01-04-11, 04:37 PM #901Originally Posted by Motor Daddy
01-04-11, 04:40 PM #902
01-04-11, 04:51 PM #903
Umm, what? That's word salad bro. There is no distance change between the top and bottom of the building. Please take a moment to read the link I provided. Light changes color when moving toward or away from a gravity source. This contradicts your notion that "as long as the distance between the objects doesn't change, the frequency doesn't either."
01-04-11, 05:02 PM #904
OK here is your equation. I'm going to rearrange it to solve for v:
v = (ct-L)/t
v = (ct/t)-(L/t)
v = (c)-(L/t)
v-c = -(L/t)
c-v = L/t
(c-v)/L = 1/t
L/(c-v) = t
t = L/(c-v)
Now, let the length of the skate be 30 centimeters. That is how far light travels in one nanosecond, so I will use nanoseconds for my unit of time. Let the velocity of the skate be 0.9c (inline with the clocks at each end of the skate). Here is what your equation tells me the elapsed time should be:
t = L/(c-v)
c = 1c
L = 1 light.nanosecond
v = 0.9c
t = 1/(1-.9)
t = 1/(.1)
t = 10 nanoseconds
Alright, so you are telling me the the time it takes for light to travel the length of the skate must be 10 nanoseconds, because its velocity is 0.9c "inline with the clocks". But let's say the train was also moving forward at 0.9c. So, during the time t=10 nanoseconds, the train moved this far forward:
And the skate also moved this far sideways to the direction of the train:
So now I can draw a right triangle with one leg 9 light.nanoseconds long, and the other leg 9 light.nanoseconds long, which makes the hypotenuse this long:
H = sqrt(9^2 + 9^2)
H = sqrt(81 + 81)
H = sqrt(162)
H = 12.7 light.nanoseconds
So, according to you, the light signal that runs up the skate must move at least this fast in the absolute reference frame:
c ?= (12.7 light.nanoseconds) / (10 nanoseconds)
c ?= 1.27c
Are you happy with your theory now? You've got light traveling at over 1.27c through your absolute reference frame. Give it up, big guy.
01-04-11, 05:10 PM #905
You do not calculate the time, you measure it.
What time did you measure from one end of the skate to the other? What is the distance between clocks on the train and what time did you measure?
You DO NOT CALCULATE TIME, you measure it! I gave you a equation and you screw it up. Tell me the times and I will do it for you, since you don;t understand it.
01-04-11, 05:13 PM #906
01-04-11, 06:02 PM #907
Originally Posted by Motor Daddy
You have a serious misconception of what measurement is.
Of course you calculate something when you measure it. Try this thought experiment:
You look at a clock at t1. Then you
a) make a cup of coffee
b) turn on the TV and watch Oprah
c) wander around aimlessly, pulling the odd hair out
After you think an interval of time has passed, you again look at the clock, at the time t2. Now you can calculate the interval which is the formula t2 -t1.
You also seem to think that once a physical constant is defined--but how is it--you can't measure it or you shouldn't, for some reason you can't divulge. This doesn't corroborate well with ongoing experiments to refine the measurements (using calculations) of physical constants such as c. Does it really stay a constant? Things like that can only be answered by measuring.
01-04-11, 06:08 PM #908
That doesn't eliminate the case requiring more time in one direction than another (as seen from outside), but as long as the velocities are the same there is no doppler shift. This is equivalent to observing an inertial frame, i.e., everything stays in one position.
01-04-11, 06:22 PM #909
If the map disagrees with the ground, the map is wrong.
If a thought experiment result disagrees with a real experiment result, the thought experiment is wrong.
People believe what they want to believe.
...they also helped me with the equations (you and Neddy Bate) for my idea, in effect strengthening my case. These are now available for everyone, and I challenge anyone to try and prove them wrong. Cheers!
You might want to look away from your navel for a moment to ponder the results of actual experiments.
01-04-11, 06:30 PM #910
Picture the box from the isolated 'elevator' experiment, and turn it on its side. The box is not moving, but the light does not travel in a straight line inside the box.
Do you see any similarity?
Last edited by phyti; 01-04-11 at 06:40 PM. Reason: clarification
01-04-11, 08:58 PM #911
The skate is 30 centimeters long. It takes 5 nanoseconds for light to travel from one end of the skate to the other end.
The absolute velocity of the skate is 239,792,458 m/s.
The train clocks are spaced 50 centimeters apart. It takes 6 nanoseconds for light to travel from one clock to the other.
The absolute velocity of the train is 216,459,124.667 m/s.
In 6 nanoseconds, the skate traveled 1.438754748 meters and the train traveled 1.298754748002 meters.
In the train frame, the skate traveled 0.139999999998 meters in 6 nanoseconds, so in the train frame the velocity of the skate is 23,333,333.333 m/s.
I checked the tracks to see if they were a zero velocity or not. They weren't, (go figure). I placed the clocks 90 centimeters apart and light took 1⁄299,792,458 of a second to travel from one clock to the other, so the absolute velocity of the tracks is 29,975,245.8 m/s. What dumb luck, eh? The tracks aren't a zero velocity. I hate it when that happens.
So in the track frame, the train is traveling 186,483,878.867 m/s, and the skate is traveling 209,817,212.2 m/s.
Do you agree with those numbers?
Last edited by Motor Daddy; 01-04-11 at 10:09 PM.
01-05-11, 08:35 AM #912
Let's look at Einstein's train thought experiment in Chapter 9. The Relativity of Simultaneity. Einstein, Albert. 1920. Relativity: The Special and General Theory.
Einstein conveniently forgot to put numbers to the thought experiment, so let's do it for him, shall we?
The observer on the train measures the time it takes light to go from the rear of the train car to the front of the train car, which is 11.9915 meters in length in the train frame. Light takes .00000004 seconds to travel the length of the train. That means the absolute velocity of the train is 4,958 m/s.
The observer on the tracks measures the time it takes light to travel the distance between two clocks on the track, which is 1 meter. It takes light .0000000033356409519815204957557671447492 seconds to travel the distance, which means the track has an absolute zero velocity.
It is 10 meters from A to B on the train in the train frame, and 10 meters from A to B on the embankment in the embankment frame. Both observers are at the midpoint between A and B in their respective frames.
Lightening strikes A and B as the two points on the train coincide with the two points on the embankment.
Light takes .000000016678204759907602478778835723746 seconds for each light from A and B to strike the embankment observer. The embankment observer was struck simultaneously from each light at precisely .000000016678204759907602478778835723746 seconds after 12:00:00. That means the strikes occurred at A and B at exactly 12:00:00.
It takes .00000001667792893852027063502108370407 seconds for light to travel from B on the train to the train observer at the midpoint. It takes .000000016678480590418212900804736688488 seconds for light to travel from A on the train to the midpoint observer on the train.
So, the train observer had the light from B impact him .00000000000055165189794226578365298441767877 seconds before the light from A impacted him.
Since the light from B impacted the train observer .00000001667792893852027063502108370407 seconds after 12:00:00 and it took light .00000001667792893852027063502108370407 seconds to travel from B to his midpoint position, the train observer concludes the strike occurred at B at exactly 12:00:00. Since the light from A impacted the train observer .000000016678480590418212900804736688488 seconds after 12:00:00 and it took light .000000016678480590418212900804736688488 seconds to travel from A to his midpoint position, the train observer concludes the strike occurred at A at exactly 12:00:00.
So both observers acknowledge that the strikes occurred at exactly 12:00:00 at A and B. The embankment observer had both lights hit him simultaneously, and the train observer had the lights hit him at different times due to his 4,958 m/s velocity.
Last edited by Motor Daddy; 01-05-11 at 09:06 AM.
01-05-11, 11:46 AM #913
Just a quick recap of the time line for those of you who missed it.
At exactly 12:00:00 lightening struck A and B on the train and the embankment.
At .00000001667792893852027063502108370407 seconds after 12:00:00 the light from B on the train impacted the train observer.
At .000000016678204759907602478778835723746 seconds after 12:00:00 the light from A and B on the embankment struck the embankment observer simultaneously.
At .000000016678480590418212900804736688488 seconds after 12:00:00 the light from A on the train impacted the train observer.
Just want to reiterate those facts, in case you missed it.
01-05-11, 02:08 PM #914
JamesR post 880
Einstein, on the other hand, gets the numbers right without any mucking around with correction factors.
Using a one dimensional example:
In the chosen/fixed/rest/static frame S, an emitter is a distance d from a mirror. The emitter-mirror setup is launched into space and passes earth at speed v. A light signal travels to the mirror and returns.
A spacetime drawing will help visualize.
The S-time is
t=2ggd/c (g=gamma for the speed v).
Applying time dilation, the A-time is
1.(note MMX calculated x-time would equal g*y-time)
A light measurement of d in the A frame (after clock synch) requires dividing the round trip A-time by two and multiplying by c.
2.(the distance is determined by the elapsed time, which by simultaneity definition is light transit time for equal path lengths)
This results in
(d'>d by definition in 2)
Dividing distance by time
Light speed is c in the A frame, but only by definition via mathematical contrivance.
Because d' for A is greater than d for S, the Lorentz transformation requires a reduction in the A measurement, i.e., d'/g=d. Contrary to popular interpretation, the distance did not contract, the measurement was too long!
The distance between A and the event, as measured in S is (x-vt). The LT states that for the moving frame A, this distance is x'=g(x-vt), which is greater than the distance in S.
The author of special relativity did in fact refer to the Lorentz transformations as "transformation of coordinates".
01-05-11, 04:21 PM #915
v = (ct-l)/t
v = (5-1)/5
v = 0.80c
Yes, that is mathematically correct. HOWEVER, I thought I demonstrated quite clearly that using this equation leads to problems in three dimensional space. All light signals must move at c through the absolute frame. Oh well, let's keep going and see what happens.
v = (ct-l)/t
v = (6-1.666)/6
v = 0.72c
OK so far, but I still don't think this is going to come out right according to the absolute frame.
d = vt
d = 0.8*6
d = 4.80 light.nanoseconds
D = Vt
D = 0.72*6
D = 4.32 light.nanoseconds
Okay mathematically, but there are still going to be problems when you look at the length of the light signals in the absolute frame. Let's look at the light signal that traveled the length of the skate. That signal traveled for 5 nanoseconds, so we have this:
d = (vt) + L
d = (0.8*5) + 1.00
d = 5.00 light.nanoseconds
D = Vt
D = 0.72*5
D = 3.60 light.nanoseconds
Now draw a right triangle with one leg equal to 5.00 and the other leg equal to 3.60. The hypotenuse of that triangle represents the length of the light path that traveled down the skate, (as viewed from the absolute rest frame).
H = sqrt(5.00^2 + 3.60^2)
H = sqrt(25.00 + 12.96)
H = sqrt(37.96)
H = 6.16 light.nanoseconds
So your light signal is traveling a distance of 6.16 light.nanoseconds in a time of 5.00 nanoseconds. Your speed of light is not c, it is:
c ?= 6.16/5.00
c ?= 1.23c
Don't you understand that space is three dimensional????? Your equations must incorporate all three dimensions if they are to ensure that the speed of all light signals in the absolute frame travel at exactly 1.00c. I already demonstrated this to you, but you just ignored it and used the same equations again. Did you think they would fix themselves? Are you waiting for me to fix them for you? It's your theory, don't you want to work on it?
01-05-11, 04:45 PM #916
Motor Daddy: I don't know if you've interpreted this correctly or not, but I would RATHER the world was the one you are describing. It makes more sense. The whole reason I started this thread is because it's difficult to accept that c is invariant as measured from any frame. In fact it still causes me some consternation.
Can you explain why, in a universe with a preferred frame and a truly absolute value for c (which is basically what you've described), gravity might cause red and blue shifting?
Originally Posted by Motor Daddy
01-05-11, 04:52 PM #917
This may sound unbelievable, but I think Motor Daddy's universe is actually more complicated than relativity! He's got light doing all sorts of odd things because of the "absolute velocities" of different reference frames. I mean, the earth is a ROTATING frame, for gosh sakes. The speed of light would vary by time of day, as well as seasonally, etc. Trying to get any physics done would be a nightmare. It would be chaos! lol
01-05-11, 05:53 PM #918
Agreed, Neddy Bate. It may be practically complex depending on whatever our absolute motion turned out to be but it would certainly be theoretically easier to stomach. Galilean velocity addition, fixed and absolute distances, etc...
Of course, as I pointed out earlier it also leads to time reversal, even when moving at speeds < c.
01-05-11, 06:21 PM #919
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