Special relativity and sameness

[Bartolomeo]

I think so (together with the length you "measure"). If that is too vague, sorry but I don't really go in for this type of calculation as I hinted above.

It seems if the clocks show the same time, the rod is at rest in our frame. If clocks are out of sync, rod is in motion. The closer velocity is to that of light, the more clocks out of sync, Another question - are all these rods the SAME rod? For example you ask me - what is velocity of the rod? - Which rod? Could you please to concertize? Tell me how it appears (what are clocks readings) and I will tell you what it's velocity is. For example, beetle pest starts eating rod simultaneously from the both ends in the rod's rest frame. In different reference frames (that are in relative motion) the remainder of the rod will look differently. From the reference frame which moves with velocity close to c the rod will be eaten to the center from the one side, but from the other side it will be almost untouched. It seems like that.

 

[Battlemage!]

It seems if the clocks show the same time, the rod is at rest in our frame. If clocks are out of sync, rod is in motion. The closer velocity is to that of light, the more clocks out of sync, Another question - are all these rods the SAME rod? For example you ask me - what is velocity of the rod? - Which rod? Could you please to concertize? Tell me how it appears (what are clocks readings) and I will tell you what it's velocity is. For example, beetle pest starts eating rod simultaneously from the both ends in the rod's rest frame. In different reference frames (that are in relative motion) the remainder of the rod will look differently. From the reference frame which moves with velocity close to c the rod will be eaten to the center from the one side, but from the other side it will be almost untouched. It seems like that.

I'm pretty sure if you waited until the beetles stopped eating, and then brought the rod to each observer, they will measure the same amount of rod eaten, and that any given frame where it looks like more of the rod is eaten than in another (take note of the when) would just be some combination of the relativity of simultaneity and length contraction/time dilation. None of them are likely going to measure the exact same times for the events, because simultaneous events in one frame are not necessarily simultaneous in another. I don't understand how this is some sort of paradox.

 

[Dale]

are all these rods the SAME rod?

What physical experiment could you perform to measure the SAMEness of a rod?

 

[Bartolomeo]

What physical experiment could you perform to measure the SAMEness of a rod?

Let’s look on the rod at the different angle.

Once upon a time there was a rod 10 miles long with marks every mile. It was like that:

(-5)==(-4)==(-3)==(-2)==(-1)==(0)==(+1)==(+2)==(+3)==(+4)==(+5)

Then beetles started eating it from both sides. Let’s say in our frame K’ which moves at 0.9 c. Then at certain moment we see surplus of the rod like that:

(-1)==(0)==(+1)

What is proper length of this rod? I don’t even know what to say.

IMHO the problem is that we can see the rod (-1) ==(0)==(+1) only from a frame K’. From the frame K’’ which moves with another relative velocity V2 the rod may appear like that:

(-3)==(-2)==(-1)==(0)==(+1)==(+2)

From the frame K’’ which moves with another relative velocity V3 the rod may appear like that:

(-5)==(-4)==(-3)==(-2)==(-1)==(0)==(+1)==(+2)=

It is simplification, sure I haven’t calculated it.

But the rod (-1)==(0)==(+1) only “exists” in a frame K’ and has exact “absolute” velocity 0.9 c. We cannot see the rod (-1)==(0)==(+1) from any other frame.

Note: part of the whole rod (-1)==(0)==(+1) has its proper length however.

Heraclitus told that we cannot enter into the same river twice because it changes with time. The same river from the different reference frames will appear different. Was it the same river or not?

I suppose that noetic rod has any velocity. But if we concertize material body to the smallest details we can say how fast and in which direction it moves.

If we want to see the “same” river from all frames, observers in every frame have to take into account their own velocity and synchronize clocks accordingly.

Maybe I am wrong,

 

[Dale]

You didn't answer my question. Again: What physical experiment could you perform to measure the SAMEness of a rod? Please answer the question.

 

[Ibix]

@Bartolomeo - The usual way to view it is that a supposedly one-dimensional rod is in fact a two-dimensional sheet in spacetime. Then you come in with your synchronisation convention and pick a one-dimensional slice to call "the rod now". So all the people with different synchronisation conventions are just taking different slices through the rod. They're viewing different parts of the same entity.

Whether that constitutes "the same rod" or not is up to you.

 

[Bartolomeo]

You didn't answer my question. Again: What physical experiment could you perform to measure the SAMEness of a rod? Please answer the question.

I think that I have to take readings of clocks on the end of the rod. At least. Or to take a photo.

 

[Dale]

I think that I have to take readings of clocks on the end of the rod. At least. Or to take a photo.

OK, so if you take a photo then how do you change that photo into a measurement of SAMEness?

 

[Bartolomeo]

@Bartolomeo - The usual way to view it is that a supposedly one-dimensional rod is in fact a two-dimensional sheet in spacetime. Then you come in with your synchronisation convention and pick a one-dimensional slice to call "the rod now". So all the people with different synchronisation conventions are just taking different slices through the rod. They're viewing different parts of the same entity.

Whether that constitutes "the same rod" or not is up to you.

OK, so if you take a photo then how do you change that photo into a measurement of SAMEness?

Pictures taken from different frames will be different I think. If the rod is covered with synchronized clocks (every spot) at each picture the readings will be different. If you will show me any chosen picture I may not agree with you. It was not the rod I wanted to see. But by appearance of the rod I can gess what frame it was photographed from.

 

[Bartolomeo]

For example you ask me. What is velocity of the rod? Which one? Any rod has any velocity. Please describe in details or show me a picture. I will tell you what is velocity of exact rod.

 

[Ibix]

No. If the cameras are co-located you will see the same clock readings regardless of frame (edit: note that this is not the same as taking identical pictures). If the cameras are not co-located you will see different clock readings regardless of frame. Remember the light speed delay.

Now, if the instantaneously co-located camera users calculate backwards from what they see then the answer will be frame dependent. But, again, they're just defining different slices of the same extended-in-space-and-time entity as the extended-in-space-only thing they call "the rod now".

 

[Dale]

Pictures taken from different frames will be different I think.

Yes. And pictures from different locations in one frame will also be different. Do you want a measure of SAMEness that behaves that way?

 

[Herman Trivilino]

Let’s look on the rod at the different angle.

Once upon a time there was a rod 10 miles long with marks every mile. It was like that:

(-5)==(-4)==(-3)==(-2)==(-1)==(0)==(+1)==(+2)==(+3)==(+4)==(+5)

Then beetles started eating it from both sides. Let’s say in our frame K’ which moves at 0.9 c. Then at certain moment we see surplus of the rod like that:

(-1)==(0)==(+1)

What is proper length of this rod? I don’t even know what to say.

If everything you've stated so far refers to frame K' then the statement that the frame moves at speed 0.9 c is meaningless because you haven't stated that speed relative to anything. If that's the case, the frame's speed is not relevant. The rod is at rest in frame K', its proper length was 10 miles, now it's 2 miles.

Note that the phrase "beetles begin eating" has meaning in only one frame. Other frames will observe that the beetles do not begin at the same time.

 

[m4r35n357]

It seems if the clocks show the same time, the rod is at rest in our frame. If clocks are out of sync, rod is in motion. The closer velocity is to that of light, the more clocks out of sync, Another question - are all these rods the SAME rod? For example you ask me - what is velocity of the rod? - Which rod? Could you please to concertize? Tell me how it appears (what are clocks readings) and I will tell you what it's velocity is. For example, beetle pest starts eating rod simultaneously from the both ends in the rod's rest frame. In different reference frames (that are in relative motion) the remainder of the rod will look differently. From the reference frame which moves with velocity close to c the rod will be eaten to the center from the one side, but from the other side it will be almost untouched. It seems like that.

Yes it is the same rod, but its "proper" age changes along its length in your rest frame. But none of this is what you see, as I said earlier. I am not a fan of "time distraction" and "length confusion" as a teaching tool. It always ends up like this thread . . .
That is why I gave a link to a demonstration of what you see, which is how you would actually measure things.

 

[Bartolomeo]

As far as I know in there are different techniques of taking pictures in Special Relativity with different outcome.

In this case an observer must take picture this way: the observer places many cameras in his reference frame close to each other. The cameras should be in immediate vicinity to the rod. Then all these cameras make click simultaneously in observer’s frame. Then this observer glues pictures into one.

It is like taking picture of long relativistic train. There are many cars in the train and each car has a clock. These clocks show the same time in train‘s frame. Observer on the platform places cameras along the platform and these cameras take picture simultaneously in his own frame. Then he glues picture into one. Due to relativity of simultaneity different clocks in different cars will show different time on the picture.

Small note about pinhole camera.

If you wish to take a picture by pinhole camera you can run into troubles, because rays from the different parts of the rod will come into aperture at different moment. If the object shines by reflected light the picture can be spoiled by optical effects, like Terrell – Penrose rotation.

However, it is possible to take a picture by pinhole camera so as to see two clocks on the ends of the rod. In this case very close to every clock we have to place a flash. Then cameraman releases these flashes simultaneously in his frame. These flashes highlight these clocks and reflected beams simultaneously come to aperture, since their patches are hypotenuses of equal triangles. The rod will appear contracted and the clocks will show different time due to relativity of simultaneity.

For example we can evaluate distortions of a square (magnitude of length contraction) by it‘s picture. In this case we place a flash at each corner of the square. Then cameraman releases flashes simultaneously in his reference frame. Since the square moves in the reference frame of a cameraman, the picture should be taken when the cameraman and the center of the square at the points of the closest approach. The square will appear distorted – it will be contracted in direction of its motion. Cameraman can measure vertical sides and horizontal ones and will measure magnitude of Lorentz contraction of the square.

But, in this case the only spatial position at which the camera can make such a picture (so as all the lamps were visible in the picture) is the point of closest approach to the square or rod.

 

[Dale]

With this definition of "same" then the rod in one frame is not the "same" as the rod in another frame. As you have defined it sameness is frame variant.

 

[Bartolomeo]

With this definition of "same" then the rod in one frame is not the "same" as the rod in another frame. As you have defined it sameness is frame variant.

Yes.

The same is the same for any observer – I mean the object, which has exact clock readings and consequently appearance. For example, objects (-1)==(0)==(+1) and objects (-1)==(0)==(+1)==(+2) are not the same, thought we call the both „THE rod“.

Are the log and a nesting box made of him the "same" tree? An artisan begins to cut it out starting from one end.

It seems to me, from different reference frames we see „different“ rods. Thus „sameness“ does not depend on frame. One million different frames – one million different „appearances“.

If we give a clear description of the physical object, this „momentarily concertized object“ has not only speed but also direction of motion. There is one and only one frame K‘ in which the rod (-1)==(0)==(+1) exists. We CANNOT make picture of THIS object from any other frame.

 

[Dale]

Note, that your concept of "same" has very little (if any) physical value. It is not conserved, it is not invariant, it doesn't predict anything else. It is just a thing you have defined for the sole purpose of saying that the object isn't the same in different frames. There is no physical relevance to any other physical quantity. It is just a convention based on the Einstein synchronization convention and the relativity of simultaneity.

 

[PeroK]

Yes.

The same is the same for any observer – I mean the object, which has exact clock readings and consequently appearance. For example, objects (-1)==(0)==(+1) and objects (-1)==(0)==(+1)==(+2) are not the same, thought we call the both „THE rod“.

Are the log and a nesting box made of him the "same" tree? An artisan begins to cut it out starting from one end.

It seems to me, from different reference frames we see „different“ rods. Thus „sameness“ does not depend on frame. One million different frames – one million different „appearances“.

If we give a clear description of the physical object, this „momentarily concertized object“ has not only speed but also direction of motion. There is one and only one frame K‘ in which the rod (-1)==(0)==(+1) exists. We CANNOT make picture of THIS object from any other frame.

I would stamp a unique serial number on the rod. Then everyone would know it's the same rod when they see it.

I'm not sure what this has to do with SR.

 

[Herman Trivilino]

The same is the same for any observer – I mean the object, which has exact clock readings and consequently appearance. For example, objects (-1)==(0)==(+1) and objects (-1)==(0)==(+1)==(+2) are not the same, thought we call the both „THE rod“.

No one I know would do that. If THE contracted rod is (-1)==(0)==(+1) then THE same rod, less contracted, is (-1)===(0)===(+1).

It certainly wouldn't be (-1)==(0)==(+1)==(+2).

 

[Bartolomeo]

I would stamp a unique serial number on the rod. Then everyone would know it's the same rod when they see it.
I'm not sure what this has to do with SR.

S/N 54321012345

It is just a thing you have defined for the sole purpose of saying that the object isn't the same in different frames. There is no physical relevance to any other physical quantity. It is just a convention based on the Einstein synchronization convention and the relativity of simultaneity.

I admit that some people consciously prefer exactly this convention. They probably enjoy that different observers would see different physical object! It gives perfect opportunity to fantasize about reality of the multidimensional worlds. To me it looks a bit queerly.

I would recommend to every observer to consider his own velocity and synchronize clicks of shutters of photo cameras in his frame accordingly. But, I am afraid to say, velocity of light in different direction will be different for every observer. They will see the same material object each and this object would have the same velocity in every observer's frame.

I do not insist, it just my personal opinion.

No one I know would do that. If THE contracted rod is (-1)==(0)==(+1) then THE same rod, less contracted, is (-1)===(0)===(+1). It certainly wouldn't be (-1)==(0)==(+1)==(+2).

Note, that your concept of "same" has very little (if any) physical value. It is not conserved, it is not invariant, it doesn't predict anything else

@MiLara asked:

Special relativity states that according to an observer at rest, a measuring stick on a moving platform will appear shorter.

Would this observer still see the measuring stick as comprising of the same amount of atoms as the observer who is at rest with respect to the measuring stick? If this is the case, would the first observer actually measure the measuring stick as being more dense since it's length is contracted?

I have a sense that this would violate conservation of mass and energy as this should still hold true regardless of the reference frame. I am no expert on relativity, so any insight as to where my logic is flawed would be greatly appreciated.

It is considered that the so-called Lorentz-invariant mass does not depend on the choice of the reference frame, as defined as the value of the square which is calculated by the formula $m^2 = E^2 / c^4-p^2 / c^2$, where E - energy, and p - momentum of the body, referred to as an invariant when changing the inertial reference systems. But to consider the value $E^2 / c^4-p^2 / c^2$ to be invariant we can only in the case if we consider the simplest objects - such as, for example, spatially extended bodies with unchanging mass or point with a changing mass. With respect to the spatially extended bodies with varying mass approval of the invariance of $E^2 / c^4-p^2 / c^2$ is incorrect.

Let‘s say, that 1 mile of mentioned above rod weights one ton.

What is proper length and proper mass of the remainder (-1)==(0)==(+1)? May we say that in reference frame K (in which a rod is “at rest”) proper length of this remainder is 10 miles and proper mass is 10 tons?

We cannot. Because of relativity of simultaneity in the reference frame K this remainder cannot be detected at any given moment in time. In the best case the remainder bounded on only one end of one of the marks (+1) can be detected.

Thus, the considered remainder of the rod limited with marks (-1) and (+1) in principle cannot be at rest and therefore does not have its own length in the conventional sense of the word.

 

[Dale]

To me it looks a bit queerly.

That is odd. It was your definition. Why did you propose a definition that you thought "looks a bit queerly"?

I would recommend to every observer to consider his own velocity

Relative to which frame?

 

[Dale]

But to consider the value E2/c4−p2/c2E2/c4−p2/c2E^2 / c^4-p^2 / c^2 to be invariant we can only in the case if we consider the simplest objects - such as, for example, spatially extended bodies with unchanging mass or point with a changing mass.

Yes, this is correct. For more general bodies we consider the full stress energy tensor instead instead of the simplified four momentum.

 

[Bartolomeo]

Relative to which frame?

Relative to any arbitrary chosen reference frame. They can call it “The Reference Frame K of Mutual Co-Operation”.

If all observers want to see the same remainder (1)==(0)==(1), they must choose the observer who will conditionally be “at rest”. This observer synchronizes clocks by Einstein in his frame and this frame will be K. Velocity of light in all directions in this frame will be c. Then every other observer synchronizes clocks considering his “proper” velocity in this frame K and adjusts clocks accordingly.

Every observer can find required synchronization of clocks (shutters of photocameras) in his frame empirically with the aim to see the same remainder (1)==(0)==(1).

Then, after these “empirical synchronizations” finished and all observers “see” the same physical object, each of them can measure velocity of light in his own frame by means of his “empirically synchronized clocks”. This way they will be able to “detect their own motion”, because measured velocity of light in different direction will be different from c.

So, if all observers employ the same synchronization convention, they will see different physical objects. If they want to see the same physical object, they must to synchronize clocks differently.

 

[Herman Trivilino]

So, if all observers employ the same synchronization convention, they will see different physical objects. If they want to see the same physical object, they must to synchronize clocks differently.

And what would be the purpose of such an arrangement? I mean, what would it accomplish?

If successful (and I'm not even sure it can be because I don't follow everything you've said and you didn't help when you didn't reply to my previous query) it would serve to demonstrate that you can pick out any inertial reference frame you want and call it special. That is the Principle of Relativity. You can do such a thing with ANY inertial reference frame. They are, in this sense, equivalent.

 

[Dale]

If all observers want to see the same remainder

But why would any of them care about it? Your definition of "same" has no physical consequences.

This way they will be able to “detect their own motion”,

Yes, they can detect their motion relative to the reference observer in any number of ways.

if all observers employ the same synchronization convention, they will see different physical objects

And why should that bother any of them? Whether they see the same or different objects doesn't matter.

 

[PAllen]

No. If the cameras are co-located you will see the same clock readings regardless of frame (edit: note that this is not the same as taking identical pictures)..

This depends on your camera abstraction. Consider each piece of film as a world tube. You have intersecting world tubes. However, the orthogonal slice for each world tube will be different, and the set of events of the rod world tube whose light reaches each film rest slice will be different. Thus, which clock readings are imaged will be different.

 

[Ibix]

This depends on your camera abstraction.

True. At the time I posted what you quoted, I was thinking of continuous ambient illumination and a pinhole camera. That was what Bartolomeo was talking about the last time we talked about this kind of thing. It's obviously not what he had in mind here...

Consider each piece of film as a world tube. You have intersecting world tubes. However, the orthogonal slice for each world tube will be different, and the set of events of the rod world tube whose light reaches each film rest slice will be different. Thus, which clock readings are imaged will be different.

I'm not sure I completely agree. First, I think it's whatever controls the light striking the film (e.g. the shutter aperture in a conventional film camera) rather than the film itself that matters. Hence all Bartolomeo's (de-)synchronised flashlamps.

Second, while I agree that the open shutter apertures are different world-tubes, I interpret "co-located" to mean that the spatial and temporal centers of the worldtubes are one event. Then, all the different states of motion just lead to different degrees and shapes of blur, plus angular aberration and Doppler, I think. But the centroid of the blurred imageof each clock ought to be the same. At least for a pinhole camera, or small conventional camera.

 

[PAllen]

True. At the time I posted what you quoted, I was thinking of continuous ambient illumination and a pinhole camera. That was what Bartolomeo was talking about the last time we talked about this kind of thing. It's obviously not what he had in mind here...
I'm not sure I completely agree. First, I think it's whatever controls the light striking the film (e.g. the shutter aperture in a conventional film camera) rather than the film itself that matters. Hence all Bartolomeo's (de-)synchronised flashlamps.

Second, while I agree that the open shutter apertures are different world-tubes, I interpret "co-located" to mean that the spatial and temporal centers of the worldtubes are one event. Then, all the different states of motion just lead to different degrees and shapes of blur, plus angular aberration and Doppler, I think. But the centroid of the blurred imageof each clock ought to be the same. At least for a pinhole camera, or small conventional camera.

I agree if it's pinhole, and the pinholes are coincident for an instantaneous opening, then you are correct. I was assuming an open aperture with continuous focus on a light sensing surface that is read at simultaneous instant for that surface. Your model provides a better abstraction of momentarily collocated cameras.

 

[Bartolomeo]

And what would be the purpose of such an arrangement? I mean, what would it accomplish?

If successful (and I'm not even sure it can be because I don't follow everything you've said and you didn't help when you didn't reply to my previous query) it would serve to demonstrate that you can pick out any inertial reference frame you want and call it special. That is the Principle of Relativity. You can do such a thing with ANY inertial reference frame. They are, in this sense, equivalent.

Please excuse me for missing your previous post. Yes, the remainder of the rod is at rest in K’ but is in motion in K at 0.9 c. I think that observers in K in no way are able to see the same remainder. They will see another remainder with different marks on it.

I wanted to say, that in Special Relativity observers from different reference frames “see” different physical objects. The clock on material body represents appearance of this spot of material body. This way you observe parts of material body of different age.

For example, from a certain reference frame an observer “sees” two identical twins of the same age, at certain distance from each other, 35 years old, healthy with rose cheeks.

Another observer from another frame sees that one of them is still sucking a pacifier and other one is in a coffin already. I think it is very funny!
Selection of a frame of reference with respect to which all observers synchronize their clocks brings some order at least. Then every observer can see the twins of the same age, 35 years old, healthy with rose cheeks.

Sure, if they need this order. Maybe they intentionally prefer havoc.

But why would any of them care about it? Your definition of "same" has no physical consequences.

Obviously, nobody will not die. I think it‘s just curious.

True. At the time I posted what you quoted, I was thinking of continuous ambient illumination and a pinhole camera. That was what Bartolomeo was talking about the last time we talked about this kind of thing.

Yes, I wrote it badly there. I wanted to say that we had to "highlight" two Einstein - synchronized clocks instantly (at equal distance from the center) by two simultaneous (in camera's frame) flashes. Two clocks on the ends of the rod i mean. The beams will come simultaneously (hypotenuses of triangles) to aperture and the shutter will make a click at this moment. It's a pity that you are were not interested. The faster the rod moves in the reference frame of the camera the shorter the rod (distance between clocks) will appear on the picture (gamma times Lorentz - contracted) and the more clocks will be out of sync on the picture.
Or we can imagine two twins at certain distance from each other. Each has a clock and a flash. Clock are Einstein - synchronized in the twins' frame. The cameraman launches flashes simultaneously in his frame.
As the distance between the observers and their transverse dimensions are gamma times less than their proper distance and dimensions, the image on the film becomes contracted, so the observers and on the photo become "thinner" and the distance between them becomes gamma times less. The clocks located close to the observers will show different time.