1

Just Chat! / The illogicality of not voting for someone who isn't leading in the polls

« on: 29/10/2016 22:48:07 »

   It is illogical to vote based upon the way everybody else is voting.  Also, it simply becomes a self-fulfilling prophecy: My candidate can't win therefore I'm not going to vote for him/her, therefore he/she won't win.  I don't understand why anyone would vote for someone just because they're leading in the polls.  I voted for Jill Stein.

2

That CAN'T be true! / Is 1 second equal to 3 seconds or 1/3 seconds?

« on: 14/08/2016 00:28:40 »

Code: [Select]

[G]         [A---1 light-second-----B]
                   ------> 0.8c
George (designated by the G) is watching his friends Sue (location A) and Sally (location B) zoom away from him at 0.8c.  His friends are located at both ends of a spaceship 1 light-second long.  From George's point of view, the ship is only 0.6 light-seconds long.  Sue sends a light beam to Sally and it only takes 1 second to get from Sue to Sally.  According to George, because Sally is running away from the light, the trip actually takes 0.6/0.2 = 3 seconds.  So 1 second for Sue and Sally is equal to 3 seconds for George.  Sally decides to send a beam of light to her friend Sue.  Once again, it takes 1 second to get from Sally to Sue.  Because Sue is racing towards the beam of light, according to George the trip takes 0.6/1.8 = 1/3 seconds.  So 1 second for Sue and Sally is equal to 1/3 seconds for George.
So which is it?
Is 1 second for Sue and Sally equal to 3 seconds for George or 1/3 seconds for George?

3

That CAN'T be true! / Can somebody be in two places at the same time?

« on: 28/04/2016 06:36:09 »

Imagine the following situation:
We have two ships that are 5 light-years in length which are moving past each other at 0.8c.
Because of length contraction each ship will view the other ship as 3 light-years in length.
5*(1-o.8^2)^0.5 = 3
We will call the ships A and B.  We are only interested in two events.  The first event is
the front of ship B passing the tail of ship A.  The second event is the tail of ship B passing
the front of ship A.
Let us see what happens and take a look shall we?
I used F for front and T for tail.
First we will look from the reference frame where A is at rest.

Code: [Select]

                             T----------F  (ship A)    (5 light-years in length)
 <---(0.8c)                      F------T  (ship  B)   (3 light-years in length)
later on.....
                             T----------F 
 <---(0.8c)                  F------T
So we see that the tail of ship B passes the front of ship A,
and then later on the front of ship B passes the tail of ship A.
Now let's look from the reference frame where B is at rest.

Code: [Select]

                             T------F    ----->(0.8c)     (ship A)   (3 light-years in length)
                             F----------T                 (ship B)   (5 light-years in length)
later on...
                                 T------F    ------>(0.8c)
                             F----------T
So we see that the front of ship B passes the tail of ship A
and then later on the tail of ship B passes the front of ship A.
Now if you look carefully you will notice that the events
are in reverse order between reference frames!
Now imagine that there is some kind of structure extending up
from the tail of ship B and there is some kind of structure extending down
from the front of ship A.  The structure could be a pole, a wall, etc., whatever.
Now the structures are going to collide when the tail of ship B passes the front of
ship A.  Now the collision of the structures will be simultaneous between both
frames of reference since ship B hitting ship A and ship A hitting
ship B occur at the same time and place.  Now at that moment where is the
front of ship B?  The front of ship B has both already passed the tail of ship A
and has not yet reached the tail of ship A as one can easily see by looking at
the diagram above.  So if you believe that relativity is true, then you must
believe that the front of ship B can be in two places at the same time!
Maybe it's the tail of ship A that's in two places at the same time or
some combination of the two!

4

That CAN'T be true! / What is the way out of this conundrum?

« on: 16/03/2016 04:52:59 »

I have already posted an unresolvable paradox on special relativity here...
http://www.thenakedscientists.com/forum/index.php?topic=59598.0

Here's another situation which I believe illustrates the absurdity and insanity of special relativity.
We have three lights in space that are equidistant from each other, forming an equilateral triangle:

Code: [Select]

                                                                    B
                                                                   / \
                                                                  /   \
                                                                 /     \
                                                                /       \
                                                               C---------A
Imagine that there is an observer at the mid-point between A and B.  An observer is at the mid-point between B and C.
An observer is at the mid-point between C and A.  The observers are at rest with respect to the triangle of
lights.  The three lights flash simultaneously in the reference frame of these observers.  There is a long ship
moving at a high rate of speed from C to A.  There is a long ship moving at a high rate of speed from
B to C.  There is a long ship moving at a high rate of speed from A to B.  So we have three different reference
frames each with a different ship.  So, A flashes then C flashes, C flashes then B flashes, B flashes then A flashes.
Wait! Wait! what? what's going on? A flashes first followed by C flashing which is followed by B flashing which
is followed by A flashing. But A flashed first, it has already flashed.  How then can it flash last?
It seems the only resolution to this conundrum is to assume that for at least one of the
ships the flashes must occur in reverse order.  For example, for the ship that is moving from
C to A, according to him, the flash at B occurs before the flash at C, whereas for the ship moving
from B to C the flash at C occurs before the flash at B.  It seems we have an impossible situation.  Thoughts anyone?                                       

5

That CAN'T be true! / Is the speed of light constant for all observers?

« on: 05/08/2015 06:09:25 »

I am going to show what I believe is evidence that the speed of light isn't constant relative to all observers.  I obviously have no proof, just evidence from an argument.  But first I need to lay out a scenario... 
     Imagine that you are standing still and you are holding a stopwatch in your hand.  There are people in a line walking swiftly past you.  They are moving at a constant speed and are equally spaced apart from one another.  You start the stopwatch when one person reaches you and you stop it when the next person reaches you.  You note how long this took and you record this time as X.  You then start walking in the same direction as the people are walking past you, but you're not walking as fast as they are.  You do the same thing again...you start your stopwatch when a person passes you and you stop it when the next person passes you.
You note two things...
1) The people are passing you more slowly.  Relative to you their speed is decreased.
2) The time between people is greater than X.
Now you start walking towards the people and again you record the amount of time between people passing you.
Again two things are noted...
1) The people are passing you more quickly.  Relative to you their speed is increased.
2) The time between people is less than X.

All of this is simple and quite obvious.  Now we are going to replace you with the Earth and the people walking past you are going to be replaced by the instant of minimum brightness from an eclipsing binary star system.  If we find that the time between the instants of minimum brightness is greater when we are moving away from the star system and less when we are moving towards the star system, the logical conclusion we should come to is that the light is passing us more slowly when we are moving away from the stars and passing us more quickly when we are moving towards the stars.
http://calgary.rasc.ca/algol_minima.htm

In May and November, the Earth is moving at "right angles" to the line to Algol. During this time we see minima happening regularly at their 2.867321 day intervals. However, during August, the Earth is rapidly moving towards Algol at about 107,229 km/hr as explained on my How Fast Are We Moving? page. (The Earth moves approximately 202 times its own size in one day.) So in 2.867321 days the Earth moves about 7,379,039 km closer to Algol. But the varying light from Algol doesn't know this(emphasis added) - its light waves left Algol 93 years ago and are travelling at a constant speed. The result - we "catch a bunch of minima early" during August as shown on Chart 2. Exactly the opposite happens during February - the Earth is moving away from Algol that fast and it takes longer for the group of minima to reach us so we see them taking longer between events. How long? 7,379,039 km divided by the speed of light 299,792.458 km/sec is 24.61382 seconds - this rough calculation explains the deviations we see in Graph 2. So in May and November when we are not moving towards or away from Algol - the period seems constant. It is our rapid movement towards or away from the events in August and February that causes the timing differences.

When we are 'running away'... the time between minimums is 2.8675875347 days.
When we are 'running towards'... the time between minimums is 2.8670608912 days.
When we are 'standing still'... the time between minimums is  2.867321 days.

I realize the analogy isn't exact because the analogy is in a line and the actual Earth and star system is in 3D space.  However, if the binary star system were in the same plane as the Earth's orbit around the Sun I think we should expect similar results.

6

That CAN'T be true! / Special Relativity falsified.

« on: 26/07/2015 11:52:37 »

[MOD EDIT: if you have been led here by a link claiming this is an unresolvable paradox, you have been misled.
                  The truth is that the original poster is unable or unwilling to understand the answers given.]


I have read that there are no 'real' paradoxes in special relativity.
However, I disagree and I believe the following scenario
poses an unresolvable paradox.  The problem concerns two balls
than are connected by a chain and they are some distance apart
from each other and they are each blocking a light source.

So here's the situation:

Code: [Select]

O2(0.99999999999999999999c)-------->
  L1 B1--------O1-------B2 L2
     M1                 M2
L1 and L2 are light sources such as a lightbulbs or perhaps a small hole in a wall with light coming out.
B1 and B2 are lead balls blocking the light sources.
B1 and B2 are separated by a distance of 10 meters.
M1 and M2 are machines holding the lead balls in place and they are underneath balls B1 and B2 respectively, ready to pull them down.
There is an observer called O1 who is located at the mid-point of the
distance between the light sources L1 and L2.
The lead balls are connected by a chain which runs between them (that's the dashed line) and
has no slack, or perhaps just a little bit of wiggle room.
At a certain time, the machines pull down the lead balls simultaneously in the reference
frame of observer O1 and the light from both sources reach observer O1 simultaneously.
The chain remains intact because the distance between the lead balls doesn't change.

The is another oberver called O2 which is moving at 0.99999999999999999999c  relative to observer O1.
Gamma is 1/(1-v^2/c^2)^.5 = 7,071,067,811.865475244
The time between events for oberver O2 is given by (gamma times the velocity times the distance)/c^2.
So we have (7,071,067,811.865475244*(299,792,458 meters/second)*(10 meters))/(299,792,458 meters/second)^2.
Or (7,071,067,811.865475244* 10 meters)/(299,792,458 meters/second.) = 235.8654 seconds.
The time between events is about 235.8654 seconds or about 236 seconds.
So for oberver O2 ball B2 is pulled down about 4 minutes before ball B1 and the chain breaks.
So we have a paradox, for observer O1 the chain is intact and for observer O2 the chain is broken.
Both situations can't simultaneously be true.

Length contraction doesn't resolve the paradox.
I posted this on another message board and the response was that the chain
for observer O1 would break because the chain was pulled tight between the balls.
This is obviously silly nonsense.  If I have a chain that's pulled tight between two
balls and the balls move simultaneously, there's no reason for the chain to break.
There could be a little bit of slack in the chain and the paradox still stands.
I honestly believe that this paradox can't be resolved.