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The Theory of Invariance
12 years 11 months ago #21332
by Cindy
Replied by Cindy on topic Reply from
Thank LB,
For every one who is interested in neutrinos travel faster than light, then faster than light gets along with the theory of invariance:
h**p://theoryofinvariance.blogspot.com/
Or you can read it in youtube video: Youtube --> The theory of invariance
For every one who is interested in neutrinos travel faster than light, then faster than light gets along with the theory of invariance:
h**p://theoryofinvariance.blogspot.com/
Or you can read it in youtube video: Youtube --> The theory of invariance
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12 years 10 months ago #24253
by Cindy
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Comparison between Special Theory of Relativity and Theory of Invariance:
1. Total Energy:
Relativity: E = gammaE<font size="1">o</font id="size1">cc
Invariance: E = E<font size="1">o</font id="size1">cc.cosh(v/c)
2. Momentum:
Relativity: p = gamma m<font size="1">o</font id="size1"> v
Invariance: p = m<font size="1">o</font id="size1"> c sinh(v/c); <b>p = dE/dv</b>
3. Mass:
Relativity: Inertial mass equal gravitational mass: m<font size="1">i</font id="size1"> = m<font size="1">g</font id="size1">
Invariance: m<font size="1">i</font id="size1"> = m<font size="1">o</font id="size1"> sinh(v/c)/(v/c)
m<font size="1">g</font id="size1"> = m<font size="1">o</font id="size1"> cosh(v/c);<b> m<font size="1">g</font id="size1"> = dp/dv</b>
4. Doppler Effect:
Relativity: f = f<font size="1">o</font id="size1">sqrt(1+v/c)/sqrt(1-v/c)
Invariance: f = f<font size="1">o</font id="size1">e^(v/c)
5. Stella Aberation:
Relativity: tan x = gamma(v/c)
Ivariance: tan x = sinh(v/c)
6. Velocity:
Relativity: v < c
Invariance: v is unlimitted.
1. Total Energy:
Relativity: E = gammaE<font size="1">o</font id="size1">cc
Invariance: E = E<font size="1">o</font id="size1">cc.cosh(v/c)
2. Momentum:
Relativity: p = gamma m<font size="1">o</font id="size1"> v
Invariance: p = m<font size="1">o</font id="size1"> c sinh(v/c); <b>p = dE/dv</b>
3. Mass:
Relativity: Inertial mass equal gravitational mass: m<font size="1">i</font id="size1"> = m<font size="1">g</font id="size1">
Invariance: m<font size="1">i</font id="size1"> = m<font size="1">o</font id="size1"> sinh(v/c)/(v/c)
m<font size="1">g</font id="size1"> = m<font size="1">o</font id="size1"> cosh(v/c);<b> m<font size="1">g</font id="size1"> = dp/dv</b>
4. Doppler Effect:
Relativity: f = f<font size="1">o</font id="size1">sqrt(1+v/c)/sqrt(1-v/c)
Invariance: f = f<font size="1">o</font id="size1">e^(v/c)
5. Stella Aberation:
Relativity: tan x = gamma(v/c)
Ivariance: tan x = sinh(v/c)
6. Velocity:
Relativity: v < c
Invariance: v is unlimitted.
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12 years 9 months ago #21337
by Cindy
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To see what is wrong in STR theorically. Let's see what Albert Einstein wrote about simultaneity:
<b>IX. The Relativity of Simultaneity - Albert Einstein</b>
UP to now our considerations have been referred to a particular body of reference, which we have styled a “railway embankment.” We suppose a very long train travelling along the rails with the constant velocity v and in the direction indicated in Fig. 1. People travelling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises:
Are two events (e.g. the two strokes of lightning A and which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A --> B of the embankment. But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A --> B on the travelling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M' in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity).
<b>IX. The Relativity of Simultaneity - Albert Einstein</b>
UP to now our considerations have been referred to a particular body of reference, which we have styled a “railway embankment.” We suppose a very long train travelling along the rails with the constant velocity v and in the direction indicated in Fig. 1. People travelling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises:
Are two events (e.g. the two strokes of lightning A and which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A --> B of the embankment. But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A --> B on the travelling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M' in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity).
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12 years 9 months ago #13706
by Cindy
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Call Lynn, who is the observer sitting at M' on the train. Einstein say Lynn doesn't see flashes coming from A and B at the same time. Was Einstein wrong?
Let's look over the experiment one more time, we have:
1. Lynn is moving away from the light source A.
2. Lynn is moving closer to the light source B.
These are equal to:
1. The light source A is moving away from Lynn.
2. The light source B is moving closer to Lynn.
Thus, light sources A and B have the same role with two stars of a binary stars. <b>Base on what we observe binary stars</b>, we can strongly confirm that <b>flashes from A and B come to Lynn at the same time.</b>
Another word,
The Special Theory of Relativity does not get along with the second postulate.
The Theory of Invariance get along with the second postulate.
Let's look over the experiment one more time, we have:
1. Lynn is moving away from the light source A.
2. Lynn is moving closer to the light source B.
These are equal to:
1. The light source A is moving away from Lynn.
2. The light source B is moving closer to Lynn.
Thus, light sources A and B have the same role with two stars of a binary stars. <b>Base on what we observe binary stars</b>, we can strongly confirm that <b>flashes from A and B come to Lynn at the same time.</b>
Another word,
The Special Theory of Relativity does not get along with the second postulate.
The Theory of Invariance get along with the second postulate.
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11 years 6 months ago #13935
by Cindy
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Hi everyone,
The theory of invariance has been posted on vixra.org by the author, with a much clearer format.
Have fun,
The theory of invariance has been posted on vixra.org by the author, with a much clearer format.
Code:
h**p://***.vixra.org/pdf/1305.0064v1.pdf
Have fun,
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