If she does this, then (according to special relativity) she will find the rod's length to be unchanged from its "basic" length (its length before it was in motion with respect to you). (We're assuming that the accelerations involved in setting the rod into motion were sufficiently gentle to not permanently distort it.) This means, for example, that if the "rod" is actually made up of a number of eggs placed end-to-end, that she will NOT observe these eggs to have been crushed.

This framework -- defining "the length of the rod" as that measured by an observer riding along with the rod (this is sometimes called a "proper observer"), is the basis for the statement you quoted above

>	But SR predicts that the moving rod (or train) does NOT "become shorter".

Now let's consider what *you* might observe. That is, you see the rod moving from left to right, and as it's moving you (somehow) measure it's length. What will you measure?

As I noted above, answering this question requires being very precise about just how the measurement is made. (It turns out that this case requires rather more precision than was the case for the proper-observer measurement.)

One obvious way for you to measure (define!) the length of the rod is for you to measure the position of the left end of the rod at some time t, and for you to measure the position of the right end of the rod at the same time t, and then for you to subtract the two positions.

[In the previous sentence I'm using "measure" in the special-relativity sense, i.e., we assume a (gedanken) infinite set of sub-observers, all at rest in your inertial reference frame, one at each spatial position. Each sub-observer has her own clock, and all these clocks are synchronized to each other via the Einstein clock-synchronization convention in your inertial reference frame.

The position of an end of the rod at a time t is then measured by that (unique) sub-observer who is coincident with that end of rod at her clock's time t. Since that sub-observer is coincident with her end of the rod, there are no light-travel-time delays from the moving rod to the sub-observer.

[It's precisely this avoidance of light-travel-time effects that's the motivation for introducing the infinite set of sub-observers.]

After making their measurements, the sub-observers then send their observations back to some central collation point for analysis.]

In relativity, "time" is observer-dependent, so we need to be specific about whose (which reference frame's) "time t" we're using. The simplest choice is to say that since *you* are making this measurement (in *your* inertial reference frame), then you should measure the left and right rod ends' positions at times which are simultaneous in *your* inertial reference frame, i.e., your sub-observers should each measure the rod ends' positions at the same clock reading (t).

That is, to recap,
* the subobserver who is coincident with the left end of the rod at her clock reading t sends a message back to the central collation point to that effect (the message includes the subobserver's position, say x = x_left, in your inertial reference frame)
* the subobserver who is coincident with the right end of the rod at her clock reading t sends a message back to the central collation point to that effect (the message includes the subobserver's position, say x = x_right, in your inertial reference frame)
* the central collation point computes L_you := x_right - x_left

This (L_you) is a reasonable operational definition of "the length of the rod as measured by you".

According to special relativity, if you do this, you will find that this length will show Lorentz contraction. Aswering *this* question is the basis of the other statement you quoted above,

>	It also predicts the moving rod WILL BE OBSERVED to "look" shorter.

It should now be clear(er) that the two apparently-contradictory statements you quoted above, are in fact the answers to two *different* questions... so it's no longer surprising or contradictory that their answers are different. -- jt]]

>	[[Mod. note -- Note that in the final paragraph above, the author is using "will be observed" in the technical sense of special relativity, i.e., this refers to observations made by a CO-LOCATED observer so that there are no speed-of-light propagation delays.]] -- jt]]

IMO physical length contraction becomes a paramount problem, considering an experiment with a set of identical rods flying or moving freely through the entire universe all with different speeds. Is there any length contraction involved? This problem involves to calculate the two positions (front and back) of each rod at regular intervals. This is specific tricky when the distances between the front and the end to a central observation point are different. (This situation is easier when all the rods move in a storage ring in which all the rods have the same speed, which an adjustable speed for all rods)

The same problem exists when you want to observe the positions of the planets around the Sun. You need in some way a set of accurate clocks which all run synchronuous to monitor the positions, such that speed of light propagation delays can be eliminated.

IMO the most important question to answer is: what is the physical explanation of length contraction. The explanation can not be in the way the rods are observed, assuming light signals are involved.

Nicolaas Vroom

17 How to test length contraction by experiment?

18 How to test length contraction by experiment?

I agree.

>

Then the simplest case is one where the observer-on-the-rod measures the length of the rod. In this case, measuring the rod's length is easy (for her): she measures the position of the left end of the rod then she measures the position of the right end of the rod and she subtracts the two positions.

The question is how exactly does she makes this measurement? Does she performs this measurement using a set of standard rods?

>	If she does this, then (according to special relativity) she will find the rod's length to be unchanged from its "basic" length (its length before it was in motion with respect to you).

I assume you mean its rest length? The question is how is this rest length measured? Is this measurement performed using a set of standard rods?

I expect if both measurements are performed in the same way (Using both standard rods) no length contraction will be measured because if there is any length contraction both her moving rod and the moving standard rods will change, making it disappear.

>	This framework -- defining "the length of the rod" as that measured by an observer riding along with the rod (this is sometimes called a "proper observer"), is the basis for the statement you quoted above

> >	But SR predicts that the moving rod (or train) does NOT "become shorter".

>

Now let's consider what *you* might observe. That is, you see the rod moving from left to right, and as it's moving you (somehow) measure it's length. What will you measure? As I noted above, answering this question requires being very precise about just how the measurement is made. (It turns out that this case requires rather more precision than was the case for the proper-observer measurement.)

I agree. I think finally both require the same precision, specific is no length contraction is observed (which could be the case in my proposal)

>	One obvious way for you to measure (define!) the length of the rod is for you to measure the position of the left end of the rod at some time t, and for you to measure the position of the right end of the rod at the same time t, and then for you to subtract the two positions.

That is what I'm proposing in my experiment. I want to measure (compare) the positions when the center of the rod is exactly in between the rod at rest. That is moment t.

>

The position of an end of the rod at a time t is then measured by that (unique) sub-observer who is coincident with that end of rod at her clock's time t. Since that sub-observer is coincident with her end of the rod, there are no light-travel-time delays from the moving rod to the sub-observer. [It's precisely this avoidance of light-travel-time effects that's the motivation for introducing the infinite set of sub-observers.] After making their measurements, the sub-observers then send their observations back to some central collation point for analysis.] In relativity, "time" is observer-dependent, so we need to be specific about whose (which reference frame's) "time t" we're using.

See above how that moment is defined in my proposed experiment.

>

The simplest choice is to say that since *you* are making this measurement (in *your* inertial reference frame), then you should measure the left and right rod ends' positions at times which are simultaneous in *your* inertial reference frame, i.e., your sub-observers should each measure the rod ends' positions at the same clock reading (t). That is, to recap, * the central collation point computes L_you := x_right - x_left According to special relativity, if you do this, you will find that this length will show Lorentz contraction. Answering *this* question is the basis of the other statement you quoted above,

> >	It also predicts the moving rod WILL BE OBSERVED to "look" shorter.

>	It should now be clear(er) that the two apparently-contradictory statements you quoted above, are in fact the answers to two *different* questions... so it's no longer surprising or contradictory that their answers are different. -- jt]]

When I understand everything correct at every moment t there is always one observer at the front end of the rod and one at the back end of each rod such that we can measure the length of two rods, one at rest and one moving in the same (inertial) frame simultaneous. What I also understand that the measured length of the moving rod is shorter than the measured length rod at rest and a function of the speed v of the moving rod. That means this demonstrates length contraction.

This leaves open the question what is the physical explanation.

What is difficult to understand the experiment, which is a thought experiment, should be in agreement, as SR predicts, while the experiment I propose, which is much simpler, is supposed to physical fail (?) IMO every experiment is very difficult to perform in reality

Nicolaas Vroom.

19 How to test length contraction by experiment?

This is really geometrical: both "length contraction" [#] and "time dilation" [#] are simple geometrical projections. No physical properties of anything are changed, but when you measure such properties from frames relative to which an object is moving, you get values different from when it is at rest.

Think about it: it must be possible for multiple observers in multiple frames to observe a given object, and they all get different values. Geometrical projections do just that, but ask yourself how any "physical change" of the object could do it....

These projections are directly analogous to "length contraction" and "time dilation", with the line's angle analogous to relative velocity of inertial frames, except the y axis is the time axis, and the geometry is hyperbolic (not Euclidean).

[#] I put them in "scare quotes" because these are not very good names -- they imply a length actually contracts (none does) and time actually dilates (it doesn't). But they are solidly established historically.

20 How to test length contraction by experiment?

21 How to test length contraction by experiment?

Personally I prefer the method described by JT in which in principle you use one (inertial) reference frame and describe (measure) the complete state of the universe simultaneous.

>	Think about it: it must be possible for multiple observers in multiple frames to observe a given object, and they all get different values.

I agree with you when a rod is considered because in all these measurements different physical time delays are involved, caused by the light travel time of the photons along different path length. However and that is important: these physical measurements don't cause any physical change within the rod. The length does not physical change.

>	Geometrical projections do just that, but ask yourself how any "physical change" of the object could do it....

But geometrical projections is mathematics ......

The problem is time dilation of moving clocks. If you want to know the time of moving clock which moves away from you, you get a time count that is always earlier than the actual time at the moment when you receive that information. The reason is simple communication time between sending and receiving any message (which is a function of distance) Sending and receiving does not cause any physical effect on the clocks in use.

What causes a physical effect is the internal operation of a clock if you compare a moving clock versus a clock at rest. The moving clock runs physical slower compared to a clock at rest. Depending about the internal operation of the clocks the mathematics which describe this physical behaviour are the Lorentz Transformations

Nicolaas Vroom

22 How to test length contraction by experiment?

Geometry, and specifically geometrical projections, can have physical consequences:

You can carry a ladder through a narrow doorway in some orientations but not others -- the geometrical projection of ladder's length onto doorway's width determines this.

Yes, collisions between ladder and doorframe are the cause, but the geometrical projection determines whether such a collision will occur.

>

The problem is time dilation of moving clocks. If you want to know the time of moving clock which moves away from you, you get a time count that is always earlier than the actual time at the moment when you receive that information. The reason is simple communication time between sending and receiving any message (which is a function of distance) Sending and receiving does not cause any physical effect on the clocks in use.

You are confusing Doppler shift with "time dilation". The usual approach to separate them is to use a set of assistants arrayed along the path of the moving clock, each at rest in the observer's frame with a clock synchronized to the observer's clock; the assistants note the time on the moving clock as it passes, and also their own clock, and they send these notes to the observer, who can then measure "time dilation" without any light delays or Doppler shift. (The moderator already mentioned this approach in this thread.)

>	What causes a physical effect is the internal operation of a clock

No. The internal operation of a clock is unaffected by its motion (relative to anything) -- that's the first postulate of SR.

The laws of physics govern the internal operation of every clock, and the first postulate says they are the same in every locally inertial frame.

>	The moving clock runs physical slower compared to a clock at rest.

Only if the first postulate is wrong. Zillions of experiments show that it is correct.

The observer will OBSERVE the moving clock to run slower than her own clock, but the moving clock ITSELF is unaffected (i.e. it does NOT run "physically slower").

Be careful what your words mean. "This clock runs slower" refers to the clock AND NOTHING ELSE, and is therefore wrong. "This observer measures that moving clock to run slower than her own clock" is a correct statement of "time dilation".

Tom Roberts

23 How to test length contraction by experiment?

24 How to test length contraction by experiment?

On Fri, 26 Jul 2019 08:57:08 PDT, "Phillip Helbig (undress to reply)" wrote:

>

What is difficult to understand is the twin paradox: After A goes away and comes back while B stays at home and they then compare clocks at rest, EVERYONE agrees that A's clock has ticked less. Recent discussion here shows that acceleration is not the "cause", since the effect depends on the length of the journey, and not on the acceleration. Since all clocks (mechanical, electronic, atomic, biological, nuclear) are equally affected, it is a) hard to imagine that some mechanism affects them all equally and b) no-one has any idea what such a mechanism could be.

>	Yes, one can adopt the "shut up and calculate" approach and calculate the strength of the effect, and it agrees with observations, but this is not the same as understanding. The question is whether such an understanding is possible.

I wonder if Mach's Law is involved. Nobody understands that either, but both issues involve an unseen reference frame of some kind. Maybe it's the same one.

25 How to test length contraction by experiment?

>	I think that everyone understands purely illusory effects: A sees B's clock running slower and vice versa.

I doubt if it is that simple. You must know the whole physical situation from start to end of this experiment, because it involves both physical (mechanical?) and optical effects.

There is a difference if A sees B's clock running slower or running behind. Running behind means that the difference in clock readings is constant during a certain period is constant. This can only be when the distance is fixed. As soon as any clock moves (which requires a force) this experiment becomes like the twin experiment with moving clocks. See below.

>	It is clear that these are only apparent effects and not "real".

Length contraction is simpler to study

>	The question of what is observed is trickier, because the finite speed of light has to be taken into account. There WAS some real confusion about this, but it was cleared up by Terrell and Penrose a long time ago.

What physicists should do is only to discuss measurements and how these measurements are made. When I see or observe something this are also measurements, but they are relatif and involve my position or my opinion.

When I observe a train which moves away from me and this train all of a sudden stops, then the first thing I will observe (measure) later is that the back of the train stops. At that moment the front is still moving away. A little later the front. This means the length of train during a small period becomes longer. This is an optical effect and not a physical effect. (later implies light-travel-time effects)

The same thing happens when a train approaches me. In that case I will observe (see, measure) that when the train suddenly stops, that the front stops first. At that moment the back is further away. A little later the back stops and during that small period the train becomes shorter, optical. This is I think what Terrell meant.

>	[[Mod. note -- Note that in the previous paragraph, the author is

using > the word "observed" NOT in the technical-special-relativity sense (of > measurements made by instantaneously co-located sub-observers so as to > eliminate light-travel-time effects), but rather in the sense of "what > a camera image would show". That is, the author's usage of "observed" > INCLUDES light-travel-time effects. > -- jt]]

When you use this method, and eliminate all light-travel-time effects we will get instantaneous measurements for all objects involved (in one reference frame) The measurements will indicate that the physical length of all moving rods will be the same (and not change).

But when identical clocks are involved free moving though space, which initially all show the same time at position (p,t), after a certain time all will be at different positions and show different clock readings. What the measurements also could show is, that the clocks do not undergo any form of length contraction.

>

What is difficult to understand is the twin paradox: After A goes away and comes back while B stays at home and they then compare clocks at rest, EVERYONE agrees that A's clock has ticked less. Recent discussion here shows that acceleration is not the "cause", since the effect depends on the length of the journey, and not on the acceleration. Since all clocks (mechanical, electronic, atomic, biological, nuclear) are equally affected, it is a) hard to imagine that some mechanism affects them all equally and b) no-one has any idea what such a mechanism could be.

To explain the above each clock should intially undergo a different force (in a different direction) or a temporarily acceleration which will change the speed of each clock differently, including its internal operation, which is based on (the direction of) the speed of the clock and (the direction of) the speed of light inside the clock. The overall result will be that the # of ticks will be different, which can be demonstrated when they meet again at one point.

Nicolaas Vroom

26 How to test length contraction by experiment?

27 How to test length contraction by experiment?

When I walk away from a building the observed height from my point decreases. This is a pure optical effect and has no physical implications i.e. the height of the buiding does not change.

> >	because it involves both physical (mechanical?) and optical effects.

>	Not in the case of purely relative motion.

> >	There is a difference if A sees B's clock running slower or running behind.

>	It runs behind because it runs slow.

Consider the following clock readings, both clocks are placed in front of me.

Case 1
Clock A    10:00   11:00  12:00   13:00  14:00
Clock B    10:00   10:30  11:00   11:30  12:00

Case 2
Clock A    10:00   11:00  12:00   13:00  14:00
Clock B     6:00    7:00   8:00    9:00  10:00

In the above two examples no motion is involved.

Case 1 can also be used as an example that there is motion involved i.e. that clock B moves away from clock A.
But the cause is now both optical (light travel time) and physical (clock B runs slower). This is the case with the twin paradox.

>	In the classic twin paradox, only one clock is accelerated.

In the classic twin paradox one clock is accelerated (deaccelerated) three times: At the start, at the midway point, at the end.

>	While it is clear that the accelerated clock runs slower (depending on the length of the journey and not on the magnitude of the acceleration!), it is less clear that acceleration "causes" this in the sense of physically affecting the operation of the clock.

The physical change in the performance (ticking rate) of the moving clock is caused by the forces enforced on the clock.
If the direction of the internal lightsignal, which creates oscillations (each cylce defines one count), is in the direction of the moving clock, then the path length will become longer and the clock rate decreases. The path length is defined as the distance between two mirrors, one in the back and one in the front.
If the lightsignal starts in the back and moves toward the front mirror, then When this mirror moves away, the time for the lightsignal to reach the mirror in front increases.
The faster the clock speed, the longer this distance. In theory when the speed approaches the speed of light, this time will reach infinite and the clock will stop counting. All this is physics.

If you drop an hourly glass, which internal operation is based on gravity, during the fall the internal glass particles will not move towards the other container, thereby increasing the cycle time i.e. the time that one container becomes empty and the hourly glass has to be turned over. As a consequence the hourly glass will run slower (just like a moving clock). All this is physics

What is also interesting is that the behaviour of a clock is influenced by the speed of light (photons) and an hourly glass by the speed of gravity (gravitons). Both are also influenced by external (temporary induced) forces.

Nicolaas Vroom.

28 How to test length contraction by experiment?

An hours glass just doesn't work when it is in free fall. While certainly a consequence of physical law, its failure has nothing to do with the relativistic slowing (in another frame) of an otherwise functioning clock. I can make the other clock stop by hitting it with a hammer. This has similarly nothing to do with relativity, but serves to emphasise that real clocks are designed to function within certain constraints, and cannot be expected to give valid results outside those constraints, with one typical constraint entailing not being hit with a hammer.

>	What is also interesting is that the behaviour of a clock is influenced by the speed of light (photons) and an hourly glass by the speed of gravity (gravitons). Both are also influenced by external (temporary induced) forces.

The behaviour of the clock in a different frame depends on the universal constant c. Light travels at that speed, but photons are not causing the relativistic effects.

I have no idea what is meant by "induced forces". It sounds like something that has been made up.

Sylvia.

29 How to test length contraction by experiment?

30 How to test length contraction by experiment?

> >

If you drop an hourly glass, which internal operation is based on gravity, during the fall the internal glass particles will not move towards the other container, thereby increasing the cycle time i.e. the time that one container becomes empty and the hourly glass has to be turned over. As a consequence the hourly glass will run slower (just like a moving clock).

>	An hours glass just doesn't work when it is in free fall.

I agree.

>	While certainly a consequence of physical law, its failure has nothing to do with the relativistic slowing (in another frame) of an otherwise functioning clock.

I comparing performance of an hourly glass with a clock in extreme. That means the hourly glass falls (with an average speed v) and the clock moves almost at the speed of light (c). Result both don't work. However if the speed of the hourly glass is 0.1*v and the speed of the clock 0.1*c both will function (operate) however as a clock each will run slower.

[[Mod. note -- In relativity, we need to (implicitly or explicitly) qualify statements of the form "X moves with speed V" with the reference frame with respect to which the speed is defined. E.g., "if the speed of the hourly [sic] glass is 0.1*v" needs to be expanded into something like "if the speed of the hour glass WITH RESPECT TO SUCH-AND-SUCH AN OBSERVER is 0.1*v".

And, suitable clocks (e.g., the decay of unstable elementary particles) work just fine when moving almost at the speed of light with respect to a laboratory. -- jt]]

> >	What is also interesting is that the behaviour of a clock is influenced by the speed of light (photons) and an hourly glass by the speed of gravity (gravitons). Both are also influenced by external (temporary induced) forces.

>	The behaviour of the clock in a different frame depends on the universal constant c. Light travels at that speed, but photons are not causing the relativistic effects.

Photons are the physical reason that a clock operates i.e. functions as an oscillator and ticks.

For an hourly glass this are the glass particles. A human is required to take care that the hourly glass becomes an oscillatot i.e. for time keeping.

>	I have no idea what is meant by "induced forces". It sounds like something that has been made up.

In order to operate properly, a clock or a hourly glass should stay at rest and not be moved.
Moving a clock (or hourly glass) will influence this normal operation. (And the clock will start ticking slower)
To move a clock (or hourly glass) you need an external induced (applied?) force. Dropping an hourly glass mimics such a force.

[[Mod. note -- Again, phrases like "should stay at rest and not be moved" are ambiguous, because they don't specify a reference frame. For example, does "stay at rest and not be moved" mean (a) "not be moved with respect to a laboratory located at the Earth's south pole"? Does "stay at rest and not be moved" mean (b) "not be moved with respect to a laboratory located on the Earth's equator"? (Since the Earth's equator moves at about 460 meters/second relative to the Earth's south pole, it's impossible to be at rest with respect to both of them simultaneously.)

And that said, some types of clocks (e.g., radioactive decay)

[That is, our "clock" is a mass of some radioactive isotope, and we measure time by counting decays/second and using the isotope's known half-life to determine elapsed time since we set up the clock.]

don't need any external forces to operate. -- jt]]

The same problem exists when you place a chronometer on board of a ship. Special design of such a chronometer is required to minimise the influence of the waves. All of that is physics.

Nicolaas Vroom

31 How to test length contraction by experiment?

I agree with you. When you try to quantify physical processes you place yourself on icy ice. The general idea is to place two identical clocks (or hourly glasses) side by side. When you move one (this always requires some force) its behaviour will change and the clock will run slower.

>	And, suitable clocks (e.g., the decay of unstable elementary particles) work just fine when moving almost at the speed of light with respect to a laboratory.

My first guess is that a clock based on the decay of unstable elementary particles (its counting rate) is not constant over a long period.

>

-- jt]]

>

In order to operate properly, a clock or a hourly glass should stay at rest and not be moved. Moving a clock (or hourly glass) will influence this normal operation. (And the clock will start ticking slower) To move a clock (or hourly glass) you need an external induced (applied?) force. Dropping an hourly glass mimics such a force. [[Mod. note -- Again, phrases like "should stay at rest and not be moved" are ambiguous, because they don't specify a reference frame. For example, does "stay at rest and not be moved" mean (a) "not be moved with respect to a laboratory located at the Earth's south pole"? Does "stay at rest and not be moved" mean (b) "not be moved with respect to a laboratory located on the Earth's equator"? (Since the Earth's equator moves at about 460 meters/second relative to the Earth's south pole, it's impossible to be at rest with respect to both of them simultaneously.)

Also here I agree with you. What exactly means at rest? In most experiments (in books) one clock stays on earth and in other clock is ejected in space with a speed of 0.3 * c. The observer on earth is considered in his reference frame at rest. If that is commonly accepted, I have no problem. Maybe it is better to consider the observer always hypothetical at the centre of the earth.

>

And that said, some types of clocks (e.g., radioactive decay) [That is, our "clock" is a mass of some radioactive isotope, and we measure time by counting decays/second and using the isotope's known half-life to determine elapsed time since we set up the clock.] don't need any external forces to operate.

I fully agree that such a clock does not need any force to operate. That is the same for a clock using lightsignals (photons) or an hourly glass (except to turn the clock when one container is empty)

Consider you have two identical clocks and one clock is moved (is influenced by an external force). The question is: does this have consequence for the performance of the moving clock versus the not moving clock. For a clock based on light signals the answer is yes, because the path length of the photons will become longer.
For an hourly glass the answer is also yes. The issue is gravity. Dropping an hourly glass a couple of times (or shaking carefully) will increase the path length.

32 How to test length contraction by experiment?

33 How to test length contraction by experiment?

Yes, "time dilation" does not affect the moving clock being observed, but it is not really "illusory", because it can have real and measurable physical consequences.

For instance, charged pions have a proper lifetime of 26 nanoseconds, which at 0.999999 c corresponds to traveling only 7.8 meters. Fermilab and CERN have had pion beamlines over a kilometer long because the "time dilation" of high-energy pions permits such long beamlines to work -- this is no "illusion".

>>	There is a difference if A sees B's clock running slower or running behind.

>	It runs behind because it runs slow.

This is just plain not true, and I wish physicists who OUGHT to know better would stop repeating such errors. The clock does NOT "run slow", it runs at its usual and natural rate -- the first postulate of SR would be violated if this were not so.

Such incorrect statements are all too likely to confuse non-experts, including some participants in this newsgroup.

Yes, in the twin paradox the traveling twin's clock accumulates less elapsed proper time than the earthbound twin's clock. But it does NOT "run slow" -- rather it travels a shorter path through spacetime.

Please remember how English works -- in the sentence "That clock runs slower than this one", only the clocks are mentioned; in SR this is simply false because the laws of physics that govern both clocks' ticking are the same in the rest frames of both clocks. To make a correct statement about this you MUST mention how its tick rate is measured. A clock moving relative to inertial frame S does not "tick slow", but S will measure it to tick slower than identical clocks at rest in S.

>>>

What is difficult to understand is the twin paradox: After A goes away and comes back while B stays at home and they then compare clocks at rest, EVERYONE agrees that A's clock has ticked less. [...] Since all clocks (mechanical, electronic, atomic, biological, nuclear) are equally affected, it is a) hard to imagine that some mechanism affects them all equally and b) no-one has any idea what such a mechanism could be.

This is an example of the confusion created by incorrect statements about "clocks running slow". The resolution is simple: the clocks do NOT "run slow", so no "mechanism" is needed.

>	In the classic twin paradox, only one clock is accelerated. While it is clear that the accelerated clock runs slower [...]

Again, stop making such erroneous and confusing claims. The clock does NOT "run slower"; rather, it accumulated less elapsed proper time (and did so while ticking at its usual and natural rate). The reason for this is geometrical: the traveling twin followed a shorter path through spacetime than did the earthbound twin.

I hesitate to call that a "cause", but it certainly is an explanation.

34 How to test length contraction by experiment?

But it CAN have physical implications. For instance, if you want to photograph the building, close up it won't fit in the camera's aperture, but far away it will.

Note this is a geometrical projection, which is directly analogous to "time dilation" and "length contraction". For them, relative motion varies the projection, while for your example it is distance that varies the projection.

Tom Roberts

35 How to test length contraction by experiment?

An hourglass is NOT a clock. Hourglass+earth is the clock, and if you change the relationship between the hourglass and the earth you have a completely different clock. Of course no hourglass is accurate enough to demonstrate any relativistic effects. Ditto for a pendulum or grandfather clock.

>

Consider you have two identical clocks and one clock is moved (is influenced by an external force). The question is: does this have consequence for the performance of the moving clock versus the not moving clock. For a clock based on light signals the answer is yes, because the path length of the photons will become longer.

No! You keep making the same mistake: not specifying the inertial frame used for reference.

If you want to discuss a clock's tick rate, you must use its own rest frame. For a light clock, in its rest frame the light path lengths are constant; regardless of which frame it is at rest in, its tick rate does not change (remember the speed of light is also the same in all inertial frames).

If you want to discuss a clock's tick rate as measured in some inertial frame other than its rest frame, you MUST say so. For any good clock, it is only when measured from an inertial frame relative to which it is moving that the observed tick rate changes.

Or when observed from a different altitude in a gravitational field.

As I keep saying, discussing relativity requires precision in thought and word. Your statements repeatedly fall short of what is required, and you MUST improve the precision of your thinking in order to understand this.

>	For an hourly glass the answer is also yes. The issue is gravity. Dropping an hourly glass a couple of times (or shaking carefully) will increase the path length.

Dropping an hourglass changes its relationship to the earth and you have a completely different clock. This is useless.

>	I don't know how constant clocks based on radioactive decay are,

There are measurements of "time dilation" using radioactive ions in a beam moving at an appreciable fraction of c, but I have lost the references. Here are references for a "clock" made from the decay rate of muons:

Bailey et al., =E2=80=9CMeasurements of relativistic time dilation for positive and negative muons in a circular orbit,=E2=80=9D Nature 268 (July 28, 1977) pg 301. Bailey et al., Nuclear Physics B 150 pg 1=E2=80=9379 (1979).

They stored muons in a storage ring and measured their lifetime. The measured decay rate is consistent with the prediction of SR.

When combined with measurements of the muon lifetime at rest this becomes a highly relativistic twin scenario (v ~0.9994 c), for which the stored muons are the traveling twin and return to a given point in the lab every few microseconds. While being stored in the ring they were subject to a proper acceleration of approximately 10^18 g (1 g =3D 9.8 m/s2).

Tom Roberts

36 How to test length contraction by experiment?

There is no material length contraction. A meter stick will return the same length after a journey. What the LT length contraction mean: The light-path-length (LPL) of a moving meter stick is predicted to be contracted by a factor of 1/gamma. This prediction by the observer is based on the assumption that the LPL of the observer’s meter stick is its material length.

37 How to test length contraction by experiment?

No. What matters is who has the shortest path through spacetime.

It is easy to set up a twin scenario in which two twins orbit a mass, one in circular orbit and one in a highly elliptical orbit, such that they meet periodically. Both have zero proper acceleration. It can be arranged so either one has the larger elapsed proper time between meetings.

One can also set up a twin scenario using clocks on the rims of rotating disks, with the disks' centers at rest in some inertial frame, arranged so they meet periodically. By varying the diameter and rotation rate of the disks, one can arrange for the clock with the larger proper acceleration to have either larger or smaller elapsed proper time between meetings. In this case, of course, it is the speed of the clock that matters, not its acceleration (both relative to the inertial frame).

Tom Roberts

38 How to test length contraction by experiment?

> > >

I doubt if it is that simple. You must know the whole physical situation from start to end of this experiment,

> >	By "purely illusory effects" I mean those which arise solely from the relative, unaccelerated motion. These are well documented, easily understood in SR, and no mystery at all.

>	Yes, "time dilation" does not affect the moving clock being observed, but it is not really "illusory", because it can have real and measurable physical consequences.

Nothing affects any physical processes to be observed, to be measured. This is not exactly physical true. If you put a thermometer in water, the water 'heats' up if the T of thermometer is higher than the T of water. As such the length of rod does not physical change being observed. The same for any clock (its physical dimensions) being observed. The problem with any clock is that its physical performances changes when a clock undergoes movement.

39 How to test length contraction by experiment?

> > >

I doubt if it is that simple. You must know the whole physical situation from start to end of this experiment,

> >	By "purely illusory effects" I mean those which arise solely from the relative, unaccelerated motion. These are well documented, easily understood in SR, and no mystery at all.

>	Yes, "time dilation" does not affect the moving clock being observed, but it is not really "illusory", because it can have real and measurable physical consequences.

Nothing affects any physical processes to be observed, to be measured. This is not exactly physical true. If you put a thermometer in water, the water 'heats' up if the T of thermometer is higher than the T of water. As such the length of rod does not physical change being observed. The same for any clock (its physical dimensions) being observed. The problem with any clock is that its physical performances changes when a clock undergoes movement.

40 How to test length contraction by experiment?

Einstein did not know better when he wrote, as this had not yet been fully understood. Pais is not a physicist and is writing for a general audience, so it is no surprise that he does not know better.

The observation mentioned is really that C2 showed less elapsed proper time than C1; the tick rates of the two clocks were NOT compared, and "running slow" was NOT actually observed.

This is merely another example of insufficiently precise wording: the notion that C2 "runs slow" compared to C1 implicitly assumes a) using the inertial frame of A (and C1), and b) the difference in final displayed times depends ONLY on the clocks' tick rates. But a) if you are talking about the tick rate of C2 you MUST use its own rest frame, and b) the final difference also CLEARLY depends on the clocks' paths.

Bottom line: C2 does not "run slow". But C2 does run slow RELATIVE TO THE FRAME OF C1. Note carefully the difference in English wording, which reflects an essential difference in meaning.

>>>	In the classic twin paradox, only one clock is accelerated. While it is clear that the accelerated clock runs slower [...]

>>

Again, stop making such erroneous and confusing claims. The clock does NOT "run slower"; rather, it accumulated less elapsed proper time (and did so while ticking at its usual and natural rate). The reason for this is geometrical: the traveling twin followed a shorter path through spacetime than did the earthbound twin.

>	See the text by Einstein above.

Einstein did not know better -- he had the excuse that this had not yet been fully understood. Today there is no such excuse.

I repeat: if a given clock actually does "run slow", then the first postulate of SR would be violated , and the entire edifice of modern physics would be overthrown. But no such violation has ever been observed.

Tom Roberts

41 How to test length contraction by experiment?

That means if Abraham Pais would have written: "and if C2 leaves A and moves along a closed orbit, then upon return to A, C2 will run slow relative to THE FRAME of C1, as often observed since in the laboratory. [...]" then everything is okay? Maybe that is what he (and Einstein) meant?

But is all off this that important? Original both C1 and C2 are resident of the same frame (state). When C2 leaves A a certain force is exerted on C2. In order to travel a closed loop more forces are exerted on C2. At the end again a force is exerted on C2 to bring him back in the frame of C1. (What this implies is that C2 during his travelling, under going accelerations, is not part of one particular frame at rest) Finally what is observed, that C2 shows less counts than C1. (relative to the frame of C1). The cause of this lies in all these extra forces, (which physical affect the internal operation of the clock) which results that the path length of C2 travelled is much larger than C1.

Nicolaas Vroom

42 How to test length contraction by experiment?

The answer lies in the details of this experiment. IMO it is important to describe this experiment in more detail. The mass involved is the earth with observer C1 at rest. For the rest the universe is considered empty. C2 orbits around the earth in one year (relative to C1). C2 will make 10 revolutions around the earth. C3 orbits around the earth in 10 years (relative to C1) C3 will make 1 revolution around the earth. After 10 years (earth time) they will all meet again on earth. This is a general rule: The duration (earth time) for all observers is the same.

When you perform such an experiment the observer closest to the earth (C2) will be subject to the largest speed*, largest acceleration*, its path length will be the longest* and its clock will run the slowest relative to the frame of C1. (*) If my calculations using Newton's law are correct i.e. r = T^(2/3)

43 How to test length contraction by experiment?

On Sunday, 4 August 2019 12:18:00 UTC+2, Tom Roberts wrote:

>	On 8/1/19 2:33 PM, Phillip Helbig (undress to reply) wrote:

> >	What matters in the twin paradox is who has the strongest acceleration.

>

No. What matters is who has the shortest path through spacetime. It is easy to set up a twin scenario in which two twins orbit a mass, one in circular orbit and one in a highly elliptical orbit, such that they meet periodically. Both have zero proper acceleration. It can be arranged so either one has the larger elapsed proper time between meetings.

The answer lies in the details of this experiment. IMO it is important to describe this experiment in more detail. The mass involved is the earth with observer C1 at rest. For the rest the universe is considered empty.
C2 orbits around the earth in one year (relative to C1).
C2 will make 10 revolutions around the earth.
C3 orbits around the earth in 10 years (relative to C1) C3 will make 1 revolution around the earth.
After 10 years (earth time) they will all meet again on earth. This is a general rule:
The duration (earth time) for all observers is the same.

[[Mod. note -- I don't know what you mean by that last sentence.

In Tom Roberts' scenario we are comparing /proper time/, i.e., readings of clocks carried by the various observers. That is, we first synchronize clocks #1, #2, and #3. Then we leave #1 on the Earth, put #2 in some one-year orbit, and put #2 in some 10-year orbit. (These orbits may be elliptical.) Then we wait a while, until (by clever arrangement of the orbits) all 3 clocks are back in the same place again and clock #1 (stationary on the Earth) says 10 years have elapsed. In this scenario the elapsed time shown by clocks #1 (= 10 years), #2, and #3 will in general all be different.

(To tie this together with the traditional "twin paradox", note that the aging of a person's body is a type of clock. It's not the most precise of clocks, but that doesn't matter for our /gedanken/ purposes.) -- jt]]

When you perform such an experiment the observer closest to the earth (C2) will be subject to the largest speed*, largest acceleration*, its path length will be the longest* and its clock will run the slowest relative to the frame of C1.
(*) If my calculations using Newton's law are correct i.e. r = T^(2/3)

44 How to test length contraction by experiment?

This experiment is not very clear. That is why I have setup a different experiment.

>

The answer lies in the details of this (each) experiment. C2 orbits around the earth in one year (relative to C1). C3 orbits around the earth in 10 years (relative to C1) C3 will make 1 revolution around the earth. C2 10 revolutions. After 10 years (earth time) they will all meet again on earth. This is a general rule: The duration (earth time) for all observers is the same. [[Mod. note -- I don't know what you mean by that last sentence.

The whole idea is that the experiment for all observers takes the same time i.e. 10 years (central mass or earth based) time.

>	In Tom Roberts' scenario we are comparing /proper time/, i.e., readings of clocks carried by the various observers. That is, we first synchronize clocks #1, #2, and #3.

The clocks for all observers are all only reset at the start of the experiment. During the trip no clock synchronization is performed. This is the only (?) correct way to study the behaviour of the individual clocks.

>	(To tie this together with the traditional "twin paradox", note that the aging of a person's body is a type of clock. It's not the most precise of clocks, but that doesn't matter for our /gedanken/ purposes.)

It is the behaviour of the (moving) clocks during the experiment to mimic the aging of the twins from the start to the end.

>

-- jt]] The purpose of this posting is to mention that any experiment which tries to demonstrate the importance or influence of acceleration is rather complex. The movement of the objects involves 'gravity' (Newton's Law or GR) and extra forces in order to control the movement of the spaceships. ONLY in order to describe the behaviour of the clocks the speed of light is involved.

The above mentioned experiment is important because it shows that identical clocks under different circumstances i.e. different path through space, different forces, different accelerations, will behave differently. In this particle setup the forces are rocket propulsion forces and gravity forces. The propulsion forces are used to bring the spaceship (clock) in orbit at the correct distance and with the correct speed. Secondly to bring the spaceship and clock back home, back to earth. The force of gravity is used to keep the speceship in orbit. The same experiment can also be done without a central mass. In that case propulsion forces have to be used to keep the spaceship in 'orbit' In both cases the result will be the same. The spaceship in the closest orbit will run the slowest, relative to the frame of the earth.