[AtoMick-u235 (OP)]

I have been using my relativity program i made in 1997 and have some ideas. Do massless particles travel along with the actual speed of the expansion of the universe close to C, , , , Is the apparent rate of the expanding universe 44.7387 mile  per second per mega parsec, and does it work out at an actual rate of 99.999997 C

My twist on Einsteins formula shows 1) how massless particles like photons can travel at light speed
2) That the apparent expansion of the universe is speeding up because the actual speed is slowing down

here is a screen shot of my program, , , Note the actual speed of .99999997, or 186243 MPS and the apparent speed at the bottom of 45.62003 MPS


EINY-  WORKS.jpg (64.05 kB . 649x329 - viewed 1858 times)

[Halc]

Your topic title is still not worded as a question, but at least this got moved to new theories.

I have been using my relativity program i made in 1997 and have some ideas. Do massless particles travel along with the actual speed of the expansion of the universe close to C

Massless particles do indeed travel at c through a vacuum.
The expansion rate of the universe is not a velocity.  It has different units.

, , , , Is the apparent rate of the expanding universe 44.7387 mile  per second per mega parsec, and does it work out at an actual rate of 99.999999 C

Since it has different units, it does not work out to a speed.  Do the algebra.  It isn't hard.  It works out to the reciprocal of the age of the universe, or at least what the age would be if the rate was not accelerating.

here is a screen shot of my program

The screen shot is only of the output, not of the program being run.

Note the actual speed of .99999997, or 186243 MPS

186243 miles/sec is about .9998c
If you want .99999997, you need more significant digits for one thing.
I'm not used to doing this in non-metric units.

and the apparent speed at the bottom of 64.30355 MPS

If an object is going at .9998c in some frame, then it is apparently going .9998c in that frame. It will apparently take 10002 seconds to get to a place 10000 light seconds away, two seconds slower than light.  That's 'apparently' .9998c.
Maybe you mean something else by 'apparent speed', but you've not defined it.

TIME DILATION IS 2896.309

The relativistic change factor at .99999997c is about 4082.  At .9998 it is about 50.  Not sure where the 2896 is coming from.

It is really unclear what you are asking or what you are proposing.

[AtoMick-u235 (OP)]

Hi Halc , , , I have checked my time dilation program results and they are correct, i may have got the light speed of 186243 miles per second a little off, but you should know that you do not have to use this in the Lorentz transformation equation, , , ,   .999 works out at 22.36649 times dilation and .8 is 1.666667, I have also checked the apparent speed value, my program works fine

i believe i am correct in my observations that there is an actual and apparent effect, and that the expansion rate of the universe has an apparent value

Mick Bartram

[Halc]

Hi Halc , , , I have checked my time dilation program results and they are correct

Being a software person myself, there is always a possibility of another bug.

.999 works out at 22.36649 times dilation and .8 is 1.666667, my program works fine

And yet .99999997c should yield a dilation of 4082.  I suspect you are not using sufficient precision on your variables to handle a value of .99999997.  A 32 bit float is insufficient.  You need at least 64 bits to compute that one.

I have no idea what your program does.  You input a time of 50, but 50 what?  Seconds? Years?  Just 50 units?
Is it time one way, round trip, and as measured back home or as measured on the ship?  None of that was clear just by looking at the output.

OK, I suspect seconds because 9312149 is 50 times your stated light speed.  Distance and time then are I imagine measured by the non-ship frame, else that distance would not be meaningful.

i believe i am correct in my observations that there is an actual and apparent effect, and that the expansion rate of the universe has an apparent value.

I just don't know what you mean by 'apparent' speed.  I cannot comment if you're wrong or right about it if I don't know to what it refers.

From the stationary point of view, the ship goes out to a location 9312149 miles away and returns and comes back, 1 minute 40 seconds (plus a smidge in the 8th digit) elapsed for a trip of 100 light seconds.  Sounds like an apparent speed of 'damn near c', much faster than an apparent speed of 64 miles per second which would have resulted in an elapsed time of ~80 hours.
You probably mean something else by 'apparent speed', but I cannot figure out what it is.

[Halc]

Actually, the screen shot says T1 is ship time (units unspecified, but I still suspect seconds).
T2 would then be about 56.7 hours, so distance traveled by the ship in Earth frame is 38 trillion miles, much more than the 9 million the program reports.
Your formula there seems to be correct, but the program doesn't compute it.  It reports 144815 'units', but not sure what unit is here.  Seconds I think, which would have been correct if the dilation factor had been correctly computed.
If a ship moves almost 56.7 light hours in 56.7 hours, it has an apparent speed of almost c.

Distance traveled in the ship frame is zero.  The ship is stationary in its own frame for the 50 second duration of the exercise.  OK, the distance traveled by Earth in ship's frame is about 9.3 million miles.  Is that what you mean by that figure?  Earth has an apparent speed of almost c in that frame.

Your 'apparent speed' seems to be computed by dividing c by the relativistic factor, which makes no sense.  There is nothing in the example that is apparently going that speed in any meaningful frame.

[AtoMick-u235 (OP)]

Hi Halc, , Like I said, my program is accurate and works out the same on paper, I suggest you try it with .999 and .8

ha ha , , ,im sorry but what im saying seems so obvious to me, im more of a philosopher than a trained physicist so maybe this will help

Einstein said - If you are moving with respect to me, we shall not agree upon the rate of flow of time, your clock runs slower than mine when you move and relative to me your speed slows down, till at C you stop "ALL PROCESSES" that change with time "INCLUDING SPEED" change at a slower rate when observed in motion, one clock does not beat out time for the whole universe, a separate clock is needed for each state of motion, , , , I remembered this from a BBC OU episode on physics in 1985, , , BOOM !!

if you were a stationary observer on earth, and i got in a space ship on a round trip to the moon and back at a speed of 87% the speed of light there would be a 2x time dilation, it would take me 3 seconds, but when i returned to earth you would say ive been gone 6 seconds. FROM EARTHS VIEW APPARENTLY THE TRIP TOOK 6 SECONDS WHEN IN FACT FROM THE SHIPS VIEW IT ACTUALY TOOK 3 SECONDS, DIFFERENT TIMES and 
SPEEDS - QED, the apparent time would have to be inserted into the S=D\T to give the apparent speed

[Halc]

Hi Halc, , Like I said, my program is accurate and works out the same on paper, I suggest you try it with .999 and .8

Did you try it on paper with .99999997?
I agree, the thing works with the small figures, but you performed no limit test on your program.

if you were a stationary observer on earth, and i got in a space ship on a round trip to the moon and back at a speed of 87% the speed of light there would be a 2x time dilation, it would take me 3 seconds, but when i returned to earth you would say ive been gone 6 seconds.

2x dilation indeed. It takes light 2.6 seconds to make the round trip so a thing going at .87c would take about 3 seconds.  That's what .87c means.
3 seconds round trip as measured on earth and 1.5 seconds elapsed on the ship clock. From Earth frame, the ship is going .87c. From the ship's point of view, the ship is stationary and the moon and  the Earth move at .87c. There is no 'apparent speed' that is something else.
If I watch the ship clock from Earth, it will run at an 'apparent' rate of much slower than half pace, and it will appear to run faster than mine as I see the ship return.  But that apparent difference is due to redshift/Doppler effect, not due to dilation.  The final time will still be 3 seconds (me) and 1.5 seconds (ship).

FROM EARTHS VIEW APPARENTLY THE TRIP TOOK 6 SECONDS WHEN IN FACT FROM THE SHIPS VIEW IT ACTUALY TOOK 3 SECONDS, DIFFERENT TIMES

Putting it in caps doesn't make it right.  If it takes 6 seconds and it takes light only 2.6 seconds, it's going only about .43c.

[AtoMick-u235 (OP)]

Yes, spot on the apparent speed is .435 and the actual speed is .87, 2 x dilation
Your getting the idea but this is wrong "The final time will still be 3 seconds (me) and 1.5 seconds (ship)".it should be 6 and 3

I can understand that it all seems counter intuitive, but its just a basic Special Relativity concept

and Yes, the input T1 should be in seconds and the elapsed time on earth is 144815 units x 50 seconds or whatever you use for T1

I shall post you my QBASIC program TXT

[Bored chemist]

I can understand that it all seems counter intuitive, but its just a basic Special Relativity concept

I'm backing Halc's track record on this.
Incidentally, another "counter intuitive" thing is that computers get the maths wrong.
But they do.

Do you actually know what precision you are calculating to?

[Bored chemist]

I shall post you my QBASIC program TXT

It can't make things less clear...

[AtoMick-u235 (OP)]

Hey halc, Hmmmm i did it on paper, you may be right

.99999997
.99999994
.00000006
.000244948

1 / 000244948 = 4082.499 not my 2896.309

Arghhhhhhhhhhh, , , i will have to debug it , , ,Eeeeek !!  it could be my calculator tho, im gona try .99999

[Bored chemist]

It may not be a problem with your code.

Computers get maths wrong.

[Halc]

Hey halc, Hmmmm i did it on paper, you may be right

.99999997
.99999994
.00000006
.000244948

1 / 000244948 = 4082.499 not my 2896.309

Arghhhhhhhhhhh, , , i will have to debug it , , ,Eeeeek !!  it could be my calculator tho, im gona try .99999

I told you, it's just beyond the precision of the variable you're using to compute it.  Use 64 bit arithmetic, or don't use an example so close to light speed.

[Halc]

Yes, spot on the apparent speed is .435 and the actual speed is .87, 2 x dilation

So if I have two ships, one faster than the other, the faster one gets back to Earth after the slow one because it has a lower apparent speed?  That's what your math implies.

Your getting the idea but this is wrong "The final time will still be 3 seconds (me) and 1.5 seconds (ship)".it should be 6 and 3

I can understand that it all seems counter intuitive, but its just a basic Special Relativity concept

I'm not 'getting the idea'.  I've been doing relativity a long time.  I'm surprised you've had a relativity program for 22 years and only now are finding out it is doing this 'apparent speed' thing which is incorrect.  Speed is not a process and the ship isn't taking its speed along with it.  Speed is a relation and the relation is .87c in this latest example and not anything else.

The trip takes 3 seconds on and Earth clock and 1.5 on ship clock.  Using that math, the faster ship does the trip in less time, just like a faster ship should.

[AtoMick-u235 (OP)]

Hey guys, I tried it at .99999 , , , by hand I got time dilation 233.6068454 and my program gave me 233.4551, just .15 in it

Yes it may be the PC maths processor, its to late to debug my program now, ill try tmora
Ive got MS DOS QBASIC, and running Win XP on one of my laptops

And halc, in your last post I think your wrong, so apparently we will have to agree to disagree, , ,

[AtoMick-u235 (OP)]

10 CLS : COLOR 2: PRINT "TIME DILATION BY MICK BARTRAM / micktbart@yahoo.co.uk"
20 COLOR 15: PRINT "AN ASTRONAUT GOES OFF IN A ROCKET AT .X THE SPEED OFF LIGHT RELATIVE TO THE     EARTH, HOW MUCH TIME HAS ELAPSED ON EARTH."
40 PRINT : PRINT " BY USING EINSTEIN'S FOLLOWING EQUATION WE CAN WORK IT OUT."
42 PRINT : PRINT "                 T1   ": PRINT "              ========="
44 PRINT "            /      V2 ": PRINT "      T2= \/ 1-  ------   OR  SP=SP*SP:SP=1-SP:SP=SQR(SP):TD=T1/SP": PRINT "                   C2"
70 PRINT : PRINT "  T1 = TIME ELAPSED ON THE ROCKET, IE.  30 YEARS."
80 PRINT "  V  = SPEED OF ROCKET, IE.  .8 C, OR 148994 MPS."
85 PRINT "  T2 = TIME ELAPSED ON EARTH. ": PRINT
90 INPUT "ENTER T1 "; T1: INPUT "ENTER V"; SP: PER = SP
95 IF SP = 1 THEN 500
100 SP = SP * SP: SP = 1 - SP: SP = SQR(SP): TD = T1 / SP
105 MPS = 186243 * PER: MPH = MPS * 60 * 60: LOCATE 18, 20: PRINT "ACTUAL SPEED ="; MPS; "MPS. OR "; MPH; " MPH"
108 DIST = MPS * T1: PRINT : PRINT "DISTANCE OF SHIP = "; DIST; "MILES"
110 FOR T = 1 TO 5000: NEXT T
120 tms = TD / T1: PRINT "TIME DILATION IS "; tms; " TIMES T1 /"; " ELAPSED TIME ON EARTH ="; TD; "UNITS"
200 APP = DIST / TD: PRINT "VIEWED FROM EARTH, THE APPARENT SPEED OF ROCKET DUE TO TIME DILATION = "; APP; "MPS"
210 INPUT "Another go (Y\N)"; go$
220 IF go$ = "y" THEN GOTO 10 ELSE END

[AtoMick-u235 (OP)]

You said = So if I have two ships, one faster than the other, the faster one gets back to Earth after the slow one because it has a lower apparent speed?  That's what your math implies.

Yes halc this is correct, if the faster ship is going fast enough, it would loose every time , ,it obviously depends on how fast and far the slow ship is moving, and how long the round trip took for it, ,

[Halc]

Yes halc this is correct, if the faster ship is going fast enough, it would loose every time , ,it obviously depends on how fast and far the slow ship is moving, and how long the round trip took for it, ,

This should be a clue that your understanding is wrong.  Nothing in special or general relativity has faster objects taking longer to get somewhere than slower objects.
I can falsify your assertion:
If this 'apparent speed' thing worked the way you say, neutrinos from supernovas would not appear before the light from the explosion does.  They have such a high dilation factor (gamma if you will) that they'd have an 'apparent speed' slower than I can walk by your computations.  Instead, they get here so fast they actually outrun the light which is slowed by needing to travel through space that is not a vacuum and thus has a refractive index of greater than 1.

Fast things get there before slow things.  Relativity does not say otherwise.

[Halc]

90 INPUT "ENTER T1 "; T1: INPUT "ENTER V"; SP: PER = SP

SP (and your other variables) are taking the default type of SINGLE, which is your problem with the tms computation.
You need to implicitly declare it as DOUBLE by using SP# or explicitly with DIM SP AS DOUBLE
A single precision just cannot hold 8 digits of precision, so the really high-V computations come out wrong.

[AtoMick-u235 (OP)]

halk, it is you who is wrong, what im saying is obvious. and it is a shame you can't grasp it

OK halc, what if I travelled near a black hole and came home, the twin paradox, , , ,it would be the same effect, my ideas explain that too

In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket or travels close to a black hole, then returns home to find that the twin who remained on Earth has aged more.

Look what it means, the 2 observers  see 2 different times and speeds, , , QED - BOOM !!, , , ha ha, i give up explaining to you, my post was meant to verify my ideas, so i'll wait for someone who can, , , ,

OK thanks, , ,I will try to define SP# as a double, and look at the DIM function and let you know