"Science vs. Religion: What Scientists Really Think"

Jan 22, 2012 | Posted by: roboblogger | Full story: Examiner.com

It is fascinating to note that atheists boast that most scientists are atheists.

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humble brother

Vanda, Finland

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#11944
Jul 29, 2012
 
polymath257 wrote:
The time dilation factor is the same from the two sources, but the number of wavefronts passed in any time interval is larger from the source in front than the source behind.
The observer observes wavefronts coming from both directions at the same speed c.
Why would there be more wavefronts coming from front? There is absolutely no reason for that.
KJV

Sioux City, IA

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#11945
Jul 29, 2012
 
-Skeptic- wrote:
<quoted text>Dude, people are:

1. Telling you you are wrong.

2. Showing you are wrong

And yet you call this babble. You don't know what you are talking about and I have no respect for you.
"I have no respect for you"

Well that ought to keep him up nights.

“Think&Care”

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#11946
Jul 29, 2012
 
humble brother wrote:
<quoted text>
The observer observes wavefronts coming from both directions at the same speed c.
Why would there be more wavefronts coming from front? There is absolutely no reason for that.
The same speed for the observer, yes. But the number of wavefronts is different simply because, from the point of view of the beacon, the spacecraft passes through more wavefronts moving towards as opposed to moving away. But the count of wavefronts does not depend on relative speed, so there are more wavefronts in a given time with the same speed for the observer going one direction than the other. This means a different frequency for that same speed.

Think of it like this: from the point of view of the beacon the wavefronts go out in a sphere. Will the incoming observer pass more wavefronts for a given time interval or will the outgoing observer? Just answer this from the point of view of the beacon first.
humble brother

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#11947
Jul 29, 2012
 
polymath257 wrote:
The same speed for the observer, yes. But the number of wavefronts is different simply because, from the point of view of the beacon, the spacecraft passes through more wavefronts moving towards as opposed to moving away. But the count of wavefronts does not depend on relative speed, so there are more wavefronts in a given time with the same speed for the observer going one direction than the other. This means a different frequency for that same speed.
In the previous case of two observers and one light source you argued that there is time dilation and length correction that cause the compressed waves of light and thus change in frequency.

Now we have only two frames. The rest frame which has the two light sources and the rest frame of the moving ship. There is no longer time dilation or length correction to account for any difference in frequency of light observed from the two sources.
polymath257 wrote:
Think of it like this: from the point of view of the beacon the wavefronts go out in a sphere. Will the incoming observer pass more wavefronts for a given time interval or will the outgoing observer? Just answer this from the point of view of the beacon first.
You are contradicting yourself now. This would be only true for the observer if they could move faster relative to the light coming from front than from back. It is not possible in the relativistic model.

We have already established the fact that two observers in different rest frames do not agree on their observations. You can not mix and match measured values between two rest frames any way you please.

You are now in the paradox that you claim:
1. the observer ship can not move faster relative to light coming from the front beacon
2. the observer ship can move faster relative to light coming from the front beacon

These claims can not both be true for the observer. You also can not combine these two into a single calculation for the observer. That is totally false.

“It's all about the struggle”

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#11948
Jul 29, 2012
 

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Aura Mytha wrote:
<quoted text>
Why professors are human , and even Einstein made mistakes?
Admitting this is a noble quality , to not is a lame arrogance.
:p

“Think&Care”

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#11949
Jul 29, 2012
 
humble brother wrote:
<quoted text>
In the previous case of two observers and one light source you argued that there is time dilation and length correction that cause the compressed waves of light and thus change in frequency.
And I corrected myself and noted that *both* the time dilation and the change in number of wavefronts in a given period of time are relevant.
Now we have only two frames. The rest frame which has the two light sources and the rest frame of the moving ship. There is no longer time dilation or length correction to account for any difference in frequency of light observed from the two sources.
Right. it is *only* the difference in counting the number of wavefronts.
You are contradicting yourself now. This would be only true for the observer if they could move faster relative to the light coming from front than from back. It is not possible in the relativistic model.
You are confusing the difference in speeds as seen from the beacon's frame and the the difference in speeds as seen by the observers.

In the beacon's frame, during a time interval t, the light moves a distance c*t and the observers move a distance v*t. The distance between a wavefront and an observer is then given by c*t-v*t or c*t+v*t depending on the direction of travel. This is NOT the same as saying the relative speeds are c-v and c+v *in the frames of the observers*. It is the speed of light in a frame that is fixed as a constant.
We have already established the fact that two observers in different rest frames do not agree on their observations. You can not mix and match measured values between two rest frames any way you please.
You are right. We cannot willy-nilly use the measurements from one frame in another. However, the count of the number of wavefronts that passes does not depend on the frame. The time intervals do; the distances moved do; but the simple count does not.
You are now in the paradox that you claim:
1. the observer ship can not move faster relative to light coming from the front beacon
No. The speed of light in any frame is the same. THAT is the statement of the constancy of the speed of light. But the c-v and the c+v are NOT the speeds of light in any frame. They are differences of speeds and come up in calculations of distances.
2. the observer ship can move faster relative to light coming from the front beacon
These claims can not both be true for the observer. You also can not combine these two into a single calculation for the observer. That is totally false.
Again, you are confusing the difference in speeds in one frame with the difference of speeds in another. In every frame, the speed of light in a vacuum is the same. But you can still calculate the distance between a wavefront and a spacecraft via c*t-v*t or c*t +v*t.
humble brother

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#11950
Jul 30, 2012
 
polymath257 wrote:
You are confusing the difference in speeds as seen from the beacon's frame and the the difference in speeds as seen by the observers.
The confusion is all in your mind.
polymath257 wrote:
In the beacon's frame, during a time interval t, the light moves a distance c*t and the observers move a distance v*t. The distance between a wavefront and an observer is then given by c*t-v*t or c*t+v*t depending on the direction of travel. This is NOT the same as saying the relative speeds are c-v and c+v *in the frames of the observers*. It is the speed of light in a frame that is fixed as a constant.
You are contradicting yourself still.

You say here that the photons from the beacon can move faster than the speed c relative to the observer ship. So the relative speed c can then be exceeded by photons after all? So it is not a constant?
polymath257 wrote:
You are right. We cannot willy-nilly use the measurements from one frame in another. However, the count of the number of wavefronts that passes does not depend on the frame. The time intervals do; the distances moved do; but the simple count does not.
You are saying that the front beacon sees light move faster than c relative to the observer ship and therefore the number of encountered wavefronts is increased.

Does the observer ship see light coming from front faster than the constant c? Answer: NO they do not.

Once more: we know that within the model observers in different rest frames disagree on measurements. So why do you bring measurements that the two frames disagree with into one calculation?

You have drowned yourself in fallacies in your paradox. You know that the two frames disagree with the measurements and still you bring them together to calculate.

You only can say within the model that from the beacon it seems that light seemingly travels faster than the constant c relative to the ship and therefore there seems to be a case of:
- relative speed of light faster than c
- increased frequency because of that speed

It is absolute nonsense to bring the frequency from that equation to the rest frame of the observer and then hypocritically slap a new speed label on it. Pure nonsense.
polymath257 wrote:
No. The speed of light in any frame is the same. THAT is the statement of the constancy of the speed of light. But the c-v and the c+v are NOT the speeds of light in any frame. They are differences of speeds and come up in calculations of distances.
And yet you have counted the encountered wavefronts for a faster than light relative speed. That's exactly what you are using in your calculations to count the wavefronts:
- faster than constant light speed for the front: c+v
- slower than constant light speed for the back: c-v

Because you change the speed like that you get different counts of wavefronts.
polymath257 wrote:
Again, you are confusing the difference in speeds in one frame with the difference of speeds in another. In every frame, the speed of light in a vacuum is the same. But you can still calculate the distance between a wavefront and a spacecraft via c*t-v*t or c*t +v*t.
I must ask you again:
Are you lying through your teeth to try and confuse other people here???

Since: Mar 12

Dubai, UAE

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#11951
Jul 30, 2012
 
humble brother wrote:
The gravity model F=(M+m)/r has a nonlinear relationship between force and calculable mass. It works perfectly for modeling gravity but the concept of mass is just different than the one you are used to.
Your model is nonsense before we even try to test it against empirical reality because its units do not stack up. The left hand side of your equation is force. Force is:

F = mass * distance * time^-2
Your right hand side = mass^2 * distance^-2

m * d * t^-2 =/= m^2 * d^-2

Your right hand side lacks any time units for starters. Your equation would only make sense...if you add a constant that balances up the units being used.

You need to add:

d^3 * m^-1 * time^-2

We will call this constant a Humble (H) in your honor. Or will we? Its the same units as G!

And funnily enough, your equation will finally make sense as a coherent hypothesis but still won't match reality.

Not only do you fail in your quest to demolish G, the only real difference left is your prediction that force will vary according to (M + m)^2, all else being equal, whereas Newton predicts it will vary by Mm, all else being equal.

Since Newton's predictions have held up as close to reality under intense scrutiny and your prediction is significantly different even after I fix your equation (and made no sense at all before)...

Your hypothesis (even after I fix your schoolboy unit errors), is falsified.
humble brother

Vanda, Finland

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#11952
Jul 30, 2012
 
Chimney1 wrote:
<quoted text>
Your model is nonsense before we even try to test it against empirical reality because its units do not stack up. The left hand side of your equation is force. Force is:
F = mass * distance * time^-2
Your right hand side = mass^2 * distance^-2
m * d * t^-2 =/= m^2 * d^-2
Your right hand side lacks any time units for starters. Your equation would only make sense...if you add a constant that balances up the units being used.
You need to add:
d^3 * m^-1 * time^-2
We will call this constant a Humble (H) in your honor. Or will we? Its the same units as G!
And funnily enough, your equation will finally make sense as a coherent hypothesis but still won't match reality.
Not only do you fail in your quest to demolish G, the only real difference left is your prediction that force will vary according to (M + m)^2, all else being equal, whereas Newton predicts it will vary by Mm, all else being equal.
Since Newton's predictions have held up as close to reality under intense scrutiny and your prediction is significantly different even after I fix your equation (and made no sense at all before)...
Your hypothesis (even after I fix your schoolboy unit errors), is falsified.
Oh you came back to show your faith.

This was already covered a long time ago but I can see that you understood none of it. From the new model you can directly derive the relation between mass and and rotational speed:
m = M/(vr - M)

From here you can already see what the concept of mass is within the new model. You poor little thing didn't understand the consequences of (vr - M). Within the new model relative mass is also a function of relative rotational speed.

You should go back to school to try and understand that you don't falsify mathematical models with units. Thank you for the laughs though.

Since: Mar 12

Dubai, UAE

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#11953
Jul 30, 2012
 
humble brother wrote:
<quoted text>
Oh you came back to show your faith.
This was already covered a long time ago but I can see that you understood none of it. From the new model you can directly derive the relation between mass and and rotational speed:
m = M/(vr - M)
From here you can already see what the concept of mass is within the new model. You poor little thing didn't understand the consequences of (vr - M). Within the new model relative mass is also a function of relative rotational speed.
You should go back to school to try and understand that you don't falsify mathematical models with units. Thank you for the laughs though.
Yeah, I do not spend seven days a week on topix. Go figure.

However, in spite of your feeble attempt at condescension, anybody can see that yet again, you have failed to align the units, which must be consistent on both sides of the equation. This is elementary stuff. You think its unnecessary?

In pure maths, 5 + 3 = 8.

But in pure maths,

5x + 3y =/= 8 anything

The algebraic statement 5x + 3y is NOT reducible to a single variable.

In physics, every variable is a quantity of a specific unit. 3 kilograms, 3 seconds, 3 somethings. Thus 3d, 3m, 3t. These are not purely numerical statements, they are algebraic of the form number, unit. 5x, not just "5".

You cannot add 5 seconds to 3 kilograms. Any more than, mathematically, you can add 5x to 3y and get "8" of anything.

You cannot do maths, and you certainly cannot do physics, with any competence. And its laughable that you try to redefine what "mass" is to suit your very silly non-unit-aligned equations. You already have a mislignment in your m = M/(vr - M) nonsense, as mass on one side includes a velocity and distance component, and lacks it on the other. So you have two entirely different qualities you are trying to call mass.

Since: Mar 12

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#11954
Jul 30, 2012
 
Velocity, of course, being a time & distance component when broken down into fundamentals.

So its time, distance and distance again that you are misaligning.

humble brother

Vanda, Finland

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#11955
Jul 30, 2012
 
Chimney1 wrote:
Yeah, I do not spend seven days a week on topix. Go figure.
However, in spite of your feeble attempt at condescension, anybody can see that yet again, you have failed to align the units, which must be consistent on both sides of the equation. This is elementary stuff. You think its unnecessary?
In pure maths, 5 + 3 = 8.
But in pure maths,
5x + 3y =/= 8 anything
The algebraic statement 5x + 3y is NOT reducible to a single variable.
In physics, every variable is a quantity of a specific unit. 3 kilograms, 3 seconds, 3 somethings. Thus 3d, 3m, 3t. These are not purely numerical statements, they are algebraic of the form number, unit. 5x, not just "5".
You cannot add 5 seconds to 3 kilograms. Any more than, mathematically, you can add 5x to 3y and get "8" of anything.
You cannot do maths, and you certainly cannot do physics, with any competence. And its laughable that you try to redefine what "mass" is to suit your very silly non-unit-aligned equations. You already have a mislignment in your m = M/(vr - M) nonsense, as mass on one side includes a velocity and distance component, and lacks it on the other. So you have two entirely different qualities you are trying to call mass.
It is obvious that you have absolutely no understanding of the nonlinear nature of relative mass in the new model.

One kilogram is a name given to a lump of a chosen size. Then mathematically it is possible to call two of those lumps two kilograms.

You can as well stop using the name kilogram and choose another name, you can start using cowabungaa as the name and choose some lump to be the fixed measure. Then you can perhaps say that one cowabungaa is 0.346333 times the lump called "kilogram".

Above you can see that the unit of mass is not a kilogram or anything else that is used your your magic model. The unit of mass in the new model is m/s. Mass in the new model should be called "mass effect" rather than just "mass".

Since: Mar 12

Dubai, UAE

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#11956
Jul 30, 2012
 
humble brother wrote:
It is obvious that you have absolutely no understanding of the nonlinear nature of relative mass in the new model.
Actually its perfectly clear that your incoherent drivel is tripping you up. Whatever it is you are trying to define, its certainly not mass.
One kilogram is a name given to a lump of a chosen size. Then mathematically it is possible to call two of those lumps two kilograms.
Yes, and if we are adding 5kg to 3kg, we get 8kg.

5x + 3x = 8x WOW!

However, you still CANNOT add 5 kg to 3 seconds.
Like 5x + 3y, they are irreducible beyond that composite statement.

And in a statement of equalities - an equation - the units must equate, as I said. If there is an expression of time on one side, there must be an expression of time on the other.
The unit of mass in the new model is m/s. Mass in the new model should be called "mass effect" rather than just "mass".
Now you accept that its NOT mass you are talking about but a new derivative quality, a function of mass and time. And now "mass effect" = mass cubed over time squared. Mass is still mass, then, the same old mass, an actual, non relative quantity just like it always was, over there on the right hand side of your "mass effect" equation, the actual quantity you have decided to cube in inventing your new, utterly useless thingamejig called "mass effect".

No way around it. To fix your equation, you have to add back in exactly the same units that comprise G: d^3 * m^-1 * t^-2

Then we just change your (M + m)^2 back to Mm, and hey presto, its Newton again and it actually works.

You doofus!

“Think&Care”

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#11957
Jul 30, 2012
 
humble brother wrote:
<quoted text>
The confusion is all in your mind.
<quoted text>
You are contradicting yourself still.
You say here that the photons from the beacon can move faster than the speed c relative to the observer ship. So the relative speed c can then be exceeded by photons after all? So it is not a constant?
No, I did NOT say that. The c+v is in the beacon's frame, not the observer's frame. The distance between the wavefront of light and the observer moving away in the beacon's frame increases at a rate of c+v, but that is not a velocity of any object nor a relative velocity in any frame.
You are saying that the front beacon sees light move faster than c relative to the observer ship and therefore the number of encountered wavefronts is increased.
The beacon sees the distance between the wavefront and the observer increase at a rate of c+v. But that is NOT a velocity in any frame. In particular, it is NOT the relative velocity between ligth and the observer *in the observer's frame*.
Does the observer ship see light coming from front faster than the constant c? Answer: NO they do not.
Right. In the observer's frame, the speed of light is always c.
Once more: we know that within the model observers in different rest frames disagree on measurements. So why do you bring measurements that the two frames disagree with into one calculation?
They disagree on distances, velocities, time durations, energies, momenta, and a great number of other things. But they agree on things like rest mass, proper time of an event (time experienced by someone in the frame of the event), and phase of a wave. So a count of the number of wavefronts passing the observer is independent of the reference frame. The Doppler effect compares two frames: that of the beacon and that of the observer. So of course we have to look at the system in both frames and compare them!
You have drowned yourself in fallacies in your paradox. You know that the two frames disagree with the measurements and still you bring them together to calculate.
They agree on number of wavefronts passing the observer.
You only can say within the model that from the beacon it seems that light seemingly travels faster than the constant c relative to the ship
No, the c+v is NOT the velocity of anything in any frame.
And yet you have counted the encountered wavefronts for a faster than light relative speed. That's exactly what you are using in your calculations to count the wavefronts:
- faster than constant light speed for the front: c+v
- slower than constant light speed for the back: c-v
Because you change the speed like that you get different counts of wavefronts.
The speed doesn't change. We are merely calculating the distance between the wavefront and the observer in the beacon's frame. The count is the same in both frames; the speed of light is the same in both frames; so the frequency differs between the frames.
I must ask you again:
Are you lying through your teeth to try and confuse other people here???
No, I am attempting to show you where you are making your mistakes so, perhaps, you will learn something. Clearly that isn't happening. Let's focus on the c+v. Is it the speed of light in any reference frame? Is it the speed of anything in any reference frame? What, exactly, does it represent?
humble brother

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#11958
Jul 30, 2012
 

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Chimney1 wrote:
<quoted text>
Actually its perfectly clear that your incoherent drivel is tripping you up. Whatever it is you are trying to define, its certainly not mass.
<quoted text>
Yes, and if we are adding 5kg to 3kg, we get 8kg.
5x + 3x = 8x WOW!
However, you still CANNOT add 5 kg to 3 seconds.
Like 5x + 3y, they are irreducible beyond that composite statement.
And in a statement of equalities - an equation - the units must equate, as I said. If there is an expression of time on one side, there must be an expression of time on the other.
<quoted text>
Now you accept that its NOT mass you are talking about but a new derivative quality, a function of mass and time. And now "mass effect" = mass cubed over time squared. Mass is still mass, then, the same old mass, an actual, non relative quantity just like it always was, over there on the right hand side of your "mass effect" equation, the actual quantity you have decided to cube in inventing your new, utterly useless thingamejig called "mass effect".
No way around it. To fix your equation, you have to add back in exactly the same units that comprise G: d^3 * m^-1 * t^-2
Then we just change your (M + m)^2 back to Mm, and hey presto, its Newton again and it actually works.
You doofus!
I have lost count of how many times I've said to you that the mass in the current gravity models is not compatible with the mass in the new model. I clearly said to you that the new concept is nonlinear.

Do you at all read what you respond to??? You are constantly trying to put your primitive linear concept of mass over the new model and then start whining like a little baby when it does not work.

OF COURSE THE NEWTONIAN GRAVITY MASS IS NOT COMPATIBLE WITH THE NEW CONCEPT.

Suddenly you act all surprised now? You should have understood this a long time ago but I can see that you were too busy with your fingers in your ears going "la-la-laa". You don't understand a word you read if its outside your existing belief system, do you..

I won't explain this to you anymore, if you still keep whining about the models not being compatible then I'll just leave you whining like a baby.
humble brother

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#11959
Jul 30, 2012
 

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polymath257 wrote:
No, the c+v is NOT the velocity of anything in any frame.
c+v is the relative speed from which you calculate how many wavefronts the observer supposedly encounters during a specific time period t.

All of you babbling above is pure nonsense. This is what you are doing:

1. From the beacon's frame you pick the relative speed v=(Vo + c) between the observer and the wavefronts. You pick this speed to calculate in the observer's frame even though the observer disagrees with this speed.

2. You then use this speed inside the observer's frame and using the observer's time you magically calculate a new frequency for the light in the observer's frame.

You just pick and choose measurements from different frames and mix them together in a soup.
polymath257 wrote:
Let's focus on the c+v. Is it the speed of light in any reference frame? Is it the speed of anything in any reference frame? What, exactly, does it represent?
c+v is exactly what would be used to calculate standard doppler effect for light dismissing the relative speed limits. In your case it is a combination of measured values that the observers do not agree on. Your case is a logical fallacy and you choose to ignore it.

“Think&Care”

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#11960
Jul 30, 2012
 
humble brother wrote:
<quoted text>
c+v is the relative speed from which you calculate how many wavefronts the observer supposedly encounters during a specific time period t.
All of you babbling above is pure nonsense. This is what you are doing:
1. From the beacon's frame you pick the relative speed v=(Vo + c) between the observer and the wavefronts. You pick this speed to calculate in the observer's frame even though the observer disagrees with this speed.
2. You then use this speed inside the observer's frame and using the observer's time you magically calculate a new frequency for the light in the observer's frame.
You just pick and choose measurements from different frames and mix them together in a soup.
<quoted text>
c+v is exactly what would be used to calculate standard doppler effect for light dismissing the relative speed limits. In your case it is a combination of measured values that the observers do not agree on. Your case is a logical fallacy and you choose to ignore it.
OK, let's try it another way. Suppose I send a light signal off to the right and the observer goes off to the left at .5*c in the beacon's frame. After one second, how far apart are the light signal and the observer in the beacon's frame?

“Think&Care”

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#11961
Jul 30, 2012
 
humble brother wrote:
<quoted text>
c+v is the relative speed from which you calculate how many wavefronts the observer supposedly encounters during a specific time period t.
All of you babbling above is pure nonsense. This is what you are doing:
1. From the beacon's frame you pick the relative speed v=(Vo + c) between the observer and the wavefronts. You pick this speed to calculate in the observer's frame even though the observer disagrees with this speed.
No, I am calculating in the beacon's frame. I am finding the number of wavefronts that pass the observer.
2. You then use this speed inside the observer's frame and using the observer's time you magically calculate a new frequency for the light in the observer's frame.
No, I use the number of wavefronts that pass the observer to calculate the frequency.
You just pick and choose measurements from different frames and mix them together in a soup.
You clearly don't understand the calculation and are not even trying to do so.
c+v is exactly what would be used to calculate standard doppler effect for light dismissing the relative speed limits.
Why is this used? Because it tells the distance between the wavefront and the observer so we can find how many wavefronts have passed the observer.
In your case it is a combination of measured values that the observers do not agree on. Your case is a logical fallacy and you choose to ignore it.
The number of wavefronts that pass the observer is not dependent on the frame.

At this point, we have two methods for finding the frequency in the observer's frame given the frequency in the beacon's frame:

1) Use the equation describing the light wave as a function of time and distance in the beacon's frame and use a Lorenz transform to find its description in the observer's frame.

2) Count wavefronts passing the observer in the beacons frame and take into account time dilation when figuring the frequency in the observer's frame.

Both methods give the same result. Both methods give results that agree with experiments. Both results keep the two frames separated and are based on a relativistic model.

YOU seem to be the only one that doesn't understand what is being said. Whether this is purposeful or not I cannot say, but clearly you do not want to learn what is wrong with your understanding of this argument.
humble brother

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#11962
Jul 30, 2012
 
polymath257 wrote:
OK, let's try it another way. Suppose I send a light signal off to the right and the observer goes off to the left at .5*c in the beacon's frame. After one second, how far apart are the light signal and the observer in the beacon's frame?
This is where the model says they disagree. In the beacon's frame the beam has made it one light second away and the ship a half light second to the other way.

In the ship's frame a lot less time has passed and that beam hasn't got nearly as far.

“Think&Care”

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#11963
Jul 30, 2012
 
humble brother wrote:
<quoted text>
This is where the model says they disagree. In the beacon's frame the beam has made it one light second away and the ship a half light second to the other way.
So how far apart are they in the beacon's frame?
In the ship's frame a lot less time has passed and that beam hasn't got nearly as far.
To be dealt with later...answer the above question first.

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