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 Please Register or Log in to view the hidden image! http://www.physforum.com/index.php?showtopic=28676&st=75
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. http://www.sciforums.com/showpost.php?p=2518556&postcount=33 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 -- \(e^{\begin{pmatrix} 0 & \ln \sqrt{\frac{c+v}{c-v}} \\ \ln \sqrt{\frac{c+v}{c-v}} & 0 \end{pmatrix} } = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\end{pmatrix} \)
Yes. Assume clock is located at x' < 0 in the primed coordinates. Assume standard configuration and time interval begins for both frame when origins same. Assume time interval ends when x' and unprimed origin same. What is this time interval in each frame. Unprimed/rest frame Apply length contraction. -x'/γ = vt. t = -x'/(vγ) Primed frame -x' = vt'. t' = -x'/v. Note in unprimed frame/rest frame clock elapsed more time in moving frame than rest frame. This verified by Lorentz Transforms. x = 0, origin at rest. t' = ( t - vx/c² )γ = tγ
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.
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. Please Register or Log in to view the hidden image! :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.
Is it not a well defined and agreed upon event when origins are the same. An event has a place and elapsed time since origins same. It is not a well defined event when x' < 0 and the unprime origin are at same place? I am not understand problem.
No clock does not move. But, unprime origin move to x'. That is the end event for primed frame. Since unprime origin moves at -vt, t = ( t' + vx'/c²)γ = t'γ This I can agree. This I can not agree. You forgot length contraction. \(-c/\gamma = v\Delta t\)
Assume start event origins same. What is distance in unprimed frame x' must travel to reach unprimed origin. What is distance in primed frame unprime origin must travel to reach x'. Then, simple (x' < 0) Δx' = ( 0 - x') = v Δt' Δx = ( 0 - x'/γ) = v Δt Δt'/Δt = γ
Of course I will write your way. I am positive you know how to show time dilation for primed origin clock starting when origins same. Just set x=vt. Plug this into Lorenntz Transforms of t' = ( t - vx/c² )γ and get t'=t/γ. Likewise for unprimed frame clock at origin primed frame calculates x' = -vt', t = ( t' + vx'/c² )γ = t=t'/γ. Many then conclude reciprocal time dilation. Just look t=t'/γ in prime frame for unprimed origin clock and t'=t/γ in unprime frame for prime origin clock. Sure looks like reciprocal time dilation. But, these are not the same events for the moving clocks since they are at different locations when motion is complete. This is apples and oranges. So, in unprimed frame, it views a clock moving toward its origin whereas this same sequence in the primed frame looks like the unprime origin is move toward a coordinate that has a clock. If we now remember that when Lorenntz Transforms calculates time, it is always the case that both frame agree with both times. This mean for this situation, both frames agree the origin of the unprimed frame elapses t = t'/γ. But, also both frame agree the clock in primed frame elapses tγ = t'. There is no dispute for correctly defined agreed events. In short there is no such thing as reciprocal time dilation since it is a comparison of two completely different end events but same start event, common origins. So here is geometry. start event O' and O at same place. . end event x' and O origin at same place. x'<0 x'-------------------------O' x'/γ-----------------------O For O to reach x' in view of primed frame, it moves left, no length contraction longer distance, same v longer time on x' clock. For x'/γ to reach O, it moves right, length contraction, shorter distance, same v shorter time on O clock. So, this means the x' clock must elapse more time than the O origin clock from start of O and O' the same place to O and x' the same place.
\((c \Delta t')^2 - (\Delta x')^2 = \left( \frac{c^4}{v^2 (c^2 - v^2)} \right) c^2 \gamma^2\) but \((c \Delta t)^2 - (\Delta x)^2 = \left(\frac{c^2}{v^2} - 1 \right) c^2 \gamma^2\) So it looks like you are not talking about the same space-time interval throughout. That's because at x'=-c, and t'=0, t is not also 0.
Thanks, invariance of space-time interval gives alternative way for me to prove my case. \((c \Delta t')^2 - (\Delta x')^2 = (c \Delta t)^2 - (\Delta x)^2\) Assume \(x'=-x\) then \(\Delta x' = x/\gamma - x'\) \(\Delta x = x - x'/\gamma\) Since \(x'=-x\) then \(\Delta x' = x/\gamma - (-x)\) \(\Delta x = x - (-x/\gamma)\) So \(\Delta x = \Delta x'\) So, based on the invariance of the space-time interval \((c \Delta t')^2 = (c \Delta t)^2\) \(\Delta t' = \Delta t\)
