The closer to the speed of light, the more length contraction in the direction of motion (SRT).

[Halc]

1 hour ago, studiot said:

The operative word being compute, not measure.

Proper length of a rigid inertial object can be directly measured.  If it's accelerating, it's arguably a computation no matter how it's done.

Proper distance between a pair of events seems to be a computation in any frame.

1 hour ago, studiot said:

You have just told me that the proper length of a ship is 100m.

Notice that I didn't need to specify a frame to say that. A statement of coordinate length would need such a specification to be unambiguous. You say it's defined only in the ship's inertial frame (if accelerating, ship frame isn't even an inertial one), implying that the statement, lacking the reference, is ambiguous. But that's your assertion. The wiki site defines the proper length in other frames (said computation above, which is defined in any frame).

2 hours ago, studiot said:

Say I have a super duper camera that is equipped with length measuring stadia.

From my observatory on a planet, where I am stationary, I monitor the ship's approach and at the instant its centre passes my telescope,

Don't know what stadia means here, but I'm find with infinite resolution telescopes that can count spider legs from 1000 LY away. The ship is approaching, so it is always centered on the telescope, not just passing in some instant. Perhaps you mean something else like it is moving from left to right or something, which would not be 'approaching'.  If you see it end-on, you cannot even tell that it has length.

2 hours ago, studiot said:

I take a photographic mesasurement of the length of the ship.

I observe 60 m

Telescopes don't measure length. They measure subtend angle, so an approaching ship appears to be getting bigger, but without knowing how far away it is, no size dimension can be inferred from a photograph.  Maybe 'stadia' is some kind of magic that lets you do this. Anyway, if it is approaching, one can only see the front, not the depth of the ship, regardless of its orientation. No size is obvious even from subtend angle.

OK, the ship has a coordinate length of 60m so it is moving in the direction of that length (and not sideways).  The telescope cannot measure that, especially since it can only see one end of the thing.  So you do not observe any length at all, nor a distance to the thing.

2 hours ago, studiot said:

in my frame the length of the ship is 60 m.

I agree with this statement of coordinate length, but not that it was something observed.  Not sure how the telescope can figure the velocity either unless one already knows say what color the lights are and what redshift is observed. I'll allow that since decent spectography can do that.  So I'll grant that one can work out (not directly measure) its approach velocity.

Not sure of the whole point of the example. We already stated the proper length, and I agree that there are ways to figure it out (with a tape measure, not a telescope) in any frame. You assert that since it might involve a computation, that it doesn't count as it having a proper length. That's your choice, but unless it has a different proper length in some other frame, the proper length is still frame invariant, which is what I'm asserting, and which the bottom line of your recent wiki quote says.

2 hours ago, studiot said:

The velocity c is invariant.

It is most decidedly not.

Totally agree there. c is not a velocity at all, let alone a constant one. It is a scalar constant, not a vector constant.  c is even a constant in alternative theories that do not posit light moving at that speed as does SR.  In some alternative theories, light speed is not frame invariant.

[studiot]

Why in the world do I bother since you do not respond to what I write but to some made up fantasy you substitute for my words.

 

I wrote

8 hours ago, studiot said:

From my observatory on a planet, where I am stationary, I monitor the ship's approach and at the instant its centre passes my telescope, I take a photographic mesasurement of the length of the ship.

I observe 60 m

How does that become a head on collision course ?

5 hours ago, Halc said:

The ship is approaching, so it is always centered on the telescope, not just passing in some instant. Perhaps you mean something else like it is moving from left to right or something, which would not be 'approaching'.  If you see it end-on, you cannot even tell that it has length.

 

I accept that this

8 hours ago, studiot said:

Since this is only defined in the that ships's own inertial frame, where we are agreed it appears at rest, and will be different in any other frame and measured as such by an observer in that frame.

But I did say

8 hours ago, studiot said:

From my observatory on a planet, where I am stationary, I monitor the ship's approach and at the instant its centre passes my telescope, I take a photographic mesasurement of the length of the ship.

I observe 60 m

So yes you can measure the length of an object in a frame moving relative to yours, but the measurement will not be the proper length.

As far as I can tell it is impossible to directly measure the proper length of something moving relative to yourself.

5 hours ago, Halc said:

Proper length of a rigid inertial object can be directly measured.

If you think that is possible please describe how it might be done.

 

5 hours ago, Halc said:

Don't know what stadia means here, but I'm find with infinite resolution telescopes that can count spider legs from 1000 LY away.

So do you know what stadia are ?

That sort of resolution is silly.

I referred to passing at close distance.

 

5 hours ago, Halc said:

Telescopes don't measure length. They measure subtend angle, so an approaching ship appears to be getting bigger, but without knowing how far away it is, no size dimension can be inferred from a photograph.  Maybe 'stadia' is some kind of magic that lets you do this. Anyway, if it is approaching, one can only see the front, not the depth of the ship, regardless of its orientation. No size is obvious even from subtend angle.

OK, the ship has a coordinate length of 60m so it is moving in the direction of that length (and not sideways).  The telescope cannot measure that, especially since it can only see one end of the thing.  So you do not observe any length at all, nor a distance to the thing.

I have spent the last 53 years since leaving university measuring distance with stadia in telescopes.

Distance or length measuring marks on the reticule of optical equipment started to come in around 200 years ago.

I will elaborate for a more polite and understanding response.

[Halc]

8 hours ago, studiot said:

Why in the world do I bother since you do not respond to what I write but to some made up fantasy you substitute for my words.

What you wrote was not understood and seemed contradictory. I was attempting a guess as to what was meant. I apparently guessed wrong.

8 hours ago, studiot said:

How does that become a head on collision course ?

The word 'approach'.  Please be more clear as to the actual trajectory. I am now imagining a ship angling in so it is both getting closer (approaching), but heading somewhere other than toward the observer such that its path crosses the fixed orientation of the telescope. I probably have that wrong as well, but a repeat of a quote isn't much of a clarification.

8 hours ago, studiot said:

I take a photographic mesasurement of the length of the ship.

I observe 60 m

A photograph takes an image of a ship  It is not a measurement of length. I cannot determine the size of a bird from a photo of it against a blank background. So kindly clarify how this 60m was 'observed' instead of just repeating the assertion.

8 hours ago, studiot said:

As far as I can tell it is impossible to directly measure the proper length of something moving relative to yourself.

OK, but it can't be done when it's stationary either. Almost all measurements are indirect, if only by the fact that involve a computation, and you seem to suggest that any computation invalidates the value computed.

8 hours ago, studiot said:

13 hours ago, Halc said:

Proper length of a rigid inertial object can be directly measured.

If you think that is possible please describe how it might be done.

I don't know what you consider to be a direct measurement. I would have glued a tape measure onto the ship, but that involves subtracting the number on the tape where the front of the ship is from the value at the back, and subtraction is a calculation which you disqualify as a direct measurement.

8 hours ago, studiot said:

So do you know what stadia are ?

OK, you've barely given enough clue to actually get me to put in search criteria that makes sense. I've never heard of stadia before.

From https://en.wikipedia.org/wiki/Stadiametric_rangefinding and my bold

"The stadia method is based upon the principle of similar triangles. This means that, for a triangle with a given angle, the ratio of opposite side length to adjacent side length (tangent) is constant. By using a reticle with marks of a known angular spacing, the principle of similar triangles can be used to find either the distance to objects of known size or the size of objects at a known distance. In either case, the known parameter is used, in conjunction with the angular measurement, to derive the length of the other side. "

Point is, if you already know the size, you can determine the distance. If you know the distance, the size can be determined. Problem is with the ship we're looking at, we know neither. Hence my utter confusion as to how any kind of telescope can be used to determine the size of something at an unknown distance.

Maybe I'm stupid and there's a way to determine these when both are unknown, but the article there didn't seem to spell that out. I also still don't know the path the ship is moving, or what the orientation of the ship is relative to either that path or relative to the line of sight of the telescope.  Is the nose of the 100m approaching rocket in front like in cartoons and Hollywood movies, or is the engine-end in front like it typically is in reality, or maybe sideways? I suppose it depends if it's planning to accelerate, and which way.

 

Edited October 22 by Halc

[KJW]

11 hours ago, studiot said:

As far as I can tell it is impossible to directly measure the proper length of something moving relative to yourself.

If you think that is possible please describe how it might be done.

You could measure the proper length of a moving object using a ruler that is moving alongside the object, bearing in mind that the object is still moving relative to you, the person doing the measurement.

 

Edited October 22 by KJW

[studiot]

3 hours ago, KJW said:

You could measure the proper length of a moving object using a ruler that is moving alongside the object, bearing in mind that the object is still moving relative to you, the person doing the measurement.

 

Since this needs just a short reply and @Halc needs a much longer one I will deal with this first.

 

Quote

That is indeed part of the definition of proper length.

However How would you manipulate that ruler since it is moving relative to you?

Perhaps you could do it by drone if the relative movement was slow enough.

But the spaceship in this example is moving at nearly 0.8c.

Also you have to take into account the time for control signals to pass back and fore.

 

 

What I mean here is that you or someone else has to straddle the spaceship (ie be comoving with it) and run the tape measure over it.

That does not have to be done at the flyby.

All you need is that someone does it and records the measurement as the proper length.

 

As the ship flys by at some point you will actually be perpendicular to its direction of motion.

That is the moment to make the measurement and with a photograph you can get both ends at once.

 

That way you will be observing in the spaceship direction of motion and so be witnessing the shortening.

 

Length or distance measurement is differenet from time measurement as a a clock will retain a permanent record of the time dilation after the eventful journey.
This has been experimentally verified with aircraft and satellites on Earth.

But the spaceship does not retain any evidence of its length contraction and measures 100m again when it returns to Earth and docks.

[Halc]

5 hours ago, KJW said:

You could measure the proper length of a moving object using a ruler that is moving alongside the object

That was my suggestion.

I thought about what was trying to be conveyed, and I think the ship is photographed when it isn't approaching, but rather moving perpendicular to the line of sight, hence exposing its contracted length. My apologies for getting thrown off by the description of it 'approaching' which made me think the motion was of an object getting closer at the time the photo gets taken when it is actually moving away when the photo gets taken.

Yes, the ruler along side (or glued to it as I had suggested) would be a nice way to almost directly measure its proper length.

The ship will need to be significantly far away (hence the need for the telescope) else the middle will be significantly closer than the two ends, effecting light travel times and making the ship not appear centered.

I presume a rectangular ship.  The thing will appear deformed, and the rear of the ship will be visible in the photograph taken when the middle of the ship is centered in the telescope view aimed at the point of closest approach. The front of the ship will not be visible, so from that asymmetry, one can likely deduce from a still shot the direction it is moving.

1 hour ago, studiot said:

However How would you manipulate that ruler since it is moving relative to you?

Engineering problem irrelevant to this topic.

1 hour ago, studiot said:

What I mean here is that you or someone else has to straddle the spaceship (ie be comoving with it) and run the tape measure over it.

That does not have to be done at the flyby.

All you need is that someone does it and records the measurement as the proper length.

Equivalently, you can just ask the ship occupants how long their ship is, even if the occupant is somebody you put there with a tape measure. Somehow that feels like a measurement in the ship frame though, a violation of what we're trying to do.

1 hour ago, studiot said:

As the ship flys by at some point you will actually be perpendicular to its direction of motion.

That is the moment to make the measurement and with a photograph you can get both ends at once.

Actually no. Say the ship is a light year away at closest approach, and I aim my telescope at that closest approach point in space (my frame). The time to take the shot is a year after the ship was there because that's when the ship will appear in the picture for a microsecond or so. You want to take the shot when it appears closest, not when it actually is closest.

1 hour ago, studiot said:

But the spaceship does not retain any evidence of its length contraction and measures 100m again when it returns to Earth and docks.

What if it's the Earth that's moving, and it's you with the telescope that needs to return back to the inertial ship that was never any proper length other than 100m?  I say this because the comment above smacks of a suggestion of a preferred frame.

 

Bell's Spaceships

The Bell spaceship scenario nicely illustrates proper distance vs proper length.

You have to small identical ships stationary in frame X and separated by a light day in X.  They're pointed east, and the east ship has a light-day string attached behind it reaching to the other ship but not attached.  At time 0 in X, both ships accelerate at a proper 1g for 1 year ship time (about 1.18 years in X) at which point they run out of fuel.

In X, the two ships are at all times separated by a coordinate distance of one light day but their proper separation (not really defined until fuel runs out) becomes larger. The string is moving in X and its contracted length no longer reaches the west ship. Were it attached there, it would break, which is the point of Bell's scenario.

In the inertial frame in which both ships eventually come to a halt, the string is stationary and is of coordinate length 1 light day, but the ships are now separated by 1.58 light days, and since the ships become mutually stationary when both their fuels run out, that is a proper separation. Point is, proper separation of ships went up (not contracted), but the coordinate length of the string went down in X (length contraction).  The string is treated as a rigid thing and its proper length never changes.

Edited October 22 by Halc
Add bell

[Mordred]

A suggestion instead of looking at physical objects its seems to me you could simply examine the problem using wavelength.

[studiot]

3 hours ago, Halc said:

Equivalently, you can just ask the ship occupants how long their ship is, even if the occupant is somebody you put there with a tape measure. Somehow that feels like a measurement in the ship frame though, a violation of what we're trying to do.

Of course it is a measurement in the ship frame which again it must be to be a proper length.

Why is that controversial ?

 

3 hours ago, Halc said:

Say the ship is a light year away at closest approach,

I didn't say anything of the sort.

So I am glad you said this

3 hours ago, Halc said:

Engineering problem irrelevant to this topic.

Because I am relying on a good quality pilot that can pass by really close (without giving a figure for now)

 

3 hours ago, Halc said:

My apologies for getting thrown off by the description of it 'approaching' which made me think the motion was of an object getting closer at the time the photo gets taken when it is actually moving away when the photo gets taken.

Accepted.

Think of Einstein's train examples.

 

Two things.

Stand on the platform and watch the approach.
At distance the train will appear to be heading straight for you.
You will only be able to see that it will pass you close by at it gets very near.

The other point is that this example enabled Einstein to mark both the train and the platform (ie both the stationary and moving systems) with one lightening flash.

Modern technology has progressed  long way from the time when we used to huddle round a surveyor's telescope, trying to keep dry at night and hoping to catch the transit of a star.
Today we can take a whole series of (perhaps digital) pictures before before and after the event at very high speed.
Then we can have a computer find the best interpolation for the exact measurement we want.

 

But you are right about the fiducial marks you require to take the measurement,
This is one of the correct terms you need to research if you want to - stadia are a form of fiducial marks.
They are used in microscopes, surveyors telescopes, photographic cameras, tacheometers, photogrammetric methods, medical science amongst others.

So your special engineers can arrange something in space that the ship can pass between them and the Earth to both appear on the photographs.

As to the photographs, I wouldn't worry about the time of flight it will be incredibly short, many orders of magnitude shorter than the time it takes for the photographic film or digital sensors to react.

 

[Halc]

16 minutes ago, studiot said:

3 hours ago, Halc said:

Say the ship is a light year away at closest approach,

I didn't say anything of the sort.

I know, but it can't be very close.  Far away removes a lot of distortion. You want to keep the light travel time from all parts to you as similar as possible, and it's not if you're far closer to the middle than you are to the ends. So no need for your pilot to get it close, especially since it is inertial and not under pilot control. We have a good telescope, so distance is not an issue.

19 minutes ago, studiot said:

Think of Einstein's train examples.

OK, but Einstein relied on timing of very explicitly defined light pulse events and not on pictures taken by observers in different frames. All the measurements of lengths (proper or otherwise) were computed, not directly measured. I still don't know how one can make a direct measurement of anything that doesn't involve a computation at some point, but you seem to disqualify a measurement of the proper length of a moving object based on the fact that the measurement involves a computation.

 

23 minutes ago, studiot said:

At distance the train will appear to be heading straight for you.
You will only be able to see that it will pass you close by at it gets very near.

Sure, but that doesn't say how its length gets measured, coordinate or proper. The train is receding when your picture is taken. Let's assume there is a red stripe mid-ship and the camesra snaps a pic just as that red stripe is centered in the frame. If you're close enough, the train end will still be getting closer when that shot is taken, but the middle and front will both already be receding at that time.

The observer can use a simple clock to determine how long it takes for the train's length to pass.  Sure, that involves a computation, but it's a measurement nonetheless.  Is that direct?  Don't know your criteria for that.

28 minutes ago, studiot said:

The other point is that this example enabled Einstein to mark both the train and the platform (ie both the stationary and moving systems) with one lightening flash.

Two flashes, each leaving a mark both on the train and the platform, each pair of marks presumed to be spatially separated by negligible distance.

31 minutes ago, studiot said:

Then we can have a computer find the best interpolation for the exact measurement we want.

OK, so you do accept computation as a valid form of measurement, meaning we do actually measure the proper distance of the ship in the frame where it's moving. all presuming we've measured its coordinate length, which has yet to be described.  The photo we took (or a series of them if that helps) didn't convey it, it only conveys one subtend angle, and that only works if the ship is quite a distance away and we know that distance, something the telescope doesn't yield.

Not saying it can't be done, just saying that you need more than a telescope which only measures angles, not distances.

37 minutes ago, studiot said:

So your special engineers can arrange something in space that the ship can pass between them and the Earth to both appear on the photographs.

What, a moving ruler?  Only works if we know it is at the ship, which is why I had suggested the glue. A ruler halfway between moving ship at ship speed will make the ship appear around 50m long, and will appear to be moving relative to the ship.

 

41 minutes ago, studiot said:

As to the photographs, I wouldn't worry about the time of flight it will be incredibly short, many orders of magnitude shorter than the time it takes for the photographic film or digital sensors to react.

I am assuming zero time shutter speeds.  But for a nearby ship, the flight time of the light can be 10x longer from the ends as it is from the middle, leading to that massive distortion in the picture. The front will appear far shorter than the rear half of the ship.  So we let it pass at a great distance where the light travel times are almost identical. That's why I suggesting it passing a light year away.

[studiot]

4 hours ago, Halc said:

4 hours ago, studiot said:

The other point is that this example enabled Einstein to mark both the train and the platform (ie both the stationary and moving systems) with one lightening flash.

Two flashes, each leaving a mark both on the train and the platform, each pair of marks presumed to be spatially separated by negligible distance.

Einstein is famous for his train analogies.

But they didn't appear in his papers.

Instead he wrote several books and you will find them in one or other of these.

I do not know if I have seen all of them but here is an extract from one

 

I see no reference to the flashes beeing separated by "negligable distance"

 

So do you have a reference for this claim ?

[Halc]

Thank you for ignoring everything that matters and finding something to nitpick about.

5 hours ago, studiot said:

I see no reference to the flashes beeing separated by "negligable distance"

So do you have a reference for this claim ?

Says the guy who doesn't back his claims.

 

I did not claim it was a quote. Throughout the small paper, each lightning strike is treated as a single event in two-dimensional spacetime. Were that not the case, the entire conclusion would be invalid.

You also only showed the first page, just before he describes the relevant bits.

 

So I had a long car ride today and used 20 minutes of mental free time to consider the case of a 100m (proper len) ship passing really close to the observer.  How big would it appear when going by at 0.8c?  Turns out it appears larger. The observer would measure 165 meters in the photo he takes when the center of the ship passes him, and if he condescends to compute anything more, he can deduce 100m proper length from that.

The middle of the ship (moving left to right) is directly in front of the camera as specified. The front half of it appears to extend 15m to the right and the rear half 150m to the left. Mind you, this is what the photo shows, not where the ship actually is when the photo is taken. The photo cannot show where the ship actually is since the camera cannot be everywhere present along its length simultaneously. The right kind of camera can, but we're using a wide angle camera which constitutes a single point of view (as does any telescope). We presume it takes a 360 degree image of everything with sufficient resolution to see what we need to.

The further away the camera is from the middle, the longer the front appears and the shorter the rear appears, approaching 30m each if the camera is infinitely distant. That's why I put my camera a light year away in my earlier example, but during the car ride I wanted to work out the limit when the ship passed arbitrarily close.

 

So your camera was quite close, and you claim it shows a 60m ship. You need to justify that claim with a description of how that is done without positing a magic device that just somehow knows. Use mathematics. I can do that with mine, although I have not yet done so since I did it all in my head.

I used an ship with length 8 (natural units) and then scaled that up to our example of 100m.

 

From a light year away, the photo shows the name of the ship printed on the rear of the rectangular prism shaped thing.  From a photo taken nearby, this name is not visible.  I think the camera needs to be at least 23 meters away for the name to be in the photo, if I computed that correctly. None of this paragraph is relevant to determining the ship's proper length except to illustrate that the photo will show a distorted image, making the task not entirely trivial.

Edited October 23 by Halc