If 0 < v < c (like I say) and if k > 0 (like you say) Event 2 has t < 0 and t' < 0, so it happened before Event 1 (where, t = t' = 0) , exactly like 1989 < 2011 so 1989 came before 2011. Why would you even try to say the opposite?
Time Dilation
- Thread starter chinglu
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If 0 < v < c (like I say) and if k > 0 (like you say) Event 2 has t < 0 and t' < 0, so it happened before Event 1 (where, t = t' = 0) , exactly like 1989 < 2011 so 1989 came before 2011. Why would you even try to say the opposite?
Ok. That all looks right.
Since the position coordinate does not change in the x frame between events 1 and 2, the time interval in the x frame is a proper time for these two events. The time interval in the x' frame is not a proper time, so we should expect it to be longer than the x time for the same two events. And it is! It's longer by a factor of $$\gamma$$.
It looks like you've got these calculations correct.
Any questions? Are you sorted now?
Yes, k < 0 for x'.
I typed it wrong.
The rest falls into place under that k < 0 assumption.
Thanks, that was the point of this thread.
If the primed frame is the moving frame, then it elapses more time then the stationary unprimed frame origin.
That is not time dilation.
Length contraction in a moving frame, as seen by an observer (that is, motion relative to their frame of reference) is a consequence of measuring say, the ends of a moving rod simultaneously.
The length is contracted as a consequence of the Lorentz transformation. Time is dilated for the same reason the length is contracted, when you invert the transformation. This is entirely consistent with transforming a frequency into a wavelength when the source is moving relative to an observer.
What would you call it?
The length is contracted as a consequence of the Lorentz transformation. Time is dilated for the same reason the length is contracted, when you invert the transformation. This is entirely consistent with transforming a frequency into a wavelength when the source is moving relative to an observer.
If the primed frame is the moving frame, events appear to take longer in that frame than the "stationary" frame. That IS time dilation, and it's relative to the observer of the moving frame.chinglu said:
What would you call it?
Nothing in the above calculation says which frame is "moving" and which is not.
The math is above. Feel free.
I have said over and over unprimed stationary. Does not matter.
If that last sentence is what you think, then what do YOU THINK it should be called?chinglu said:
Feel free (I know I do).
t = (-k/v)/γ
and
t' = (-k/v)
It does not matter which frame is stationary. t' is always larger.
I am not sure what to call this mathematical fact.
Well if you aren't sure, it's probably acceptable to call it "Lorentz contraction".chinglu said:
Seeing that it's a consequence of the Lorentz transformation of moving coordinates relative to a stationary observer.
Hell, you could get carried away and call it "a principle of relativity of simultaneity", but that's harder to type.
I suppose you could call it R of S. In fact, this is a consequence of it.
I am not sure about the "Lorentz contraction", because that is something completely different.
To what?chinglu said:
The contraction of length and the dilation of time (together) is a consequence of the transformation. Ergo, distance and time in moving coordinate systems are both transformed (in parallel) relative to local coordinate systems
I doubt we communicate.
The point of this thread is that a clock moving toward the origin between the start event of both origins being same to the end event of moving clock at same place as origin will beat faster than the clock of the stationary origin.
That is what alll the valid math in this thread has proven.
So because an observer sees a clock beat faster, that can't be time dilation?
If another observer is with the 'faster' clock, won't the other clock seem to be slower than their one?
I think so. Any other possibility just doesn't seem logical, Jim.
If another observer is with the 'faster' clock, won't the other clock seem to be slower than their one?
I think so. Any other possibility just doesn't seem logical, Jim.
You're right. It doesn't matter which frame you call "stationary", because an observer at rest in either frame can consider himself stationary.
t' is not always larger. It depends on which events you're looking at, obviously.
The point of this thread is that a clock moving toward the origin between the start event of both origins being same to the end event of moving clock at same place as origin will beat faster than the clock of the stationary origin.
That is what all the valid math in this thread has proven.
That is what all the valid math in this thread has proven.
In step 3 #117, you claimed (x',t') = (-k, k/v) when t'=t=0 with k > 0.
Assume a clock is located at -k in the primed system.
Now, let's shoot a light beam at that clock when the origins are the same.
Hence, while light moves c from the origin, x' moves toward the light at v cutting the distance light must travel.
ct = vt + k/γ
t = k/(γ(c-v))
x = -k/γ
According to LT,
t' = ( t - vx/c² )γ
Plug in the numbers.
t' = ( k/(γ(c-v)) - v(-k/γ)/c² )γ
t' = k/(c-v) + vk/c²
But, this assume the clock at (x',t') = (-k, k/v) had t'= 0.
We add k/v to the result above and have
t'failed = k/(c-v) + vk/c² + k/v.
Thus, you contradict LT.
This is a bit impenetrable.chinglu said:
The stationary origin doesn't make sense unless there is a clock there. But you say "the start event of both origins being same", but there is a moving clock, which can't be at the origin (or it would be stationary), then the moving clock is "at the same place", eventually.
If there are two clocks in your scenario they can only be in the same stationary frame if they aren't moving relative to each other. But you say one of them is moving toward the origin, so it can't be at the origin and it can't be in a stationary frame.