Discussion in 'Physics & Math' started by MacM, Oct 11, 2004.

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## Do You Agree or Disagree that the conclusions of James R's thread are misleading?

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1. ### MacMRegistered Senior Member

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Just a note to readers. I had posted this with three vote choices. Apparently James R has amended my vote choices to include "Don't Care".

Well, I don't really care either that he has done that but I find it in poor taste for him to have done so without comment or permission. That is not unlike altering the text of my post.

There is currently a thread running which advocates a principle of Relativity for which the stated conclusion is deliberately misleading.

The issue has to do with the very requirement by Relativity for something called reciprocity in cases of constant relative velocity and/or equal acceleration. That means whatever happens to A happens to B.

In the diagrams given in this case the calculations are made for only one direction of a light signal, frankly the light signal is a complexiety which isn't required to properly analyze the problem.

After running a one hour test under such conditions at 0.9c relative velocity the following would be true statements:

Clock A reads 3,600 seconds but percieves Clock B as reading 1,569.2 seconds.

Clock B reads 3,600 seconds but percieves Clock A as reading 1,569.2 seconds.

Upon returning the clocks to a common rest frame they will both still read 3,600 seconds. There is no physical shift of time. The illusion is created by only calculating one half of the actual physical situation.

This mathematical trick is obvious if you consider this simplified demonstration.

If I tell you that X = Y<sup>2</sup>. You can plot that relationship and you would see a curve. You would claim that this is reality and that X and Y have this non-linear relationship.

However, if you are forced to apply reciprocity you would have to also claim that Y = X<sup>2</sup>.

Substituting:

X = (X<sup>2</sup>)<sup>2</sup> = X<sup>4</sup>
Y = (Y<sup>2</sup>)<sup>2</sup> = Y<sup>4</sup>

X<sup>4</sup> = Y<sup>4</sup>

X = Y

So a non-linear illusion of reality is created by looking at only half of the problem. The reality is it is linear.

Likewise, in Relativity one sees T2 = T1(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>.5</sup> as a non linear time function between clocks and it is claimed that flying around the universe at relavistic speeds will cause you to return younger than your twin you left at home.

Well, the truth is that no such thing happens as a consequence of relative velocity. Upon return your watches (and ages) may only differ a small amount due to General Relativity involving acceleration and gravity but no changes due to long term relative velocity.

This is made abundantly clear if one actually computes the total reality of the situation and not calculate only one half of the reality.

Last edited: Oct 11, 2004

3. ### James RJust this guy, you know?Staff Member

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It does not advocate a principle of relativity. It derives results mathematically and rigorously from one particular set of principles of relativity.

It is left to the judgment of readers to agree or disagree with the validity of the principles. Readers are also urged to examine the arguments to satisfy themselves that the stated conclusions do actually follow from the given postulates.

I also note that real-world experiments which test the real world against the mathematically derived results provide strong evidence in favour of the truth of the postulates. Relativity appears to provide an excellent description of our physical world.

There is no such principle in relativity. This must be the "misleading science" MacM is trying to get at here.

Nothing in relativity says "what happens to A happens to B". In fact, in the example given in the linked thread, A emits light, while B does not. B receives a light signal; A does not.

Note that MacM has given no alternative way to "analyse the problem".

And indeed they do, if the equation is correct and X and Y are variables.

MacM here is introducing a concept of "reciprocity" which has nothing to do with relativity, making his analogy useless from this point onwards.

MacM here explicitly avoids the implication of his two equations, which show that X=Y=0 or X=Y=1 are the only solutions to his "reciprocal" pair.

That means that the relation X=Y^2 is not a relation between two variables, but between two constants.

The reality in MacM's example is that X and Y are not related, linearly or in any other way. They are simply constants, equal to either 1 or 0.

Actually, for constant v, the given function is linear, not non-linear as MacM says. The variables are T2 and T1.

The twin paradox is a well-known example which cannot be solved using the given equation alone, since the spaceship must accelerate to return home.

In fact, large differences due to GR effects are possible. These are countered by SR effects. In combination, they result in a net time difference.

MacM doesn't know how to calculate any part of the reality.

5. ### Paul TRegistered Senior Member

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This is not illusion but stupidity! Only in MacM's relativity such a stupid test exist. Based on standard SR, we don't compare clock reading on A and B (a non-symmetrical system) that way. MacM compared only time component but ignored length component. It is therefore invalid test as time and length (or space) are interconnected.

MacM is proposing a new kind of relativity called "MacM's relativity" so if it looks silly, that's the way it is.

A school kid would have done a better math than this. The two equations, X = Y<sup>2</sup> and Y = X<sup>2</sup>, if plotted on cartesian coordinate displaying two parabolas respectively open to the right and up. They have two intersection points, as shown by James R on the other thread. X and Y equal only on that two intersection points, nothing else. It tells nothing more than that except for someone suffering from serious delusion.

Sound poetic, but this is just bullshit as pointed above.

It is again very stupid. Relationship between T1 and T2 in that equation are linear since (1 - v<sup>2</sup>/c<sup>2</sup>) with v constant is yes...constant. So, it is T2=T1*C, linear relationship...only a fool would think it is non-linear!

Wild speculation. The above statement will fall apart when scrutinized closely.

Last edited: Oct 11, 2004

7. ### MacMRegistered Senior Member

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10,104
Yes please do and note that he calculates only one view not the recipocal view using the same relavistic principles. Relativity states wither A or B may be considered at rest.

If you calculate the view of the observer at rest he "Sees" differences in the time flow of the moving clock. But he doesn't then show you that under the same rules of Relativity you should also consider the other clock at rest and calculate his view as though he were at rest and the other clock is in motion.

That is what relativity is. A<---- 0.9c------>B is 0.9c to either clock regardless of which clock is placed into any particular frame, including a rest frame whire v = 0.

It is noted that indeed his mathematics are correct, howver, that he is still only showing the "Perception" based on one view not both which Relativity applies equally to both views.

Do it correctly and these functions vanish.

Indeed I concur that mathematically Relativity has value but that doesn't alter the conclusion that many of its attributes are "Perception" and not physical reality.

For example the H&K Atomic Clock test which claimed to have recorded time dilation between clocks being flown around the earth vs a clock stationary on earth, proving Relativity; actually violates Relativity.

The clock on earth did not produce the slowing reading expected (according to Relativity) by the pilot flying the plane. In terms of only relative velocity each clock should have shown an equal amount of slowing and there should be no measureable time differance.

You need to ask yourselves "How is it that we are claiming the observation by a remote party (indeed trillions of remote parties, at trillions of different relative velocities) can alter your watch?" It can't and it doesn't. Not even according to relativity. What Relativity actually shows is "Perception" of your watches "Reality".

Nice try James R. Now explain to us how you can say that relative velocity between two clocks is not the same velocity? If I say B is receeding from A with a relative velocity of 0.9c, are you going to tell us that B does not see A as receeding at 0.9c?

I think not. That SIR is reciprocity and while it is not made clear (and for good reason) it is a fact of Relativity and must be considered.

8. ### PeteIt's not rocket surgeryRegistered Senior Member

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Mac,
If you'd bother to read my posts, perhaps you'd understand by now.
Remember [post=692998]that post[/post] you've been avoiding like the plague?
Why not just admit you don't understand it, and leave it at that?

9. ### MacMRegistered Senior Member

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10,104
READERS: Please note the quality of this post. Does he actually adres the issue? No. Does he point out any actual flaw? No. Does he run off at the mouth and cast innuendo and slander? Yes. that should tell you a lot about his worth.

Check it out. He either doewn't understand relativity or wishes to seep this under the rug by simply denying that if A is moving 30 Mph relative to B that B must also be moving 30 Mph relative to A. That isn't my theory it is simple fact and is an integral part of Relativity.

Which is it you don't understand Relativity or you can't overcome this issue and choose to try and mask the jroblem with BS?

Someone is suffering from dillusion but it isn't I unfortunately for you. Your objection here is meaningless. Address the issue of inherent reciprocity in Relativity.

Yes I have pointed out your bullshit above.

Only a fool would not understand the relationship I am speaking of is the non-linear time dilation affect of the formula for various velocities. That damn sure is not linear. Twit.

We are waiting. The clock is ticking (pun intended).

10. ### PeteIt's not rocket surgeryRegistered Senior Member

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And yet there is no frame in which both A and B are moving at 30Mph...
If you choose a frame where one is moving at 30Mph, then the other is stationary (or moving at 60Mph).

If you want to put the same number into an equation for both A's velocity and B's velocity, then that number must be 15 (and must be negative for one of them)... but so what?

Mac's concept of "reciprocity" is ill-defined, and doesn't seem to give any useful results (unlike the well-defined concept of symmetry).

11. ### MacMRegistered Senior Member

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10,104
Well, I haven't avoided your ost. I did read it but I noted two things.

1 - First it is not the test I am refering to and gets aways fromthe issue at hand which is reciprocity of clocks in constant relative motion.

2 - that you are trying to confuse simultaneity with reciprocity. Please note that the time span onyour time sacles are equal.

3 - What is the respective time differential between these events for each observer?

Want to try again?

12. ### James RJust this guy, you know?Staff Member

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MacM:

Wrong. Both observers' views are taken into account in the given derivation, as they msut be to come to any conclusion. Relativity is, after all, all about the comparison of different reference frames.

Two points here.

1. The given example deals not with what observers "see", but with the actual times and locations of events in their reference frames.
2. The reference frames chosen correspond to the "rest" frames of the two cars. Both frames are considered. That is made quite explicit.

I doubt if MacM has actually read the post. Or understood it.

This is just unclear. Who knows what MacM's crazy notation means? A<---0.9c--->B? Huh?

Why?

Sure, the given postulates are for ONE theory only - special relativity. Change the postulates and you'd get different results.

I note here that in the past MacM has agreed that the postulates correctly reflect reality. Which means he must agree that SR is a correct description of relativity, provided he uses logic.

Let's wait for the inevitable flip-flop.

Wrong. It is made quite clear in the thread that both views must be considered to derive the relativistic result.

Empty waffle.

You already agreed I did it correctly. Quote: "Indeed his mathematics are correct."

Poor confused MacM.

This is MacM's imagination. He has no relevant data to support this.

Which is the whole point. What relativity says and what MacM says it says are completely different.

The results are the opposite, in the situation shown. To produce that result, refer to the thread in question. Simply swap the words "blue" and "green" everywhere, and replace v with -v everywhere. The rest of the argument is the same, and so is the mathematics.

Let's hope we don't hear anything more about the twin paradox or the H&K experiment, then.

13. ### MacMRegistered Senior Member

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How do you confuse yourself and come up with 60Mph? ONe stationary certainly infact please calculate time dilation for each assuming each to view the other from a rest postion (since each always sees themselves at rest) but change 30 Mph to 0.3c.

How much does A see B clcok slow down? ___________

How much does B see A clcok slow down ____________

Please compute the time differential (according to Relativity) between A and B clocks upon their return to a common rest frame and are compared:________

Not so. You can take the view of either A or B as at rest. If we convert these numbers into %c instead of Mph, so that we can do some actual relavistic calculations:

At 0.15c each clock would see the other clock running only 0.988689 as fast.

At 0.3c (correct view to calculate A or B views) the number is 0.953939.

You are in error on your calculations. You have done it wrong. The 15 Mph view would only be valid to a third observer at rest relative to both A and B.

Mac's concept of "reciprocity" is ill-defined, and doesn't seem to give any useful results (unlike the well-defined concept of symmetry).[/QUOTE]

I would agree you don't get results. That is you don't get time dilation. Which rather is the point now isn't it?

14. ### PeteIt's not rocket surgeryRegistered Senior Member

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I'm trying to do no such thing, since both are made up concepts that obscure the simplicity of the issue at hand. I'm addressing what relativity says directly, without laying any obscuring misinterpretations on top.

Why is that a problem?

I deliberately kept quantities out of it. The different order of events in each frame should be sufficient.

That post was actually hauled up from an eariler thread in which the focus was the symmetry of length contraction. I will revamp it to better focus on time dilation.

15. ### MacMRegistered Senior Member

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I note that you still refuse to calculate the issue as though each car is considered at rest and light beams are sent both directions. Why is that do you suppose? Could it be that you know that to do that you will show that your affect is perception and not reality?

Again lets do this correctly and not bias the result by using a single light beam in one direction. Shall we?

Really. You can pretend to be so dumb at times. You only forestall the inevitable. I will explain and then you will be forced to actually address the issue.

If A is considered at rest then B is moving away at 0.9c. If B is considered at rest then A is moving away at 0.9c. That really wasn't that hard now was it?

[quoteWhy?

Sure, the given postulates are for ONE theory only - special relativity. Change the postulates and you'd get different results.[/quote]

Nobody has changed any postulates. they have applied the postulates to the entire problem and not quit halfway thorugh the relavistic relationship as you have.

WHOA. I have never agreed to any such thing. I challenge your assumption that you know and understand the nature of light and its "Apparent" invariance.

But accepting it as stated the problem of reciprocity remains inherent in Relativity when calculated properly and not just one sided views.

You must work for the Bush campaign. Think you say it enough people will believe it. I have been rock steady on this issue of 1 1/2 years. You however have been all over the map trying to find reasons that the calculation cannot be done or that it must include things which it doesn't have to. Any excuse but to directly facr the consequences of Relativity's mandate that such reciprocity be applied if you want to view the reality and not just the perception.

Only if you apply your light signal unidirectionally. Apply it the other direction and recompute the relationship with the same relative velocities. What answer do you get?__________________

Correct as far as computaional results but not correct as far as being complete.

One must have data to suggest that relative velocity means relative velocity?. That the velocity between two observers is the same to each observer. Gee, I would like to see you prove otherwise, then I can ask why you would use the same figures had you choosen to compute A instead of B.

Bull. Arew you suggesting that Relativity says that the relative velocity is different to each observer? I should hope not.

I have read all such threads with equal amazement as to the gullability and short sightedness of some peole.

No lest do it my way and wapr the direction of the light beam.

You can hope but until you properly address this issue it will not go away.

16. ### MacMRegistered Senior Member

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So you see pointing out that relative velocity between two observers is equal in each observers eyes is "obscuring"?

Not a problem. that is they way it should be and as I have been pointing out that it is. Your discussion addresses simultaneity and time differentials or offsets. You should just have noticed that the affect is recipocal (equally) and that if you consider the issue of time dilation it is going to cancel.

Do that, you might find it interesting but be sure and do both views.

Last edited: Oct 11, 2004

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18. ### Paul TRegistered Senior Member

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Bahhh. MacM, you still think you understand relativity??? Invariant velocity is the basic assumption in relativity. Have you ever seen in relativity term g '= 1/sqrt(1-(v'/c)<sup>2</sup>)???? May be in MacM's relativity. Address the issue, stop making bullshit statement.

Done many times. But everytime you say: "I skipped the math" and talk something else. We already know that you just too silly to understand the issue and therefore keep coming back to the same question.

Very poor math ability. T2=T1*0.8, say....isn't this a linear relationship. Think with your brain MacM, not with your butt.

19. ### Paul TRegistered Senior Member

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It was done in the other thread:

A is on earth and C is on Pluto, which is 4.95 light hours away from earth. B in a spacecraft moving at constant velocity 0.9c from A to C. Assumed that both A and C are at rest, a fair assumtion considering the spacecraft velocity involved.

According to A, how long is the journey from A to C? The answer should be straight forward, require very little knowledge of relativity. It is (4.95c hours)/(0.9c)=5.5 hours.

According to B, how long is the journey from A to C? Use equation (t' = g [ t - (b/c)x]), insert in g=2.2941578, t=5.5 hours and x=4.95c hours. It is 2.2941578*(5.5-0.9*4.95)=2.3973949 hours. Note that this result can be also obtained using time dilation formula, that is 5.5/2.2941578=2.3973949 hours.

We should now make the computation from B perspective. Which equation should we use? Equation (t = g [ t' + (b/c)x']) and the parameters should be: g = 2.2941578, t' = 2.3973949 hours (not 5.5 hours, do you know why?) and x' = 0 (because the event is B reaches C). The result is certainly t=5.5 hours!​

Read it carefully mister. If don't understand ask. You don't have to send email to "Ask Physicist". There are so many physicists here who understand the issue and able to help you to understand.

20. ### MacMRegistered Senior Member

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Sorry you have become so blinded but there is no MacM's Relativity there is only Einstien's Relativity and IT has a problem which you rather not address.
Just where inb the HELL do you see "g" in my presentations? If it is there then you know as well as anyone that it is a typo. So get back to the issue.

Not done once yet. ALL attempts to sweep this issue under the rug are obvious failures. Now address the issue and knock off the meaningless personal attacks. You make yourself look impotent. Well may you are. Hmmm.

"v" is the variable asshole. Stop trying to confuse people with your bullshit responses.

21. ### James RJust this guy, you know?Staff Member

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MacM:

To do it, simply swap the words "green" and "blue" in my post, and replace v everywhere with -v.

Done.

Let's clear this up, shall we?

Respond.

The speed of light is invariant, for all observers. True or false?

Can you give a straight answer?

More lies. This is getting tiresome. Not only can the calculation be done, but I
have done it, in my special relativity derivation thread.

Where can we find your calculation? There isn't one, is there?

Same as before.

Wapr?

As previously pointed out, the quoted response refers to a completely different scenario.

MacM switches scenarios so often, he can't keep track of them all, and often gets them confused.

22. ### (Q)Encephaloid MartiniValued Senior Member

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Mac

JamesR has nothing to do with your misunderstanding of relativity. Of course, your misunderstandings have led to your misleadings of science, which is with whom this poll should ultimately be encapsulated.

23. ### MacMRegistered Senior Member

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10,104
First your stattmentis an unsupported innuendo. I have no misunderstadnings of Relativity. It is hollow words only.

You are now obligated to take the following Challenge:

[post=693326]Challenge[/post]

Either perform or apologize.