SR is that nothing can move faster than light

[nieuwenhuizen]

Can somebody explain my error to me?

1. The base of SR is that nothing can move faster than light, c + v == c,
c - v = c

2 The next step many authors do is proving non-existance of simultaneity by
on observer at the platform versus one in a fast train. Flash from the
front, to be received at the tail, which is. they say "racing toward the
rays". Conclusion: Speed of approach is $c+v$ so that
$$ ( c + v ) \cdot \Delta t = L $$
Combination with a forward flash [ c - v ] leads to

$$ \frac{1}{c+v} + \frac{1}{1-v} = \frac{1}{c^2 - v^2 } $$

Why does this not contradict the base - statement 1 ?

Thanks to the one that does.

Nieuwenhuizen, J.K.
2009-02-16T15:36

 

[nieuwenhuizen]

Can somebody explain my error to me?

1. The base of SR is that nothing can move faster than light, c + v == c,
c - v = c

2 The next step many authors do is proving non-existance of simultaneity by
on observer at the platform versus one in a fast train. Flash from the
front, to be received at the tail, which is. they say "racing toward the
rays". Conclusion: Speed of approach is $c+v$ so that
$$ ( c + v ) \cdot \Delta t = L $$
Combination with a forward flash [ c - v ] leads to

$$ \frac{1}{c+v} + \frac{1}{1-v} = \frac{1}{c^2 - v^2 } $$

Why does this not contradict the base - statement 1 ?

Thanks to the one that does.

Nieuwenhuizen, J.K.
2009-02-16T15:36

I am new to this and this is only to find how a reaction is handled. Sorry

 

[ZikZak]

Can somebody explain my error to me?

1. The base of SR is that nothing can move faster than light, c + v == c,
c - v = c

2 The next step many authors do is proving non-existance of simultaneity by
on observer at the platform versus one in a fast train. Flash from the
front, to be received at the tail, which is. they say "racing toward the
rays". Conclusion: Speed of approach is $c+v$ so that
$$ ( c + v ) \cdot \Delta t = L $$
Combination with a forward flash [ c - v ] leads to

$$ \frac{1}{c+v} + \frac{1}{1-v} = \frac{1}{c^2 - v^2 } $$

Why does this not contradict the base - statement 1 ?

Thanks to the one that does.

Nieuwenhuizen, J.K.
2009-02-16T15:36

No body can travel at greater than c, but there's no problem with the distance between two moving bodies closing at more than c. Two bodies approaching each other, each at nearly c in the observer's frame, will close distance at nearly 2c. Of course, in the frame of reference of one of the bodies, the distance is closing at nearly c, not nearly 2c, because all of the motion is in the other body, which cannot travel faster than c.

 

[bernhard.rothenstein]

Can somebody explain my error to me?

1. The base of SR is that nothing can move faster than light, c + v == c,
c - v = c

2 The next step many authors do is proving non-existance of simultaneity by
on observer at the platform versus one in a fast train. Flash from the
front, to be received at the tail, which is. they say "racing toward the
rays". Conclusion: Speed of approach is $c+v$ so that
$$ ( c + v ) \cdot \Delta t = L $$
Combination with a forward flash [ c - v ] leads to

$$ \frac{1}{c+v} + \frac{1}{1-v} = \frac{1}{c^2 - v^2 } $$

Why does this not contradict the base - statement 1 ?

Thanks to the one that does.

Nieuwenhuizen, J.K.
2009-02-16T15:36

Consider please the following experiment performed in an inertial reference frame in the limits of Newton's mechanics. A source of light S located at the origin O emits successive light signals in the positive direction of the x-axis at constant time intervals t(e). The light signals
illuminate a target that moves with speed u in the positive direction of the x axis. When the first signal is emitted the target is located in front of the source. We impose the condition that the second emitted signal illuminates the target at a time t(r) i.e.
c[t(r)-t(e)]=ut(r).
where from
t(r)=ct(e)/[c-u]
Special relativity is not involved so far. Is c-u the result of a classical sddition law of velocities? That is the way in which the classical Doppler Effect formula is derived.