
010611, 10:20 PM #1001
Here's what Motor Daddy's equations look like for the 3D problem from two pages back:
t(x) = 1.201 seconds
v(x) = (ct  l) / t
v(x) = (1.201  1.000) / 1.201 = 0.167c
t(y) = 1.336 seconds
v(y) = (ct  l) / t
v(y) = (1.336  1.000) / 1.336 = 0.251c
t(z) = 1.482 seconds
v(z) = (ct  l) / t
v(z) = (1.482  1.000) / 1.482 = 0.325c
And once again, the problem is that the numbers should be:
v(x) = 0.100c
v(y) = 0.200c
v(z) = 0.300c
which shows that the accuracy of Motor Daddy's equations goes down even more for the 3D problems.

010611, 10:22 PM #1002

010611, 10:33 PM #1003
...and it looks like in the 3D numbers that the greater the speed the greater the accuracy. My numbers are correct by the definition of distance and time!

010611, 10:35 PM #1004
No, I'm not comparing anything to Einstein. I am approaching this as if there is an absolute rest frame where light must travel at c in that frame. The reason your equations lose accuracy is because your equations were derived from a onedimensional case. We never derived any multidimensional equations, we just started using your equation "asis" in the multidimensional cases.
What you are doing is applying your equations seperately to x, seperately to y, and seperately to z. At no time do your equations acknowledge that x and y affect z, y and z affect x, and x and z affect y. My equations do include that effect, but it makes them much more complicated to use.

010611, 10:40 PM #1005

010611, 10:49 PM #1006
Here's a 2D example that might help show what's going on. If I told you this time:
t(x) = 1.155 seconds
You would automatically assume there was motion along the xaxis, and you would do this:
t(x) = 1.155 seconds
v(x) = (ct  l) / t
v(x) = (1.155  1.000) / 1.155 = 0.13c
But it turns out that there was no xcomponent to the velocity at all. The reason the light took 1.155 seconds to travel the xside of the square was actually because of the ycomponent of the speed. The real velocity was like this:
v(x) = 0.00c
v(y) = 0.50c
The fact that the square was moving up the yaxis made it take longer for the light to travel the xside of the square. If you saw this from the absolute frame, you would see the square moving up, and the light ray would appear as a long diagonal line.

010611, 10:54 PM #1007

010611, 11:11 PM #1008

010611, 11:36 PM #1009
I think the bigger problem is that many of the explanations are themselves confused and conflicting. Compounding the matter is the miscommunications, so that a great deal of time is spent arguing against a position that no one holds. Then more time is spent trying to correct the miscommunication, and the problem worsens.
It's now well beyond the point of no return. Some people in this thread will learn from the process of developing their own arguments, but I doubt that anyone is learning anything from others.

010711, 07:28 AM #1010

010711, 07:40 AM #1011
Right, so say the primary corner of the box is at (0,0,0), and the length of the box is 1 light second long along all sides.
If the box moves in the y direction only for 1 second at .5c, the primary corner is .5 light seconds from (0,0,0) in the y direction, and still 0 in the x direction. From that you know where all the other corners are because you know the size of the box. So the primary corner is at (0,.5,0), the y corner is (0,1.5,0), and the x corner is at (1.0,.5,0), right?
H = sqrt(1^2 + .5^2)
H = sqrt(1 + .25)
H = sqrt(1.25)
H = 1.1180339887498948482045868343656
So the x corner is 1.12 light seconds from the origin of the primary corner.
Assuming that is correct, I assume the z corner is originally at (0,0,1), now at (0,.5,1) and the distance from the origin of the primary corner is:
H = sqrt(1^2 + .5^2)
H = sqrt(1 + .25)
H = sqrt(1.25)
H = 1.1180339887498948482045868343656
So the x and z corner are the same distance from the primary corner's origin after 1 second of motion.
The x and z corners are:
H = sqrt(1^2 + 1^2)
H = sqrt(1 + 1)
H = sqrt(2)
H = 1.4142135623730950488016887242097
1.41 light seconds apart from each other in the box frame.
After 2 seconds of motion, the x corner is at (1,1,0) and the z corner is at (0,1,1). That means the x and the z corners are each
H = sqrt(1^2 + 1^2)
H = sqrt(1 + 1)
H = sqrt(2)
H = 1.4142135623730950488016887242097
1.41 light seconds from the primary corner's origin.
So at exactly 2 seconds of motion at the velocity of .5c, the x and z corners are the same distance from the origin as they are from each other in the box frame.
If the box were to continue to travel, the x and z corner's distances from the primary corner's origin would be increasing at the same rate as each other, hence the distances would remain equal to each other at all times, and those distances would be greater than the distance between the two corners in the box frame.
So after 10 seconds of travel, the x corner is at (1,5,0) and the z corner is at (0,5,1), so the two corners are:
H = sqrt(1^2 + 5^2)
H = sqrt(1 + 25)
H = sqrt(26)
H = 5.0990195135927848300282241090228
5.1 light seconds away from the primary corner's origin.
An observer at the primary corner's origin would see the distance between himself and the x and z corners as increasing. It would appear to the observer that the distance between the two corners of the box was decreasing, in effect making it appear to him that the object is shrinking, but we know that isn't true, right? It's an illusion created by fact that the angle between the lines drawn from the primary origin to the corners is decreasing due to the fact that the object is getting further away from the observer.
Hence, the moon appears smaller to an observer on the Earth than it does for an observer on a ship closer to the moon.Last edited by Motor Daddy; 010711 at 11:03 AM.

010711, 08:27 AM #1012
If Einstein put numbers into that thought experiment, and if he knew the speed of light that precisely, then he would say that the train observer measures the light to take 0.000000039999 seconds to travel the length of the train car 11.915 meters long.
Absolute simultaneity!!!
No, in Einstein's model it is not.
Experiments show that Einstein's model is closer to reality than your model.Last edited by Pete; 010711 at 08:34 AM.

010711, 08:50 AM #1013

010711, 11:19 AM #1014
 Posts
 3,742
And back we go to page 1.

010711, 12:36 PM #1015
 Posts
 211
Your understanding of absolute space and time is fine, but you are at a disadvantage because you didn't move on to learn other factors that are needed for this type of analysis.
There are velocities, which are just speeds with a specific direction, like the Neddy Bate examples.
You could imagine these as arrows with specific lengths. For n different length arrows, you place them end to end, with each pointed in its proper direction. The resulting motion is a new arrow that begins at the free end of arrow 1 and ends at the free end of arrow n. For a material object, this implies that the object moves along all arrows in the same amount of time. It's here we make an exception for light. Light having one speed requires different times to travel different length arrows.
There is time dilation, which shouldn't be difficult for you since you accept constant light speed.
Imagine a light clock that sends a signal vertically to a mirror that reflects it back to a detector that flashes, representing one time tick. If the clock moves horizontally at some speed v, you say the light would travel farther to complete its trip. Wouldn't that mean the interval between flashes gets longer?
A good tutor could refine you math abilities and eliminate the excessive number crunching you perform. It's confusing sorting through all those numbers.
I have no doubt you can do it if you want to.

010711, 01:44 PM #1016

010711, 04:49 PM #1017
Neddy Bate,
I'm still racking my brain on the 3D stuff.
Let's use your example:
The problem gives us three times. The first time is 1.201 seconds, so the PRIMARY corner of the cube would be located at (0.120, 0.240, 0.360) as you can see from these equations:
x = 1.201(0.1) = 0.120
y = 1.201(0.2) = 0.240
z = 1.201(0.3) = 0.360
And from that, it follows that the corner which used to be at (1,0,0) is now located at (1.120, 0.240, 0.360) because all I had to do was add one to the xcoordinate.
If I want to know exactly how far away that corner is from the origin, I can use this equation:
d = sqrt(1.120^2 + 0.240^2 + 0.360^2)
d = 1.201
And that happens to be how far the light sphere is from the origin, because the time is (1.201 seconds) and the light sphere expands at one lightsecond per second. So we know the light is just now reaching that corner of the cube.
I know the length of the box is 1 light second from the primary corner to the x corner. I know the time you give is 1.201, so I know that distance from the x corner to the origin of the primary corner. I can find the distance of the other side of that triangle by using A^2+B^2=C^2, so I know all three distances of that triangle, plus all the sides of the other two triangles in question. Is there any way I can know the accurate positions of the corners by knowing all the sides of all the triangles at the given times?Last edited by Motor Daddy; 010711 at 04:59 PM.

010711, 06:07 PM #1018
In 3D problems, you should use the 3D version of A^2+B^2=C^2. That equation looks like this:
d^2 = x^2 + y^2 + z^2
So, for example, if the time is 1.201 seconds, and the primary corner of the cube is located at (0.120, 0.240, 0.360), then we know that the corner which used to be at (1,0,0) is now located at (1.120, 0.240, 0.360).
The distance from that corner to the origin (0,0,0) is:
d = sqrt(1.120^2 + 0.240^2 + 0.360^2)
d = 1.201
I don't think there is any way to make your equation work for 3D, but it's good that you are trying to find a way. The only thing I can do right now is start with an assumed velocity with known x, y, and z components, and then find the light travel times from that. I don't know how to start with light travel times and then calculate the velocity, (although "trial and error" should work pretty well).Last edited by Neddy Bate; 010711 at 06:17 PM.

010811, 05:24 AM #1019
If we assume the track is stationary, then we can conclude that the strikes occurred simultaneously.
If we assume the strikes occurred simultaneously, then we can conclude that the track is stationary.
Neither of which relates directly to the train observer's measurement of time.
You clearly don't understand Einstein's model.Last edited by Pete; 010811 at 05:35 AM.

010811, 07:26 AM #1020
 Posts
 211
MD;
Here's what you are leaving out in post 970
skate length d = 30cm
c = 3*10^8 m/s
light travel time t0 = d/c = 1/(10)^9 sec
front skate clock reads .1 sec
with absolute simultaneity, this is 10^8 times as long as t0,
an extremely high gamma,
thus skate speed is extremely fast!
t1 is front clock reading which is dilated (slow)
absolute time t = g(t1) = t0/(1v)
g=gamma for v, v is speed as fraction of c
using g(t1) = t0/(1v) results in
v=[(t1)^2(t0)^2]/[(t1)^2+(t0)^2]
v = [(10^161)/(10^16+1)]c
a mosquito whisker short of c!Last edited by phyti; 010811 at 07:28 AM. Reason: correct equation
Similar Threads

By Paul W. Dixon in forum Astronomy, Exobiology, & CosmologyLast Post: 123010, 10:07 AMReplies: 1953

By Scaramouche in forum Physics & MathLast Post: 011810, 05:37 PMReplies: 69

By Scaramouche in forum Physics & MathLast Post: 010510, 07:19 AMReplies: 33

By dkane75 in forum Science & SocietyLast Post: 010608, 03:57 AMReplies: 34

By noahfor in forum General PhilosophyLast Post: 040606, 02:17 PMReplies: 9
Bookmarks