OK Relavists

Mac, the length of the rim of the merry-go-round does not change.Again, know one said that this is the case. The diameter changes.
Look again at the link provided by Janus58. It has to do with the observer on the merry-go-round being in an accelerated frame of reference.
 
And stop using the ruler as your measurement instrument. Use a laser or something. You are being severly limited to properly visualise the problem because you are doing it wrong.
 
Wrong Answer

ryan,

Look again.

If the observer applies his standard measuring-rod (a rod which is short as compared with the radius of the disc) tangentially to the edge of the disc, then, as judged from the Galileian system, the length of this rod will be less than 1, since, according to Section XII, moving bodies suffer a shortening in the direction of the motion. On the other hand, the measuring-rod will not experience a shortening in length, as judged from K, if it is applied to the disc in the direction of the radius. If, then, the observer first measures the circumference of the disc with his measuring-rod and then the diameter of the disc, on dividing the one by the other, he will not obtain as quotient the familiar number = 3.14 …, but a larger number, 2 whereas of course, for a disc which is at rest with respect to K, this operation would yield exactly.
Click to expand...


From the link you just mis-quoted. The radius (hence diameter) does NOT change. The circumference (in the vector of motion changes).

And why not use rulers and/or rods as measuring devices. Old Al did, Brian Greene did and this link does and you have agreed they all know what they are talking about. So why now are you reversing Relativity in opposition of what your mentors have said and want to introduce lasers, etc which didn't even exist when old Al made this mistake?

PS: Going into these discussion, I hope you are prepared and capable of producing bonfide responses (which you haven't to date) because appeal to authority isn't going to cut it.

Now tell me again why the merry-go-round, made of measuring rods, ends up with a different number of rods in its circumference while rotating (you see it is saying the same thing, since the circumference is made of these rods, the number of rods must change to yield a different measurement).

Relativity affects are in the vector of motion, not along the radius. You have it backwards.

You do of course understand when they say dividing 3.14 (Pi) by the radius (2 or diameter) becoming larger infers that the circumference either became larger or as they state the measuring rod became smaller don't you. I mean this isn't even algebra yet and we are having trouble?
 
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Mac

you are ****

READ THE **** ARTICLE.

The answer to this question cannot be achieved within the framework of SR. The observer on the **** merry-go-round experiences an acceleration away from the centre of the merry-go round which proportional to the distance from the centre of the merry-go-round.

The only way to answer this question correctly is via the use of non-euclidian spaces and differential geometry.

Anyone who has a faint idea of these topics Knows that in general the ratio of the circumference of a circle to its radius is only Pi in flat regions of space. You can even visulise this as I said before by drawing larger circles on a sphere and measuring its circumference and its radius, where the radial vector is constrained to the surface of the sphere.

I have said this all before and you still refuse to listen. The question posed in B.Greenes book is simply a qualitative example to show a point. Like a **** like you is ever going to understand differential geometry and non-euclidian geometry, so shut **** up because you do not have a **** clue as to what is going on.

Moderator edit: Personals insults and profanity removed.
 
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Again

ryans,


Yes, but I interpret in a totally different way that you do.
Do the calculation, it will take about 2 minutes.

From this measurement of Pi, the observer will be able to determine that he is an accelerated frame of reference. When the merry-go-round is travelling clockwise, Pi would be smaller than if the merry go round was at rest. Likewise for the anti-clockwise case.

From this he plots a graph of Pi versus angular velocity and finds the value of Pi for a flat region of space-time to be the maximum of this distribution.
Click to expand...


Reference to your above post. Your attitude still fails to answer my question. You can call it ryans geometry. I know of these different geometries but believe me that is not the answer.

Remember my merry-go-round is made of these measuring rods. If the measuring rod changes length so does the merry go round. It will still measure the same. Pi does not change. No geometry you use seperates the construction of the merry-go-round from the rods measuring it. If one changes they both change and the measurement stays the same.

You have more than once stated that it is the diameter that changes not the circumference. That contridicts every article ever written, even those posted here.

Try again.
 
Are you telling me Mac that there is no length contraction due to accelerated frames of reference? there is no accelaration tangential to the circumference of the merry-go-round.

Have you tryed the Pi thing for circles on spheres yet buddy.
You will never understand this.
 
Shorter

ryans,

Let me shorten this paragraph even more to focus in on what has been said.




If the observer applies his standard measuring-rod (a rod which is short as compared with the radius of the disc) tangentially to the edge of the disc, then, as judged from the Galileian system, the length of this rod will be less than 1, since, according to Section XII, moving bodies suffer a shortening in the direction of the motion. On the other hand, the measuring-rod will not experience a shortening in length, as judged from K, if it is applied to the disc in the direction of the radius.
Click to expand...



The only acceleration here is the centrifugal force produced by the rotation. That is not motion, the motion is the rotation.


I don't want to beat a dead horse here because it really doesn't matter since I have already said I accept Lorentz Contraction.

What I differ with is these examples. Pi simply cannot change because whatever shrinks so does the ruler and the measurement stays the same. Hence so does Pi. That is my point.

I do think the issue needs to be re-visited, perhaps amended.

After all gravity is supposed to effect time dilation, suppose that the centrifugal force (gravity equivelent per Einstien) shrinks the radius and length contracts the circumference due to motion.

Now Pi not only stays the same but you don't get the struts sticking out of the rim.

[Even shorter should there be an attention span problem here]]:


*************************************************
On the other hand, the measuring-rod will not experience a shortening in length, as judged from K, if it is applied to the disc in the direction of the radius.
*************************************************


PS: In most books I have read this illustration is accompanied by drawings and they show the ruler and man being smaller measuring the perimeter and standard length measuring the radius. So this has nothing to do with reading and interpreting words.
 
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Read the **** article again.

The author says that this calculation is not valid, acceleration needs to be taken to account.

What's your problem with acceleration.

Moderator edit: Personal insults removed.
 
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Still

ryans,

This is unbelievable and you expect to discuss something more complicated?

Number one you mis-state the issue. Number two you ignore the fact that it doesn't matter about motion or acceleration.

Whatever affects the ruler affects the the item being measured and there is no way anything even ryans geometry seperates the two from any purported affects. The measurement will be the same and Pi doesn't change.

Now if you still disagree on this let me propose that you answer the question and stop referring to "What the man said (or you say he said)".

I told you appealing to authority is not the answer and you can't prove Relativity using Relativity. That should be evident even to a kindergartner. Lets try to rise up out of your hole you've had your head stuck in and give a suitable answer.

How do you propose to seperate the ruler from the object in rotation you are measuring such that you get a change in Pi?

I'll repeat since it is obvious that you are not paying attention.

If Relativity alters the ruler, it also alters the rotating object hence no measurable change. That is my position. Show me wrong. That doesn't mean quote somebody because anybody that says anything other than what I have just said is simply wrong, just as you are wrong. Period.

Give me a suitable explanation of how you seperate the affects of Relativity from the rotating objects such that it only changes the ruler or conceed failure. Thank you.
 
Just the way you gave me an answer on how the quantum teleporter works. You were to stupid to answer that, but you still appeal to its existence by misquoting professional researchers.

I have given you the answer to the problem, if you don't like, that's not my problem.

Again Mac, have you drawn the circles on the sphere. If not, fuck off, if so then you would have seen the answer.
 
Any Supporters

To Other Members:

Does ryan have any supporters out there that thinks drawing circles on spheres answers the question?


THE QUESTION: "How do you propose to seperate the affect of Relativity such that it alters the ruler but not the merry-go-round, in a manner that causes the Pi ratio calculation to change."

I'm serious. If you have that answer I would love to see it.
 
Yes, Mac -- ryans is correct -- but not just due to drawing circles on spheres. ryans is really just trying to show you something you don't yet know -- but, as usual, you already think you know everything. You freely admit that you have never done math beyond algebra, and this severely limits your ability to comprehend the majority of the physics you want to understand. It does not matter how logical or how philosophically skilled you think you are -- you are suffering greatly from a lack of mathematical ability.

Here's why pi is different on a merry-go-round (nb for others reading this, I am purposefully being sloppy with terminology to help MacM come to grips with this problem):

Step 1) The acceleration 'produced' by the rotation is indistinguishable from (and this equivalent to) a gravitational field pushing things outward toward the edge of the merry-go-round. This is the Einstein equivalence principle: accelerations and gravitational fields are the same thing.

Step 2) Most systems involving accelerations cannot be expressed in special relativity theory alone. Special relativity deals with Minkowski (flat) spacetime, and only deals with accelerations indirectly as integrals. The more complete formalism is called general relativity. General relativity describes space as an arbitrary manifold of Lorentz signature, and postulates that gravitation is merely a result of bodies following the straightest possible lines in such curved spacetime.

Step 3) The rotating merry-go-round, with its radial acceleration, can be thought of as a radially symmetric gravitational field. It does not matter that the merry-go-round actually exists under your feet -- just imagine a system in which everything gravitates toward the edge. The math is, of course, the same. In a system where everything gravitates radially toward the edge, a metric (think of it as a general-purpose ruler set) is found that represents the space as being curved. Remember, any system with acceleration (gravitation) is expressed as curved spacetime in general relativity theory.

Step 4) As ryans was trying to indicate, the value of pi depends upon the curvature of space. The spacetime in the relativistic merry-go-round is curved. Pi is not 3.1415926535... anymore.

You've also said about a hundred times that "since the ruler and the merry-go-round experience the same length contraction, pi must be always the same." You are making the incorrect assumption that spacetime is at all times flat in this statement. I'm not just trying to bullshit you into shutting up, Mac. You really are making this assumption. Think about it for a while.

The bottom line, Mac, is that you really do not know very much math. All of the steps 1-4 that I outlined above are readily and rigorously described by differential geometry. Diff geo is, of course, a rather advanced topic. You seek to understand things on a qualitative level, in words alone, and this will often get you into serious problems. Words are not rigorous enough, which is why physicists generally try to avoid people like you.

You are unable, due to lack of education (and presumably willpower) to understand the mathematics that describe the rotating merry-go-round. Therefore, you will literally just have to take the word of those who do understand the math fully, until you are capable of understanding on your own.

The things you say, you realize, are like a child having just learned addition trying to tell someone how to do college algebra. You would do well to either begin educating yourself properly, or begin listening to those who have already made the commitment.

- Warren
 
MacM,

I agree that Pi can change if spacetime curves, but I don't think that acceleration can curve spacetime.

Here is where chroot makes the mistake:

Step 1) The acceleration 'produced' by the rotation is indistinguishable from (and this equivalent to) a gravitational field pushing things outward toward the edge of the merry-go-round. This is the Einstein equivalence principle: accelerations and gravitational fields are the same thing.
Click to expand...

It is true that accelaration can be indistinguishable from a gravitational field, but that doesn't mean that they are both the result of the curvature of spacetime. The force in a gravitational field is the result of curved spacetime, or the exchange of particles, while the force of acceleration is the result of inertia.

Note: By the way, let me remind everyone that it has never been proven that spacetime actually curves or bends.

Tom
 
Note: By the way, let me remind everyone that it has never been proven that spacetime actually curves or bends.
Click to expand...

Light is bent.

but that doesn't mean that they are both the result of the curvature of spacetime
Click to expand...

Relative. To an observer not undergoing rotation, viewing the spinning object, space time will not appear to be bent.
 
ryans,

Light is bent.
Click to expand...

The path of an electron bends when it goes through a magnetic field, or a non-uniform electric field. Does that mean that magnetic and electric fields bend space as well??

Relative. To an observer not undergoing rotation, viewing the spinning object, space time will not appear to be bent.
Click to expand...

OK. I see your point. Light will appear to act to an accelerating observer the same way it would if the observer was stationairy in a gravitational field. So even though spacetime isn't really curved for the accelerating observer, the observer sees it as being curved.

Tom
 
Light will appear to act to an accelerating observer the same way it would if the observer was stationairy in a gravitational field. So even though spacetime isn't really curved for the accelerating observer, the observer sees it as being curved.
Click to expand...

What is really? Its all relative.

The path of an electron bends when it goes through a magnetic field, or a non-uniform electric field. Does that mean that magnetic and electric fields bend space as well??
Click to expand...

The electric field can be uniform. Yes they bend space but in a different way. It is through the quantity called action. Action is defined as

S=T-V

where T is kinetic energy and V is potential energy. The path taken by a particle in a potential field is one which minimises the quantity action. Thus it can be seen that the path of an electron is a geodesic which minimises S. Lagrangian formulation of mechanics.
 
I Agree Inpart

chroot,

Good writting as usual but there are several things wrong with your post.

1 - Although you (and others here) like to make assumptions which are baseless; such as Old Mac doesn't know anything. The fact is I know Einstiens equivelence principle and I had already considered the affect of its "gravity" on bending time-space.

2 - You still have to subject the ruler to the same time-space however, bent, if bent and therefore the measurement remains the same. Hence the same Pi.

3 - I have said it several times here. My arguement is not with Lorentz Contraction perse but the examples being given and claims being made.

4 - Further the affect of velocity at the rim (in the direction of motion) is being ignored here. The rim should contract due to velocity and the strut could contract by "gravity" (acceleration) and again there is still no "Meaaureable" change or change therefore in Pi.

5 - Common guys. Read the posts. I said in the beginning I accept Lorentz Contraction. Stop saying I don't know or don't accept what I have already told you I accept and therefore must know.

Look at what you are saying and why it is simply wrong. We are not in disagreement as to what theory says about contraction we are in disagreement about what relavist say in support of that concept.

PS: Nice to have a civil conversation with you ch. Please keep it that way since James r., has already given you notice about coming back here and doing otherwise. I would like to see you remain here. Your knowledge is is respected. But unfortunately you have let your assumptions about Old Mac cause you to miss the point once again.

No change in Pi fellows.

"Show me how the invocation of any geometry ryans or ch's can change Pi. That is affect only the ruler and not the rotating object being measured" You haven't done that yet.
 
To Any Intelligence Out There

I just hope you can see that this selfappointed genius has failed to answer the question and that all the insults leveled at me are intended to infer that it is I that am wrong.

It is a very weak tactic by a very weak mind. His cutting and running under this cover should give you my answer.

"There is no change in Pi."

James R.,

Are you out there? What is your take on this issue. I respect your level head and knowledge. The arguement is not if Lorentz Contraction exists but if relavist are using a false example to try and convey the concept.

How does one, using whatever geometry, have Relativity to affect only the measuring rod and not the rotating object such that Pi becomes affected?
 
Originally posted by Prosoothus
It is true that accelaration can be indistinguishable from a gravitational field, but that doesn't mean that they are both the result of the curvature of spacetime.
Click to expand...
If acceleration and gravity are indistinguishable, then their source is also indistinguishable. It is nonsense to say that one is created by one mechanism and the other created by a different mechanism.

- Warren
 
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