Concerning the Muon Experiment:

The muon doesn't see time dilation in it's own frame nor does the Earth. So both their clocks are ticking along at the same rate in their respective frames but they see the other's clock tick slower. The muon is explained by saying the Earth sees the muon's clock slower and the muon, with a normal tick rate in it's own frame, sees the Earth move over a smaller distance. This argument seems inadequate to me because isn't the reverse true if neither frame is preffered. That is, the muon sees the Earth's clock slower while the Earth sees the muon move a smaller distance. Can anyone point out where this thinking is flawed? I have not been able to resolve this issue and would like to!

Consider time for the muon, in order for no light to reach it beyond the edge of the known universe as defined in the earth frame (156b ly) and the muon frame (9.6b ly), then the time accumulated on the muons clock (if it could exist forever, consider the time accumulated by an observer in the muons frame) would have to be less than that accumulated on the Earth clock. That statement does not agree with the statement that each clock ticks normally in their own frame and each sees the other clock as running slower.

I don't see any problem, the observers on Earth will also observe the muon to move a shorter distance. If you consider the muon to be embedded in a fictional crystal, then the size of the crystal as observed from Earth will be shorter if the Earth is moving towards the muon (i.e. the muon is moving towards the Earth), so you can say that the muon has moved a shorter distance

I don't see any problem, the observers on Earth will also observe the muon to move a shorter distance. Impossible and Nonsensical. If the Earth observer saw the muon move over a shorter distance, how does the muon reach the Earth's surface? I am not sure you are completely familiar with the Muon Experiment. See here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html)


If you consider the muon to be embedded in a fictional crystal, then the size of the crystal as observed from Earth will be shorter if the Earth is moving towards the muon (i.e. the muon is moving towards the Earth), so you can say that the muon has moved a shorter distance In who's frame? I already think I mentioned that the Earth moves a shorter distance from the muon's perspective.

The resolution of your difficulty is that in both frames the determining factor is the perceived thickness of the Earth's atmosphere. In order to get the symmetric situation you are looking for you would have to collide two Earths.

The resolution of your difficulty is that in both frames the determining factor is the perceived thickness of the Earth's atmosphere. I stated this. I already know know how the muon experiment is explained. Read my post carefully. The difficulty arrises from this explaination, NOT, this explaination resolves the difficulty.


In order to get the symmetric situation you are looking for you would have to collide two Earths. I am not looking for such a symmetric situation. I merely posed the question, why does the symmetry not exist if neither frame is preferred. Nothing more nothing less.

You may have missed my point. Let me restate it this way. The muon is created in the Earth's upper atmosphere, traversed it, and then decayed. The Earth wasn't created in the muon's "upper atmosphere", didn't traverse it, and didn't decay.

The factor that controls both frames is soley in the Earth's frame.

Or perhap's I'm not understanding your question?

You may have missed my point. Let me restate it this way. The muon is created in the Earth's upper atmosphere, traversed it, and then decayed. The Earth wasn't created in the muon's "upper atmosphere", didn't traverse it, and didn't decay.

The factor that controls both frames is soley in the Earth's frame.

Or perhap's I'm not understanding your question? The Earth created the muon so the Earth's viewpoint superceeds the muon's viewpoint. Thanks! That is tremendously helpful. :rolleyes:

Anyway, the muon is a point, how do you want to contract a point?

The Earth created the muon so the Earth's viewpoint superceeds the muon's viewpoint. Thanks! That is tremendously helpful. :rolleyes:
Aer, Kevinalam, Lucas et al you miss the point of science,

The muons are created not at one altitude, not at two altitudes, but a spectrum of altitudes. At each measured point then, of the number created there, 1/2 were created 600 meters higher up and so on.

Prove to us, you who believe, but are not experimentally saavy, that 1/2 of the muons counted on the surface of the earth weren't created 600 meters up, and 1/4 created 1200 meters op, and 1/8 created 1800 meters up and 1/16 created 2400 meters up on ad infinitum.

I don't recall seeing any reference to "1/2 life" in this thread, or 1/2 life significance. I only see three brainless ones discussing special realtivity idiocy.

Prove the source of atoms detected on the surface originated higher than 600 meters for instance. I bet a clever fellows like you can find us an Einstein quote to solve this ticklish problem can't you? or something devilishly clever to resolve the "where created" problem can't you?

The three of you ought to combined your brain power and come close to unity, on this I am sure.

Why not analyze the problem from first principles {if you aware of these things, that is], instead of assuming you singularly know the answer? As to this points I am only able to determine collective stupidity, as well meaning as it is projected in this thread, that is.

Geistkiesel :cool:

Concerning the Muon Experiment:

The muon doesn't see time dilation in it's own frame nor does the Earth. So both their clocks are ticking along at the same rate in their respective frames but they see the other's clock tick slower.

Right.

The muon is explained by saying the Earth sees the muon's clock slower and the muon, with a normal tick rate in it's own frame, sees the Earth move over a smaller distance.

Come again? What do you mean the "Earth move[s] over a smaller distance"? In the muon frame the Earth (and it's atmosphere) are length contracted, and since the muon originates somewhere in the atmosphere the surface reaches it before it (the muon) decayes.

This argument seems inadequate to me because isn't the reverse true if neither frame is preffered. That is, the muon sees the Earth's clock slower while the Earth sees the muon move a smaller distance. Can anyone point out where this thinking is flawed? I have not been able to resolve this issue and would like to!

I think your problem might originate in the "smaller distance" thing. I'm not sure what you are referring too, but I assume that you're thinking of length contraction. An important invariant is that both the muon and the Earth will observe the other to be moving at the same speed, relative to the observer's rest frame.

On second thought, I think I know exactly where the flaw lies: I see what you mean with "smaller distance"; the distance from the point in the atmosphere the muon originates to the surface of the Earth is shorter in the muon frame than the Earth frame, so for the Earth's surface to reach the muon it only has to travel a short distance, which is possible within the decay time of the muon (from the muon frame). You want this situation to be equally true from the point of the Earth, but the equivalence (which is real) does not let you conclude that the muon cannot reach the Earth. Why? Because the two frames will not agree on time. Think about it...

Consider time for the muon, in order for no light to reach it beyond the edge of the known universe as defined in the earth frame (156b ly) and the muon frame (9.6b ly), then the time accumulated on the muons clock (if it could exist forever, consider the time accumulated by an observer in the muons frame) would have to be less than that accumulated on the Earth clock. That statement does not agree with the statement that each clock ticks normally in their own frame and each sees the other clock as running slower.
I'm not sure what you mean that it does not agree. Near relativistic particles could indeed pass over the observable universe (as viewed from earth) in quite small proper time. I fail to see the inconsistency, because the observable universe of the muon would be quite different. Remember that time doesn't make sense without space, so the fact that the muon and the Earth would disagree about these and other things (such as the age of the universe) doesn't make it inconsistent.

Concerning the Muon Experiment:

The muon doesn't see time dilation in it's own frame nor does the Earth. So both their clocks are ticking along at the same rate in their respective frames but they see the other's clock tick slower. The muon is explained by saying the Earth sees the muon's clock slower and the muon, with a normal tick rate in it's own frame, sees the Earth move over a smaller distance. This argument seems inadequate to me because isn't the reverse true if neither frame is preffered. That is, the muon sees the Earth's clock slower while the Earth sees the muon move a smaller distance. Can anyone point out where this thinking is flawed? I have not been able to resolve this issue and would like to!

The resolution is based on the conditions of the experiment. How much time passes for both the Muon and Earth clock between two Events? These two events are the creation of the muon in the upper atmosphere, and the muon reaching the surface of the Earth. Both of these events must occur for both frames in order for the experiment to be complete.

Thus in the Earth Frame, the Muon can not be seen as moving a shorter distance because the conditions of the experiment require it to cross the depth of the atmosphere, and the atmosphere is in the same frame as the Earth and thus this distance does not contract in the Earth frame.

That is not to say that you could not devise a different experiment in which the reverse is true, where the Earth measures the muon as traveling the shorter distance, but this would be a different experiment with different conditions. For instance, you could have two muons, one created after the other and compare how much time passes between the first muon striking the Earth and the second striking the striking the Earth as measured by the second muon and an Earth clock.

geist:

As to this points I am only able to determine collective stupidity

Geist you fucking asshole. Back to the pseudoscience forum and let people discuss actual science.

If you don't want to read all this, they use balloon lofted instruments and know very well the muon fluxes in the upper atmosphere.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html

http://www.cosmicrays.org/

http://www.lbl.gov/abc/cosmic/SKliewer/Cosmic_Rays/Muons.htm

http://pandora.physics.lsa.umich.edu/schubnel/papers/PhysRevD_70_092005.pdf

http://people.roma2.infn.it/~aldo/A51prl_cap94_mu.pdf

And the explanation for why muons with a HALF-LIFE of 2.2us reach the ground, is that from our frame, the muon lasts much longer. From the muon's frame an earth clock would also appear to run slower. So, as the muon ticks normally, it decays when it reaches the ground (say). It will always have seen the earth clock as much slower. It never changes frames, it just dies. Time and space are different for earth and muon.

Anyway, the muon is a point, how do you want to contract a point? What?

Come again? What do you mean the "Earth move[s] over a smaller distance"? In the muon frame the Earth (and it's atmosphere) are length contracted, and since the muon originates somewhere in the atmosphere the surface reaches it before it (the muon) decayes. Perhaps this is the source of the issue. How is what you are saying any different from what I said?

I think your problem might originate in the "smaller distance" thing. I'm not sure what you are referring too, but I assume that you're thinking of length contraction. An important invariant is that both the muon and the Earth will observe the other to be moving at the same speed, relative to the observer's rest frame. How is the "smaller distance" not length contraction.

On second thought, I think I know exactly where the flaw lies: I see what you mean with "smaller distance"; the distance from the point in the atmosphere the muon originates to the surface of the Earth is shorter in the muon frame than the Earth frame, so for the Earth's surface to reach the muon it only has to travel a short distance, which is possible within the decay time of the muon (from the muon frame). You want this situation to be equally true from the point of the Earth, but the equivalence (which is real) does not let you conclude that the muon cannot reach the Earth. Why? Because the two frames will not agree on time. Think about it... I am not convinced by this explaination - I think the issue is just cloudy for one or the other of us.

I'm not sure what you mean that it does not agree. Near relativistic particles could indeed pass over the observable universe (as viewed from earth) in quite small proper time. I fail to see the inconsistency, because the observable universe of the muon would be quite different. Remember that time doesn't make sense without space, so the fact that the muon and the Earth would disagree about these and other things (such as the age of the universe) doesn't make it inconsistent. I am not talking about the time issue. I made up an entirely different issue which I have not yet come to the conclusion yet as being valid/invalid. That is the light that the muon sees compared to the light the Earth sees.

The resolution is based on the conditions of the experiment. How much time passes for both the Muon and Earth clock between two Events? These two events are the creation of the muon in the upper atmosphere, and the muon reaching the surface of the Earth. Both of these events must occur for both frames in order for the experiment to be complete. Agreed.

Thus in the Earth Frame, the Muon can not be seen as moving a shorter distance because the conditions of the experiment require it to cross the depth of the atmosphere, and the atmosphere is in the same frame as the Earth and thus this distance does not contract in the Earth frame. This seems like an ex post facto argument. How might you reach this conclusion using Special Relativity laws only a prior?

That is not to say that you could not devise a different experiment in which the reverse is true, where the Earth measures the muon as traveling the shorter distance, but this would be a different experiment with different conditions. For instance, you could have two muons, one created after the other and compare how much time passes between the first muon striking the Earth and the second striking the striking the Earth as measured by the second muon and an Earth clock. Are you sure this predicts what you say it will predict?

Geistkiesel :cool: [/indent] I do not wish to go into a discussion with you on any topic related to science or otherwise. Not because I think you are an idiot or anything, just personal preference :cool:

Concerning the Muon Experiment:

The muon doesn't see time dilation in it's own frame nor does the Earth. So both their clocks are ticking along at the same rate in their respective frames but they see the other's clock tick slower. The muon is explained by saying the Earth sees the muon's clock slower and the muon, with a normal tick rate in it's own frame, sees the Earth move over a smaller distance. This argument seems inadequate to me because isn't the reverse true if neither frame is preffered. That is, the muon sees the Earth's clock slower while the Earth sees the muon move a smaller distance. Can anyone point out where this thinking is flawed? I have not been able to resolve this issue and would like to!

Consider time for the muon, in order for no light to reach it beyond the edge of the known universe as defined in the earth frame (156b ly) and the muon frame (9.6b ly), then the time accumulated on the muons clock (if it could exist forever, consider the time accumulated by an observer in the muons frame) would have to be less than that accumulated on the Earth clock. That statement does not agree with the statement that each clock ticks normally in their own frame and each sees the other clock as running slower.

I don't see any problem with the lack of symmetry in your gedankenexperiment, as long as the experiment agrees with the laws of Special Relativity. Show me what law is violated, and then we can discuss (after all, the theory is called Relativity, that is different observers measure different times, sizes, etc)

With respect to your response in the third post, yes I knew that the distance of the muon as seen from the Earth is the same, but you're assigning extra importance to the fact of the creation of the muon in the athmosphere. You should try to imagine a fictional frame of reference centered in the muon. As observed from earth, this frame of reference would contract if the muon is moving, so in a sense, the movement of the muon between two points of its frame of reference is shorter as seen from Earth than the distance between those points if the muon was stationary (as seen from Earth). This is the same as the case as say, as seen from the muon, in a frame of reference centered in the surface of Earth, you choose the two points to be the surface of Earth and the upper part of the athmosphere

I don't see any problem with the lack of symmetry in your gedankenexperiment, This is not a thought experiment. It is an analysis of the explaination given to the Muon Experiment.

as long as the experiment agrees with the laws of Special Relativity. Show me what law is violated, and then we can discuss (after all, the theory is called Relativity, that is different observers measure different times, sizes, etc) You need to read my post. I am analysising the explaination of the laws of Special Relativity, not saying they are violated.

With respect to your response in the third post, yes I knew that the distance of the muon as seen from the Earth is the same, but you're assigning extra importance to the fact of the creation of the muon in the athmosphere. This makes no sense. "distance of the muon as seen from the Earth is the same"? What frames are you talking about? I never put any importance to the creation of the muon in the atmosphere. Would you like to explain this process in detail?

You should try to imagine a fictional frame of reference centered in the muon. As observed from earth, this frame of reference would contract if the muon is moving, so in a sense, the movement of the muon between two points of its frame of reference is shorter as seen from Earth than the distance between those points if the muon was stationary (as seen from Earth). This is the same as the case as say, as seen from the muon, in a frame of reference centered in the surface of Earth, you choose the two points to be the surface of Earth and the upper part of the athmosphere You are just reiterating the analysis I already provided.

This makes no sense. "distance of the muon as seen from the Earth is the same"? What frames are you talking about?


i mean,seen from the Earth frame of reference, when the muon is created its distance to the surface of Earth is the same if the muon is created stationary with respect the frame of Earth, or moving with respect to the same frame

i mean,seen from the Earth frame of reference, when the muon is created its distance to the surface of Earth is the same if the muon is created stationary with respect the frame of Earth, or moving with respect to the same frame No it's not.

This seems like an ex post facto argument. How might you reach this conclusion using Special Relativity laws only a prior?
It is not a conclusion, it is a intial condition of the test. The muon crosses the depth of the Earth's atmosphere as measured in both frames.

Are you sure this predicts what you say it will predict?

I didn't predict anthing about about the results of this experiment.

It is not a conclusion, it is a intial condition of the test. The muon crosses the depth of the Earth's atmosphere as measured in both frames.

I didn't predict anthing about about the results of this experiment.
It appears we are approaching the problem from different perspectives. The only condition related to special relativity in the experiment is the fact that the muon is created and decays and at all points in time (from the muon's perspective or the Earth's perspective) the muon is travelling at .998c. Is this correct? the fact that the muon reaches the Earth's surface is not a condition of the experiment or special relativity, it is a result.

Concerning the Muon Experiment:

The muon doesn't see time dilation in it's own frame nor does the Earth. So both their clocks are ticking along at the same rate in their respective frames but they see the other's clock tick slower.

Yes.

The muon is explained by saying the Earth sees the muon's clock slower

Yes.

and the muon, with a normal tick rate in it's own frame, sees the Earth move over a smaller distance.

Yes.

This argument seems inadequate to me because isn't the reverse true if neither frame is preffered. That is, the muon sees the Earth's clock slower while the Earth sees the muon move a smaller distance.

Can anyone point out where this thinking is flawed? I have not been able to resolve this issue and would like to!


No.

The muon sees all distances and lengths that have relative velocity to it as contracted.

The earth sees all distances and lengths that have relative velocity to it as contracted.

In our rest frame, the muon is created 10000m up and must traverse that distance. In the muon's rest frame it is created 600m up and need only traverse that distance.

Consider time for the muon, in order for no light to reach it beyond the edge of the known universe as defined in the earth frame (156b ly) and the muon frame (9.6b ly), then the time accumulated on the muons clock
...
would have to be less than that accumulated on the Earth clock.

Assume muon v = 0.79c Distance to traverse in earth frame = 15.6b ly. When we meet:

On the earth, we would see the muon clock at 12.0by with our clock at 19.7by.

Now, when the muon is created the distance it must travel in its frame is only 9.6bly and will take it 12by on it's own clock.

So, the muon would read our clock at 7.3by with its clock at 12by

That statement does not agree with the statement that each clock ticks normally in their own frame and each sees the other clock as running slower.

Yes, it does.

Hopefully this helps?

That is, the muon sees the Earth's clock slower while the Earth sees the muon move a smaller distance.



No.

The muon sees all distances and lengths that have relative velocity to it as contracted.

The earth sees all distances and lengths that have relative velocity to it as contracted.

In our rest frame, the muon is created 10000m up and must traverse that distance. In the muon's rest frame it is created 600m up and need only traverse that distance. Ah ha! This is more of the answer I was looking for. However, from the muon's perspective, the Earth travels 600m, as the muon doesn't consider itself "traversing". Anyway, thank you superluminal. I'm still not sure that the other issue has been resolved.

Anytime. And of course you are right, the muon is happy in its rest frame and all else moves about it.

Which was the other issue? The accumulated time thing?

Assume muon v = 0.79c Distance to traverse in earth frame = 15.6b ly. When we meet:

On the earth, we would see the muon clock at 12.0by with our clock at 19.7by. That is all well and good, but what would the muon see when we meet?

Now, when the muon is created the distance it must travel in its frame is only 9.6bly and will take it 12by on it's own clock. I think the issue of who is considered travelling is confused here.

So, the muon would read our clock at 7.3by with its clock at 12by This is of course assuming the muon's frame is length contracted and on second thought, I am not sure about my agreement with you on the post above.



Hopefully this helps? Hopefully I'll be able to reconsider the issue from what you have said and express my thoughts in a more clear manner.

Hmmm... Ok

Sorry about the "who is travelling" confusion. I have trouble sometimes yanking myself out of my own rest frame and into another. :D :m:

Aer:

This is of course assuming the muon's frame is length contracted and on second thought...

From the muons POV (travelling at .79c or whatever) the universe exists in a contracted state. What appears to us to be 15.6bly is, to the muon, only 9.6bly. The earth only travels (:)) that distance to the muon.

Geist you fucking asshole. Back to the pseudoscience forum and let people discuss actual science.

If you don't want to read all this, they use balloon lofted instruments and know very well the muon fluxes in the upper atmosphere.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html

http://www.cosmicrays.org/

http://www.lbl.gov/abc/cosmic/SKliewer/Cosmic_Rays/Muons.htm

http://pandora.physics.lsa.umich.edu/schubnel/papers/PhysRevD_70_092005.pdf

http://people.roma2.infn.it/~aldo/A51prl_cap94_mu.pdf

And the explanation for why muons with a HALF-LIFE of 2.2us reach the ground, is that from our frame, the muon lasts much longer. From the muon's frame an earth clock would also appear to run slower.
Sl, One paper you provided (see link below) show muon creation over three orders of altitude magnitude.


While I am in the process of matching the use of d/m² terms to determine altitude
this paper indicates variation (experimental data) of muon fluxes with altitude. (http://people.roma2.infn.it/~aldo/A51prl_cap94_mu.pdf)

Notice that these data indicate muons created at altitudes nearing the planet surface. I have seen no information in this forum other than SRT claims that time dilation accounts for the muons reaching the planet surface. Obviously these data in the paper cited above show experimental contradictions, not that this means anything to a believer in special relativity theory. As was pointed out earlier if one ignores these data the reliance on pure special relativity theory turns out to be just another pile of pure scientific bullshit.

Geistkiesel :cool:

From the muons POV (travelling at .79c or whatever) the universe exists in a contracted state. What appears to us to be 15.6bly is, to the muon, only 9.6bly. The earth only travels (:)) that distance to the muon. The distance the Earth travels was not the issue here, it was the time on the muon's clock. In order for light to not reach the muon in it's contracted universe, the total time accumulated in the muon's frame must be less than that accumulated in the Earth frame. However, this is not what Special Relativity claims.

geist:

I have seen no information in this forum other than SRT claims that time dilation accounts for the muons reaching the planet surface

Well big G, since this thread is about accepted science and how we explain muon fluxes at the surface, then I guess you better star another thread dedicated to why that ain't so. Whaddya think?

The distance the Earth travels was not the issue here, it was the time on the muon's clock. In order for light to not reach the muon in it's contracted universe, the total time accumulated in the muon's frame must be less than that accumulated in the Earth frame. However, this is not what Special Relativity claims.

Ok let me think. I'll get back to it tomorrow.

Actually, I don't understand the problem as you state it.

It is not a conclusion, it is a intial condition of the test. The muon crosses the depth of the Earth's atmosphere as measured in both frames.

I didn't predict anthing about about the results of this experiment.

Janus58 there are some data needing explaining by someone with familiarity with SRT as these data are contradictory. See what you can do with this, ok?what does special realativity theory say about muons created at various levels of the atmosphere? (http://people.roma2.infn.it/~aldo/A51prl_cap94_mu.pdf)I have seen nothing from SRT on this. How, or why does SRT ignore these kinds of experimental data?
I suppose it is, not by ignoring the data, rather the data is not sought out, is it Janus 58?

Why do you appear on this forum professing to offer anything other than what is expected from a robot?

Geistkiesel :cool:

Janus58,

If It were me, my answer to:

Why do you appear on this forum professing to offer anything other than what is expected from a robot?

Would be "Fuck you geistkiesel, and the fucking relativistic jackass you rode in on."

Just my opinion though.

by superluminal:

"The muon sees all distances and lengths that have relative velocity to it as contracted.

The earth sees all distances and lengths that have relative velocity to it as contracted.

In our rest frame, the muon is created 10000m up and must traverse that distance."
================================================== =============

Do you guys really not see the mistake here? 'The earth sees all distances and lengths
that have relative velocity to it as contracted.' Then an UNCONTRACTED distance is
stated, 10000m.

Surely you guys know that all you have to do to skew Special Theory results is to start the gedanken from the 'wrong frame'. Example:

The muon is created at 10,000 meters from the Earth through decay. (varies,of course). The muon sees the Earth clock as beating slower than its own. The muon lives a MEAN life (not half-life, a different thing) of 2.2 microseconds in its 'rest' frame,
and 'sees' the Earth's clock accumilate only a fraction of this time, about .13 microseconds, before it decays. The Earth 'sees' the distance to the muon as length
contracted, instead of 10,000 meters, the Earth observer sees the muon created at
about 600 meters. Special Theory blows up when starting the exercise from the 'wrong'
frame, even though STR states there are no preferred frames.

2inq:

There's no mistake in superluminal's post. Both the muon and the Earth observer agree that the muon originates in the atmosphere right next to the (imaginary) altitude balloon with "Height: 10km" on it. The muon doesn't agree that there's 10km to the surface of the earth, however.

Your gedanken fails not because you start from the "wrong" frame, but because your setup doesn't match the conditions of the muon experiment: If the muon originates 10,000 muon meters away from the earth, it'll originate far outside the atmosphere, because the earth (and it's atmosphere) is length contracted.

The starting condition, however, was that the muon originated 10,000 Earth meters up in the atmosphere, not 10,000 muon meters. To interchange the two is to mix length measures from different frames, and that's a no-no.

Uh, no funkstar. When beginning the exercise from the muon's frame of reference,
the muon sees the Earth clock running slow. The muon DOES NOT see distance
contracted from this frame. Changing to the Earth's frame of reference, the distance
to the muon seems contracted. It is exactly opposite of starting from the Earth's frame
of reference, in which the Earth observer sees the muon's clock running slow, then
changing to the muon's frame of reference, the muon sees the distance contracted.

2inq:

When beginning the exercise from the muon's frame of reference,
the muon sees the Earth clock running slow. The muon DOES NOT see distance contracted from this frame

Of course it must. Maybe we should talk about this in the "Visual SRT 1" thread? It addresses this very issue.

2inq, I'm sorry, but you're mistaken.

In one gedanken the muon originates 10,000 Earth meters from the surface of the earth. In the other gedanken the muon originates 10,000 muon meters from the surface of the earth. This is not the same thing. Can you see why? Think about the altitude balloon...

Also, of course the muon sees the Earth as length contracted. After all, in the muon frame the Earth is moving and so must be length contracted. And therefore the Earth's atmosphere, moving with the Earth, is also length contracted.

Of course it must. Maybe we should talk about this in the "Visual SRT 1" thread? It addresses this very issue.

When beginning from the Earth observer's frame of reference, the Earth observer sees the muon's clock running slow. Does this Earth observer also see the distance to the muon contracted?

When beginning from the muon observer's frame of reference, the muon observer sees the Earth's clock running slow. Does the muon observer also
see the distance to the Earth contracted?

The muon is at rest in its frame of reference anf the Earth is approaching it.
The Earth is at rest in its frame of reference and the muon is approaching it.

2inq:

When beginning from the Earth observer's frame of reference, the Earth observer sees the muon's clock running slow. Does this Earth observer also see the distance to the muon contracted?

No.

When beginning from the muon observer's frame of reference, the muon observer sees the Earth's clock running slow. Does the muon observer also see the distance to the Earth contracted?

Yes.


The muon is at rest in its frame of reference and the Earth is approaching it.
The Earth is at rest in its frame of reference and the muon is approaching it.

Yes.

I could be wrong about everything. I haven't slept for 26hrs. :eek:

The first two statements were exactly symmetrical. Why a 'no' to the first and a 'yes'
to the second statement?

2inq,

Can you please go and look at the "Visual SRT 1" OP? and read it because this is exactly why I posted it.

2inq,

Can you please go and look at the "Visual SRT 1" OP? and read it because this is exactly why I posted it.

OK, I read and responded to your other thread. That thread does not answer the question I asked of you in this thread. Why a yes and a no to summetrical situations? There is no acceleration phase (non-inertial phase) in either situation. Why can these two inertial frames not be treated identically?

2inq:

I've told you twice, already: The muon start 10,000 Earth meters up in the atmosphere - there's no disagreement on this!. Both the Earth observer and the muon agree that there's a big "10 clicks" sign right where the muon pops into existence. However, due to length contraction of the Earth and it's atmosphere, because the Earth is moving in the muon frame, the muon sees those 10,000 Earth meters as (say) just 600.

Now, you're asking why the Earth observer doesn't see the 10,000 meters as contracted. Why would he? The 10,000 meters is a distance measurement in the Earth frame. Nothing about a relativistic muon can make that contract for a Earth observer.

The reason the situation is not symmetrical, is that you cannot say that the distance from the Earth's surface to the muon is invariant between frames.

Think of it this way: You start out with an implicit measurement in the Earth frame, namely that the muon originates 10,000 Earth meters up in the atmosphere. This cannot be reversed to say that the Earth is 10,000 muon meters away in the muon frame! To do so would be using a measurement from one frame directly in a different frame, and you can't do that in str: You have to use the Lorentz transforms to "translate" between frames.

D'you see?

2inq,

OK, I read and responded to your other thread. That thread does not answer the question I asked of you in this thread. Why a yes and a no to summetrical situations? There is no acceleration phase (non-inertial phase) in either situation. Why can these two inertial frames not be treated identically?

The muon is a massive particle and is created and accelerated to 0.987c with enormous energy. The earth does nothing. So in the animation, the muon is A and the earth is B.

Yes?

funkstar:

Nothing about a relativistic muon can make that contract for a Earth observer.

Right.

This is from the Visual SRT 1 thread. I was trying to address that same issue with this statement.


Me:

A note. The postulate regarding preferred frames (there are none) does not imply symmetry outside of that frame. It simply means that you in your frame are at rest and Newtonian physics rules.

2inq:

I've told you twice, already: The muon start 10,000 Earth meters up in the atmosphere - there's no disagreement on this!. Both the Earth observer and the muon agree that there's a big "10 clicks" sign right where the muon pops into existence. However, due to length contraction of the Earth and it's atmosphere, because the Earth is moving in the muon frame, the muon sees those 10,000 Earth meters as (say) just 600.

Now, you're asking why the Earth observer doesn't see the 10,000 meters as contracted. Why would he? The 10,000 meters is a distance measurement in the Earth frame. Nothing about a relativistic muon can make that contract for a Earth observer.

The reason the situation is not symmetrical, is that you cannot say that the distance from the Earth's surface to the muon is invariant between frames.

Think of it this way: You start out with an implicit measurement in the Earth frame, namely that the muon originates 10,000 Earth meters up in the atmosphere. This cannot be reversed to say that the Earth is 10,000 muon meters away in the muon frame! To do so would be using a measurement from one frame directly in a different frame, and you can't do that in str: You have to use the Lorentz transforms to "translate" between frames.

D'you see?

No, funkstar, you and superluminal still don't get it. I am starting the exercise FROM THE MUON'S FRAME OF REFERENCE, not the Earth observers.

More detail. A proton sees the Earth approaching it at approximately .987c.
The proton interacts with the Earth's atmosphere and decays into high-energy pions, they also see the Earth approaching at appox. .987c. The pions quickly decay into a high-energy muon and two neutrinos, which also
see the Earth approaching at .987c. The muons are not accellerated TO .987c, they are created at that velocity relative to the approaching Earth. The muon will see the distance to the Earth as 10,000 meters, its meters. The muon will see the Earth clock as ticking much slower than its own clock. By its own clock, the muon will live 2.2 microseconds before decaying, the muon will see only a small fraction of that time elapse on the Earth clock before it dies due to decay. The muon believes the Earth will see a contracted distance between them due to the Earth's velocity, still measuring 10,000 meters but those meters are contracted, equal to only 600
of the muon's meters. Understand? I started and completed the exercise from the muon's point of view. According to the muon's point of view, the Earth's clock will only tick off about .13 microseconds while its own proper time will elapse 2.2 microseconds and it will decay before the Earth reaches it. That is what I mean by beginning the exercise in the 'wrong' frame.

2inq,

When the muon is created it will see the distance to the earth as 600m. It will tick off its 2.2us life and decay as it hits the surface. The end.


2inq:

The muon will see the distance to the Earth as 10,000 meters, its meters.

Ok. It decays at 2.2us when it hits the earth, 10000m away. That's an effective speed of:

10000/2.2e-6 = 454 x 10⁷m/s = 15 times the speed of light. Still think you're right?

by superluminal:

2inq,

When the muon is created it will see the distance to the earth as 600m. It will tick off its 2.2us life and decay as it hits the surface. The end.
================================================== =============

As previously stated by funkstar, both frames of reference agree there are 10,000 meters to the Earth's surface when the muon is created. The Earth is the frame moving in the muon's rest frame. Why should the muon see a contracted distance when it is not the one moving?
----------------------------------------------------------------------------------

by superluminal:


Ok. It decays at 2.2us when it hits the earth, 10000m away. That's an effective speed of:

10000/2.2e-6 = 454 x 107m/s = 15 times the speed of light. Still think you're right?
================================================== ============

I am following the Special Theory method of the assuming the observer is the one at rest and the other frame is the one moving. I just started from the muon's frame instead of the Earth's frame as is explained in STR examples. When beginning the example, the observer in the first frame of reference always 'sees' the clock ticking slowly in the other frame, the frame which is moving. Then, STR switches to the second frame of reference and states this observer 'sees' the distance as contracted, again the 'other' frame is the one in motion. You ask me if I think I'm right? Can you not keep track of the posts, superluminal? I said early on that Special Theory blows up when the exercise is begun in the 'wrong' frame of reference. This is what I am illustrating.
Now to your assertion that the muon will see the distance to the Earth as 600 meters
when it is created and it will hit the surface of the Earth in 2.2 microseconds. Remember, light travels one meter in 1/299,792,458 seconds in ANY frame according to Special Theory. That means a muon's meter is 16.7 times as long as an Earth meter. No 'shorter' meters there! There are FEWER meters. Think about it, it also means the speed of light is not constant, but is frame dependent.

The only way that makes sense is if the muon clock is the one beating slower from EITHER frame of reference, due to the fact that the muon IS the one with greater velocity. Just as in General Relativity, the slower clock keeps its status as the clock beating slower and the faster clock (Earth's) is always seen as the one beating relatively faster. But again, the speed of light would have to be frame dependent here also. The speed of light IS frame dependent in General Relativity, a light clock in low Earth orbit will beat relatively slower than a light clock in high Earth orbit.

Janus58,

If It were me, my answer to:



Would be "Fuck you geistkiesel, and the fucking relativistic jackass you rode in on."

Just my opinion though.
SL,

I posted a paper with experimental data included. A paper that you referenced in your previous post. You haven't a clue what you are doing do you SL? You didn't even know what the paper was you submitted with your gross condescending bullshit? Who gave you the references you post, your little propaganda research team? Most of the links you provided were just advertisements, did you know that?

You respond with obscenities. But then what else is to be expected? What's the matter, a little unsure of yourself that you have to wave off a scientific discussion using an all too familiar belligerent mouth [hand]?

Muons are not created at "10,000" meters, muons are created throughout the total volume of the atmoswphere.

How do you and your robot friend you defend, who isn't able to answer for itself, handle actual experimental results, data that is, in your religious discussions? I get it, simply ignore it right? Like Aer. If it suggests a threat to SRT crunch the messenger. That's the drill your've picked here isn't it SL, you and your friend that is too personally wounded with its big bad emotional boo boo to respond with anything even approaching sensibility?


Geistkiesel :cool:

by superluminal:

2inq,

When the muon is created it will see the distance to the earth as 600m. It will tick off its 2.2us life and decay as it hits the surface. The end.
================================================== =============

As previously stated by funkstar, both frames of reference agree there are 10,000 meters to the Earth's surface when the muon is created. The Earth is the frame moving in the muon's rest frame. Why should the muon see a contracted distance when it is not the one moving?
----------------------------------------------------------------------------------

I never said that! In fact, I repeatedly emphasized to you that the 10,000 meters is in the Earth frame, and that due to the relative velocity of the Earth frame wrt. the muon frame, those 10,000 Earth meters become 600 muon meters, because of length contraction!

Don't you get it yet? You can't use a distance measurement from the Earth frame directly in the muon frame!

I never said that! In fact, I repeatedly emphasized to you that the 10,000 meters is in the Earth frame, and that due to the relative velocity of the Earth frame wrt. the muon frame, those 10,000 Earth meters become 600 muon meters, because of length contraction!

Don't you get it yet? You can't use a distance measurement from the Earth frame directly in the muon frame!

by funkstar:

"I've told you twice, already: The muon start 10,000 Earth meters up in the atmosphere - there's no disagreement on this!. Both the Earth observer and the muon agree that there's a big "10 clicks" sign right where the muon pops into existence. However, due to length contraction of the Earth and it's atmosphere, because the Earth is moving in the muon frame, the muon sees those 10,000 Earth meters as (say) just 600."
================================================== ========

I repeat: I am beginning the exercise FROM THE MUON'S FRAME OF REFERENCE. Don't you get it? The Earth observer or his measurement of distance has not even entered the exercise yet. No 'Earth meters'. The muon is at rest in its frame and the Earth is approaching THE MUON. Consider an example of two particles in open space approaching each other at .987c. When the particles are 10,000 meters apart, does one proclaim that the distance is only 600 meters in its frame? WHICH ONE? No, both will measure the distance as 10,000 meters according to Special Theory. You are choosing a preferred frame by insisting the distance is only contracted in the muon's frame of reference. That is a no-no.

================================================== ========

I repeat: I am beginning the exercise FROM THE MUON'S FRAME OF REFERENCE. Don't you get it? The Earth observer or his measurement of distance has not even entered the exercise yet. No 'Earth meters'. The muon is at rest in its frame and the Earth is approaching THE MUON. Consider an example of two particles in open space approaching each other at .987c. When the particles are 10,000 meters apart,
As measured from which Frame?
does one proclaim that the distance is only 600 meters in its frame? WHICH ONE?The one which isn't the one that measures the distance as 10,000 meters.
No, both will measure the distance as 10,000 meters according to Special Theory. You are choosing a preferred frame by insisting the distance is only contracted in the muon's frame of reference. That is a no-no.

No, you are the one trying to infer a prefered frame of reference when you said above that the two muons are 10,000 meters apart at some point without specifying which frame that measurement will be made. Try thinking of it this way: Each muon has a measuring tape that it carries with it and extends 10,000 meters ahead of it (as measured in its own frame). At the instant a muon is opposite the end of the other muon's tape, where is the other muon with respect to the first moun's tape according to the first muon? Do this for both Muons.

Aer, I don't know where I didn't make myself clear in the other posts, but why do you still say that light from the edge of the universe reaches the muon? It doesn't. In the muon's own frame, it's life is very short (the normal length). So, just when it reaches the earth, it dies. Any light, which isn't seen as reaching it in the earth frame, doe not reach it in it's own frame either. How can you not understand this. I'l give my earlier diagram here also.
Length contraction accounts for both the ability of the muon to reach the earth (in the muon frame), <I>and</I> the <I>inability</I> of light from beyond to reach the muon. It is clear enough from the diagram.

<Img src= "http://www.geocities.com/virusmakermad/Muon.GIF"></Img>

Yellow: Source; Red: Photon; Brown: Muon; Black platform: Earth;

In fig. B. instead of showing the earth and the source move towards the Muon, as really happens in that frame, I've superimposed two images from t=0 and t=t on top of each other in such a way that the Muon appears to move.

by funkstar:

"I've told you twice, already: The muon start 10,000 Earth meters up in the atmosphere - there's no disagreement on this!. Both the Earth observer and the muon agree that there's a big "10 clicks" sign right where the muon pops into existence. However, due to length contraction of the Earth and it's atmosphere, because the Earth is moving in the muon frame, the muon sees those 10,000 Earth meters as (say) just 600."
================================================== ========

I repeat: I am beginning the exercise FROM THE MUON'S FRAME OF REFERENCE.

So did I.

Don't you get it? The Earth observer or his measurement of distance has not even entered the exercise yet. No 'Earth meters'.

Yes, there are. It is undisputably so, because there's an imaginary altitude balloon at Earth height 10 km with "Height 10km" on it right where the muon comes into existence (with relativistic speed wrt the Earth immediately; no acceleration necessary.)

The muon is at rest in its frame and the Earth is approaching THE MUON.

I never disputed this. In fact, I wrote it explictly several times.

Consider an example of two particles in open space approaching each other at .987c. When the particles are 10,000 meters apart, does one proclaim that the distance is only 600 meters in its frame?

Yes, because that measurement of 10,000m is frame dependent! Don't you know that distances are relative?

Do the math as per Janus58's example. It should clear it up for you, as well as enlightening you to the relativity of simultaneity (it's the pole-in-the-barn "paradox" in light disguise.)

WHICH ONE? No, both will measure the distance as 10,000 meters according to Special Theory.

No, they won't. There's no way in hell they can both measure the distance as being the same if they are in relative movement, without violating the basic postulates of str.

You are choosing a preferred frame by insisting the distance is only contracted in the muon's frame of reference. That is a no-no.
Absolutly not! All I'm saying is that the Earth meters are contracted in the muon frame which is perfectly in line with str. You, on the other hand, are assuming a frame invariant distance between the muon and the Earth which is in direct contradiction with str. Not once have I claimed that the Earth observer won't see the muon meter as length contracted! However, that is a vastly different claim than what you are saying must happen! Do you understand why it is nonsensical to claim that because the muon is moving in the Earth frame, the Earth meter must contract for the Earth observer?

Again, you can't switch frames in str except via the Lorentz transforms. Your use of a measurement from the Earth frame in the muon frame without using the transforms is not allowed. And that measurement most definitely comes from the Earth frame, because we know that muons originate in the atmosphere, right next to that damned balloon!

OK, after careful examination I have come to the conclusion that you are ALL wrong (in one regard or another).


Surely you guys know that all you have to do to skew Special Theory results is to start the gedanken from the 'wrong frame'. Example:

The muon is created at 10,000 meters from the Earth through decay. (varies,of course). This is the Earth frame, you know this right?

The muon sees the Earth clock as beating slower than its own. The muon lives a MEAN life (not half-life, a different thing) of 2.2 microseconds in its 'rest' frame,
and 'sees' the Earth's clock accumilate only a fraction of this time, about .13 microseconds, before it decays. The Earth 'sees' the distance to the muon as length
contracted, instead of 10,000 meters, the Earth observer sees the muon created at
about 600 meters. No. This is wrong.

In one gedanken the muon originates 10,000 Earth meters from the surface of the earth. In the other gedanken the muon originates 10,000 muon meters from the surface of the earth. A meter is a meter is a meter. A meter is not redefined from one frame to another.


When beginning from the Earth observer's frame of reference, the Earth observer sees the muon's clock running slow. Does this Earth observer also see the distance to the muon contracted? This is ambiguous, are you talking about what the Earth observer must conclude the muon sees?


When beginning from the muon observer's frame of reference, the muon observer sees the Earth's clock running slow. Does the muon observer also
see the distance to the Earth contracted? Again, I think you are confusing frames.




When beginning from the Earth observer's frame of reference, the Earth observer sees the muon's clock running slow. Does this Earth observer also see the distance to the muon contracted?

No.


When beginning from the muon observer's frame of reference, the muon observer sees the Earth's clock running slow. Does the muon observer also see the distance to the Earth contracted?

Yes.


How can you say No to one and Yes to the other? You are assuming your own interpretation to each question differently than the other.




The first two statements were exactly symmetrical. Why a 'no' to the first and a 'yes'
to the second statement?
This is precisely the problem they are having with my question as well. :D

When the muon is created it will see the distance to the earth as 600m. Where are you pulling these numbers? This is what initially confused me in your very first response. It should be 2000m, not 600m. (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html)


I never said that! In fact, I repeatedly emphasized to you that the 10,000 meters is in the Earth frame, and that due to the relative velocity of the Earth frame wrt. the muon frame, those 10,000 Earth meters become 600 muon meters, because of length contraction! You must mean Earth Meter = distance as measured in meters in the Earth rest frame and Muon Meter = distance as measured in meters in the Muon rest frame? Again, why 600m? Is my source wrong?


When the particles are 10,000 meters apart, does one proclaim that the distance is only 600 meters in its frame? WHICH ONE? No, both will measure the distance as 10,000 meters according to Special Theory.
The other must 'see' the other's spatial coordinates as contracted.


Aer, I don't know where I didn't make myself clear in the other posts, but why do you still say that light from the edge of the universe reaches the muon? It doesn't. In the muon's own frame, it's life is very short (the normal length). So, just when it reaches the earth, it dies. Any light, which isn't seen as reaching it in the earth frame, doe not reach it in it's own frame either. How can you not understand this. I'l give my earlier diagram here also.
Length contraction accounts for both the ability of the muon to reach the earth (in the muon frame), and the inability of light from beyond to reach the muon. It is clear enough from the diagram. As I said, your explaination is inadequate as I see it. I merely started a new thread to get other explainations from other people - sorry if this is offensive to you.

OK People, here is the math - we are going to first work in the Earth frame and conclude what the Muon must see in it's own frame according to Special Relativity.

Then we are going to start in the Muon frame from our conclusions above and we are going to predict what the Earth must see in it's own frame - again, according to Special Relativity.

First the Earth frame: The average distance from the surface of the Earth to where a muon is created is 10,000 m, from the Earth frame. The muon travels at a velocity of .98c for it's entire existence. Thus to reach the Earth's surface, the average life of the muon as seen in the Earth frame is t = x/v = 3.4e-5 s. This is too long of a life for the average muon - luckily, Special Relativity says that the muon's time will be dilated as seen from the Earth frame. The following MATLAB code will be used for the calculations to see what Special Relativity says of the Muon Frame:

----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Earth frame:');
Lo=10e3
To=Lo/v

disp('Muon Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Earth frame:
Lo = 10000
To = 3.4014e-005
Muon Predictions:
L = 1.9900e+003
T = 6.7686e-006
----------------------------------

Thus, in the Muon's frame, the length to the Earth's surface is 2000m upon when the muon was created. The average life of the muon is 6.8e-6 s, well within the life of the average muon based on it's half-life.



Now for analysis from the Muon's frame: The average distance from the surfaceof the Earth to where a muon is created is 2,000 m, from the Muon frame. The Earth travels at a velocity of .98c for the entire time the muon exists. Thus to reach the Muon, the time between which the muon is formed and the time the Earth's surface hits the Muon in the Muon's frame is t = x/v = 6.8e-6. Now to see what Special Relativity predicts the Muon will say the Earth frame should see:


----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Muon frame:');
Lo=2e3
To=Lo/v

disp('Earth Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Muon frame:
Lo = 1.9900e+003
To = 6.7686e-006
Earth Predictions:
L = 396.0000
T = 1.3469e-006
----------------------------------

Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m upon when the muon was created. The time on Earth is 1.3e-6 s.

But alas, the Earth predictions from the Muon frame according to Special Relativity do not match what we actually know happens in the Earth frame. - Where is this analysis wrong?

Aer,

"Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m"

If muons are observed by balloon instruments in the earth frame, to detect muon creation at 10000m, how can you say this. I haven't analyzed your post yet, but you are making an invalid assumption somewhere.

As funkstar has pointed out several times, the speed of a muon cannot affect distances in the earth's frame.

If muons are observed by balloon instruments in the earth frame, to detect muon creation at 10000m, how can you say this. I haven't analyzed your post yet, but you are making an invalid assumption somewhere. Assumption? I only assumed Muons start at 10000m above Earth and then only applied Special Relativity - notably, the reciprocity effect of Special Relativity. However, you may be correct - I could have errored somewhere in my analysis of special relativity.

Aer:

How can you say No to one and Yes to the other? You are assuming your own interpretation to each question differently than the other.

Because, as I have stated before, the muon has been created in the earth's frame and then accelerated by the enormous energy of the gamma ray to 0.988c, while the earth has not. The earth has not changed it's motion through spacetime.

The motion of a particle cannot effect distances in other frames. You are misinterpreting the symmetry of the "no preferred frames" statement of relativity.

The motion of a particle cannot effect distances in other frames. You are misinterpreting the symmetry of the "no preferred frames" statement of relativity. So you are saying the muon neccessarily undergoes acceleration. So therefore it is inappropriate to analyze the situation from the Muon's perspective, as Special Relativity would say you can if the Muon did not undergo acceleration?

No. Acceleration has nothing to do with it.

I think I see the problem with all of this. A problem of scale. Let's try this:

P1 and P2 are identical planets with 10000m thick atmospheres. They are some distance away from each other and approaching each other at 0.987c.

P1 sees P2 length contracted, including seeing it's atmosphere as 600m thick and vice-versa.

When P1 and P2's atmospheres first touch, P1 sees P2's atmosphere touching his at 10000m up, (P2's still appears 600m thick) and vice-versa. Completely symmetrical.

Now, shrink P2 down to the size of a muon.

P1 still sees it hitting the atmosphere at 10000m, and the little muon people still see P1's atmosphere as 600m thick.

So, what I'm saying is that the space between objects in different frames does not change. However the distance between objects that are moving in the same frame appears contracted.

So, my animation in the "Visual SRT 1" needs to be corrected as it is misleading. I shall correct it.

No. Acceleration has nothing to do with it.
I don't think you followed my post - the explaination you just gave does nothing to fix the reciprocity issue here.

Edit: That is, from the muon's perspective, shouldn't less time appear to elapse in the Earth frame? If so, then there should equally be length contraction in the Earth frame. And it appears 2inq has one thing right - as long as you don't assume the wrong frame, then everything in special relativity is just hunky-dory.

Change in velocity is all that matters so, when I say acceleration has nothing to do with it, I mean that the fact that change in velocity through spacetime is what's important (and of course this requires acceleration) but the magintude of the acceleration is irrelevant.

does nothing to fix the reciprocity issue here.

I think it does.

Let's take this one step at a time. Do you agree with the P1, P2 planet scenario above?

Let's take this one step at a time. Do you agree with the P1, P2 planet scenario above? Can you repeat the explaination but include calculations along the way. I still don't know where 600m comes from.

Edit: That is, from the muon's perspective, shouldn't less time appear to elapse in the Earth frame? If so, then there should equally be length contraction in the Earth frame. And it appears 2inq has one thing right - as long as you don't assume the wrong frame, then everything in special relativity is just hunky-dory.

Ok, less time passes on the earth clock. In the muon's frame 2.2us elapses as the surface travels the 600m. Yes? And the muon will see say, 0.3us on the earth clock. Ok?

In the earth's frame say 1000us elapses as the muon travels the 10000m. Yes? And the earth will see 2.2us on the muon clock. Ok?

Problems with this?

Problems with this? Did you analyze my post dealing with the math thoroughly? For the muon to see less time on the Earth clock, it must also see the Earth's frame length contracted.

Can you repeat the explaination but include calculations along the way. I still don't know where 600m comes from.

I'm just using representative numbers here. The actual numbers don't really matter as long as we have the right order of magnitude. I think math details at this point just confuse things.

The 600m was one quote for the atmospheric depth in the muon's frame, from some website based on some earth frame altitude that I don't remember.

Did you analyze my post dealing with the math thoroughly? For the muon to see less time on the Earth clock, it must also see the Earth's frame length contracted.

Yes. I completely agree (with the dilation effects)

Gotta go now. Be back later.

Did you analyze my post dealing with the math thoroughly? For the muon to see less time on the Earth clock, it must also see the Earth's frame length contracted.

So, yes, and the earth sees the muons frame contracted. If it had a 10000m thick atmosphere we would see it as 600m (or whatever).

Remember, a frame is just a set of comoving coordinates. So, the earth and the molecules of air are comoving and according to the muon, they are all contracted. The only object comoving with the muon is the muon, and we see that contracted.

Yes?

Remember, a frame is just a set of comoving coordinates. Yes.

So, the earth and the molecules of air are comoving and according to the muon, they are all contracted. So it should seem.

The only object comoving with the muon is the muon, and we see that contracted. We? Who's frame are you in now?

Yes? When analyzing things from the Earth frame, we never refer to anything as seen by the muon, so why are we doing this when analyzing everything from the Muon frame? - That is, refering back to the Earth frame.

We? Who's frame are you in now?

Earths

When analyzing things from the Earth frame, we never refer to anything as seen by the muon, so why are we doing this when analyzing everything from the Muon frame? - That is, refering back to the Earth frame.

Ok fine. If i say the earth sees the muon thusly and the muon sees the earth thusly, Im not mixing frames, just shifting perspective.

Im not mixing frames, just shifting perspective. That is precisely what I did with my example - and the muon see's the Earth frame contracted to 396m or whatever, no?

I'm workin' on it...

Ok I found your error. Give me a while...

Ok I found your error. Give me a while... On a scale of 1-10, how bad was it? Other than the acceleration issue - I still don't see any possible counter argument. Anyway, I await your response.

OK People, here is the math - we are going to first work in the Earth frame and conclude what the Muon must see in it's own frame according to Special Relativity.

Then we are going to start in the Muon frame from our conclusions above and we are going to predict what the Earth must see in it's own frame - again, according to Special Relativity.

First the Earth frame: The average distance from the surface of the Earth to where a muon is created is 10,000 m, from the Earth frame. The muon travels at a velocity of .98c for it's entire existence. Thus to reach the Earth's surface, the average life of the muon as seen in the Earth frame is t = x/v = 3.4e-5 s. This is too long of a life for the average muon - luckily, Special Relativity says that the muon's time will be dilated as seen from the Earth frame. The following MATLAB code will be used for the calculations to see what Special Relativity says of the Muon Frame:

10000 earth-frame meters.


----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Earth frame:');
Lo=10e3
To=Lo/v

disp('Muon Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Earth frame:
Lo = 10000
To = 3.4014e-005
Muon Predictions:
L = 1.9900e+003
T = 6.7686e-006
----------------------------------

A-ok.

Thus, in the Muon's frame, the length to the Earth's surface is 2000m upon when the muon was created. The average life of the muon is 6.8e-6 s, well within the life of the average muon based on it's half-life.

2000 muon-frame meters. (calculated)

Now for analysis from the Muon's frame: The average distance from the surface of the <strike>Earth</strike> muon to where the <strike>muon</strike> Earth is created is 2,000 m, from the Muon frame. The Earth travels at a velocity of .98c for the entire time the muon exists. Thus to reach the Muon, the time between which the muon is formed and the time the Earth's surface hits the Muon in the Muon's frame is t = x/v = 6.8e-6. Now to see what Special Relativity predicts the Muon will say the Earth frame should see:

Ok? (if we're being strict with our frame perspectives).

----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Muon frame:');
Lo=2e3
To=Lo/v

disp('Earth Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Muon frame:
Lo = 1.9900e+003
To = 6.7686e-006
Earth Predictions:
L = 396.0000
T = 1.3469e-006
----------------------------------

Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m upon when the muon was created. The time on Earth is 1.3e-6 s.

No.

In special relativity, an event that occurrs, occurrs in all frames, it's just the timing that is in question (simultaneity). So, in the earth frame the muon creation event occurred at T=0 at 10000m (given). This can never change in the earth frame.

So far so good.

The only given at the beginning is that the muon was IN FACT created at 10000m in the earth frame.

You then go on to correctly calculate what the muon would see in its frame.

You then proceed to use the CALCULATED muon-frame distance (Lo=2e3m) to back-calculate and see what the earth should see. This is incorrect, and you can see that this process will lead to ever decreasing distances as you calculate each frame!

You are violating the initial conditions of the system. All you can ever know is what is given as FACT at the start of the exercise.

So, a new problem:

An earth is created 2000m (a given FACT) above the surface of the muon, travelling at 0.98c. We calculate, correctly, that the earth will see us (muon) as 396m away in its frame.

You cannot then use the CALCULATED value to deduce the original given FACTS.

But alas, the Earth predictions from the Muon frame according to Special Relativity do not match what we actually know happens in the Earth frame. - Where is this analysis wrong?

The upshot is that you need to pick a frame and stay there. You cannot calculate anything for the original frame from a calculated frame.

You can calculate what the muon will see, but what we see is etched in reality and can't change. There is only one primary frame in any calculations.

You then proceed to use the CALCULATED muon-frame distance (Lo=2e3m) to back-calculate and see what the earth should see. This is incorrect, and you can see that this process will lead to ever decreasing distances as you calculate each frame! Yes it would.


You are violating the initial conditions of the system. All you can ever know is what is given as FACT at the start of the exercise. I never assumed anything other than what special relativity tells me.


So, a new problem:

An earth is created 2000m (a given FACT) above the surface of the muon, travelling at 0.98c. We calculate, correctly, that the earth will see us (muon) as 396m away in its frame. And you don't see a problem with this? That is, we used this exact same analysis to determine that the muon was 2000m away when the Earth was created as you describe it. Get what I am saying?


You cannot then use the CALCULATED value to deduce the original given FACTS. But it leads to contradiction - are you imposing that we cannot first assume the Muon frame? What if that was the frame in which the experiment was determined and the 2000m and 6.8e-5s were the FACTS as you put them.

Now to see what Special Relativity predicts the Muon will say the Earth frame should see:


----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Muon frame:');
Lo=2e3
To=Lo/v

disp('Earth Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Muon frame:
Lo = 1.9900e+003
To = 6.7686e-006
Earth Predictions:
L = 396.0000
T = 1.3469e-006
----------------------------------

Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m upon when the muon was created. The time on Earth is 1.3e-6 s.

But alas, the Earth predictions from the Muon frame according to Special Relativity do not match what we actually know happens in the Earth frame. - Where is this analysis wrong?
You're not being meticulous enough in your analysis. While the Earth observer will see 1990 Muon meters as 396 Earth meters, he will not agree that this is the Muon's height. Why is this so, when you've calculated the length contraction? It's because you're treating space as independent of time.

You have to use the Lorentz transforms (http://en.wikipedia.org/wiki/Lorentz_transformation_equations):

In keeping with the link's notation let S be the Muon's frame, S' be the Earth frame, moving along the x-axis with speed 0.98c wrt S, such that at time t=t'=0, x=x'=0. Assume gamma=5.025.

Now, the muon pops into existence in muon coordinates (x,t)=(1990,0). This means that at time 0 for the muon, there's 1990m to the Earth surface located at x'=0 at all times, and specifically at t=t'=0, when x=x'=0.

We now use the Lorentz equations to translate this to the Earth frame, and we get (x',t')=(10,000,3.27e-5). Note here that the Earth observer and the muon do not agree on when the muon came into existence! This is extremely important to note, because it shows the frame dependence of the initial condition! In other words, because time and space are not independent, from the earth frame, saying that the muon has height 10,000 at time 0, is not the same as what this thought experiment states.

At muon spacetime coordinates (x,t)=(1990,6.50e-6) the surface hits the muon. Again, using the Lorentz transforms this translates to the Earth frame as (x',t')=(0,1.32e-4). That is, even though the muon only saw the Earth travel 1990m, the Earth observer saw the Muon travel 10,000.

In other words, switching frames does not mean ever decreasing lengths, as is implieed in your argument.

No I don't see a problem and it is not a contradiction. And I did assume the muon frame in my post, but it's NOT the same physical situation anymore.

So, a new problem:

An earth is created 2000m (a given FACT) above the surface of the muon, travelling at 0.98c. We calculate, correctly, that the earth will see us (muon) as 396m away in its frame.

You cannot then use the CALCULATED value to deduce the original given FACTS.

They are two mutually exclusive and valid physical situations.

funkstar - yes.

The mutual length contraction is just as weird as mutual time dilation, but it does not lead to any inconsistencies in str*.



*In theory, inconsistencies are impossible, because str is nothing more than R^4 with the Minkowski metric.

Note here that the Earth observer and the muon do not agree on when the muon came into existence! I figured simultaneity had to be a part of this somehow - length contraction and disagreement on events being simultaneous go hand in hand. Neither of the two have been proven by experiment - but they haven't been disproven either. I think the goal here should be to determine an experiment that is practicle to conduct which would confirm it one way or the other. If you want to argue on this experimental proof point - then I will ask you to first point to an experiment to back up your arguments. I'd love to read it.

*In theory, inconsistencies are impossible, because str is nothing more than R^4 with the Minkowski metric. I thought there had to be some explaination - which is why I asked the question seeking out where the analysis went wrong. :D Some others on this forum need to learn this trait before they go around claiming a theory to be invalid (uh-HUM - not going to name names)

Just for illustration purposes, let's try these animations. The first shows events in the same frame as the Earth and the other in the smae frame as the Muon. Each has a measuring rod with which it measures distances. The Earth's measuring rod reaches from surface to the top of the atmosphere (10,000m). The muon measuring rod is also 10,000m as measured in its own frame. The muon is shown as a yellow dot. The measuring rods exists thorughout both animations but the Muon doesn't appear until the left end of the Muon's rod reaches the upper edge of the Earth's atmosphere.

In order to make things easy to see, the thickness of the atmosphere is not to scale, and I reduced the relative velocity to .866 c.

The first animation shows things from the Earth frame.

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/muon1.gif

The muon rod approaches from the left, and due to length contraction, is half the length of the Earth Measuring Rod. At the instant the left end of the rod hits the atmosphere the muon is created. At this instant, the Moun is at the End of the Earth measuring rod, and the Earth measures the distance as 10,000m (note at this point, the right end of the Muon's rod has not yet reached the Surface of the Earth.)

It continues on until it hits the surface of the Earth.

In the second Animation:
http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/muon2.gif

We see things from the same frame in which the Muon is in. In this case, the Earth rushes towards the left, and undergoes length contraction.) Again, the muon appears when the atmosphere reaches the left end of the Muon measuring rod. In this frame however, the Muon measuring rod hits the Earth's surface before the muon is formed. Thus when the Muon measures its distance to the Earth at the instant of its creation, it finds that it is less than 10,000 m away from the surface.( note however that, at that same exact instant, it is at the left end of the Earth measuring rod, which we know is 10,000m long as measured from the Earth.

Again, the Earth continues on until it hits the Muon.

Perfect symmetry of length contraction, The muon sees everything in the Earth frame as contracted and the Earth sees everything in the muon frame as contracted by the same factor. and we conclude that when the muon is created it measures its distance from Earth as less than what the Earth measures the distance to the muon to be when the muon is created.

Nice!

I agree, very nice illustration, Janus58!

Now a couple of questions. In your first illustration, drawn from the Earth observer's frame of reference, the Earth observer sees the muon created at 10,000 earth-meters
above the surface. This same observer sees the muon's meterstick (rod) contracted by
50% due the the relative velocity of .866c. According to the Earth observer, he thinks the muon must transverse 20,000 of the muon's meters to reach the surface, since the
muon's meters are half as long as his. This is also illustrated by the animation. 10,000 Earth meters equal 20,000 muon meters according to the Earth observer.

In your second illustration, the muon observer's frame of reference, when the muon is created at the top of the approaching Earth's atmosphere, part of his meterstick has already passed throught the atmosphere, and the muon observer sees 5,000 of his
(the muon observer's) meters left until the Earth's surface strikes him. The muon observer, according to your illustration, thinks 5,000 of his meters are equal to 10,000 of the Earth's meters, again because the Earth's meters are 50% the length of his own meter due to the .866c velocity of the approaching Earth.

Question. Since the relative velocities are identical in both illustrations, the defining event that leads to different measurements and relativity of simultaneity is the creation point of the muon, which is in the same x,y,z and t coordinates in both reference frames. Without this event, the distances in both frames of reference would be symmetrical, correct? Is this not the same as using a 'third' frame of reference to
obtain correct results?

Edit: By the way, we can fill-in the time component to this illustration at a later time. The suspense is great, isn't it?

I agree, very nice illustration, Janus58!



Question. Since the relative velocities are identical in both illustrations, the defining event that leads to different measurements and relativity of simultaneity is the creation point of the muon, which is in the same x,y,z and t coordinates in both reference frames. Without this event, the distances in both frames of reference would be symmetrical, correct? Is this not the same as using a 'third' frame of reference to
obtain correct results?


Without two events, The creation of the muon and the muon striking the Earth, we have no time interval to be measured in either frame, and no way to compare clocks between frames, and thus no experiment.

The creation of the muon is a spacetime event, it is not a reference frame by any usage of the term. So no, this is not the same as using a third reference frame.

I could introduce a third reference frame, say, one in which the Earth and Muon are approaching each other from the right and left at equal velocites, but it would come up with the same results as to the respective readings on the muon and Earth clock when the muon hits the surface. I could introduce an infinite number of reference frames, but they would all come up with the same final result.

Perhaps you didn't notice it, but your illustration does not illustrate perfect symmetry.
I pointed it out when I filled in the numbers.

The first illustration, calculated from the Earth's frame of reference, has the muon created at 10,000 Earth meters. The second illustration, calculated using the muon's
frame of reference as the beginning frame, agrees with this number, 10,000 Earth meters. This number is symmetric and agreed upon in both illustrations.

The first illustration, calculated from the Earth's frame of reference, has the 10,000 Earth meters equal to 20,000 muon meters. The Earth observer states the muon has
20,000 of its OWN meters (the muon's) to transit to reach the surface. The second illustration, calculated using the muon's frame of reference as the beginning frame, has the Earth surface located 5,000 of the muon's meters away when the muon is created due to relativity of simultaneity. This is equal to 10,000 of the Earth's meters. Using the Earth as the beginning frame, the muon must travel 20,000 muon meters to reach the surface. Using the muon frame as the beginning frame, the Earth must travel 10,000 Earth meters to reach the muon. These frames are not symmetric. The defining event, the creation of the muon at the x,y,z coordinates, must be used to explain this asymmetry. Its coordinates constitute a third frame of reference without
which both scenarios in your illustrations WOULD be symmetric, i.e. each would measure 10,000 of their own meters and state the moving frames meters were contracted to equal 20,000 of their own.

There is no problem assigning time intervals to the illustrations, we have the velocity of the muon/Earth and the distances to the muon/Earth in both illustrations. Yes, I know why you balk at assigning time intervals to the illustrations. Everyone can think about this for awhile, see what you can come up with that makes sense.

Animations zoomed in and with time indexes added. Note that I do not have the Muon clock starting until the muon is created. The time indexes are not in any standard time unit, just note that each clock's tick rate would be the same if all the clocks were put int he same frame. The Earth's clock's (one at each end of the Earth's measuring rod) are sychronized in the Earth frame. and are indexed such that the left clock reads zero at the instant the muon is created next to it. (so that for convenience, both clocks next to the muon at the moment of creation read zero. )

The animations pause at the begining, when the Muon is created, when the right end of the muon measuring rod reaches the surface of the Earth and when the muon reaches the center of the Earth so that you can read and compare the clock readings.
To be technically correct, the second animation should be half the length of the first, but I time expanded it so that it would be easier to keep track of the clocks.

I could have extended the Muon measuring rod out to 20,000 km in length and in that way shown that it would hit the surface of the Earth when the Surface clock read zero in both frames, but it would have meant not being able to zoom in as much to get it all in,(as it is, I let the upper atmosphere end of the Earth measuring rod exit the frame before the second animation ends.) and it would not be consistant with the first animation.

From the Earth Frame

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/muon3.gif


From the Muon frame
http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/muon4.gif

Excellent, Janus58.

This is getting to be fun!

OK, now time to get serious.
Janus58, in your very first illustration, from the Earth frame of reference:
c = 300 meters/microsecond, in all frames according to Special Theory. I will hereafter
use the symbol 'm' to refer to a meter and the symbol 'us' to refer to a microsecond.
You stated the muon traveled at .866c in your illustrations. That equals 259.8 m/us.
10,000 m divided by 259.8 m/us = 38.49 us, the time according to the Earth observer in the Earth's frame of reference for the muon to reach the surface.
This same Earth observer in Earth's frame of reference will calculate the muon 'sees' 20,000 meters to the Earth surface due to length contraction of the muon's meter in the muon's frame of reference (again, as calculated from Earth's frame)
This same Earth observer in Earth's frame of reference will state the muon's clock is ticking at half the rate of Earth's clock due to time dilation of the muon's clock in the muon's frame of reference. Thus the Earth observer in Earth's frame of reference will calculate the muon travels 20,000 muon-meters in 19.25 microseconds of the muon's clock, equal to 1038.96 meters/microsecond. That is 3.46c. Whoops!

Now, do you realize why I kept saying 'Earth observer in the Earth's frame of reference? Because in your second illustration, you are using an Earth observer TRANSFORMED to the muon's frame of reference. The Lorentz transforms are not noted in your illustration, but they are there nontheless. That is why the 'muon' only 'sees' 5,000 meters to the Earth's surface. If the second illustration truly started in the muon's frame of reference with NO PRIOR HISTORY, the muon would see the Earth approachin at .866c and at a distance of 10,000 meters away. A meter is defined by the speed of light and cannot contract in the muon's own frame of reference. The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference. No 'contracted' atmosphere if STARTING from the muon's frame of reference. And 'relativity of simultaneity' does not come into play yet with no prior history in the frame.

This is getting to be fun!

OK, now time to get serious.
Janus58, in your very first illustration, from the Earth frame of reference:
c = 300 meters/microsecond, in all frames according to Special Theory. I will hereafter
use the symbol 'm' to refer to a meter and the symbol 'us' to refer to a microsecond.
You stated the muon traveled at .866c in your illustrations. That equals 259.8 m/us.
10,000 m divided by 259.8 m/us = 38.49 us, the time according to the Earth observer in the Earth's frame of reference for the muon to reach the surface.
This same Earth observer in Earth's frame of reference will calculate the muon 'sees' 20,000 meters to the Earth surface due to length contraction of the muon's meter in the muon's frame of reference (again, as calculated from Earth's frame)
This same Earth observer in Earth's frame of reference will state the muon's clock is ticking at half the rate of Earth's clock due to time dilation of the muon's clock in the muon's frame of reference. Thus the Earth observer in Earth's frame of reference will calculate the muon travels 20,000 muon-meters in 19.25 microseconds of the muon's clock, equal to 1038.96 meters/microsecond. That is 3.46c. Whoops!
That figure has no physical significance. In order to do a proper transform, you have to decide which event you are going to transform from (The Surface clock reading Zero and being opposite the 20,000m mark of the Muon measuring rod, or the muon appearing at the 10,000 m mark of the Earth rod and its clock rading Zero.)

In the first case, when you transform to the Muon frame at this instant the left end of the measuring rod is 20,000m from the Earth surface and the time in that frame is -57.74us before the Muon is created. the muon takes 19.24us to cross the rest of the distance for a total time of 76.98us. 20,000/76.98 = 259.8m/s = .866c

In the second case the muon time =0 and the muon finds itself 5000m from the earth surface it takes 19.24us seconds to cross that 5000m. 5000/ 19.24 =259.8m/s =.866c. [quote]


Now, do you realize why I kept saying 'Earth observer in the Earth's frame of reference? Because in your second illustration, you are using an Earth observer TRANSFORMED to the muon's frame of reference. The Lorentz transforms are not noted in your illustration, but they are there nontheless. That is why the 'muon' only 'sees' 5,000 meters to the Earth's surface. If the second illustration truly started in the muon's frame of reference with NO PRIOR HISTORY, the muon would see the Earth approachin at .866c and at a distance of 10,000 meters away.
This is complete gobbly-gook. the second animation is completely independent from the earth observer, it only based on what an observer traveling in the same frame as the muon woul determine. You have managed to completely confuse yourself.
A meter is defined by the speed of light and cannot contract in the muon's own frame of reference. The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference. No 'contracted' atmosphere if STARTING from the muon's frame of reference.
[quote]That is patently ridiculous. Length contraction happens to anything that is moivng relative to your frame. The formula for that contraction is:
L = L' * sqrt(1-v²/c²
where L' is the lenght as measured from the the relatively moving frame.
Is the Earth moving relative to the Muon?
Yes.

Does the atmosphere move with the Earth?
Yes.

How deep is the atmosphere as measured in the Earth frame?
10,000m

What length does this transform to as measured in the Muon frame using the above formula for v=.866c?
5000m.

How deep is the Earth atmosphere then according to the Muon frame?
5000m

Where does the muon form?
At the upper limit of the atmosphere

How far is the muon from the Surface of the Earth when its forms as measured in its own frame?
5000m.

This is true no matter what frame you start from!

And 'relativity of simultaneity' does not come into play yet with no prior history in the frame.

Relativity of simultaneity comes into play anytime you compare clocks in relative motion which are separated by a distance that is measured parallel to the line of relative motion, "prior history" or not.

A meter is defined by the speed of light and cannot contract in the muon's own frame of reference. The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference. No 'contracted' atmosphere if STARTING from the muon's frame of reference.
The formula for that contraction is:
L = L' * sqrt(1-v²/c²
where L' is the lenght as measured from the the relatively moving frame.
Is the Earth moving relative to the Muon?
Yes.

Does the atmosphere move with the Earth?
Yes.

How deep is the atmosphere as measured in the Earth frame?
10,000m

What length does this transform to as measured in the Muon frame using the above formula for v=.866c?
5000m.

How deep is the Earth atmosphere then according to the Muon frame?
5000m

Quick sequential set of questions:

Is the Muon moving relative to the Earth?


Does the atmosphere move with the Earth?


How deep is the atmosphere as measured in the Muon frame?


What length does this transform to as measured in the Earth frame using the above formula for v=.866c?


How deep is the Earth atmosphere then according to the Earth frame?


Where does the muon form?


How far is the muon from the Surface of the Earth when its forms as measured in its own frame?


This is true no matter what frame you start from! Then start with the muon frame - with the questions above, knowing no information from the Earth frame - that is, you are an observer in the Muon frame taking measurements - you have no prior knowledge of any measurements in the Earth frame.

I'm not saying you are wrong - I would just like to see how you apply special relativity.

This is getting to be fun!

It is, isn't it? :)

This'll be a long post, but I beg your patience because it shows an important point.

OK, now time to get serious.
Janus58, in your very first illustration, from the Earth frame of reference:
c = 300 meters/microsecond, in all frames according to Special Theory. I will hereafter
use the symbol 'm' to refer to a meter and the symbol 'us' to refer to a microsecond.
You stated the muon traveled at .866c in your illustrations. That equals 259.8 m/us.
10,000 m divided by 259.8 m/us = 38.49 us, the time according to the Earth observer in the Earth's frame of reference for the muon to reach the surface.
This same Earth observer in Earth's frame of reference will calculate the muon 'sees' 20,000 meters to the Earth surface due to length contraction of the muon's meter in the muon's frame of reference (again, as calculated from Earth's frame)

This is wrong. The 10,000 m in the Earth frame will be seen in the muon frame as 5,000 m. Length contraction, remember. Also, the muon will see no contraction of it's own frame, of course.

This same Earth observer in Earth's frame of reference will state the muon's clock is ticking at half the rate of Earth's clock due to time dilation of the muon's clock in the muon's frame of reference. Thus the Earth observer in Earth's frame of reference will calculate the muon travels 20,000 muon-meters in 19.25 microseconds of the muon's clock, equal to 1038.96 meters/microsecond. That is 3.46c. Whoops!

Again, not 20,000 m but 5,000 m making it... 0.866c. (And, in the muon frame the Earth is the one travelling.)

I'll explain, but first a bit of notation:

Let "ee" stand for "Earth meter as viewed from the Earth frame".
Let "em" stand for "Earth meter as viewed from the muon frame".
Let "mm" stand for "Muon meter as viewed from the muon frame".
Let "me" stand for "Muon meter as viewed from the Earth frame".

(The first letter indicates in which frame some length is in, and the second letter indicates which frame the length is being viewed from.)

The argument you are making is the following: At the time the Earth observer sees the muon pop into existence, there's 10,000ee to it. The Earth observer calculates that the muon will see these 10,000em as just 5,000mm in the muon frame. Now, the Earth observer sees 5,000me as just 2,500ee, and concludes that the 10,000em must be seen as 20,000mm in the muon frame!

Now, of course, there is a glaring inconsistency in what the Earth observer predicts the muon as seeing the distance as being (5,000mm? 20,000mm?), meaning that something is horribly wrong.

The error you make lies in the part in red. The problem is that you suppose that because the Earth observer sees the muon frame as contracted wrt. the Earth frame, then so will the muon. However, as we know, this is not the case! You're using a length contraction of the muon frame that the Earth observer sees in the Earth frame, directly in the muon frame, and that's not allowed (because you can't change frames without the Lorentz transforms). The muon will not see the Earth as length dilated as your analysis claims.

This is also suggested by the special notation: The dilation of 10,000em to 20,000mm is based on a relationship that relates 2,500ee to 5,000me. The units are different, meaning that such a dilation is a non sequitur; it is unjustified (though it doesn't seem so at first glance). Length contraction has the following consequence quantitatively when speaking of distances: ee to me means multiply the number by gamma, me to ee means divide by gamma. Similarly mm to em means multiply by gamma, and (the pertinent case) em to mm means divide.

When relating 10,000em to 20,000mm you multiplied by gamma where you should have divided. Simple error, paradox solved.



Now, do you realize why I kept saying 'Earth observer in the Earth's frame of reference? Because in your second illustration, you are using an Earth observer TRANSFORMED to the muon's frame of reference. The Lorentz transforms are not noted in your illustration, but they are there nontheless. That is why the 'muon' only 'sees' 5,000 meters to the Earth's surface. If the second illustration truly started in the muon's frame of reference with NO PRIOR HISTORY, the muon would see the Earth approachin at .866c and at a distance of 10,000 meters away.

The distance to the Earth from the muon is frame dependent. That's really all there is to it. If two frames agree on any (non-zero) distance at any time, then they must be the same frame (modulo translation and rotation and such) due to light speed being constant in all frames. That means no relative velocity, which is certainly not the case here.

A meter is defined by the speed of light and cannot contract in the muon's own frame of reference.

Exactly the same holds true in the Earth (as you know), yet you proceed to claim

The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference.

No. Length contraction is what happens to objects at rest in the other frame when viewed from your own frame (in which they are moving). Nothing about the muon moving at relativistic speed relative to the Earth can make the meter contract in the Earth frame - how could that possibly be so? (Especially considering that there are infinitely many reference frames...)

No 'contracted' atmosphere if STARTING from the muon's frame of reference. And 'relativity of simultaneity' does not come into play yet with no prior history in the frame.
2inq, saying that the distance from the Earth to the muon is 10,000m in both frames when they have a relative velocity is incompatible with the basic postulates of str, so if you insist on this then you are not doing an str analysis. In the notation I introduced above: Either there's 10,000mm from the muon to the surface of the Earth, or there's 10,000ee, but not both. Str forbids it.

Aer essentially asked my question of Janus58. I will respond later according to how Janus58 answers.
------------------------------------------------------------------------------

by funkstar:
"This is wrong. The 10,000 m in the Earth frame will be seen in the muon frame as 5,000 m. Length contraction, remember. Also, the muon will see no contraction of it's own frame, of course."
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Yes, length contraction. That's very ambigeous, the reason I introduced the '20,000 meter' figure to open discussion. Seems simple to say the 10,000 meters are contracted to 5,000 meters in the muon's frame of reference, as seen by the Earth observer. The 'distance' is cut in half, correct? Half of 10,000 meters IS 5,000 meters, 5,000 of THE SAME LENGTH METERS. If the 'meter' is contracted by 50%, it will take
20,000 of them to equal 10,000 Earth frame meters. So, contrary to common belief, the meter itself is not contracted, the contracted DISTANCE consists of FEWER meters, correct? Can't have it both ways. If the meter is contracted by 1/2 PLUS the distance is contracted by 1/2, then you end up with 10,000 'half-length' meters in a distance that is contracted by 1/2. The muon can't transverse that distance in any shorter period of time UNLESS the muon clock beats at half the rate of an Earth clock. That has been my contention all along, length contraction is due to a slower clock only, the distance SEEMS less because the muon can transverse 10,000 meters in half the time relative to the Earth clock. Time beats slow in the relativistically moving muon's frame of reference only.
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by funkstar:

"The argument you are making is the following: At the time the Earth observer sees the muon pop into existence, there's 10,000ee to it. The Earth observer calculates that the muon will see these 10,000em as just 5,000mm in the muon frame. Now, the Earth observer sees 5,000me as just 2,500ee, and concludes that the 10,000em must be seen as 20,000mm in the muon frame!"
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No, funkstar, that is not my arguement. You have '10,000 meters/Earth frame' on the brain. BEGIN the exercise in the MUON frame of reference, not as transformed from an Earth frame. Think the muon will measure the Earth's atmosphere as 5,000 of its meters? That's fine, no problem. NOW, do your Lorentz transformations TO the Earth frame of reference. The number in this FIRST transform becomes 2,500 meters. See?
When beginning in the muon frame of reference, the muon sees the EARTH frame as contracted, half his own measurement FROM HIS REST FRAME. The Earth is in motion, the universe is in motion, relative to the muon's rest frame. ALL is contracted relative to the muon's own meter. The muon's frame of reference does not consist of a tiny 'box' that he is in, it includes the whole universe. This 'whole universe' rest frame of reference is relative to the approaching 'whole universe' frame of the Earth. Or do you believe the muon's rest frame only encompasses the tiny 'point' of its existance? This misconception arises because when beginning in the Earth frame, THE MUON ONLY is the one moving relative to the Earth and the rest of the universe.

Do you remember my post where I explained this?