Please respond if my claim is not correct.

Hi Ziaeden,

ZS: You are right, we already very well cleared the relation between an observer and his own frame. We already agreed that the observer is not moving relative to his own frame. When I say "observer's frame is moving" it means both observer and its frame are moving relative to a second frame/observer. I have not said anything which disregard this assumption. I also tried to simplify this by introducing Bob and Alice instead of the two particles to make sure I definitely assume that observer and its frame are the same. Hope this is clear now.

Yes, that's great.

ZS: What I have claimed is that relativistic physicists usually go for the first option below but they allow themselves switch to the second option.

You have to be careful when reading stuff intended for the public, they tend to be very careless about how they express things in order to make it more understandable.

1. Bob (OMG particle) can claim he is stationary and it is Alice (second particle) that is moving with a very high speed, v, towards him. In this case the time is calculated to be one year.

That is fair but to be a bit more rigorous, a scientist might say "In Bob's frame, after passing an object at rest one light year away from Bob, it will take only slightly more than a year for Alice to reach him."

2. Bob can also assume that Alice is stationary and he is moving with the speed of v towards Alice.

"Bob" isn't a person, it is the anthropomorphism of the frame, so "he" can't say anything, the velocity in the coordinate system "tells us" how long it will take, but in the sense of a numerical result from an equation, not words from a person.

What the scientist can say is "In Alice's frame, after passing an object at rest one light year away from Alice, it will take only slightly more than a year for Alice to reach him." Note this is not the same distant object mentioned in point 1. The scientist could also say "Alice would note the object mentioned in point 1 passed her followed almost immediately by Bob." due to length contraction.

You are taking the metaphors people use too literally, the maths is robust.

Dear George

The question in this forum is based on a well-known scientific paper, as referenced in the bottom of the main question. Any other references are all based on journal papers and well known textbooks. Concentrating on the subject, if “the mathematics is robust” then can you please mathematically show why both answers are right according to your first answer in this forum.

GD - “Both times are correct but in the frame of the particle, the distance is reduced by length contraction so we see the particle approaching at nearly c, the particle sees us approaching at the same speed. The whole thing is just a geometric projection.”

Hi Ziaeden,

I take it you mean this paper:

http://physics.gmu.edu/~rubinp/courses/122/readings/AJP000342.pdf

Which part of it do you think conflicts with what I said?

Dear George

I said “if the mathematics is robust then can you please mathematically show why both answers are right”. How come both answers, one year and zero, are correct?

Dear Christian~

Where does the number 0.999,999,999,999,999,999,999,995,1c come from? Of course, people do not question about it unless they want to defend SR. My question is based on the speed of OMG particle by those who observed it.

According to SR there is no preferable inertial reference frame (IRF). It means no observer in any IRF knows its absolute speed. However, SR gives every IRF observer the right to assume its own velocity is zero, first breakage of SR principle. When it is necessary, such as in the Mount Washington experiment, the observer is given the right to assume some considerable speed, second breakage of SR principle.

I thus want to highlight this contradiction in SR freehand way of analysis.

PS: noted that the presentation material is not shown but the reference to the paper for the experiment is given in the question.

The answer is simple: Time is not a frame-independent parameter, the 'Proper Time' is.

Dear Xiaohao Cui

Suppose you want to go to Andromeda galaxy which is about 2.5 million-light-year away from earth. You travel with the constant speed of OMG particle, or for simplicity the speed of light. It should take at least 2.5 years to get there, according to the proper time as you measure time on board of your space ship. However, according to theory of length contraction the distance is shrunk almost to nothing for you on board. Thus, with no distance to travel you are supposed to be there in no time. This is the paradox.

I shall observe the particle to take almost a year to cross the distance, but the particle " will observe" itself to require an extremely small time. v/c=1-4.9e-25.

In fact, the light, from its own frame of reference, travels from one place to other instantaneously, directly from principles of SR