Neddy Bate
Valued Senior Member
Motor Daddy, can you tell me when this happens :
In your scenario of y and z times being 0.832 seconds?
In your scenario of y and z times being 0.832 seconds?
Do you agree I was correct?
Motor Daddy, can you tell me when this happens :
In your scenario of y and z times being 0.832 seconds?
v(x)=0.79927794245177367361794866830634 c
t(x)=2.4910067488714731588930217979394 seconds
v(x)=0.79927794245177367361794866830634 c
t(x)=2.4910067488714731588930217979394 seconds
Like I asked you, how does relativity resolve the issue that it takes ~2.49 seconds to travel the length of .5 light seconds in the cube frame?
0.27788925112852684110697820206067 seconds
Thank you.
v(x)=0.799
t(x)=2.491
Is pretty darn close to my answers, v(x)=0.8 and t(x)=2.5.
Relativity says that your diagram is correct for the reference frame through which the cube is moving. You drew your diagram with the cube moving relative to the reference point from which the light expands. But relativity says things look differently in the cube frame. In the cube frame, relativity says the reference point from which the light expands would always be at the center of the cube.
After the light flashes in front, and before the light in the rear flashes, he covers it up with the bag, thereby blocking the light. Now, how do we explain this from the point of view of the embankment observer? There is no passage of time between the two flashes of light. So we seem to have a paradox.
Gosh, I was hoping we were past this difference between MD's Light Box coordinate system and SR's coordinate system and its Lorentz transformations.
Two different animals. Each with its own logic. You can believe one or the other but not both.
Close but no cigar.
Well then relativity would be wrong, because the light sphere doesn't expand relative to the source, it expands relative to the point in space the light was emitted, regardless of what the source does.
My diagram was built on the measured times to the receivers in the cube, nothing else.
From those measured times at the receivers I am able to calculate the absolute velocity of the box.
You say relativity can do that too?
Tell me the distances and times in the cube frame from the source to the receivers of the .832 second example.
There is no changing times and distances in the cube.
The only way you know the times is to measure them,
If you think the y receiver can be at two different coordinates at the same time then tell me both.
Can relativity tell me the absolute velocity of the box?
Does relativity acknowledge that it takes .65 seconds for light to travel from the source to the y and z receivers?
Well then relativity would be wrong, because the light sphere doesn't expand relative to the source, it expands relative to the point in space the light was emitted, regardless of what the source does.
My diagram was built on the measured times to the receivers in the cube, nothing else. From those measured times at the receivers I am able to calculate the absolute velocity of the box.
Zeno said:After the light flashes in front, and before the light in the rear flashes, he covers it up with the bag, thereby blocking the light. Now, how do we explain this from the point of view of the embankment observer? There is no passage of time between the two flashes of light. So we seem to have a paradox.
Neddy Bate said:If the bag goes over the lamp before it flashes, then nobody sees it flash. In the embankment frame, the flashes would still be simultaneous, but there would be a bag hiding one of the flashes, that's all. No paradox.
Motor Daddy:
Ok. We're assuming light is emitted at t=t'=0 in the embankment and train frames. The y receiver is initially located at (x,y,z)=(x',y',z')=(0,0.5,0), where the distances are in light seconds.
You say that the light reaches the y receiver after 0.65 seconds. I assume that is in the embankment frame using embankment clocks.
In 0.65 seconds, light travels a distance of 0.65 light seconds.
For light to hit the y receiver, the y receiver must have travelled a distance:
$$d = \sqrt{(0.5)^2 + x^2} = 0.65$$
Here, x is the x coordinate of the y receiver when the light hits it.
Solving for x we find x = 0.42 light seconds.
So, the coordinates of the y receiver when the light hits it in the embankment frame are (x,y,z)=(0.42,0.5,0).
In the box/train frame, things are much easier, since the y receiver never moves. The coordinates of the y receiver when the light hits it are (x',y',z')=(0,0.5,0).
Summary of y receiver coordinates when light hits it:
Embankment frame: (x,y,z)=(0.42,0.5,0)
Train frame: (x',y',z')=(0,0.5,0)
Since it takes 0.65 seconds for the box to move 0.42 light seconds in the x direction, the speed of the box in the embankment frame is
speed = 0.42/0.65 = 0.65c.
That is, the train is moving at 65% of the speed of light.
No. Absolute velocity doesn't exist.
See calculation above, which is in accordance with relativity. Based on your calculations, do you agree that the speed of the box is 0.65c? If so, then we agree. If not, then one of us is wrong.
No. you only calculate the velocity of the box relative to the embankment.
James is correct from the perspective of SR.
And have you done any calculations of motion in more than one axis? These new receptors should help.
SR is incorrect so James is incorrect. It is impossible for the times to be .5 c in the cube frame in my pic. It is impossible to measure the speed of light to be c in the cube frame.
Measured times to receivers:
x time: .761972 seconds
y time: .761972 seconds
z time: .92 seconds
Component velocities:
v(x) = .2300c
v(y) = .2300c
v(z) = .4022c
Coordinates of source at center of cube at .761972 seconds: (0.17525356,0.17525356,0.3064651384)
x .761972(.2300c) 0.17525356
y .761972(.2300c) 0.17525356
z .761972(.4022c) 0.3064651384
Coordinates of source at center of cube at .92 seconds: (.2116,.2116,.370)
x .92(.2300c) .2116
y .92(.2300c) .2116
z .92(.4022c) .370
Coordinates of source at center of cube at 1.0 seconds: (.23,.23,.4022)
x 1.0(.2300c) .23
y 1.0(.2300c) .23
z 1.0(.4022c) .4022
Distance center of cube traveled from start coordinates (0,0,0) to (.23,.23,.4022) in 1 second is 155,072,655.74 meters, so the absolute velocity of the center of the cube is 155,072,655.74 m/s.