Yes, the clock.chinglu1998 said:Stationary with respect to the observer in WIKI. Does WIKI have something else in the problem?
Originally Posted by chinglu1998
This same set of light beams mapped by LT do not measure a constant distance from the light emission point in the moving frame.
These are mathematical facts.
DaleSpam said:Right here:
No, I already disproved this, and it is contrary to the second postulate. If you believe otherwise then post your derivation.
Sorry, I don't know what you're looking for. I thought you wanted to know how to calculate time dilation as shown on the WIKI page, but with the frames reversed? Realize that the 'time dilation' formula relates the proper time measured by some clock with the time measured from a frame in which that clock is moving.chinglu1998 said:Light aberration is absolute I was very specific.
Are you going to calculate how the time dilation works for this WIKI example?
I am curious how you will convert absolute light aberration to reciprocal time dilation.
I was hoping you would calculate this.
No, you were trying to use the light sphere to show that one frame involves "Euclidean space" and the other involves "Minkowski space". But in the "stationary" frame you can have both a light sphere and a light ellipsoid depending on whether you consider simultaneous or non-simultaneous events in that frame, and likewise in the "moving" frame you can have both a light sphere and a light ellipsoid depending on whether you consider simultaneous or non-simultaneous events in that frame. So, I still don't see how the labels "stationary" and "moving" have any significance beyond an arbitrary verbal distinction.chinglu1998 said:Sure you are correct. Except, we were considering the light sphere, remember?
JesseM said:Similarly, if your initial set of events was such that when you transform into the "moving frame", you get a bunch of events that are simultaneous in the moving frame, then their positions form a sphere. If your events in the moving frame are non-simultaneous, then they may form some other shape like an ellipsoid. And if you are looking at all events on the worldlines of the light beam in the moving frame, they form a 4D cone. So, I still can't make any sense of your distinction between "Euclidean space" in the stationary frame and "Minkowski space" in the moving one, still seems like a totally incoherent distinction.
I meant that if you pick events that are simultaneous in the moving frame, their positions form a sphere in the moving frame. Obviously their positions form an ellipsoid in the stationary frame.chinglu1998 said:No, I have an equation from the context of the "stationary frame" frame such that for all light beams that strike this object in the stationary frame, the LT calculates the same t'. It is not a sphere BTW. Do you have this math?
JesseM said:I have no idea what you mean by "sees a sphere". If you think of yourself as an actual physical observer at rest in some frame (as opposed to adopting the omniscient perspective of someone reading a problem in a textbook), then you understand that you can't actually visually "see" a set of simultaneous events in your frame at a single moment, right? Since you are at different distances from different points in space, what you see visually at a single moment will be light from a bunch of events at different times in your frame. Statements about what was happening at a single t-coordinate can only be made in retrospect, like if in 2010 I receive a signal from an event E1 10 light-years away in my frame, and in 2020 I receive a signal from an event E2 20 light-years away in my frame, and I conclude retroactively that they both happened simultaneously at the t-coordinate of 2000 in my frame. So the "light sphere" is every bit as much of a retroactive reconstruction as the "light cone", both involve charting the coordinates of a bunch of events that I didn't become aware of until various later times.
Well, he only "calculates a sphere" if he happens to pick events that are simultaneous in his frame, but there is no physical reason why he must use simultaneous events as the initial data which he plugs into the Lorentz transformation, he could equally well use a set of events which are non-simultaneous in his frame. So he has a totally arbitrary choice of what events to pick, if he makes one choice then the positions of the events will form a sphere in his frame but an ellipsoid in the moving frame, if he makes a different choice then the positions of the events will form an ellipsoid in his frame but a sphere in the moving frame. Do you disagree? If you agree with the above, it seems there is no coherent sense in which the stationary frame inherently involves "Euclidean space" and the moving frame inherently involves "Minkowski space".chinglu1998 said:Calculates a sphere.
JesseM said:"Stationary" is meaningless unless understood to mean "stationary relative to" some object or frame. Certainly an observer (or any other object) is stationary relative to their own frame, but moving relative to other objects and frames.
To the position coordinates of your rest frame.chinglu1998 said:I do not know how to answer this. Let's ask you a question. Assume you are in a rocket in space without acceleration. You want to use SR from your view. What are you stationary wrt?
JesseM said:In their own frame yes, but the observer is perfectly capable of understanding that they would be seen as "moving" in other frames, unless they are an idiot who doesn't understand the LT.
It was you who introduced the idea of what an observer "thinks" when you said "That observer does not move and thinks all other objects move", I was just responding to that. If you want to drop all notion of what an observer "thinks" and just talk about the math that's fine with me, it's plain to see that there's nothing in the math about needing to pick one frame as "stationary" when doing calculations.chinglu1998 said:Where can I find this in the axioms of SR? This is a math theorynot a human theory.
Doc Al said:Sorry, I don't know what you're looking for. I thought you wanted to know how to calculate time dilation as shown on the WIKI page, but with the frames reversed? Realize that the 'time dilation' formula relates the proper time measured by some clock with the time measured from a frame in which that clock is moving.
What do you mean by 'reciprocal' time dilation?
Why don't you describe exactly what you're looking for? You started out this thread wondering if the WIKI description of the time dilation formula was wrong. Many posters, including myself, have shown that the WIKI page is just fine. Are you still confused about the meaning of the 'time dilation' formula?
Yes, but this has nothing to do with the "stationary" frame being treated as "Euclidean" and the "moving" frame being treated as "Minkowski". After all, it is just as true that if you take a set of simultaneous events in the "moving" frame which form a sphere, the LT says these events form an ellipsoid in the "stationary" frame (likewise if you start with an ellipsoid of non-simultaneous events in the 'stationary' frame, then under the LT they translate to a Euclidean light sphere in the 'moving' frame).chinglu1998 said:Sure you can pick any Newtonian/Euclidean equation that works in one frame, it works in any frame. This is true by the relativity postulate. Everyone knows that.
But, if you take the Euclidean light sphere in a frame, LT does not translate it to a Euclidean light sphere. Again, everyone knows that, do you?
You mean, the one where he says it's just a verbal distinction? The one where he doesn't say a damned thing about "stationary" having any relevance whatsoever to the math? That one?chinglu1998 said:I posted over and over Einstein's distinction for the "stationary" system.
"Stationary wrt to the universe" is meaningless. Are you just asking for a calculation which first shows the time t between the events of hitting the top and bottom mirror in the light clock frame, then switches to the observer frame and calculates the time t' between these same two events based on the aberrated path of the light beam in this frame? If that's not it, you need to be more specific about what sort of calculation you're looking for.chinglu1998 said:I am having trouble calculating t' = t γ using the light aberration argument of WIKI and the clock frame as stationary wrt to the universe. Can you help me?
The observer isn't stationary wrt clock, the observer is moving relative to the clock (or the clock is moving relative to the observer, in relativity these two statements are equivalent).chinglu1998 said:However, now let's calculate time for each frame when observer taken as stationary wrt clock.
And you are still wrong. What don't you understand about that derivation? That's pretty basic stuff.chinglu1998 said:At first, my reading of WIKI article indicate their conclusion incorrect.
I still stand by that.
"Clock stationary with respect to observer" and "observer stationary with respect to clock" are equivalent statements. There's only one clock described on the WIKI page. It happens to be at rest in the unprimed frame.However, now let's calculate time for each frame when observer taken as stationary wrt clock.
WIKI already did this and no one disagrees.
Then, let's calculate time for each frame when clock taken as stationary wrt observer.
JesseM said:No, you were trying to use the light sphere to show that one frame involves "Euclidean space" and the other involves "Minkowski space". But in the "stationary" frame you can have both a light sphere and a light ellipsoid depending on whether you consider simultaneous or non-simultaneous events in that frame, and likewise in the "moving" frame you can have both a light sphere and a light ellipsoid depending on whether you consider simultaneous or non-simultaneous events in that frame. So, I still don't see how the labels "stationary" and "moving" have any significance beyond an arbitrary verbal distinction.
Fine.I meant that if you pick events that are simultaneous in the moving frame, their positions form a sphere in the moving frame. Obviously their positions form an ellipsoid in the stationary frame.
Well, he only "calculates a sphere" if he happens to pick events that are simultaneous in his frame, but there is no physical reason why he must use simultaneous events as the initial data which he plugs into the Lorentz transformation, he could equally well use a set of events which are non-simultaneous in his frame. So he has a totally arbitrary choice of what events to pick, if he makes one choice then the positions of the events will form a sphere in his frame but an ellipsoid in the moving frame, if he makes a different choice then the positions of the events will form an ellipsoid in his frame but a sphere in the moving frame. Do you disagree? If you agree with the above, it seems there is no coherent sense in which the stationary frame inherently involves "Euclidean space" and the moving frame inherently involves "Minkowski space".
Doc Al said:And you are still wrong. What don't you understand about that derivation? That's pretty basic stuff.
"Clock stationary with respect to observer" and "observer stationary with respect to clock" are equivalent statements. There's only one clock described on the WIKI page. It happens to be at rest in the unprimed frame.
JesseM said:The observer isn't stationary wrt clock, the observer is moving relative to the clock (or the clock is moving relative to the observer, in relativity these two statements are equivalent).
chinglu1998 said:At first, my reading of WIKI article indicate their conclusion incorrect.
I still stand by that.
However, now let's calculate time for each frame when observer taken as stationary wrt clock.
WIKI already did this and no one disagrees.
Then, let's calculate time for each frame when clock taken as stationary wrt observer.
What does that mean? As has been pointed out endlessly in this thread, stationary with respect to what?chinglu1998 said:So, are you going to calculate the times in both frame if the clock is taken as stationary?
Meaningless, unless "stationary" is understood as an arbitrary label.chinglu1998 said:Assume frame stationary.
Again, meaningless--what does "views" mean? If you just mean "if we decide to pick a set of events which are simultaneous in that frame, then we get a sphere". That's true, but of course this is true of the frame you call "moving" as well.chinglu1998 said:It views a light sphere.
Meaningless again, the LT doesn't "view" anything, it just translates arbitrary sets of events from one frame to another.chinglu1998 said:Does LT? No.
I am able to distinguish them, I just don't know what the hell this has to do with the choice of which frame we label "stationary" and which we label "moving". You agree that we can take a light sphere in the moving frame and use the LT to get a light ellipsoid in the stationary frame, right? So to me this means the situation is totally symmetrical, and "stationary" and "moving" has no relevance beyond an arbitrary label...you have yet to provide any coherent argument as to why you disagree.chinglu1998 said:That is the distinction. If you are not able to distinguish a light spherefrom a non-light sphere, I do not know what to do.
JesseM said:Well, he only "calculates a sphere" if he happens to pick events that are simultaneous in his frame, but there is no physical reason why he must use simultaneous events as the initial data which he plugs into the Lorentz transformation, he could equally well use a set of events which are non-simultaneous in his frame. So he has a totally arbitrary choice of what events to pick, if he makes one choice then the positions of the events will form a sphere in his frame but an ellipsoid in the moving frame, if he makes a different choice then the positions of the events will form an ellipsoid in his frame but a sphere in the moving frame. Do you disagree? If you agree with the above, it seems there is no coherent sense in which the stationary frame inherently involves "Euclidean space" and the moving frame inherently involves "Minkowski space".
Why do you think I need to "learn this"? Isn't this just a mathematical illustration of exactly what I just said in words, namely "if he makes a different choice then the positions of the events will form an ellipsoid in his frame but a sphere in the moving frame"?chinglu1998 said:OK I will learn you this.
Equation
(x-vrγ/c)²/(rγ)² + y²/r² + z²/r² = 1
Intersect the light postulate in the unprimed frame with this equation at every point and the primed frame will conclude t' is a constant r/c in all directions ie, the light sphere in the moving frame.
chinglu1998 said:OK I will learn you this.
Equation
(x-vrγ/c)²/(rγ)² + y²/r² + z²/r² = 1
Intersect the light postulate in the unprimed frame with this equation at every point and the primed frame will conclude t' is a constant r/c in all directions ie, the light sphere in the moving frame.
I assume you can do the math since I gave you everything.
You can't calculate things from the "context" of only one frame if you want an equation that relates the time in both frames like t' = t γ. The wiki article uses physical arguments to find the time from the perspective of both frames, as you can see. Do you just mean you want to start by using physical arguments to find the coordinates of the events in the clock frame, then just use the Lorentz transform to find the coordinates in the observer's frame?chinglu1998 said:I thought this is all just terminology.
So, use the context of the clock frame and do all the calculations.
Doc Al said:What does that mean? As has been pointed out endlessly in this thread, stationary with respect to what?
In the WIKI page, the clock is stationary in the unprimed frame, thus the proper time between 'clicks' is Δt'. The page gives the standard derivation of how the time between clicks according to the primed frame will be Δt = γΔt'.
You can't 'view the clock as stationary in the primed frame', unless you introduce a second clock that is at rest in the primed frame.
espen180 said:What does "intersect the light postulate" mean? Please adopt mainstream terminology.
I think he basically means to find a light cone such that every position on the ellipsoid is also the position of some event on the light cone, then translate the positions and times of these events into the primed frame using the LT. Of course it's easier to work backwards--assume a light cone in the primed frame starting from x'=y'=z'=t'=0, consider the set of events at t'=r/c which all satisfy x'2 + y'2 + z'2 = r2, then translate these events to the unprimed frame and show their positions form an ellipsoid.espen180 said:What does "intersect the light postulate" mean? Please adopt mainstream terminology.
JesseM said:You can't calculate things from the "context" of only one frame if you want an equation that relates the time in both frames like t' = t γ. The wiki article uses physical arguments to find the time from the perspective of both frames, as you can see. Do you just mean you want to start by using physical arguments to find the coordinates of the events in the clock frame, then just use the Lorentz transform to find the coordinates in the observer's frame?
In that case there is only one frame to consider, the frame where both the clock and observer are at rest, not a primed frame and an unprimed frame. Unless you are adopting some ridiculous definition of "wrt" that means something like "the observer decides to label the clock as 'stationary' even if the clock is moving in his own rest frame"--if you want to use the language of every physicist in the world rather than making up your own private language, "X is stationary wrt Y" means "X and Y share the same rest frame".chinglu1998 said:Geez, please take the clock as the stationary frame wrt to the observer.
JesseM said:I think he basically means to find a light cone such that every position on the ellipsoid is also the position of some event on the light cone, then translate the positions and times of these events into the primed frame using the LT. Of course it's easier to work backwards--assume a light cone in the primed frame starting from x'=y'=z'=t'=0, consider the set of events at t'=r/c which all satisfy x'2 + y'2 + z'2 = r2, then translate these events to the unprimed frame and show their positions form an ellipsoid.
JesseM said:In that case there is only one frame to consider, the frame where both the clock and observer are at rest, not a primed frame and an unprimed frame. Unless you are adopting some ridiculous definition of "wrt" that means something like "the observer decides to label the clock as 'stationary' even if the clock is moving in his own rest frame"--if you want to use the language of every physicist in the world rather than making up your own private language, "X is stationary wrt Y" means "X and Y share the same rest frame".
Geez, that's trivial. In the WIKI example, the time observed from that frame would be Δt'. Obviously a clock observed from a frame in which it is at rest just reads the proper time. No need for time dilation, as the clock isn't moving in that frame.chinglu1998 said:Geez, please take the clock as the stationary frame wrt to the observer.
Doc Al said:Geez, that's trivial. In the WIKI example, the time observed from that frame would be Δt'. Obviously a clock observed from a frame in which it is at rest just reads the proper time. No need for time dilation, as the clock isn't moving in that frame.
JesseM said:You can't calculate things from the "context" of only one frame if you want an equation that relates the time in both frames like t' = t γ. The wiki article uses physical arguments to find the time from the perspective of both frames, as you can see. Do you just mean you want to start by using physical arguments to find the coordinates of the events in the clock frame, then just use the Lorentz transform to find the coordinates in the observer's frame?
I don't know what you mean by "calculate from the clock frame", that's why I asked the questions. To me, this section of the wiki already did calculate the time from the clock frame:chinglu1998 said:You people are so willing to accept the WIKI calculations and yet so reluctant to calculate from the clock frame. Why?
Do you agree that here they have calculated [tex]\Delta t[/tex] "from the clock frame"? If so, are you asking that somehow we also calculate [tex]\Delta t'[/tex] "from the clock frame"? That seems completely nonsensical, since [tex]\Delta t'[/tex] inherently refers to the perspective of the observer frame, not the clock frame. Of course we could obtain [tex]\Delta t'[/tex] by starting from the coordinates of the events in the clock frame and then just performing a LT on them, but I don't know what else it could possibly mean to calculate [tex]\Delta t'[/tex] "from the clock frame", you'll have to explain if you don't want your question to be seen as incoherent gibberish.In the frame where the clock is at rest (diagram at right), the light pulse traces out a path of length 2L and the period of the clock is 2L divided by the speed of light:
[tex]\Delta t = \frac{2L}{c}[/tex]
"Light aberration" is only needed if you want to calculate the time in some other frame, one in which the clock is moving. WIKI shows all the math you need.chinglu1998 said:That is not how the WIKI article made their conclusions.
The article used light aberration.
Where is your light aberration math to calculate the times?
JesseM said:In that case there is only one frame to consider, the frame where both the clock and observer are at rest, not a primed frame and an unprimed frame. Unless you are adopting some ridiculous definition of "wrt" that means something like "the observer decides to label the clock as 'stationary' even if the clock is moving in his own rest frame"--if you want to use the language of every physicist in the world rather than making up your own private language, "X is stationary wrt Y" means "X and Y share the same rest frame".
Are you trolling now? Unless you have absolutely no ability to keep track of the context of statements, I think you should know perfectly well I didn't say the wiki article said "the clock is at rest with the observer", I was responding to your request to "please take the clock as the stationary frame wrt to the observer." (that's, y'know, why I quoted that exact statement before making the comment above) To me and every other physicist in the world, "take the clock as the stationary frame wrt to the observer" means exactly the same thing as "the clock at rest with the observer". Do they mean something different to you? If you see any difference between those two statements, then as I said above you must be using some ridiculous nonstandard definition of "stationary frame wrt to the observer".chinglu1998 said:Where in the WIKI article is the clock at rest with the observer? You make many mistakes.
Doc Al said:"Light aberration" is only needed if you want to calculate the time in some other frame, one in which the clock is moving. WIKI shows all the math you need.
JesseM said:Are you trolling now? Unless you have absolutely no ability to keep track of the context of statements, I think you should know perfectly well I didn't say the wiki article said "the clock is at rest with the observer", I was responding to your request to "please take the clock as the stationary frame wrt to the observer." (that's, y'know, why I quoted that exact statement before making the comment above) To me and every other physicist in the world, "take the clock as the stationary frame wrt to the observer" means exactly the same thing as "the clock at rest with the observer". Do they mean something different to you? If you see any difference between those two statements, then as I said above you must be using some ridiculous nonstandard definition of "stationary frame wrt to the observer".
That is simply the equation for all events on the light cone in the unprimed frame. If we determine which of these events also satisfy (x-vrγ/c)²/(rγ)² + y²/r² + z²/r² = 1, and find the corresponding time for each position, then we can use the LT to translate these positions and times into the primed frame and show that in the primed frame they all occur at the same time and for a sphere.chinglu1998 said:No, I mean t = √(x²+y²+z²)/c.
And how do you want to prove t'=r/c in the context of the primed frame, except by first finding the positions and times of all the events on the ellipsoid in the unprimed frame, then translate those events into the primed frame using the LT? If this is not the procedure you have in mind, you need to either explain more clearly or just show the full derivation you have in mind.chinglu1998 said:We are operating from the context of the unprimed frame proving t' = r/c in the context of the primed frame.
You're not making any sense!chinglu1998 said:Are you saying SR cannot use light aberration to check times from the clock frame? That would be a CPT violation.
Sowhat are the times when the clokc frame is taken as stationary.
Why are all you people terrified of answering this?
JesseM said:That is simply the equation for all events on the light cone in the unprimed frame. If we determine which of these events also satisfy (x-vrγ/c)²/(rγ)² + y²/r² + z²/r² = 1, and find the corresponding time for each position, then we can use the LT to translate these positions and times into the primed frame and show that in the primed frame they all occur at the same time and for a sphere.
And how do you want to prove t'=r/c in the context of the primed frame, except by first finding the positions and times of all the events on the ellipsoid in the unprimed frame, then translate those events into the primed frame using the LT? If this is not the procedure you have in mind, you need to either explain more clearly or just show the full derivation you have in mind.
If you have the context, why did you accuse me of the "mistake" of saying that the wiki article said "the clock at rest with the observer"? The context made clear I was talking about your statement, not the wiki article. If you're not a troll, please acknowledge that this was an unfounded accusation.chinglu1998 said:I have the context.
Then it's simple, in the observer's frame the light source is at rest so there is no aberration, therefore the light just travels on a vertical path of length L from the bottom to the top, so the time is [tex]\Delta t = \frac{L}{c}[/tex] to go from bottom to top.chinglu1998 said:I want to see your calculation with the clock at rest wrt to the observer.
All I'm asking is whether the method I suggested to prove it was the one you had in mind. If not, I don't understand what type of proof you're looking for. In any case, are you ever going to respond to [post=3072773]post #224[/post], where I asked at the end why you think I need to do this exercise, given that it would only be a mathematical demonstration of what I already said in words?chinglu1998 said:You can prove this yourself. At least you did not have to find it like I did, so it is easier for you.