Mac's Special Relativity

[przyk]

You must be pragmatic in your assumptions and to assume the entire universe shifts gears just because you expend energy and change veloicty simply fails the laugh test. You must remeber the balance of the universe IS (are) nothing but other clocks in the relative veloicty equation so to suggest the station does anything as a consequence of your induced motion is to suggest the entire univers e responds .

Only as seen by the accelerating twin/shuttle/whatever, and it's simply a result of one inertial coordinate system "morphing" into another. Since the use of these coordinate systems in which the laws of physics take the same form is simply a matter of convenience, your real issue with STR is "the laws of physics just can't be Lorentz covariant".

Now you deviate from SR to GR and try to claim time dilation is a GR function related to relative veloicty.

It's an effect with more than one contributing factor. STR deals with situations in which only one of these - relative velocity - is significant.

The bottom line issue is that ONLY the accelerated clock shows a loss of time and bothclocks experienced the same relative velocities

Be careful with this assumption, especially in your own home-made versions of relativity. How velocities transform is dependent on how space and time coordinates transform from one reference frame to another, and I've never seen you postulate your own coordinate transformation to replace the one used in STR.

(even during the acceleration period).

I'm not so sure about this. I plan on taking my own look at accelerating frames in the near future, but I don't have time for it now.

But the acceleration magnitude and period do not correspond to changes in accumulated time. Only the period of differential velocity does so.

There will be an inverse relationship between the magnitude and period of acceleration for fixed initial and final velocities, so the magnitudes of the effect is approximately the same. This is why it doesn't matter if you arrange for the acceleration period to occupy 1% or 0.001% of the trip. What does have an important effect (as seen by the accelerating twin) is the distance between the two twins at the time of the acceleration, which is directly dependent on the "period of differential velocity".

 

[MacM]

Aha, so here is the basis of your argument then no?
Perhaps you could show the math for this, because I do not beleive you.

-Andrew

That shouldn't be to difficult.

Case 1:

Step 1 - Two clocks are mounted aboard rockets and a flight course is laid out. The start is at a common location and all at relative rest with a schedule having equal acceleration and comoving causing both rockets to accelerate and coast inertially side by side throughout the test.

Step 2 - Some distance away there is a stationary space bouy (Clock "A") to mark where rockets carrying clocks "B" & "C" will stop accelerating and go inertial at 0.8c and where all clocks set to "zero" and are started.

Step 3 - A second bouy 6,000 light seconds from station "A" marks where the rocket "B" stops his clock.

Step 4 - A third bouy 60,000 light seconds away marks where rocket "C" stops his clock.

It is obvious that in steps 1 and 2 the clocks are subjected to the same period and magnitude of acceleration and achieve identical higher inertial velocity.

At 0.8c t "B & C" = 0.6t "A" for both rockets, such that when the rockets cross the 2nd bouy it took 7,500 seconds according to "A" but "B" stops his clock at 0.6 * 7,500 = 4,500 seconds.

When "C" crosses the 3rd bouy A now reads 75,000 seconds and "C" stops his clock at 45,000 seconds.

The difference in accumulated time is a direct function of inertial velocity and duration in an absolute universal sense.



Case 2:

Step 1 - Two clocks are mounted aboard rockets and a flight course is laid out. The start point is shifted but is at a common relative rest with a schedule having accelerations to cause the rockets to cross a start bouy simultaneously and comoving causing both rockets to accelerate and coast inertially throughout the test but at different higher inertial velocities.

Step 2 - Some distance away there is a stationary space bouy (Clock "A") to mark where rockets carrying clocks "B" & "C" will stop accelerating and go inertial at 0.6c and 0.8c respectively and where all clocks set to "zero" and are started.

Step 3 - A second bouy 6,000 light seconds from station "A" marks where both rockets stops their clocks.

It is obvious that in steps 1 and 2 the clocks are subjected to different magnitude of acceleration but for a common period in a universal sense and achieve different higher inertial velocities.

At 0.6c "B" crosses the 2nd bouy after 10,000 seconds according to "A" but his clock reads 8,000 seconds.

At 0.8c when C crosses the 2nd bouy after 7,500 seconds according to "A" but his clock stops at 4,500 seconds.

The difference in accumulated time is a direct function of inertial velocity and distance in an absolute universal sense. t = d/v.

Acceleration affects the terminal velocity but the terminal velocity dictates tick rate and time loss is a function of tick rate and universal duration of the test.

 

[(Q)]

Correct and the latest study of that shows an ansitropy and using the ansitropy of muon decay they correctly computed the earth's absolute motion relative to the CMB. The conclusion being that the "Correct" "V" to use in the time dilation formula is an absolute velocity relative to the CMB and not relative velocity to the earth. Totally inconsistant with the SR view.


http://redshift.vif.com/JournalFiles/Pre2001/V03NO2PDF/V03N2MON.PDF

"Significance of this research
It is of interest to know the absolute velocity of the
solar system for a number of reasons: The fact that a
solar system velocity can be observationally determined
is further evidence for the existence of absolute space, a
stationary ether, or a preferred zero velocity frame of
reference, and it provides evidence against relativity
theories. The measured value of the absolute velocity of
the solar system, after subtracting off the velocity due to
the general Milky-Way galactic rotation and
translation, might provide an indication of dark
neighbors to the solar system."

Mac, the linked article makes assumptions of absolute velocities and then immediately confirms the assumptions without explaining anything. Nonsense.

 

[MacM]

"Significance of this research
It is of interest to know the absolute velocity of the
solar system for a number of reasons: The fact that a
solar system velocity can be observationally determined
is further evidence for the existence of absolute space, a
stationary ether, or a preferred zero velocity frame of
reference, and it provides evidence against relativity
theories. The measured value of the absolute velocity of
the solar system, after subtracting off the velocity due to
the general Milky-Way galactic rotation and
translation, might provide an indication of dark
neighbors to the solar system."

Mac, the linked article makes assumptions of absolute velocities and then immediately confirms the assumptions without explaining anything. Nonsense.

You can choose to ignore the ansitropy measured and utilized to compute the absolute motion of our planet but that doesn't alter the fact it is there.

Go ahead it's OK to cry.

 

[(Q)]

You can choose to ignore the ansitropy mesured and utilized but that doesn't alter the fact it is there and the signifigance that it can be used to compute absolute velocity of motion.

Go ahead it's OK to cry.

No one is ignoring anything, Mac. The fact of the matter is that the article takes an assumption and makes it a fact, without any explanation and regardless of measurements.

 

[andbna]

That shouldn't be to difficult.

Alas it was: that was not mathematics, that was some measured values and assertions of a thought experiment. Look I can do that to:
Relativity is wrong because if I have a cookie traveling at 0.8c then its mass will increase 40 times and thus if I eat it, I will violate the laws of conservation of mass because I ate 40 cookies because 40*1=40. Go ahead, show some empirical observational data that if I ate a cookie travelling at 0.8c I would not have gained 40 cookies of mass.
You need to prove your assertions.

How did you derive the reasult of those thought experiments mathematicly? Use agreed upon axioms, for instance Lorentz equations, not MacM assertions.

-Andrew

 

[MacM]

Alas it was: that was not mathematics, that was some measured values and assertions of a thought experiment. Look I can do that to:
Relativity is wrong because if I have a cookie traveling at 0.8c then its mass will increase 40 times and thus if I eat it, I will violate the laws of conservation of mass because I ate 40 cookies because 40*1=40. Go ahead, show some empirical observational data that if I ate a cookie travelling at 0.8c I would not have gained 40 cookies of mass.
You need to prove your assertions.

How did you derive the reasult of those thought experiments mathematicly? Use agreed upon axioms, for instance Lorentz equations, not MacM assertions.

-Andrew

My error I assumed you knew how to compute time dilation and do basic mathematics (using SR not Mac's relativity).

BTW: 0.8c doesn't equate to 40 cookies only 1.6 cookies. Of course that isn't a problem either since you are traveling with the cookie and it has "0" relative velocity to you. - Shessh.

 

[andbna]

Obviously, if your computations utilize relativistic mathematics, they will be dead on with relativities predictions. So either your predicting that which the theory you are trying to discredit will predict (thus making your argument meaningless), or your doing something wrong with the mathematics. If it's the first, then theres no point in disscussion, if it's the second, than showing your work will point out the problem, either way showing you work will tell us which of the two it is.

My error I assumed you knew how to compute time dilation and do basic mathematics (using SR not Mac's relativity).

I do, it is actualy you who appears not to.
Looking at case 1:
Since each clock is in an inertial reference frame while they are on, according to Relativity, the Lorentz factor is all the time dialation we have during the test. And thus, while the buoy will observe one rocket ship to be dialating, and a few thousand ticks behind, the rocket will observe the buoy to be equally behind, and will stop it's clock at 7,500 ticks, just like the buoy, and upon comparison they will see the dialation was only observed.
This is what STR predicts.

Case 2: same as case 1.

Acceleration affects the terminal velocity but the terminal velocity dictates tick rate and time loss is a function of tick rate and universal duration of the test.

Observed time loss that is. Actual time loss only happens during acceleration. If you think I am wrong, this is why I asked you to show your work. Tell me what equations you are using to derive that resault.
Clearly, your thought experiments are not going anywhere.

BTW: 0.8c doesn't equate to 40 cookies only 1.6 cookies. Of course that isn't a problem either since you are traveling with the cookie and it has "0" relative velocity to you. - Shessh.

I'll merely note that none of the above is rebuttal data.
And I'm not travelling with the cookie, it would be very hard to eat if our distances remained relativly fixed...

-Andrew

 

[MacM]

It's an effect with more than one contributing factor. STR deals with situations in which only one of these - relative velocity - is significant.

I would agree as to the result but not the theory. SR does not qualify that only ONE relative velocity is significant, it specifies that both views are equally valid.

Be careful with this assumption, especially in your own home-made versions of relativity. How velocities transform is dependent on how space and time coordinates transform from one reference frame to another, and I've never seen you postulate your own coordinate transformation to replace the one used in STR.

That is because there are none. In my view Lorentz Spatial Contraction does not occur. I maintain an absolute view of clock tick rate. That is the accelerated clock (which ultimately emperically demonstrates a loss of time) IS physically ticking slower than the resting clock.

Time and distance are physical enities and velocity is just a calculated ratio v = d/t. Such that when t1 = 10 and t2 = t1/gamma = i.e: 0.5t1 = 5 and d = 200.

v1 = d/t1 = 200/10 = 20 ; d1 = v1t1 = 10*20 = 200

v2 = d/t2 = 200/5 = 40; d2 = v2t2 = 5 * 40 = 200

d1 = d2

That is relative velocity in an absolute sense is a function of the tick rate of the clock used to measure the trip and retaining that measurement standard distance cannot change.

So velocity is frame dependant, not distance. SR is invalid (my opinion) because they claim a reduced tick rate for a clock (i.e. - t2 = 0.5t1) but then turn around and claim because the trip took less time at the same veloicty distance was shorter. But velocity is nothing more than the calculated d/t and disregarding that t2 is universally different produces the requirement for contraction. If t2 were not physically different then it would not physically lose time.

For example if point A to point B which are in the same inertial rest frame is 60 miles and I drive two cars over the course.

Car "A" has a good speed-o-meter, odometer and clock but car "B" speed-o-meter is broken and his watch batteries are low such that his watch ticks only once for each tick of car "A" clock and we drive side by side at 60 Mph according to "A"; upon arrival when they discuss the trip "B" would say look we were driving 120 Mph because we went 60 miles in 30 minutes!

He would not say you went 60 miles but I only went 30 miles. It is really just that simple.

I'm not so sure about this. I plan on taking my own look at accelerating frames in the near future, but I don't have time for it now.

Just keep in mind that every instantaneous moment in time is an inertial velocity because there is no change in velocity in an instant. The affect is an integration of velocity affects over the acceleration schedule.

There will be an inverse relationship between the magnitude and period of acceleration for fixed initial and final velocities, so the magnitudes of the effect is approximately the same. This is why it doesn't matter if you arrange for the acceleration period to occupy 1% or 0.001% of the trip. What does have an important effect (as seen by the accelerating twin) is the distance between the two twins at the time of the acceleration, which is directly dependent on the "period of differential velocity".

I think we are getting into symantics here and I would not argue over it.

I have claimed time loss is a function of tick rate at the elevated velocity times the period at that velocity.

You seem to be claim the acceleration magnitude and duration set the tick rate and times period determines time loss.

I would agree with that because the tick rate and the induced velocity are directly and inversely linked by the time dilation formula.

 

[Pete]

What they see is only relevant as to the difference in the empirical reality. What you see is NOT what you get.

Good.

When the flash arrives is not at issue. That is the reason for using tick rates and not accumulated time. It is ONLY the time intervals measured between flashes once they have started to arrrive that counts.

Good again.

He sees 36%, then calculates (predictes) what the accumulated time is on the other clock...

That's what I thought. Like I said, you still have the same misunderstanding of SR that you've had since I've known you.

From the OP:

The station observer notes that according to his observation of his clock and the shuttle tick rate that the shuttle clock is only ticking at 36% the rate of his clock. Not the anticipated 60%

The researchers haven't done their predictions correctly.
They should have anticipated that he station observer would see the shuttle clock ticking at 33.3% the rate of his clock.

The prediction of SR is that each observer will see the flashes from the other's clock arrive at 1/3 of their own clock's tick rate - not 60%.



Here's a question for you.
During the approach (ie before the shuttle crosses the light trigger, but after it has finished accelerating), at what rate would the station observer and shuttle pilot see the flashes from each other's clocks arive? And what would they anticipate seeing, according to SR?

 

[James R]

MacM:

I can only assume you missed my long post where I took apart your thought experiment. Either that, or you're willfully ignoring it. The only response I've had from you was for you to dishonestly misquote me, in relation to a tangential post I made concerning you ability to understand special relativity.

Go back to that post and read my post two or three posts above it. Then respond, if you have a response.

 

[MacM]

Good.


Good again.



That's what I thought. Like I said, you still have the same misunderstanding of SR that you've had since I've known you.

From the OP:

The researchers haven't done their predictions correctly.
They should have anticipated that he station observer would see the shuttle clock ticking at 33.3% the rate of his clock.

The prediction of SR is that each observer will see the flashes from the other's clock arrive at 1/3 of their own clock's tick rate - not 60%.



Here's a question for you.
During the approach (ie before the shuttle crosses the light trigger, but after it has finished accelerating), at what rate would the station observer and shuttle pilot see the flashes from each other's clocks arive? And what would they anticipate seeing, according to SR?

You are correct. I screwed this one up because I was not applying doppler and therefore should not have used the light signal. I'll re-formulate and re-post to acheive what I was really wanting to do.

To answer your question SR = 300%

 

[MacM]

MacM:

I can only assume you missed my long post where I took apart your thought experiment. Either that, or you're willfully ignoring it. The only response I've had from you was for you to dishonestly misquote me, in relation to a tangential post I made concerning you ability to understand special relativity.

Go back to that post and read my post two or three posts above it. Then respond, if you have a response.

I'll pass on your post because I screwed up my presentation and it doesn't do what I wanted. I'll re-think the presentation and re-post.

 

[przyk]

That is because there are none. In my view Lorentz Spatial Contraction does not occur. I maintain an absolute view of clock tick rate. That is the accelerated clock (which ultimately emperically demonstrates a loss of time) IS physically ticking slower than the resting clock.

Then you do have a coordinate transformation with respect to your "absolute" frame and it is (for motion along the x axis):

$$t' = \frac{1}{\gamma(v)} t $$​

$$x' = x - v t$$​

If you define $$u = \frac{\mathrm{d}x}{\mathrm{d}t}$$ and $$u' = \frac{\mathrm{d}x'}{\mathrm{d}t'}$$, then your MacM velocity transformation formula is:

$$u' = \gamma(v) ( u - v )$$​

You can use this to directly calculate your v[sub]1[/sub] and v[sub]2[/sub]. Incidentally, this transform has a variant c.

So velocity is frame dependant, not distance. SR is invalid (my opinion) because they claim a reduced tick rate for a clock (i.e. - t2 = 0.5t1) but then turn around and claim because the trip took less time at the same veloicty distance was shorter.

No, STR assumes that measuring instruments contract in the direction of their motion and that clocks which appear to be synchronized in one frame will not be synchonized in another. And it's got good reasons for drawing both these conclusions.

You seem to be claim the acceleration magnitude and duration set the tick rate and times period determines time loss.

As I said before, the "acceleration due to time dilation" is simply a result of the gradual transition from one inertial frame to another. It's completely analagous to a spatial rotation. A stationary or slow moving object will appear to move a great distance at a great speed if we rotate the coordinate system. The faster we rotate and the farther the object is from the centre of rotation, the more pronounced the effect is. It's basically the same story with STR: inertial frames are related by (hyperbolic) "rotations" of the space and time coordinates, so as you switch from one inertial frame to another, you expect a change in the time coordinate of any event that increases with distance and the rate of transition between the two inertial frames. The universe doesn't "change gears" as a result of the observer's acceleration.

 

[halo07guy]

Can anyone explain the last three pages in english?

 

[MacM]

Then you do have a coordinate transformation with respect to your "absolute" frame and it is (for motion along the x axis):

$$t' = \frac{1}{\gamma(v)} t $$​

$$x' = x - v t$$​

If you define $$u = \frac{\mathrm{d}x}{\mathrm{d}t}$$ and $$u' = \frac{\mathrm{d}x'}{\mathrm{d}t'}$$, then your MacM velocity transformation formula is:

$$u' = \gamma(v) ( u - v )$$​

You can use this to directly calculate your v[sub]1[/sub] and v[sub]2[/sub]. Incidentally, this transform has a variant c.

Thanks. When I said "no" I meant that I don't transform time into space and vice-versa. I don't have time-space as in SR.

 

[MacM]

Having reconsidered what it was I was trying to show is masked by doppler. So I'll merely ask the following:

v = 0.8c; t = 100

Time dilation is 60%. It is argued that observers with relative velocity will "See" the others clock ticking at it's time dilated rate (60%)

t' = t * (1 - v^2/c^2)^0.5

t' = 100 * ( 1 - (0.8^2)/1^2)^0.5

t' = 100 * (1 - 0.64)^0.5

t' = 100 * (0.36)^0.5

t' = 100 * 0.6 = 60 = 3/5

Yet doppler for the same velocity indicates that what you will see is only 1/3 the true tick (pulse) rate. That appears in conflict with the arguement about what observers "See" of others clocks.

Further the question then becomes, since we know empirically that the accelerated clock t' IS only actually ticking at 0.6t, the pulses viewed by doppler would suggest what you actually see would only be 1/5 or 20%.

The only way around that would seem to be to do what they do in the first instance "which is ignore the absolute differential in clock tick rate" when computing travel time and then suggest that distance changed.

Here you can only get 1/3 if you ignore the fact that the clock IS losing time by ticking at 60% when you compute doppler because you still use 100% in the formula..

 

[Pete]

Time dilation is 60%. It is argued that observers with relative velocity will "See" the others clock ticking at it's time dilated rate (60%)

What do you mean by "See"?
Do you mean "Measure after accounting for any signal delays"?

Further the question then becomes, since we know empirically that the accelerated clock t' IS only actually ticking at 0.6t, the pulses viewed by doppler would suggest what you actually see would only be 1/5 or 20%.

How do you figure that? My calculations suggest that you'd actually see 1/3.

 

[MacM]

What do you mean by "See"?
Do you mean "Measure after accounting for any signal delays"?


How do you figure that? My calculations suggest that you'd actually see 1/3.

OK. I guess I need to clarify this. The 1/3 doppler figure is the relavistic doppler shift which is actually non-relavistic doppler / time dilation.

f' = f * (1- v/c) / (1 - v^2/c^2)^.0.5

f' - f * 1 - .8) / (1 - .8^2)^.5

f' = f * .2 / (.36)^.5

f' = f * (.2 / .6) = 0.333333

 

[Pete]

OK. I guess I need to clarify this. The 1/3 doppler figure is the relavistic doppler shift which is actually non-relavistic doppler / time dilation.

f' = f * (1- v/c) / (1 - v^2/c^2)^.0.5

f' - f * 1 - .8) / (1 - .8^2)^.5

f' = f * .2 / (.36)^.5

f' = f * (.2 / .6) = 0.333333

I agree... like I said, the calculation indicates that you'd actually see a 1/3 tick rate from a clock actually ticking at at rate of 0.6 (ie time dilated from a rate of 1.0) and moving away at 0.8c.

So how do you figure that "what you actually see would only be 1/5 or 20%"?

And you didn't answer the first question: What do you mean by "See", when you put it in quotes?
I think you mean "Measure after accounting for any signal delays." Is that right?