If this conversation is happening on two forums at once, I suggest you keep it in one forum. Please choose one.
I can not know why posters in other forums are typing about me and are afraid to come here.
This is where I am going to have this discussion and no other forums.
If they do not come here that means they are afraid.
OK This I couldn't help. I have giggles
I don't know your aim for one. Two, I hope fear is not the word you really want to use. To say "statement of fear" is absurd.
I'm not going over everything that's been said. I have not even gone over rpenner's retort once, forget thrice!
So, if I can be forgiven for my laziness, what is it exactly you are saying?
I don't know what he's saying. But in this thread alone, he has demonstrated the absolute necessity to not just talk about intervals of space or intervals of time when doing relativity because you need to quantify the exact same interval of space and time.
These are the points I wish to convey:
- Special Relativity is a description of space and time as an inseparable whole -- the concepts of "place" and "when" don't mean much without each other.
- Length contraction you only fairly derive between co-moving time-like world-lines of endpoints because it is an effect caused by relativity of simultaneity.
- Time dilation only happens between two events on the world-line of the clock under consideration.
- Relativity of simultaneity means that even though clocks might be synchronized at one event, those clocks need not be synchronized at another event.
- And the full Lorentz transformations are homogeneous, meaning they work just as well on an interval of coordinate differences as they do on sets of coordinates.
but I have not seen that the poster managed to engage any of them.
//Edit --
Fun fact --
$$e^{\begin{pmatrix} 0 & \ln \sqrt{\frac{c+v}{c-v}} \\ \ln \sqrt{\frac{c+v}{c-v}} & 0 \end{pmatrix} } = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\end{pmatrix} $$
Event's have a place and a time.I quantified the intervals precisely.
How long in rest frame does it take a clock located at x' to reach the rest origin when everything starts when origins are same?
Start event: origins same
End event: x' same with rest origin.
Assume standard configuration and time interval begins for both frame when origins same.
Assume clock is located at x' < 0 in the primed coordinates.
Assume time interval ends when x' and unprimed origin same. What is this time interval in each frame.
I was actually looking at what you replied to me with. It wasn't hard to restrain myself from posting cause it was actually enjoyable to have no idea what you typed. Hence, a mystery was afoot and tedious deciphering to come - in between whatever else of more importance I was doing.
It could have been fun, but then you had to go and respond to rpenner and suddenly I didn't feel so special.
:bawl:
Event's have a place and a time.
Your start event is O, when you need to pick exactly one of $$P_{t'=0}$$ or $$P_{t=0}$$ if you want to answer the full question.
When you start at O and go to Q, you are only talking about a clock which is not moving in S (what you call the rest frame) and is moving in S'.
In other words, assume the origins of the x and x' frames are co-located at t=t'=0.
i.e. Assume the clock doesn't move in primed frame, so it always has x'=-c, for c a positive constant. So at t'=t=0, the clock is at x'=-c. At this time, the clock's location in the x frame is:
$$x=\gamma (x' + vt') = -c\gamma$$
If x' frame moves in positive x direction with speed v, then the time interval ends when the x coordinate of the clock is zero. The clock moves at speed v in the x frame (along with x' frame). In the x frame, we have distance covered = speed multiplied by time taken, so:
$$c\gamma = v\Delta t$$
which gives the time interval in the x frame. i.e.
$$\Delta t = \frac{c\gamma}{v}$$
The corresponding time in the t' frame is for the event at x=0 is:
$$\Delta t' = \frac{c\gamma^2}{v}$$
On further evaluation, however, your question appears a little bit too textbook and given the extraordinary measures to explain mathematically you have done yourself a tremendous disservice by demanding only the math.
It could be so simple as explaining your dissension in mere words. Just starting a post with saying "This is what my textbook says and this is what I don't understand." Rather than jump in with any incorrect mathematical conclusion and leaving others' bewildered in how you reached it.
Read my post again, please.
In other words, assume the origins of the x and x' frames are co-located at t=t'=0.
i.e. Assume the clock doesn't move in primed frame, so it always has x'=-c, for c a positive constant. So at t'=t=0, the clock is at x'=-c. At this time, the clock's location in the x frame is:
$$x=\gamma (x' + vt') = -c\gamma$$
If x' frame moves in positive x direction with speed v, then the time interval ends when the x coordinate of the clock is zero. The clock moves at speed v in the x frame (along with x' frame). In the x frame, we have distance covered = speed multiplied by time taken, so:
$$c\gamma = v\Delta t$$
which gives the time interval in the x frame. i.e.
$$\Delta t = \frac{c\gamma}{v}$$
The corresponding time in the t' frame is for the event at x=0 is:
$$\Delta t' = \frac{c\gamma^2}{v}$$
$$(c \Delta t')^2 - (\Delta x')^2 = \left( \frac{c^4}{v^2 (c^2 - v^2)} \right) c^2 \gamma^2$$ but
$$(c \Delta t)^2 - (\Delta x)^2 = \left(\frac{c^2}{v^2} - 1 \right) c^2 \gamma^2$$