I've been at sciforums for over a year debating relativity, and yet it still confuses me. I have conjured up a thought experiment, and would appreciate any feedback regarding the results of the experiment according to SR. Maybe this will help me understand relativity a little better.

Here's the thought experiment (see attachment):

There are two atomic clocks: clock A and B. Both of them are off, and their times are set to 0, at location s0. The two clocks begin accelerating. They both reach a speed of .90c before they reach point s1. As a result, when they reach s1 they are both travelling at a speed of .90c and ARE NOT accelerating. When they reach point s1, the beam of laser 1 hits both of the clocks. The clocks detect the signal using photocells. As soon as the two clocks detect the laser beam, they switch on and start to tick. They both continue to travel at .90c, and continue to tick, until they reach lasers 2a and 2b. When the clocks detect the laser beams from lasers 2a and 2b, they switch off (they record the time and stop ticking). Finally, the clocks are brought together so that an observer can see their recorded time. What time will the observer see on each clock?

Note:The clocks are only ticking when they are travelling at a constant velocity (not accelerating). Also, there are no gravitational fields in the area that can influence the clocks.

Edit: I changed the distance measurements in the illustration to make calculations easier. Attachment is in next post.

Tom

Attachment:

Answer: The time will be 1 second on both clocks.

James,

No time dilation???

From the view point of each observer moving with their respective clock (at 0.9c), they do not travel 270 000 km but
117 690 km. Thus the time elapsed of each clock is 0.436 sec.

ryans,

From the view point of each observer moving with their respective clock (at 0.9c), they do not travel 270 000 km but 117 690 km. Thus the time elapsed of each clock is 0.436 sec.

Relative to what? Clock A, clock B, or the lasers?

Relativity dictates that relative to the lasers, both clocks are moving at .90c. However, relative to clock A, clock B is moving at .99448c and relative to clock B, clock A is moving at .99448c. As a result, relative to clock A, clock B should be ticking slower, but relative to clock B, clock A should be ticking slower.

You have three frames of reference (lasers, clock A, and clock B) which relativity claims are all valid. Unfortunately, each clock can only show one time at any given moment. What is the preferred frame of reference for the clocks, and why?

Tom

The distance specified as 270 000 km was assumed to be that measured by an observer at rest with the lasers i.e. the starting and the finishing point.

You have three frames of reference (lasers, clock A, and clock B) which relativity claims are all valid. Unfortunately, each clock can only show one time at any given moment. What is the preferred frame of reference for the clocks, and why?

There are no preferred frames of reference, it's all relative.:o

ryans,

There are no preferred frames of reference, it's all relative.

If each clock can only show one time, then the combination of the two times on the two clocks can only show one of the three frames of reference (or perhaps a fourth). Which frame of reference will the clocks show?

Tom

Have you remebered that from clock A's perspective, the length travelled by clock B is no longer 270 000 km.

ryans,

Have you remebered that from clock A's perspective, the length travelled by clock B is no longer 270 000 km.

You're right, but why does that matter? From clock A's frame of reference, clock B was travelling at .9944c. Because of this, clock B should be ticking slower relative to clock A. Regardless of the distance that clock B is travelling relative to clock A, whether it's 1 km or 10 million kilometers, the time on clock B can't catch up to the time on clock A relative to clock A.

Tom

Yes, but you are assuming from clock A's frame of reference, clock B will pass its laser at the same time Clock A passes its own. This will only be true for the observer at the starting point, who sees both clocks travelling at 0.9c. From clock A's perspective, clock A and clock B do not pass their repective markers at the same instant. Simultanity just flew out the door my friend.:)

ryans,

From clock A's perspective, clock A and clock B do not pass their repective markers at the same instant. Simultanity just flew out the door my friend.

Clocks A and B do pass their markers at the same instant, even though clock A sees clock B pass its marker after clock A passes its marker. This delay is not caused by time dilation or length contraction, but is due to the finite speed of light.

I hope that your not claiming that delays caused by the finite speed of light are needed in order to preserve relativity.

Tom

in a motionless universe then each clock would have the same time measurement.
but in ours they will have diffrent times due to the motion of the universe it's self as
if speed relitive with rotation or expansion then Sa= r or e + or - and Sb= r or e + or -
example plane leaves ny at 600mph heading west plane a
leaves ny at 600mph heading east plane b which returns first assumeing no stops.

with respect :cool:

Originally posted by Prosoothus
ryans,



Clocks A and B do pass their markers at the same instant, even though clock A sees clock B pass its marker after clock A passes its marker. This delay is not caused by time dilation or length contraction, but is due to the finite speed of light.

I hope that your not claiming that delays caused by the finite speed of light are needed in order to preserve relativity.

Tom

No, this does not occur due to light signal delay, but due to time and length dilation.

Consider the situation from A's perspective. Due to length contraction, A will measure the distance between laser 1 and laser 2A as 117690 km. he will measure the same distance between lasers 1 and 2B.

At .9c, A will measure the time it takes to cross this distance as .4359 sec. Now he will also measure laser 2B's velocity as .9c relative to himself. He will also measure clock B's velocity as .994475c relative to himself, for a relative velocity difference of 0.094475c. (He will see clock B move at .094475c relative to lasers 1 and 2B. ) By A's clock this means that it will take 4.15241 sec for clock B to intersect laser 2B, after factoring out the light siganl delay. (light signal delay would add another 235380/300000 = .7846 sec)

Thus clock A will stop before clock B from A's perspective(A stopped after .4359 sec, and B after 4.15241 sec, as measured by A) . If we now apply time dilation to clock B as seen by A, we see that clock B will appear to run slow by a factor of sqrt(1-.994475c²/c²) = .10497

4.15241 * .10497 = .4358 sec. Thus clock A will see clock B read .4358 sec as it crosses laser 2b, which is the same time that clock A reads when it crossed laser 2A. What Clock B sees with respect to clock A will be exactly the same.

So when you bring both clocks together, They both agree that they should read the same time.

Oops. Sorry. ryans and Janus58 are right.

The time displayed by the clocks is 0.436 seconds. The time displayed by a clock at rest with respect to laser 1 is 1 second.

There is no signalling involved here, Tom, so there are no effects from light delays. All effects are due to time dilation and length contraction. All observers agree on the final times displayed.

Don't tell me that after so long on this board, you're still making the same elementary mistakes as MacM. (?)

James R.,

Don't tell me that after so long on this board, you're still making the same elementary mistakes as MacM. (?)



REPLY: Actually I prefer that you call them "Choices" than "Mistakes" but I'm sure you won't agree.:D

MacM:

We are discussing relativity here. If you ignore the relativity of simultaneity in a relativity problem which requires it then you are making a mistake.

James R.,


I guessed you might say that.:D

Janus58,

Thanks for the explanation. I see now that SR is a mathematically consistent theory. Your explanation can be used to explain MacM's three clock problem as well.

However, I still don't believe that the math of relativity matches physical reality. I still have a problem with understanding how space can be contracted and not contracted at the same time. For example you state:

A will measure the distance between laser 1 and laser 2A as 117690 km. he will measure the same distance between lasers 1 and 2B.

At .9c, A will measure the time it takes to cross this distance as .4359 sec.

I may agree that clock A measures a contracted distance. However, I don't agree with your assumption that the measured distance is the actual distance that clock A must traverse. I do, however, understand that you must accept that the measured distance is the actual distance if you accept the principle of invariance of light. I still don't believe that the principle of invariance of light is valid. I think more experiments need to be done in order for me to be convinced of its validity (experiments done far away from the strong gravitational field of the Earth).

Tom

Originally posted by Prosoothus
Janus58,



However, I still don't believe that the math of relativity matches physical reality. I still have a problem with understanding how space can be contracted and not contracted at the same time.

Tom

That's because you keep wanting to treat space as an absolute.

An analogy:

A long time ago, the prevailing model of the Earth was Flat. As such "down" was considered an absolute and universal direction.

When the spherical Earth Model was introduced, one of the objections raised to it was that the people on the "underside" would fall off. The answer to this, of course, is that Down is always towards the center of the Earth. Some still had trouble with this, because how could "down" be many different directions at once? They couldn't shake the idea of "down" being absolute.

This is a similar problem to the one you are having. You are having trouble making the paradigm shift needed to let go of the outmoded concept of absolute space.

Janus58,

Let's, for the sake of argument, assume that a certain distance is different for different observers moving at different speeds. Here are the only two possible explanations:

1) A certain distance can have multiple values for multiple observers at the same time.

In this case, distance must be aware of the observer and his/her speed. Otherwise, the distance won't be able to show the observer the correct value. Also, the distance must somehow show two different observers, travelling at different speeds, two different values even if one observer is directly behind the other.

2) A certain distance has only one absolute value, but different observers will see it as having different values if they are travelling at different speeds.

In this case, the observer is seeing a kind of shadow, or an illusion, as a result of his/her motion. Because of this, the observer can't use the values that he/she sees in scientific formulas because they may not be the absolute value of the distance being observed.


Do you see the dilemma that I am facing? Relativity claims that the values that the observer measures are the actual values. This means that example 2 can't be correct since relativity allows the observed values to be used in scientific formulas.

But how can example 1 be correct? How does a certain piece of space know how fast an observer is travelling. Example 1 implies that there must be a form of communication between space and an observer. How does the space know what value to show to an observer if the space has no way of knowing how fast the observer is moving? If that's not bad enough, how can the same space show two different values to two observers travelling at different speeds, if one observer is directly behind another? This seems impossible.

To summarize, if an observing is measuring a certain value, and the value is changing as the observer's speed is changing, then there can only be two explanations: Either the value is changing at the source, or it is constant at the source and only appears to change for the observer. If the value changes at the source, then the source must be aware of the speed of the observer. If the value at the source remains constant, then the values that the observer measures SHOULD NOT be used in scientific formulas because they are not absolute values.

The two examples I've listed above seem to be the only two explanations that are possible (of course it can also be a combination of the two). If there is a third explanation that I'm not aware of, that explains length contraction and time dilation in relativity, then please share.

Which explanation do you believe is correct, or do you have a third explanation? I would also like to hear from James, Crisp, chroot, and anyone else who may have alternate explanations.

Tom

Tom:

<i>To summarize, if an observing is measuring a certain value, and the value is changing as the observer's speed is changing, then there can only be two explanations: Either the value is changing at the source, or it is constant at the source and only appears to change for the observer.</i>

A distance has no "source". Neither does a time interval. These are things which are measured by observers. They have no physical meaning apart from an observer.

Originally posted by Prosoothus
Janus58,

Let's, for the sake of argument, assume that a certain distance is different for different observers moving at different speeds. Here are the only two possible explanations:

1) A certain distance can have multiple values for multiple observers at the same time.

In this case, distance must be aware of the observer and his/her speed. Otherwise, the distance won't be able to show the observer the correct value. Also, the distance must somehow show two different observers, travelling at different speeds, two different values even if one observer is directly behind the other.
Tom

No, No more than an object has to be "aware" of how far it is away from an Observer in order to "show" the Observer the correct angular size. (Note that I said "angular size" which is a direct measurement, and not apparent linear size, which is just subjective.)

Let's not get caught up now in a discusion about illusions and "real" distances and size. Tom, it is called relativity for a reason, everything is relative. An oserver who measures the size of an object, gets a figure which is no less or no more real than another observer moving relative to him. There is no such thing as absolute in relativity, it's all relative.

James,

A distance has no "source". Neither does a time interval. These are things which are measured by observers. They have no physical meaning apart from an observer.

Even if distance and time have no physical meaning, they must be linked to something physical in order to have values. And perhaps the physical properties that distance and time are linked to should be used instead of time and distance in scientific formulas.

Tom

Janus58,

No, No more than an object has to be "aware" of how far it is away from an Observer in order to "show" the Observer the correct angular size. (Note that I said "angular size" which is a direct measurement, and not apparent linear size, which is just subjective.)

Angular size, as James stated for distance and time, has no physical meaning. However, angular size is linked to something physical (distance). If you know the size of an object (distance from the bottom to the top of the object), and you know how far the object is away (the distance between the object and the observer), you can calculate the angular size.

It seems that all relative measurements must be linked to one or more physical properties of an object either directly or indirectly. If distance and time are relative measurements, as James stated, then they must also be linked to something physical and more fundamental. If this is the case, then I don't think that distance or time should be used in scientific formulas simply because they are relative properties. Using distance and time in scientific formulas would be as wrong as using an object's shadow to calculate its gravitational field.

Tom

THERE IS NOTHING WRONG WITH THE EQUATIONS.

The equations predict perfectly what will happen in the relativistic limit of particle velocity. You are again trying to impose absolutness into a theory which explicity forbids it. Distance and time DO have physical meaning. Go measure something. That's how long it is, this is how much it weighs, this is how long it exists for. Look in the world around you Tom, isn't it obvious that they have physical meaning. But James points out correctly that they are not necessarily the same for different observers. A good anology given by Einstein.

"When you are at dinner with a lovely girl, an hour seems like a minute. When you are standing on hot coals, a minute seems like an hour, That's relativity!"

ryans,

There is no such thing as absolute in relativity, it's all relative.

It's not that simple. Relativity appears to have introduced a new type of measurement to science (space and time). For example, relativity suggests, through time dilation and length contraction, that distance and time aren't physical properties, but then it uses them in scientific formulas as if they are physical properties.

If time and distance are relative properties, then they must be linked to more fundamental physical properties. And these physical properties should be used in scientific formulas, and not the relative properties of distance and time.

Honestly, I'm still not sure what type of properties distance and time are. I always believed that distance was a physical property of space. For example, GR says that gravity is the result of curved spacetime. Doesn't spacetime have to have the physical property of distance in order to curve? How do you curve something that doesn't have the physical property of distance?

I believe that relativity has been taken too far. I don't know if the overgeneralization was done by Einstein or the scientists after him. For example:

v=v2 - v1
v=(v2+a) - (v1+a)

Both equations above give the same result. Even though "a" in the second equation is superfluous, it doesn't mean that "a" doesn't exist. Relativity assumes that absolute motion doesn't exist just because it is currently superfluous. I don't agree with this type of logic.

I believe that in many cases, absolute distance and absolute time may be superfluous as well. However, that doesn't mean that absolute distance and abslolute time don't exist. I think it would be much easier to use absolute time and absolute distance in scientific formulas for the entire universe instead of constantly having to switch frame's of reference. If absolute distance and time do exist, then relativity would have to explain the mechanics of how they are transformed into relative space and time. I guess it would be easier for relativists to just assume that absolute time and distance don't exist.

Tom

For example, GR says that gravity is the result of curved spacetime.

NO NO NO

Gravity IS curved space time.

Originally posted by Prosoothus
Janus58,

I see now that SR is a mathematically consistent theory. Your explanation can be used to explain MacM's three clock problem as well.
[...]
Tom

holy crap, i don t believe my eyes.

Originally posted by lethe
holy crap, i don t believe my eyes.
Too early, lethe. i think Tom is now contemplating the idea to seperate physics&maths together from reality.:D

everneo,

i think Tom is now contemplating the idea to seperate physics&maths together from reality.

Too late, Einstein beat me to it. :D

Tom

Tom,
He is the one to show us the real reality thro' physics&maths..

everneo,

Einstein twisted reality so much, I don't know what's real and what isn't anymore. Hell, I'm still trying to figure out if length contraction and time dilation are real, or just a bunch of illusions a moving observer experiences (that is, if length contraction and time dilation occur at all).

Tom

Tom,
You are angry with Einstein for changing the absoluteness..!
Accumulating evidences show that Time-dilation/Length Contraction are not just illusion to a moving observer. they are his reality and effects of TD & LC remain even after he comes back to intial state. No illusion leaves lasting physical effects.

Don't see Einstein as a fraud any more. Just consider him as fellow crackpot (for your convenience) and go through, once again, SR & GR, sure you will realiase them to be ingenious.

ryans,

NO NO NO
Gravity IS curved space time.


ANS: I'll be taking exception to this in the very near future. Just giving you time to bone up for a challenge.:D

Hit me Mac. Anything I say you won't understand/accept anyway. I will debate with you, but you must accept my mathematical arguements

ryans,

Gravity IS curved space time.

I make a distinction between gravity and a gravitational field. Gravity is the force resulting from interaction inside a gravitational field.

So to clarify, a gravitational field would be curved spacetime in GR. While gravity would be the result of curved spacetime.

Tom

ryans,

Hit me Mac. Anything I say you won't understand/accept anyway. I will debate with you, but you must accept my mathematical arguements


ANS: Why would it be necessary that I (or anybodyelse) accept your mathematical arguements, unless they represent physical reality.

Mathematics can describe reality but mathematics cannot create reality. That is part of the problem today is that we operate from this system of algorithms with no physical underpinning and claim it is reality.

everneo,

Accumulating evidences show that Time-dilation/Length Contraction are not just illusion to a moving observer. they are his reality and effects of TD & LC remain even after he comes back to intial state. No illusion leaves lasting physical effects.

But how do you explain a property that is real for a moving observer but nonexistant for a stationairy observer? Do you call it semi-real, or quasi-real??

Isn't science based on the rule that something isn't real unless all observers can measure it? If there are two people in a room, and one of them sees a vase on the table, but the other doesn't see a vase on the table, is the vase "real", or is it only real for the observer that sees it?

Tom

Prosoothus,


I make a distinction between gravity and a gravitational field. Gravity is the force resulting from interaction inside a gravitational field.

So to clarify, a gravitational field would be curved spacetime in GR. While gravity would be the result of curved spacetime.

Tom



ANS: While straining to accept this view keep one thing in mind.

The view of the gravity well where a bowling ball stretches a rubber membrane and you roll a BB along the curved surface.

They say that curvture is gravity and that the BB will naturallly spiral down the well toward the bowling ball.

That theory supplies no impetus for the BB to do anything but follow a circular path or an elliptical path forever. Without some underlying force field there is nothing there to cause a spiraling in and down reaction.

For the BB to spiral in requires energy be absorbed., a static field doesn't get the job done.

In otherwords for the curved membrane to be gravity still requires gravity.

It isn't the answer I am afraid.

MacM,

That is part of the problem today is that we operate from this system of algorithms with no physical underpinning and claim it is reality.

Nicely said.

MacM,

That theory supplies no impetus for the BB to do anything but follow a circular path or an elliptical path forever. Without some underlying force field there is nothing there to cause a spiraling in and down reaction.

I agree. I brought the same question up a few months ago on sciforums. The curved spacetime model can explain why the path of an object curves when travelling through a gravitational field, but it doesn't explain why a stationairy object begins to move in a gravitational field.

Is the object attracted to curved spacetime? If so, doesn't this require another force besides simply the curvature of spacetime?

Tom

prosoothus,

Precisely. But remember the object need not be stationary. A BB rolling aound the curve will continue an elliptical path over and over, down into and back out of the well unless there is some change in its energy along the path.

Yeh, it some of this exotic stuff called matter.

Satelite falls from the sky, hits earth, is stopped.

I sorry, but how does your theory simplify things?

Gravity IS curved space time

So what are "gravitons" and why are we looking for them?

Dee Cee

Originally posted by Prosoothus

But how do you explain a property that is real for a moving observer but nonexistant for a stationairy observer? Do you call it semi-real, or quasi-real??

Are you asking why it is happening or how it is happening..? Science can answer the later.

Isn't science based on the rule that something isn't real unless all observers can measure it?


If all observers can measure a property that must be real. Even any one of them can measure it with proven scientific method then it is real. Their measurement must be same if all of them are doing the measurement under the 'same conditions'. Measurement is relative to under what condtions the observer measures.

If there are two people in a room, and one of them sees a vase on the table, but the other doesn't see a vase on the table, is the vase "real", or is it only real for the observer that sees it?

If one of them see a white ghost and another see a grey ghost then can a ghost be a 'real' entity..? Its not seeing but measuring, with a proven scientific method, decides whether its real or not. If both take photograph of the top of the table at the same time then we can say whether the vase is there or not.

Originally posted by DeeCee
So what are "gravitons" and why are we looking for them?

Dee Cee

gravitons are quanta of gravitational energy. we are not looking for them.

Tom:

Once again, your lackof knowledge is shining through. Why talk about things as if you understand them, when clearly you haven't even bothered to find out about them?

<i>The curved spacetime model can explain why the path of an object curves when travelling through a gravitational field, but it doesn't explain why a stationairy object begins to move in a gravitational field.</i>

Yes, it does.

An object in curved spacetime will naturally follow a geodesic in spacetime. That means that if it is released from rest, it will fall under gravity.

<i>Is the object attracted to curved spacetime?</i>

No, spacetime is the background against which the object moves. It naturally follows a geodesic, unless some force prevents it from doing that. (Remember: gravity is not a force).

It's late - very late, but I think what's being asked is this...

But using the BB on the rubber analogy, what makes the BB go down.

Gravity/space-time curving = the reason it dips, but what is making the BB go down.

If you curved a rubber sheet in space and put a bb on it, it wouldn't spiral or fall, it'd just stay there.

Therefore that analogy is flawed.

If I knew the answer, I'd give it - I'm not pretending to understand.

As far as I can see what people are saying is this...

"the curve in space-time can be called gravity, and it's gravity that's making the ball fall/spiral towards the centre"

I dont see that as true. You cant say "the shape is square because of it's squareness", not on these forums anyway ;)

Did I just discover some "5th dimensional force" that's pulling everything to what we see as "down" in 4d space...I doubt it. So what's going on?

EDIT: James R, you say "No, spacetime is the background against which the object moves. It naturally follows a geodesic, unless some force prevents it from doing that. (Remember: gravity is not a force)."
You obviously know a LOT more about this than me, but to me it sounds like "Things are naturally pulled downwards because of their orientation", which I think, people said a very long time ago...

shutupandshave:

In the general relativistic picture of gravity, the "natural" state of motion for objects is free-fall. Things fall unless a force stops them doing that. Gravity itself is not a force, but the curvature of spacetime which determines which way things fall.

lethe,

gravitons are quanta of gravitational energy. we are not looking for them.

ANS: Now that is news. However, I think you should amend your response. You and other relavists are not looking for them.

A whole lot of people are and BTW gravitons would fit nicely into my view of gravity as well.

Originally posted by MacM
lANS: Now that is news. However, I think you should amend your response. You and other relavists are not looking for them.
They aren't looking for them as defined by most scientists.

A whole lot of people are and BTW gravitons would fit nicely into my view of gravity as well.
You'd need anti-gravitons actually... which would be missed in the search for gravitons, regardless of how exactly you detect gravitons and not the gravity that they are claimed to cause (as claimed by most people looking for them)

James R,

In the general relativistic picture of gravity, the "natural" state of motion for objects is free-fall. Things fall unless a force stops them doing that. Gravity itself is not a force, but the curvature of spacetime which determines which way things fall.


ANS: So James what happened to F=ma. If an object accelerates in free fall, F=ma must be a component part.

Also let me correct a prior post. Amazed nobody caught it and pounced me.

A BB set in motion on a rubber membrane in space (absent the Force we call gravity) will not follow a continuous circle or elliptical path. It would change direction under a change of vector momentum due to the curvature of its motion being imposed by the membrane curvture but at the first opportunity would eject itself out of the well in a linear vector never to return.

Persol,

Now at least we are in partial agreement. You think gravity is curved space and I think it must be something like the graviton concept which involves energy transfer.

MacM:

<i>So James what happened to F=ma. If an object accelerates in free fall, F=ma must be a component part.</i>

F=ma still holds. An object in free fall doesn't accelerate in the GR picture.

<i>A BB set in motion on a rubber membrane in space (absent the Force we call gravity) will not follow a continuous circle or elliptical path. It would change direction under a change of vector momentum due to the curvature of its motion being imposed by the membrane curvture but at the first opportunity would eject itself out of the well in a linear vector never to return.</i>

Right.

This shows you why the rubber membrane analogy is not exactly right as a description of curved spacetime, doesn't it? For example, a bowling ball orbiting the Earth does follow a continuous circular path. It follows a geodesic in the local curved spacetime.

James R.,

MacM:

So James what happened to F=ma. If an object accelerates in free fall, F=ma must be a component part.

F=ma still holds. An object in free fall doesn't accelerate in the GR picture.

A BB set in motion on a rubber membrane in space (absent the Force we call gravity) will not follow a continuous circle or elliptical path. It would change direction under a change of vector momentum due to the curvature of its motion being imposed by the membrane curvture but at the first opportunity would eject itself out of the well in a linear vector never to return.

Right.

This shows you why the rubber membrane analogy is not exactly right as a description of curved spacetime, doesn't it? For example, a bowling ball orbiting the Earth does follow a continuous circular path. It follows a geodesic in the local curved spacetime.


ANS: They do seem to be proficient at giving incorrect analogies.

Or the bowling ball circles the earth in a pattern that balances its centrifugal force with the centriputal force of a dynamic energy based gravity.

Relativity is not so exclusive in this area.


Question: If an object in free fall gains kenetic energy and displays increasing velocity, how is it that the feature of "a" as in F=ma can be denied. ?




__________________

MacM:

<i>They do seem to be proficient at giving incorrect analogies.</i>

An analogy is only good as long as you know how far it can be applied. A full understanding includes knowing where an analogy is not exact, but analogous ... obviously.

<i>Or the bowling ball circles the earth in a pattern that balances its centrifugal force with the centriputal force of a dynamic energy based gravity.

Relativity is not so exclusive in this area.</i>

Pardon me, but I believe we were talking about relativity.

You're right, of course. For example, Newtonian mechanics describes orbits and free fall and so on pretty well (though not totally accurately, as it turns out).

<i>Question: If an object in free fall gains kenetic energy and displays increasing velocity, how is it that the feature of "a" as in F=ma can be denied. ?</i>

This is a very important question: <b>Who is viewing the velocity as increasing?</b>

If an observer is also free-falling, they see no increase in velocity and no acceleration. If they are not free falling, they are not in an inertial reference frame any more and they will see accelerations which they might incorrectly attribute to a "force" of gravity.

Kinetic energy is a relative quantity in relativity.

James R.,
This is a very important question: Who is viewing the velocity as increasing?

If an observer is also free-falling, they see no increase in velocity and no acceleration.


ANS: We agree fully.


If they are not free falling, they are not in an inertial reference frame any more and they will see accelerations which they might incorrectly attribute to a "force" of gravity.


ANS: We agree because of the word "might"



[/quote]Kinetic energy is a relative quantity in relativity.[/quote]


ANS: I would think the same applies in any case, not just relativity. No relativity velocity , no kenetic energy.


__________________

Or the bowling ball circles the earth in a pattern that balances its centrifugal force with the centriputal force of a dynamic energy based gravity

The centrifual force is not a pure force, it's a psuedo force felt do to an object being in circular motion. It is related to the inertia of a body.
The orbit will only be stable if the tangential velocity squared, times the mass of the body divided by the radius of its orbit, is equal to the radial force due to gravity acting in the direction of the centre of gravity of the body this object is rotating around, namely

mv^2/r=GMm/(r^2)

There is no such thing as a centrifugal force, as an observer outside this frame of reference will see that it is the rotating body inertia which alters its motion. Same goes for the coriolis force.

James,

No, spacetime is the background against which the object moves. It naturally follows a geodesic, unless some force prevents it from doing that. (Remember: gravity is not a force).

In the general relativistic picture of gravity, the "natural" state of motion for objects is free-fall. Things fall unless a force stops them doing that.

It seems that you switched gravity and inerta around. It used to be that an object would continue to move at a constant speed and in a straight line unless acted upon by an outside force. Now you claim that an object will only move at a constant speed and in a straight line if acted upon equal external forces, and that gravity is the result of unequal external forces.

Either, way you still have to have a force present. You just replaced the force of gravity on an object with the forces of empty space on that object. I don't see how this curved spacetime model is any simpler than the classical model for gravity.

Tom

A body in free fall does not "feel" gravity. Gravitational force is manifest through constaints. One particular constraint is that we are constrained to the surface of the earth. If we wish to be no longer constrained to the surface of the earth, we must supply a force greater than that of gravity, in the direction opposite to it.

ryans,




The centrifual force is not a pure force, it's a psuedo force felt do to an object being in circular motion. It is related to the inertia of a body.

The orbit will only be stable if the tangential velocity squared, times the mass of the body divided by the radius of its orbit, is equal to the radial force due to gravity acting in the direction of the centre of gravity of the body this object is rotating around, namely

mv^2/r=GMm/(r^2)

There is no such thing as a centrifugal force, as an observer outside this frame of reference will see that it is the rotating body inertia which alters its motion. Same goes for the coriolis force.



ANS: So you disagree with Einstein and his equivelence principle.

Seems like a lot of hollow talk and words about "Constraints" etc., and no useful physical underpinnings. What causes these constraints that moves bodies around without using force?

I am not disagreeing with einstein, but agreeing with him.
And Mac do you regard the centrifugal force as a pure force which is manifested through the exchange of what particles or fields. I'd like to see you to come up with field equations for the centrifugal force.

For those of you who are interested, the centrifugal force is simply due to an observer being in a frame of reference which has angular momentum. An observer who is not on this spinning object, will derive that the accelearation of the observer towards the periphery of the spinning object is simply due to the observer being in a non-inertial frame of reference. You can derive this force by considering newtons equations in spherical polar co-ordinates. Coincidently, this also leads to the 2 observers measuring different values of Pi, c.f. the relativistic
merry-go round.:D

Sorry to answer for you ryan... correct me if I'm wrong:).

Originally posted by MacM
ANS: So you disagree with Einstein and his equivelence principle.
No. Nothing ryan said contradicted the principle. There are various ways of looking at the system which will make the results easier to observe depending on what you are looking for. I'm not 100% sure on this, but the equation listed most likely is from a simplified force balance which treated gravity as a force (because the result is the same).

Persol,

His intentions are not my concern. My concern is he states CF is not gravity. Einstein says they are equivelent. CF is a function of F=ma or inertial mass subjected to a force of acceleration. A feature that GR you claim isn't involved.

[QUOTE]Originally posted by MacM
His intentions are not my concern. My concern is he states CF is not gravity.
Centrifual is not gravity. It is an imaginary force which acts away from the center of rotation, in the rotating frame. Imagine you are spinning a box around on a rope and someone is in the box. They feel a force against the back wall (centrifual force). When the guy in the box lets go of an object it flies out on a tangent... but to the guy in a box it looks like it is accelerating due to centrifual force. This does not happen if the 'string' is gravity. If a satalite releases a piece, and doesn't impart force too it, the piece does not fly away from earth at a tangent... but stays in orbit.

Einstein says they are equivelent.
If he said this then he would indeed be wrong in this case. However, I don't know if he did.

CF is a function of F=ma or inertial mass subjected to a force of acceleration.
Centrifual force is an imaginary force created in the rotational frame of reference.

Persol,

Centrifual force is an imaginary force created in the rotational frame of reference.


ANS:The above is your absolute statement. The folloiwng is from the link chroot provided for my to "learn" Relativity.


These forces are even today sometimes called fictitious forces, but according to Mach's principle they must be gravitational forces. In this sense the centrifugal force, for example, is not any more or less fictitious than gravity.

ANS: Perhaps I should provide you the link.

I am at lunch and don't have tiime to find a direct statement from Einstein but I will look later.

Originally posted by MacM
These forces are even today sometimes called fictitious forces, but according to Mach's principle they must be gravitational forces. In this sense the centrifugal force, for example, is not any more or less fictitious than gravity.
Exactly... gravity is technically not a force either... but is represented as one (or as acceleration) for simplicity.

I don't have the link, so I'm not sure how he goes about this... but if you keep reading it will probably go into Mach's fixed stars and that this is an 'appearance of force'. Or it may use the other teaching route and explain circular motion as having a dynamic frame of reference.

Persol,

I deliberately didn't post the entire paragraph since the issue was the "Equivelence Principle". They do state that both CF and Gravity are viewed as "Ficticous Forces".

I won't argue over that view but I don't agree with it. I know it is the GR view but seems to me pretty silly since weight is a force and it is the result of gravity.

The point here was that you want to break that equivelence principle and GR says you shouldn't. I find it also a bit confusing when people keep making different claims.

In "Relavistic Pi" ryans argued (rightfully) that the radius would undergo contraction due to GR gravity due to acceleration.

He was wrong about Pi changing since both the radius and ruler would undergo the same affects of GR.

But now James R., says there is no acceleration and you say that observers inside a rotating frame could tell they were rotating.

Let me suggest that is not correct. Obsevers inside a rotating frame would not know they were rotating (unless they had windows. They would feel gravity only. This assume they are in a small cabin and not free to roam and notice the "Zero" g's and reversal of forces toward the ceiling on the other side.

Originally posted by MacM
I deliberately didn't post the entire paragraph since the issue was the "Equivelence Principle". They do state that both CF and Gravity are viewed as "Ficticous Forces".
Then please don't take the sentence out of context.

The part with cf and gravity being the same is more complicated then I (and probably the book) make it out to be. It basically has to due with the cause of inertia possibly being gravity. This would then relate cf to gravity because gravity->inertia->cf. It is said that while Einstein knew of this theory, he did not use it.

I won't argue over that view but I don't agree with it. I know it is the GR view but seems to me pretty silly since weight is a force and it is the result of gravity.
You can consider gravity a force, but it is a different way of loking at the system which doesn't answer all the questions.

The point here was that you want to break that equivelence principle and GR says you shouldn't.
There are different ways of examining different situations (and the same situation). It doesn't really matter how, as long as it is consitent and gives the right answer.

Regardless, the equivalence principle does not have to do with centrifigual force as the equivalence principle does not take rotating reference frames into account.


He was wrong about Pi changing since both the radius and ruler would undergo the same affects of GR.
Sorry, I'm not aware of the exact circumstances here... but it is possible for pi to change.

But now James R., says there is no acceleration and you say that observers inside a rotating frame could tell they were rotating.
One is a rotating frame, one is not. They are different situations.

Let me suggest that is not correct. Obsevers inside a rotating frame would not know they were rotating unless they had windows. They would feel gravity only.
Observers in a rotating frame CAN know they are rotating. All they need to do is drop a ball. Now observers in constant and straight acceleration will not know. I explained this earlier using my guy in a box who releases an object. In gravity it continues to orbit. In forced rotation it flies off on a tangent.

Persol,

The following extract is from one of several link I have to Relativity.

*************** Extract **********************
Principle of Equivalence

Experiments performed in a uniformly accelerating reference frame with acceleration a are indistinguishable from the same experiments performed in a non-accelerating reference frame which is situated in a gravitational field where the acceleration of gravity = g = -a = intensity of gravity field.
**************************************************
**


? Do you interprete the term "uniformly" to exclude rotary motion? I don't. You can have uniform rotation producing uniform acceleration. If your position is the correct one then I think these sites should state "Linear uniform" acceleration.

The above does not differentiate between linear or angular acceleration.

Are you saying ryans was incorrect stating that GR causes contraction of the radius?

And finally my arguement was that neither SR at the rim nor GR along the radius could result in a change in Pi since a ruler in motion is subject to the same affects. If the radius contracts then so does the ruler. No measurement change hence no change in Pi.

******************** Extract *********************
One way of stating this fundamental principle of general relativity is to say that gravitational mass is identical to inertial mass. One of the implications of the principle of equivalence is that


since photons have momentum and therefore must be attributed an inertial mass, they must also have a gravitational mass.




Thus photons should be deflected by gravity. They should also be impeded in their escape from a gravity field, leading to the gravitational red shift and the concept of a black hole. It also leads to gravitational lens effects. Index
**************************************************
*


ANS: I brought this forward in that I noticed they claim an inertial mass and gravitational mass for a photon.

Is not "Inertial Mass" and "Rest Mass" one in the same? This runs counter to everything I have seen about photons.

Originally posted by MacM

Do you interprete the term "uniformly" to exclude rotaary motion? I don't. You can have uniform rotation producing uniform acceleration.
I disagree. Acceleration is a vector quantity. In 'rotary motion' the vector is changing direction.

Are you saying ryans was incorrect stating that GR causes contraction of the radius?
As I said... I don't know the conditions/

And finally my arguement was that neither SR at the rim nor GR along the radius could result in a change in Pi since a ruler in motion is subject to the same affects. If the radius contracts then so does the ruler. No measurement change hence no change in Pi.
pi is determined along 2 axis. If only 1 axis on 'contracts' then pi changes .

Is not "Inertial Mass" and "Rest Mass" on in the same?
'rest mass' is now just called 'mass'.
'inertial mass' refers to relatevistic mass

'mass' is a constant. 'relativistic mass' depends on velocity


This runs counter to everything I have seen about photons.
It shouldn't. I'm sure you knew that photons were influenced by gravity... as per lensing/blackholes.

Persol,

pi is determined along 2 axis. If only 1 axis on 'contracts' then pi changes .

Is not "Inertial Mass" and "Rest Mass" on in the same?

'rest mass' is now just called 'mass'.

'inertial mass' refers to relatevistic mass

'mass' is a constant. 'relativistic mass' depends on velocity

ANS: I had never seen this switch. It used to be that F = ma was a function of inertia or inertial mass.


This runs counter to everything I have seen about photons.

It shouldn't. I'm sure you knew that photons were influenced by gravity... as per lensing/blackholes.


ANS: Yes but the arguement was curved space and I've seen arguements agains photon mass being the cause of the bending.


Running counter was considering photon to have rest mass, which was the way I was readding this based on my antiquated terminology.

Tom:

<i>It seems that you switched gravity and inerta around. It used to be that an object would continue to move at a constant speed and in a straight line unless acted upon by an outside force.</i>

No, I haven't switched anything. Objects in curved spacetime move at constant speeds in a straight line (i.e. on geodesics) unless acted on by an outside force.

<i>Now you claim that an object will only move at a constant speed and in a straight line if acted upon equal external forces, and that gravity is the result of unequal external forces.</i>

Please don't put words in my mouth. If you don't understand something, ask.

<i>Either, way you still have to have a force present. You just replaced the force of gravity on an object with the forces of empty space on that object.</i>

Wrong.

<i>I don't see how this curved spacetime model is any simpler than the classical model for gravity.</i>

It's not. It's more complex by far. But it is also much more accurate.

MacM:

<i>They do state that both CF and Gravity are viewed as "Ficticous Forces".

I won't argue over that view but I don't agree with it. I know it is the GR view but seems to me pretty silly since weight is a force and it is the result of gravity.</i>

What you are saying here is: "It seems pretty silly to me because I believe in Newtonian physics." Well, experiments show us that Newton was wrong and Einstein is right, MacM. The universe doesn't care what you think is silly. GR is the way it works.

<i>In "Relavistic Pi" ryans argued (rightfully) that the radius would undergo contraction due to GR gravity due to acceleration.</i>

Yes, and he was right, although it could have explained it more clearly.

<i>Let me suggest that is not correct. Obsevers inside a rotating frame would not know they were rotating (unless they had windows.</i>

Have you ever been on a ferris wheel, or a merry-go-round, or in a car going around a corner. Can you tell you're rotating without looking out the window? I think you can.

<i>You can have uniform rotation producing uniform acceleration.</i>

No. The acceleration in a rotating frame is continuously changing direction. It is not uniform for that reason.

<i>If your position is the correct one then I think these sites should state "Linear uniform" acceleration.</i>

"Linear" is what is meant by "uniform" in this context.

<i>And finally my arguement was that neither SR at the rim nor GR along the radius could result in a change in Pi since a ruler in motion is subject to the same affects. If the radius contracts then so does the ruler. No measurement change hence no change in Pi.</i>

That's what you've been repeating over and over again, without bothering to look elsewhere for actual facts. You are wrong, MacM. I'm not going to argue with you on this point because frankly I don't think you'll understand the explanation. If ryans wants to take it further it is up to him.

No I do not wish to take it further. Go to the thread and have a look at my arguements, they still hold. If I take it further, I will be simply hitting the same brick wall known as Mac's "intuition"

James R.,

And finally my arguement was that neither SR at the rim nor GR along the radius could result in a change in Pi since a ruler in motion is subject to the same affects. If the radius contracts then so does the ruler. No measurement change hence no change in Pi.

That's what you've been repeating over and over again, without bothering to look elsewhere for actual facts. You are wrong, MacM. I'm not going to argue with you on this point because frankly I don't think you'll understand the explanation. If ryans wants to take it further it is up to him.

ANS: Neither do I care to pursue the issue further. I do however want to make one thing clear. You accuse me of blindly repeating the same thing over and over without any evidence.

The fact is that happens because replies here have repeated the same thing over and over without any evidence. That is to say when it is all said and done on this topic, none of you have addressed the root question.

There is and was no question regarding SR, GR, MR , RR or JR views.

The question was and is (left unanswered) to explain how any affect of any theory causing contraction at the rim, at the radius and/or both could cause a change in calculated Pi based on measurement by a ruler on the moving frame.

That is explain how such a process changes either the rotating platform and not the ruler or changes the ruler and not the platform, such that there becomes a measurement change.

If the affect changes the platform it must alsochange the ruler and there can be no measureable change, hence no change in Pi.

The response to this question is to state simply "You just don't understand" but appartently nobody here can answer this simple question over a simple process without appealing to authority and saying because Relativity says so. The fact is this issue goes beyond Relativity, that is has nothing to do with Relativity.

It has to due with if the m-g-r is constructed of rulers, it cannot be denied that both the m-g-r and the measuring ruler change equally and there can be no measureable change hence no change in Pi.

Don't bother responding if you don't have a direct answer.

Thank you.

That's right Mac, when the entire world is shouting "you just don't get it, do you? God, it's so simple!" it must be that the entire world is wrong.

- Warren

chroot,

Could be. You know the whole world was wrong about Newton once too.

Wow.


Talk about allusions of grandeur.

Isn't length contraction at relativistic speeds easily explained with SR? Its all in the frame of reference.

Blackholesun,

Talk about allusions of grandeur.

ANS: This has nothing to do with illusions of grandure. It has to do with something pretty simple of which members here have refused to acknowledge.


The mis-information and description of reality when it comes to Relativity.

I did not and do not in this question challenge Relativity. I challenged the statements of those making erroneous presentations.

It has to do with the fact that Sr affecting the rim and GR affecting the radius of a rotating frame, cannot result in a measurable change in Pi.

The reason has nothing to do with a flw in Relativity but a flaw in the presentation arguement.

READ THIS SLOWLY AND STOP AND THINK BEFORE ENGAGING MOUTH:

"Whatever theory by whatever affect alters the circumference and/or radius of a rotating frame must also by like amounts and like fashion alter any ruler used to make the measurement; hence no means to measure a change in Pi."

On radioactive Waves,


Isn't length contraction at relativistic speeds easily explained with SR? Its all in the frame of reference.


ANS: It has to do with something pretty simple of which members here have refused to acknowledge.

The mis-information and description of reality when it comes to Relativity.

I did not and do not in this question challenge Relativity. I challenged the statements of those making erroneous presentations.

It has to do with the fact that SR affecting the rim and GR affecting the radius of a rotating frame, cannot result in a measurable change in Pi.

The reason has nothing to do with a flaw in Relativity but a flaw in the presentation arguement.

************************************************** **
"Whatever theory by whatever affect alters the circumference and/or radius of a rotating frame must also by like amounts and like fashion alter any ruler used to make the measurement; hence no means to measure a change in Pi."
************************************************** **

This is the issue not Relativity perse.


Thanks.



report | quote | edit

Originally posted by MacM
READ THIS SLOWLY AND STOP AND THINK BEFORE ENGAGING MOUTH

I cannot believe you just told someone else to stop and think before engaging their mouth.... I'm rolling on the damn floor over here!

- Warren

chroot,

That is called turn about. That is what I'm being told when I screw up.

He was making irrelevant accusations about illusions of grandure.

Not much grandure in knowing if a rotating frame contracts and the ruler contracts by the same amount you will get the same measurement as though it were at rest.

Originally posted by MacM
Not much grandure in knowing if a rotating frame contracts and the ruler contracts by the same amount you will get the same measurement as though it were at rest. They probably went into this, but it depends on what you mean by 'ruler'.

Persol,

The arguement started because I made the statement I did not like Brian Greenes "the elegant universe" because he used incorrect analogies in presenting Relativity.

He didn't say ruler but used the term measuring rod. He gave the illustration of a man crawling along the circumference of a merry go round would get a different measurement of its circumference using the same rod than he did if it was not in motion.

He further went on to say that that caused the calulation of Pi to change hence evidence of different geometry.

After that there were several other examples of the same arguements being made in the presentation of Relativity.

ryan made one of them by introducing GR affecting the radius and changing the calculation of Pi.

My whole issue was really quite simple and had nothing to do with Relativity perse.

It was the fact that any affect by any theory on the circumference and/or radius would also affect the measuring rod in the same manner and magnitude, hence the measurement would not change and hence Pi would calculate the same.

MacM:

The circumference of a merry-go-round moves faster than the axis. Do you agree?

The circumference moves faster than a point half way between the circumference and the centre. Do you agree?

Then, according to relativity, rulers at the circumference shrink by a different factor than rulers at the centre or half way between the centre and the edge. Do you agree?

Do you therefore agree that, just maybe, rulers on the m-g-r might conceivably measure different distances depending on where they are?

James R.,

The circumference of a merry-go-round moves faster than the axis. Do you agree?


ANS: Yes.

The circumference moves faster than a point half way between the circumference and the centre. Do you agree?



ANS: Yes



Then, according to relativity, rulers at the circumference shrink by a different factor than rulers at the centre or half way between the centre and the edge. Do you agree?


ANS: Yes.


Do you therefore agree that, just maybe, rulers on the m-g-r might conceivably measure different distances depending on where they are?

ANS: Absolutely not. The difference in velocity as you transgress the radius, changes the amount of contraction of BOTH the m-g-r and the ruler by like amounts.


__________________

MacM:

If you're standing at the centre of the m-g-r watching rulers on the edge, even though those rulers are rotating with the m-g-r you still see them as contracted.

James R.,




MacM:

If you're standing at the centre of the m-g-r watching rulers on the edge, even though those rulers are rotating with the m-g-r you still see them as contracted.


ANS: That was not the presentation. The presentation and the objection is that they show men crawling along the circumference and crawling along the radius.

I am well aware if you look upon a moving object you can induce observational changes but even in this case the observation would include the contraction of the circumference by amounts equal to that of the ruler. Still no measureable change.

Infact imagine that instead of multiple rulers you attach a flexable tape measure to the circumference. Now tell me the tape measure contracts but the circumference doesn't.

You know watching you Mac, is like watching a really bad daytime soapie (drama). You can miss 2 months of the show, but still know exactly what's going on.

If you are so brilliant, what are you doing here. All the time you've spent here could have been used to further the mathematics for your theories. That is what makes you a super crackpot.

The m-g-r pi problem was solvec about a month ago by myself, move on, I and everybody else argueing with you is right.

ryans,

The m-g-r pi problem was solvec about a month ago by myself, move on, I and everybody else argueing with you is right.



ANS: That is pathetic. You now want to claim victory saying you gave a correct answer?

You said Pi changes. Explain how that is supposed to happen when any relavistic affect on the m-g-r must not only affect the ruler but the m-g-r as well.

Both change, no measurement change, hence no change in calculated Pi.

Trying to slip in the back door and say your answer was valid from months ago don't cut it.

Post an explanation of how your "Pi Changes" response occurs and I indeed will acknowledge you are correct. As of this date you have been incorrect and simply refuse to acknowledge it.

Originally posted by MacM
ANS: That is pathetic. You now want to claim victory saying you gave a correct answer?
MacM, I'd just like to point out that you have done the same thinbg numerous times. The only difference is that ryans has demonstrated the math/knowledge on the subject and proved it. You just claim victory without showing any proof... just things you consider inconsistant, which really aren't.

You said Pi changes. Explain how that is supposed to happen whn any relavistic affect on the m-g-r must not only affect the ruler but the m-g-r as well.
The ratio of circumference to diameter is not fixed, so pi changes.

In general relativity, space and spacetime are non-Euclidean geometries. The ratio of the circumference to diameter of a circle in non-Euclidean geometry can be more or less than pi.

Pi the constant does not change, but pi the measurement does. The definition of Pi the constant is based on Euclidean geometries, but the geometries in questoin may not be Euclidean.

Persol,

I am not even going to cut and paste your responses because quite simply they are double talk.

Different geometries certainly but neither you nor ryans have yet shown that a moving frame alters the m-g-r and not the ruler in the same fashion and by the same amount.

That being the case there is no measurement change, as claimed by the presenters of Relativity, which was my objection. Not that Relativity doesn't or may not infact alter dimension due to motion.

Once again by what justification do you or anybodyelse claim that motion of the m-g-r alters the circumference and/or radius by SR and GR respectively that would not have the same SR and GR affect also alter the ruler, whereby the measurement would always be the same, hence no change in Pi.?

You really don't seem to understand. It does not matter what geometry you invoke. There is no change in Pi because the affects of Relativity alter the m-g-r and ruler the same..

Geez.

Length only changes in the direction of motion (circumference). The radius does not change.

You measure radius, and the ruler is the same length (although different width) as a stationary ruler. When you measure circumference, the ruler is a different length. Then you calculate pi, and won't get 3.14~.

It's the same argument that is used for near c length contraction. The rotating frame is an accelerating frame.

Persol,


Geez.

Length only changes in the direction of motion (circumference). The radius does not change.

You measure radius, and the ruler is the same length (although different width) as a stationary ruler. When you measure circumference, the ruler is a different length. Then you calculate pi, and won't get 3.14~.

It's the same argument that is used for near c length contraction. The rotating frame is an accelerating frame.

ANS: Untrue. Ryans claims GR also affects the radius. I believe James R., agreed with that. So the radius doesn't stay the same.

No wonder you get a wrong Pi. You just turned the ruler sideways. You measure the circumference in the direction of motion, with the ruler turned lengthwise.

And still whatever Relavistics affects there are they affect the ruler and the m-g-r exactly the same. Try again.

O.K. Consensus reached. Pi changes. Good to see that you agree Mac. I knew I was right.:)

ryans,

O.K. Consensus reached. Pi changes. Good to see that you agree Mac. I knew I was right.



ANS: Nice try but I never argued against GR at the radius and you never explained how it and SR at the circumference affect the m-g-r without affecting the ruler.
;)

What?

You just said that you agreed with the variable Pi arguement. Read your post Old Man Mac.......

ryans,

If you are referring to this then you should read more slowly. As I hve said before I am not contesting contraction affect of SR (circumference) or GR (radius) and I agree SR and GR will not produce equal change hence a different geometry must be used BUT the arguement has been and still IS that any such affects affect the ruler in equal and like amounts and as presented cannot be measured and Pi calculates the same.


************************************************** *
ANS: Untrue. Ryans claims GR also affects the radius. I believe James R., agreed with that. So the radius doesn't stay the same.

No wonder you get a wrong Pi. You just turned the ruler sideways. You measure the circumference in the direction of motion, with the ruler turned lengthwise.

And still whatever Relavistics affects there are they affect the ruler and the m-g-r exactly the same. Try again.

Oh my god, it's sucking my will to live:

"Relavists"
"Relavistic"

etc.

- Warren

As I said Mac we agree. I think you're just to used to argueing with me. Yes the radius changes and so does Pi. Well done Mac:)

ryans,

As I said Mac we agree. I think you're just to used to argueing with me. Yes the radius changes and so does Pi. Well done Mac

ANS: As usual ryans you are half right. We have always agreed on the properties of contraction.

But as far as "The measurement of the dimensions, Pi does not change. Sorry, you lose. Guess you will never admit it but everybody that has followed this crap knows you just can't find the guts to say "Mac, you are right".

I may fight to the end but when I'm whipped, I step up to the jplate and admit it. When you going to find the guts to do that?