Clocks, rulers... and an issue for relativity

[michel123456]

Again

 

The muon, in its own frame, measures its own proper length L0 and is at rest while 2km of atmosphere comes through it.

 

As seen from Earth, the atmosphere is 10km thick at rest and the muon of length L1 travels through it.

 

You are saying that L1 is less than L0

I say that L0 is the contracted value of L1

[swansont]

Again

 

The muon, in its own frame, measures its own proper length L0 and is at rest while 2km of atmosphere comes through it.

 

As seen from Earth, the atmosphere is 10km thick at rest and the muon of length L1 travels through it.

 

You are saying that L1 is less than L0

I say that L0 is the contracted value of L1

 

 

 

You are wrong.

[michel123456]

 

 

 

You are wrong.

You are saying that L1=1/5 of L0

[swansont]

You are saying that L1=1/5 of L0

 

 

Yes.

[michel123456]

 

 

Yes.

 

You are wrong.

[swansont]

 

You are wrong.

 

 

 

Show me the physics. We've seen this effect (not with point particles, for what I hope is an obvious reason) in accelerators. You have to account for the length contraction of the moving particle when colliding nuclei with each other

 

http://wl33.web.rice.edu/research.html

 

"As illustrated below, fast-moving Pb nuclei are initially compressed into a pancake shape due to the Lorentz contraction."

 

So, please explain how I am wrong.

[michel123456]

 

 

 

Show me the physics. We've seen this effect (not with point particles, for what I hope is an obvious reason) in accelerators. You have to account for the length contraction of the moving particle when colliding nuclei with each other

 

http://wl33.web.rice.edu/research.html

 

"As illustrated below, fast-moving Pb nuclei are initially compressed into a pancake shape due to the Lorentz contraction."

 

So, please explain how I am wrong.

10 km corresponds to M1 muon length as observed in S

 

2km correspond to M0 muon length as observed in S'

 

The value 2km is the path as seen from the muon's frame and is the value contracted by a value of γ=5 wrt the 10 km atmosphere.

 

The value 2km is the path contracted. Whatever corresponds to the 2km value is the result of the equations of Relativity. It is accepted physics.

 

This value of 2km corresponds to M0 which is the length of the muon in its own rest frame. This value M0 is the value already contracted. It is the proper length of the muon.

 

What you are proposing is the contraction of an already contracted value.

What you should propose is that the value M1, as measured in S from Earth, divided by γ=5 gives you the value M0 contracted. Unfortunately the accurate value of M1 is null which doesn't help much, but you have the path and can make the calculation 10km/5=2km which gives you the contracted value of the path.

[swansont]

10 km corresponds to M1 muon length as observed in S

 

2km correspond to M0 muon length as observed in S'

 

The value 2km is the path as seen from the muon's frame and is the value contracted by a value of γ=5 wrt the 10 km atmosphere.

 

The value 2km is the path contracted. Whatever corresponds to the 2km value is the result of the equations of Relativity. It is accepted physics.

 

This value of 2km corresponds to M0 which is the length of the muon in its own rest frame. This value M0 is the value already contracted. It is the proper length of the muon.

 

What you are proposing is the contraction of an already contracted value.

What you should propose is that the value M1, as measured in S from Earth, divided by γ=5 gives you the value M0 contracted. Unfortunately the accurate value of M1 is null which doesn't help much, but you have the path and can make the calculation 10km/5=2km which gives you the contracted value of the path.

 

 

The length of the muon is not the length of the path that the muon traveled. 2 km is the contracted length of the path, i.e. the length of travel of the earth in the muon's frame. It is not the size of the muon itself, which is what "length of the muon" implies. These are distinct values.

 

The size of the muon has no impact on the analysis of the experiment. For all practical purposes it is zero. It can be ignored.

[michel123456]

 

 

The length of the muon is not the length of the path that the muon traveled. 2 km is the contracted length of the path, i.e. the length of travel of the earth in the muon's frame. It is not the size of the muon itself, which is what "length of the muon" implies. These are distinct values.

 

The size of the muon has no impact on the analysis of the experiment. For all practical purposes it is zero. It can be ignored.

Yes, I agree it is well understood.

In my posts, L0 is the length of the path in muon's frame (2km) and M0 is the proper length of the muon. These are distinct values.

 

-----------------

But they are tied together: when the path is contracted, so is the muon. The same γ factor apply to both values.

Edited September 9, 2016 by michel123456

[swansont]

 

But they are tied together: when the path is contracted, so is the muon. The same γ factor apply to both values.

 

 

Not in the same frame. Never in the same frame. Only moving things are contracted. The muon is stationary in its own frame.

[michel123456]

 

 

Not in the same frame. Never in the same frame. Only moving things are contracted. The muon is stationary in its own frame.

Well understood. The muon is stationary in its own frame.

 

Here below:

 

On the left side, the muon is at rest and the atmosphere goes from down to up. L0=2km

The contraction of the path L & the muon M is shown.

 

Do you have any problem with that?

 

--------------

oops the labeling does not correspond to previous posts. Working on it. Done

Edited September 9, 2016 by michel123456

[imatfaal]

Well understood. The muon is stationary in its own frame.

 

Here below:

muon-rev1.jpg

 

On the left side, the muon is at rest and the atmosphere goes from down to up. L0=2km

The contraction of the path L & the muon M is shown.

 

Do you have any problem with that?

 

--------------

oops the labeling does not correspond to previous posts. Working on it. Done

 

 

"Do you have any problem with that?" Yes.

 

The path is contracted and the muon is unaffected in the frame of the muon. The path is unaffected and muon (if it were not a point particle) is contracted in the frame of the earth.

 

Nothing is contracted nor dilated within its own frame

[michel123456]

 

 

"Do you have any problem with that?" Yes.

 

The path is contracted and the muon is unaffected in the frame of the muon. The path is unaffected and muon (if it were not a point particle) is contracted in the frame of the earth.

Do you mean something like this?

It does not correspond to the equations, it makes no sense. It is wrong.

What really happens is that the contraction of the path from 10km to 2 km takes the muon with it. How else to say?

 

You cannot state that the path is contracted and the muon is unaffected. It is wrong. The contraction of the path takes the muon with it.

Edited September 9, 2016 by michel123456

[Mordred]

I don't know how many times we have to repeat the same statement.

A muon has no physical dimensions. Point-like literally means indeterminate volume.

If it was any object with a spatial volume then yes it would be affected. A muon doesn't have a physical volume to affect.

 

no length to affect

 

if this causes too much confusion you may be better off just examining a normal everyday falling object rather than a point-like particle with no determinant dimensions.

Edited September 9, 2016 by Mordred

[swansont]

Do you mean something like this?

muon-rev2.jpg

It does not correspond to the equations, it makes no sense. It is wrong.

What really happens is that the contraction of the path from 10km to 2 km takes the muon with it. How else to say?

 

You cannot state that the path is contracted and the muon is unaffected. It is wrong. The contraction of the path takes the muon with it.

 

 

 

No, it's not wrong. It is your misunderstanding of relativity. As I, imatfaal and probably others have repeatedly said, nothing is contracted in its own frame.

I don't know how many times we have to repeat the same statement.

 

A muon has no physical dimensions. Point-like literally means indeterminate volume.

 

If it was any object with a spatial volume then yes it would be affected. A muon doesn't have a physical volume to affect.

 

no length to affect

 

Even ignoring this, the concept is wrong, since the muon is not moving. Replace it with some other composite particle that has a size, and it will not be length contracted in its own frame.

[Mordred]

yeah I hadn't gotten to "own frame" that has also been endlessly repeated. Thankfully I noticed the cross post between edits.

Edited September 9, 2016 by Mordred

[michel123456]

 

 

 

No, it's not wrong. It is your misunderstanding of relativity. As I, imatfaal and probably others have repeatedly said, nothing is contracted in its own frame.

 

Even ignoring this, the concept is wrong, since the muon is not moving. Replace it with some other composite particle that has a size, and it will not be length contracted in its own frame.

Sure, it is not length contracted in its own frame.

But for the frame S, it is contracted together with its path. If the path itself was a solid object, you would say that it is contracted.

Here below the formula of contraction is applied to the path:

When the equation gives a value of 2km instead of 10km, it means that the muon also is affected by the same equation. There is no equation A for the path and equation B for the muon, as far as I know.

[Tim88]

With all my respect maybe you should start to think by yourself instead of repeating other's thoughts.

 

Fact 1: The atmosphere is at rest and 10 km thick. You are observing the muon going through 10km.

 

Fact 2: A contracted muon goes through a 2km atmosphere, not through a 10 km atmosphere.

 

Fact 3: what we call a "contracted muon" is the muon's proper length.

 

Is anything wrong in the above?

Because if nothing is wrong, it comes out that a contracted muon does not go through a 10 km atmosphere, IOW that this observation is impossible.

 

-----------------

edited. Strikethrough useless comment

 

The "proper" or "rest height" of the dense part of the atmosphere is ca. 10 km.

As I (and others) already explained, 2 and 3 are wrong (and no need to care about irrelevant distractions concerning "point particles").

 

BTW you could have noticed from the foregoing discussions that I do my own thinking.

 

PS I will now search back the post where I started to explain the error that has nevertheless been repeated over and over since, and add it here.

 

OK according to me it was my post #131, just before I added the muon link - but possibly someone else clarified it before me:

 

"Objects are predicted to contract if they change velocity on the way from A to B. [i meant: if their speed increases] Obviously their contraction, as measured on Earth, cannot contract for example the distance between London and Paris! We even would have different contractions at the same time of that distance if objects with different speeds were flying in-between.

 

New emphasis added.

 

I don't know why that (and many follow-ups) did not "click"...

Edited September 9, 2016 by Tim88

[swansont]

Sure, it is not length contracted in its own frame.

But for the frame S, it is contracted together with its path. If the path itself was a solid object, you would say that it is contracted.

How can the path be considered to be moving, in the earth's frame? When a car or bicycle drives by you, is the road moving? Because that's what you're insisting is true.

 

Here below the formula of contraction is applied to the path:

Screen Shot 09-09-16 at 09.24 PM.JPG

When the equation gives a value of 2km instead of 10km, it means that the muon also is affected by the same equation. There is no equation A for the path and equation B for the muon, as far as I know.

No, that's not what it means. It means the muon sees the atmosphere — which is moving along with the earth, is 2 km thick. The muon, which is not moving, does not see itself as contracted.

 

The muon would also see the earth's clock as running slow, but it's ignored since it's irrelevant to the experiment.

[michel123456]

 

The "proper" or "rest height" of the dense part of the atmosphere is ca. 10 km.

As I (and others) already explained, 2 and 3 are wrong (and no need to care about irrelevant distractions concerning "point particles").

 

BTW you could have noticed from the foregoing discussions that I do my own thinking.

 

PS I will now search back the post where I started to explain the error that has nevertheless been repeated over and over since, and add it here.

 

OK according to me it was my post #131, just before I added the muon link - but possibly someone else clarified it before me:

 

"Objects are predicted to contract if they change velocity on the way from A to B. Obviously their contraction, as measured on Earth, cannot contract for example the distance between London and Paris! We even would have different contractions at the same time of that distance if objects with different speeds were flying in-between.

 

However you are almost right: it's just a matter of picking the right reference frame. According to measurements in a moving frame, the distance between London and Paris is contracted. Thus, in the example with muons from space: according to our measurements their radioactive "clock" is time dilated, so that some can reach the Earth before falling apart. As reckoned from their rest frame, instead the height of the atmosphere is strongly reduced. And then your calculation works: v = less time / less distance."

 

New emphasis added.

 

I don't know why it did not "click"...

I have emphasized some part in blue color,

v = less time / less distance."

 

So we have a "less distance" as the result of Relativity. And a "less time"

Now, the muon at rest, that sees this "less distance" passing through it, has a proper length M0, it is not contracted, in its own frame S'.

 

Everything is as usual for the muon, nothing special happens.

 

So, you have some equations that give you the transform of this situation in Earths frame:

 

1.It is "less time" multiplied by γ = more time (34μs)

 

2. It is "less distance" multiplied by γ = more distance (10km)

 

and

 

3. you say that we have M0 divided by γ = less length. Why?

[studiot]

 

michel123456

Everything is as usual for the muon, nothing special happens.

 

So, you have some equations that give you the transform of this situation in Earths frame:

 

1.It is "less time" multiplied by γ = more time (34μs)

 

2. It is "less distance" multiplied by γ = more distance (10km)

 

I hesitate to renter this discussion since Michel is not talking to me but this appears to be the basis of misunderstanding to me.

 

Perhaps someone will explain simply when you should divide by gamma and when you should multiply.

[swansont]

I have emphasized some part in blue color,

So we have a "less distance" as the result of Relativity. And a "less time"

Now, the muon at rest, that sees this "less distance" passing through it, has a proper length M0, it is not contracted, in its own frame S'.

 

Everything is as usual for the muon, nothing special happens.

 

So, you have some equations that give you the transform of this situation in Earths frame:

 

1.It is "less time" multiplied by γ = more time (34μs)

 

2. It is "less distance" multiplied by γ = more distance (10km)

 

and

 

3. you say that we have M0 divided by γ = less length. Why?

 

 

Multiplication and division just depend on how you write the equation down If z=xy, x = z/y (or y = z/x). Reverse the order in 1 and 2, to be consistent in going from one frame to the other.

 

1. The time in the Earth's rest frame divided by γ gives you the time in the moving frame

 

2. The travel length in the Earth's rest frame divided by γ gives you the travel length in the moving frame.

 

3. The length of the moving object as observed in the Earth's rest frame is the object's rest frame length divided by γ.

 

Moving things are contracted: divide by γ

Moving clocks runs slow: divide by γ

 

There's no inconsistency here. But you need to keep careful track of what frame you're in.

[Celeritas]

 

I hesitate to renter this discussion since Michel is not talking to me but this appears to be the basis of misunderstanding to me.

 

Perhaps someone will explain simply when you should divide by gamma and when you should multiply.

 

In supplement to swansont's post ...

 

Note 1 ... any length is measured between 2 simultaneous events per the measurer, one event co-located with one end-point of the considered length, the other event co-located with the other end-point of the considered length.

 

Note 2 ... a stationary length need not be considered by 2 simultaneous events at each end, although it may be and usually is. If the 2 events located at each end of the considered length exist at different moments in time (per he who co-moves with this length), their spatial separation is the very same no matter the duration between those 2 events ... because every point of said stationary length (including its endpoints) possess spatial locations that are independent of time.

 

 

1) wrt SPATIAL LENGTHs ...

 

PROPER LENGTH ... If a length measured by an observer does not move over duration, it is a stationary length. This is referred to as a proper length, and is the longest length measurable for the said length. He divides his measured stationary-length by gamma if he wishes to know what an observer in relative motion would measure that length as (ie its contracted-length, per POV).

 

Since the atmosphere is stationary wrt the earth, earth-frame rulers directly measure the atmosphere's thickness at its PROPER LENGTH (10km). Earth-frame rulers cannot measure the atmosphere's contracted-length directly, because the atmosphere does not move wrt the muon. The contracted-length measurable directly by the muon's ruler must be calculated by earth-frame observers ... contracted-length = proper-length / gamma = 10km / 5 = 2km.

 

CONTRACTED LENGTH ... If a length measured by an observer moves over duration, it is a moving length. A moving length is referred to as a contracted-length, and is always shorter than the measure of its proper-length. He mulitplies his (directly measured) moving-contracted-length by gamma if he wishes to know what an observer who co-moves with the moving length would measure that length as (ie its proper length).

 

Since the atmosphere moves wrt the muon, the muon's ruler directly measures the atmosphere's thickness at its contracted-length (2km). The muon cannot measure the atmosphere's proper length directly, because the muon is not at rest with the atmosphere. The proper-length measurable directly by earth-frame rulers must be calculated by muon-frame observers ... proper-length = contracted-length x gamma = 2km x 5 = 10km.

 

 

2) wrt TIME INTERVALs ...

 

PROPER TIME ... If the clock exists at both events, that clock measures the proper time for the time interval between the 2 events . This is the shortest duration measurable for the defined time interval. The carrier of that clock multiplies his measured duration by gamma if he wishes to know what an observer in relative motion would measure for the duration between those same 2 events (ie its coordinate time, per POV).

 

Since the muon's clock is AT both events (EVENT 1 ... muon enters atmosphere here, EVENT 2 ... muon exits atmosphere there), it directly measures the PROPER TIME for the interval (6.8 μs). The muon cannot directly measure the coordinate-time measured directly by earth-frame clocks, because the muon is AT both events (it does not move wrt itself). The coordinate-time measurable directly by earth-frame clocks must be calculated by the muon as follows ... coordinate-time = proper-time x gamma ... 34 μs = 6.8 μs x 5.

 

COORDINATE TIME ... If the clock (even virtual) cannot exist at both events, the frame co-moving with that clock measures coordinate time between the 2 events. Coordinate time varies per v, and is always a greater duration than the proper time. The carrier of that clock divides his measured duration by gamma if he wishes to know what an observer AT both events would measure for the duration between those events (ie its proper time).

 

Since earth-frame clocks are not AT both events (EVENT 1 ... muon enters atmosphere here, EVENT 2 ... muon exits atmosphere there), they directly measure the COORDINATE TIME for the interval (34 μs). Earth-frame clocks cannot directly measure the proper-time measured directly by the muon's clock, because earth-frame clocks are not AT both events (they move wrt the muon). The proper-time measurable directly by the muon's clock must be calculated by earth-frame observers ... proper-time = coordinate-time / gamma ... 6.8 μs = 34 μs / 5.

 

Best regards,

Celeritas

Edited September 10, 2016 by Celeritas

[VandD]

-

Edited September 9, 2016 by VandD

[Celeritas]

... I can understand for the question: How much time did the travelling clock lose? But what if the question is: How did the travelling clock lose time? What is the answer (with certainty) to that question?

 

robinpike,

 

So the LTs explain the relative measure of time for any 2 observers moving relatively. Per your study of that math, you do not yet understand WHY the LTs do what they do. A derivation of the LTs from scratch (on your own) would help you considerably in that regard. Usually, if one understands the derivation then one envisions the mechanism, and your answer lies therein. Also, the fine spacetime diagrams presented thus far in this thread, have apparently not gotten you there. I would recommend you spend a good week studying spacetime figures, if you haven't already. Often, they are the ticket for a more expedient and complete understanding of WHY the relativistic effects exist, as they present the abstract math per the LTs both geometrically and visually.

 

Ok, that said, I will try and attempt answering your question using only a 2-spacetime figure (with time implied), which most folks are plenty familiar with. Read this carefully, think about it before responding, and let me know if it helps any ...

 

 

Best regards,

Celeritas

Edited September 10, 2016 by Celeritas