On Einstein's explanation of the invariance of c

Ok, so tell me what the skater inside the skate box saw. How far did the light travel, in how much time? Please don't use too many decimal places, it gets me dizzy!

 

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I need to use so many decimal places to be accurate, sorry.

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Inside the skate box.

v = (ct-l)/t

The distance between the clocks is .21198528000038323887394410859085 meters. It takes light .000000001 seconds to travel from one clock to the other clock.

The skate box is traveling at an absolute velocity of 87,807,177.99961676112605589141 m/s.

The observer taking the measurements sees that light traveled .21198528000038323887394410859085 meters in .000000001 seconds, so he measures light to travel at the rate of 211,985,280.00038323887394410859 m/s in the box. He knows the speed of light is 299,792,458 m/s, and that he measured light to only travel 211,985,280.00038323887394410859 m/s in the box, so he therefore knows that the absolute velocity of the box must be 87,807,177.99961676112605589141 m/s.

 

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OK, I am going to try to sort through that. In the meantime, if your equations are working so well, why don't you try this problem. I've converted the numbers to meters, since you seem to prefer meters over lightseconds:


Say I have a cube that is 299,792,458 meters in length, per side. I place that cube at the origin point (0,0,0) of the absolute frame. Let the units of length be meters, so that these points represent corners of the cube while it is at rest:

(299792458, 0, 0)
(0, 299792458, 0)
(0, 0, 299792458)

Next, I choose a random absolute velocity for the cube, and tell you that it takes light the following times:

1.201 seconds to reach the corner which used to be at (299792458, 0, 0).
1.336 seconds to reach the corner which used to be at (0, 299792458, 0).
1.482 seconds to reach the corner which used to be at (0, 0, 299792458).

Now all you have to do is tell me the absolute velocity of the cube.

 

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Lets start with the first corner, in the x direction.

If light travels 1.201 seconds, light traveled 360,050,742.058 meters. Since the box is 299792458 meters in length in the x direction, the box traveled 360,050,742.058 - 299,792,458 = 60,258,284.058 meters in 1.201 seconds, right? That means the box has a 50,173,425.527060782681099084096586 m/s velocity in the x direction, correct?

Using my equation v = (ct-l)/t finds the same absolute velocity in the x direction.

For the y direction I get 75,396,905.604790419161676646706587 m/s
For the z direction I get 97,503,350.037786774628879892037787 m/s

 

So your equations are telling you this:
x-component = 50,173,425 m/s = 0.167c
y-component = 75,396,905 m/s = 0.251c
z-component = 97,503,350 m/s = 0.325c

And, unfortunately, my equations are telling me this:
x-component = 0.10c
y-component = 0.20c
z-component = 0.30c

I don't see how your equations can possibly work for three dimensional space. You were using them earlier, and you were getting speeds of light around 1.23c. You haven't changed them at all since then, so they are not just going to start being correct.

 

The incorrect times didn't work, it wasn't that my equation was bad. I showed you the proper times with the hypotenuse.

Let's talk about your .1c in the x direction.

How far do you think light traveled in 1.201 seconds?

 

The light would have traveled 1.201 lightseconds.

And, at time 1.201 seconds, 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

 

Right, which is 360,050,742.058 meters, correct? So if the box is 299,792,458 meters in length in the x direction,and light traveled 360,050,742.058 meters to get to the corner of the box, then the box must have traveled 360,050,742.058 - 299,792,458 = 60,258,284.058 meters in 1.201 seconds, right?

Well if the box traveled 60,258,284.058 meters, the PRIMARY corner is located at .201, not 1.20.

 

You're the one who says it traveled 60,258,284.058 meters. You got that by subtracting one distance from another. But the box is actually moving in three dimensions. You can't just subtract a couple of numbers like that.

When the primary corner of the cube is located at (0.120, 0.240, 0.360) then the corner which used to be at (1,0,0) would be located at (1.120, 0.240, 0.360) because all I had to do was add one to the x-coordinate.

How far away is that corner form the origin (0,0,0)?
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, too. So we know the light is just now reaching that corner of the cube.

 

This is how it happens in the the x and y direction, agreed? Forget the 3rd dimension for now. Do you agree that this is correct for the x,y dimensions?

Light traveled 29,979,245.8 meters in .1 seconds along the hypotenuse.

a^2+b^2=c^2

The distance light traveled in .1 seconds is c in the theorem, so c^2=898,755,178,736,817.64 meters


a^2=449,377,589,368,408.82 meters
b^2=449,377,589,368,408.82 meters

a=21,198,528.000038323887394410859085 meters
b=21,198,528.000038323887394410859085 meters

In the a and b frames (train and skate frames), light is measured to travel 21,198,528.000038323887394410859085 meters in .1 seconds.

In the a and b frames (train and skate frames), light travels at the velocity of 211,985,280.00038323887394410859085 m/s.

That means the train and skate are traveling an absolute velocity of 87,807,177.9996167611260558914092 m/s (299,792,458 - 211,985,280.00038323887394410859085 = 87,807,177.9996167611260558914092).

If two clocks are .21198528000038323887394410859085 meters apart from each other in the train and skate frames, it takes light .000000001 seconds to travel the distance between the clocks. That means in the train and skate frames, light travels at 211,985,280.00038323887394410859085 m/s. Why? Because the frames have an absolute velocity of 87,807,177.9996167611260558914092 m/s, so it takes light more time (in this case) to travel a length in that frame than it would if the frame had a zero velocity. Again, 211,985,280.00038323887394410859085 m/s + 87,807,177.9996167611260558914092 m/s = 299,792,458 m/s

So again, using v = (ct-l)/t

The distance between the clocks is .21198528000038323887394410859085 meters. It takes light .000000001 seconds to travel from one clock to the other clock. Of course c is 299,792,458 m/s.

The absolute velocity of the train and skate is 87,807,177.9996167611260558914092 m/s.

H = sqrt(.21198528000038323887394410859085^2 + .21198528000038323887394410859085^2)
H = sqrt(0.044937758936840881999999999999638 + 0.044937758936840881999999999999638)
H = sqrt(0.089875517873681763999999999998638)
H = 0.29979245799999999999999999999773

0.29979245799999999999999999999773 meters in .000000001 seconds is 299,792,458 m/s
Light travels at 299,792,458 m/s along the hypotenuse.

My equation is correct.

 

Why did you give up on the cube problem? Do you want me to make a two dimensional cube problem?

 

I told you I'm not a mathematician. I just gave you a two dimensional solution.

Give me another two dimensional problem and I'll do it. I don't know how to do a 3 dimensional one yet, but I'm sure my equation can do it, as it handles 1 and 2 dimensional problems just fine.

 

Here is a 2-dimensional cube problem. There is no z-axis, so we can just call it a square.

Say I have a square that is 299,792,458 meters in length, per side. I place the primary corner of that square at the origin point (0,0) of the absolute frame. Let the units of length be meters, so that these points represent corners of the cube while it is at rest:

(299792458, 0)
(0, 299792458)

Next, I choose a random absolute velocity for the cube, and tell you that it takes light the following times:

1.724 seconds to reach the corner which used to be at (299792458, 0, 0).
1.395 seconds to reach the corner which used to be at (0, 299792458, 0).

Now all you have to do is tell me the absolute velocity of the square.

 

I'll even give you the answer in advance:

I have chosen a velocity such that the two components are as follows:
x-component = 0.4c
y-component = 0.2c




The first time given is 1.724 seconds, so the primary corner of the cube would be located at (0.689, 0.344) as you can see from these equations:
x = 1.724(0.4) = 0.689
y = 1.724(0.2) = 0.344
And from that, it follows that the corner which used to be at (1,0) is now located at (1.689, 0.344) because all I had to do was add one to the x-coordinate.

If I want to know exactly how far away that corner is from the origin, I can use this equation:
d = sqrt(1.689^2 + 0.344^2)
d = 1.724
And that happens to be how far the light sphere is from the origin, because the time is 1.724 seconds.



The next time given by the problem is 1.395 seconds, so the primary corner of the cube would be located at (0.558, 0.279) as you can see from these equations:
x = 1.395(0.4) = 0.558
y = 1.395(0.2) = 0.289
And from that, it follows that the corner which used to be at (0,1) is now located at (0.558, 1.289) because all I had to do was add one to the y-coordinate.

If I want to know exactly how far away that corner is from the origin, I can use this equation:
d = sqrt(0.558^2 + 1.289^2)
d = 1.395
And that happens to be how far the light sphere is from the origin, because the time is 1.395 seconds.

 

I can't figure it out now.

 

But how do you figure it out without knowing the velocity? I did it before but I worked so many numbers now I can't think straight.

 

-not really, the train speed is .07c, the skate speed is .1c, it takes a longer path. The light follows the skate!

 

If you drop all the numbers after the decimal point, you won't sacrifice accuracy.

 

Ok,
v = (ct-l)/t

x component
v=125,898,920 m/s
0.41c



y component
v=84,887,470 m/s
0.28c

 

Yes, that's pretty much what I get using your equations, too.

t = 1.724 seconds
v = (ct - l) / t
v = (1.724 - 1.000) / 1.724 = 0.42c

t = 1.395 seconds
v = (ct - l) / t
v = (1.395 - 1.000) / 1.395 = 0.28c


The problem is that the numbers should be
v(x) = 0.40c
v(y) = 0.20c
The reason your equations are pretty close is because these speeds are not too terribly close to c. If I had chosen higher speeds, your equations would have been even less accurate.

I haven't figured out how to do that with my equations. My equations are more complex than yours, so it's not as easy. But I'll keep working on that.