It has been explained to you several times, the start event is not x'=k, t'=0.
The start event is x'=0, t'=0 and x=0, t=0.
Time Dilation
- Thread starter chinglu
- Start date
It has been explained to you several times, the start event is not x'=k, t'=0.
The start event is x'=0, t'=0 and x=0, t=0.
Your 316 is a summation of the same failed issue over and over.
You have failed to refute that the clock at k from the view of the unprimed frame is the moving clock and elapses -k/v in the view of both frames. Whereas, in the view of both frames, the clock at the unprimed origin elapses -k/(γv).
These calculations stand unrefuted.
Therefore, in the view of the unprimed frame, the moving clock at k is not time dilated relative to the unprimed origin clock from the start event to the end event.
You agreed that looking at the interval starting at x'=0 doesn't tell us about the primed clock at x'=k.
Looking at your interval only tells us about the elapsed time on the unprimed clock at x=0.
And all frames agree that the unprimed clock at x=0 elapses dt=-vk/γ between the two specified events.
If you want to determine the elapsed time on the primed clock at x'=k, you need to consider a different interval, such as one of these:
For the interval beginning at x'=k, t'=0 (x=kγ,t=vk/c^2), and ending at x'=k, x=0 (t'=-k/v, t=-vk/γ), all frames agree that the primed clock at x'=k elapses dt'=-k/v
For the interval beginning at x'=k, t=0 (x=k/γ,t'=-vk/c^2) and ending at x'=k, x=0 (t'=-k/v, t=-vk/γ), all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2
You seem to be unwilling to consider those intervals, chinglu. Why?
Don't you want to know the time elapsed on the primed clock at x'=k?
Chinglu, your basic contradiction is this:
Do you want to make claims about the elapsed time on the clock at x=0? Great!
What do you claim about that clock?
Do you want to make claims about the elapsed time on the clock at x'=k?
Then you're looking at the wrong interval.
- You want to make a claim about the time elapsed on the clock at x'=k
- You are looking at an interval attached to the clock at x=0
Do you want to make claims about the elapsed time on the clock at x=0? Great!
What do you claim about that clock?
Do you want to make claims about the elapsed time on the clock at x'=k?
Then you're looking at the wrong interval.
chinglu fails once again to respond to my challenges. For reference, chinglu must address one or both of the following issues before any further discussion will occur:
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
My bet: chinglu will do anything to avoid answering these challenges, especially number 2.
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
My bet: chinglu will do anything to avoid answering these challenges, especially number 2.
I'm fairly convinced that chinglu is just one more internet troll who has no desire to learn anything. All he wants is to keep reeling in the line for as long as he can. This is why I terminated our interactions earlier. I thought I'd give him a second chance when he reopened the topic, but that's looking like a mistake. So, no more interaction with him until he responds to my challenges.
That is fine with me. I am not interested in your nose hair challenges. They have nothing to do with this thread. It is amazing to me that you come into this thread and issue off topic challenges and call me a troll when I have provided all the math for this thread.
Here it is again. if you can't refute the math, then your are forced to accept it or admit it is beyond you.
I have made this simple. There is a clock at k in the context of the primed frame with k < 0.
The question is from the time the origins are the same to the time k and the unprimed origin is the same, what will be the time on the clock at k and what will be the time on the unprimed origin clock.
You then introduce some clock at k/ γ in the unprimed frame when that clock is not even in the problem.
I assume you understand when the origins are the same, t'=t=0.
Next, from the view of the primed frame, the unprimed origin is a distance k from the clock at the unprimed origin when the two origins are the same.
Hopefully, you can understand, the clock at k will elapse -k/v for the unprimed origin to reach k. Note all measurements are in the context of the primed frame so this should be simple with not disagreements.
Next, the final question is what is the elapsed time on the unprimed origin clock?
1) From the view of the primed frame, this is standard time dilation so, the primed frame concludes the unprimed origin clock elapses -k/(γv)
2) From the view of the unprimed frame, since the clock at k is measured in primed frame coordinates, then when the origins are the same, the unprimed frame concludes primed clock at k is a distance k/γ from the unprimed origin. Since the distance is k/γ to the unprimed origin and the speed is v, then the elapsed time from the view of the unprimed origin on the unprimed origin clock is -k/(γv).
Note how both frames agree on the time of the unprimed origin clock. There is no disagreement.
So far, we have both frame agree the unprimed origin clock will elapse -k/(γv) from the time the origins are the same to the time the clock at k meets the unprimed origin.
We have the primed frame claiming the clock at k will elapse -k/v.
Last, we need the conclusion of the unprimed frame for the clock at k. This is standard LT. k us measured in light seconds and v is measured in terms of c.
t = -k/(γv)
x = 0
t' = ( t - vx/c² )γ =( -k/(γv) - 0)γ = -k/v.
Let's makes sure we have the correct x'.
x' = ( x - vt )γ = ( 0 - v(-k/(γv))γ = k. Yes, we have the correct x'.
Therefore, both frames agree the clock at k will elapse -k/v.
Both frames agree the clock at the unprimed origin will elapse -k/(γv).
Finally, we take the view of the unprimed origin and the moving clock at k coming toward the origin beats time expanded as agreed on both frames. And frame agreement is necessary because of the invertibility of LT.
To summarize your error, you consider a clock in the unprimed frame at the same location as k when the origins were the same and placed some time on it. That clock at that location is not in any way involved in this problem and has no affect on the conclusions. We are only concerned with two clocks, the clock at the unprimed origin and the clock at k in the primed frame.
1) I agreed the start event is as I said. I could care less what is on the k clock at the start event. If you do, then you are confused.
2) "And all frames agree that the unprimed clock at x=0 elapses dt=-vk/γ between the two specified events."
Good, we have agreement here on the unprimed origin clock.
Here is your error.
If you want to determine the elapsed time on the primed clock at x'=k, you need to consider a different interval, such as one of these:
Whenever you deviate from the start and end interval as above, you are required to include a simultaneity shift based on the realativity of simultaneity. This proves you do not understand SR.
You mix apples and oranges.
The start and end intervals are as I gave them and none other. Learn to follow directions.
Here is your failure.
In SR, you have an agreed upon start event.
Mine is x'=0 t'=0, x=0 t=0. Now if this is false, prove it.
The end event when the unprimed origin and k clock are at the same place.
We then need to calculate the time for each frame.
If you believe the frames cannot agree on the event then prove it.
If you cannot prove these start and end events are false, you are forced to accept them.
Your opening post chinglu:
You are making a claim about the time elapsed on the primed clock at x'=k moving in the unprimed frame. But knowing the time it elapsed since it read zero doesn't tell you the time it elapsed since the start event, unless it read zero at the start event. So, you do need to care whether that clock read zero at the start event.
Yes chinglu, the reason that I suggested two alternative intervals is because of the simultaneity shift. I was avoiding describing it in that way, because you seemed to be confused by the concept earlier in the thread, calling it 'spooky action at a distance' ([post=2688731]post 241[/post]). Also because it can be considered as a failure of synchronization rather than a failure of simultaneity without affecting the resulting analysis.
Do you understand that the "spooky action at a distance desynchronization" you didn't like in James's [post=2665420]post 117[/post] is just the simultaneity shift?
So anyway, that's why I suggested two intervals. One starts simultaneously with the origins meeting in the primed frame, the other starts simultaneously with the origins meeting in the unprimed frame.
And as I said before ([post=2694156]post 311[/post]), the simultaneity shift is exactly why you can't talk about time elapsed on the x'=k clock in the unprimed frame when your interval starts at x'=0. You mix apples and oranges.
Here is where you implicitly use a different start event:
Wrong.chinglu said:
The proper time interval for the primed clock at x'=k obviously does not start at x'=0.
In the unprimed frame, the clock at x'=k is obviously at x=k/γ,t=0,x'=k,t'=-vk/c^2) when the origins meet at t=0.
Note the simultaneity shift.
The simultaneity shift means that the frames do not agree on the elapsed time for the primed clock during the given interval.
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So, the following events are simultaneous in the primed frame:
E1: (x',t')=(0,0); (x,t)=(0,0)
E2: (x',t')=(k,0)
For now, I omit the (x,t) coordinates of event E2, since you can't understand those.
In the primed frame, the unprimed origin travels at speed -v, so it takes the time you say in the primed frame: -k/v.
The end event is therefore:
E3: (x',t')=(k,-k/v)
The time interval between events E2 and E3 is -k/v in the primed frame. Similarly, the time interval between events E1 and E3 is -k/v in the primed frame.
The time interval between events E1 and E3 is DIFFERENT than the time interval between events E2 and E3 in the unprimed frame, even though these intervals are the same in the primed frame.
That depends on whether you want the elapsed time between events E1 and E3 or between E2 and E3.
In that case, you have calculated the time interval between events E1 and E3 in the unprimed frame. In that case, this is proper time interval in the unprimed frame, but it is not a proper time interval in the primed frame.
Since the proper time interval is expected to be shorter than the other time interval, we expect the time interval to be shorter in the unprimed frame. It is, so this is in accordance with special relativity.
This is one place you go wrong, since you now introduce event E2: (x',t')=(k,0), which says the clock is at x'=k in the primed frame at time t'=0.
Using the Lorentz transform on that event, we obtain:
E1: (x,t) = (γk,γkv/c^2).
So, when the origins are the same in the primed frame, the primed clock is at distance γk from the unprimed origin, and not k/γ as you claim. However, this does NOT occur at the time that the origins are the same in the unprimed frame.
The elapsed time depends on the start and the stop time in the unprimed frame. The start and end events in the unprimed frame are:
E1: (x,t)=(0,0)
E3: (x,t)=(0,-k/γv)
The time interval you have calculated is correct for this pair of events, even if your reasoning is wrong.
For events E1 and E3 the time intervals are:
$$\Delta t' = -k/v, \Delta t = -k/\gamma v$$
There is, as you say, no disagreement for that pair of events.
I note that all you have done here is apply the Lorentz transform to event E3, and you have done that correctly for that event.
We agree that the time interval between events E1 and E3 is longer in the primed frame than in the unprimed frame.
When you talk about the "moving clock" you are making an assumption about which clock is moving. Between events E1 and E3, the relevant clock is moving according to the primed frame, but stationary according to the unprimed frame. We can see this by observing that the x' coordinate of that clock changes between the two events, while the x coordinate remains the same. So, for these two events, it is best to regard the clock as "moving" in the primed frame.
To repeat: the primed frame watches the clock moving from x'=0 to x'=k. The unprimed frame has a stationary clock that remains at x=0 at all times.
Special relativity says that stationary clocks measure the proper time, so the proper time in this case is measured in the unprimed frame. Special relativity also states that the proper time should be shorter than the time in any other frame. This is true for this particular pair of events.
So, once again we conclude that there is no problem for special relativity here - only problems with chinglu's understanding of special relativity.
No. That would be to consider event E2. As you can see quite clearly from the current post, I have used the interval you specified: the one between events E1 and E3.
You, on the other hand, seem to have trouble distinguishing between events E1 and E2. You flip-flop from discussing one to discussing the other, without taking any care to make sure you know which one you're talking about at any given time.
Then you should not introduce it into the problem.
Are you now happy that when we use only events E1 and E3 we get the correct answers, in accordance with what special relativity predicts? Yes or no?
More specifically, we are only concerned with two events: E1 and E3, and we're concerned with the time interval between those two events as measured in both the primed and unprimed frames. We must be careful that if we're comparing time intervals we are always comparing the time interval between the same pair of events.
I end by repeating my challenges to chinglu, since they are still unanswered:
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
My bet: chinglu will do anything to avoid answering these challenges, especially number 2. So far, this is 100% confirmed.
A further small summary that combines my points and Pete's (hopefully) may help.
Diagram from the point of view of the primed frame (which sees the unprimed frame moving in the negative x' direction at speed v):
Note that "g" in the diagrams means $$\gamma$$, and that $$k<0$$, so that the second diagram shows a later time than the first.
From the initial situation the events we can immediately write down are:
E1: (x',t')=(0,0)
E2: (x',t')=(k,0)
Note that events E1 and E2 occur simultaneously in the primed frame. Also notice that the primed frame sees the distance $$\gamma k$$ between events E1 and E2 contracted to k, which makes sense because moving distances are seen as contracted and the primed frame sees the unprimed frame moving.
From the final situaton we immediately write down:
E3: (x',t')=(k,-k/v)
----
The Lorentz transformations of these events are:
E1: (x,t)=(0,0)
E2: $$(x,t)=(\gamma k, \gamma kv/c^2)$$
E3: $$(x,t)=(0,-k/\gamma v)$$
Here's the picture in the unprimed frame:
Things to note:
chinglu said:
Diagram from the point of view of the primed frame (which sees the unprimed frame moving in the negative x' direction at speed v):
Code:
t'=0:
-------k--------0------>x' t'=0
-------gk-------0------>x
E2 E1
t'=-k/v:
-------k--------0------>x' t'=-k/v
----------------0------>x
E3Note that "g" in the diagrams means $$\gamma$$, and that $$k<0$$, so that the second diagram shows a later time than the first.
From the initial situation the events we can immediately write down are:
E1: (x',t')=(0,0)
E2: (x',t')=(k,0)
Note that events E1 and E2 occur simultaneously in the primed frame. Also notice that the primed frame sees the distance $$\gamma k$$ between events E1 and E2 contracted to k, which makes sense because moving distances are seen as contracted and the primed frame sees the unprimed frame moving.
From the final situaton we immediately write down:
E3: (x',t')=(k,-k/v)
----
The Lorentz transformations of these events are:
E1: (x,t)=(0,0)
E2: $$(x,t)=(\gamma k, \gamma kv/c^2)$$
E3: $$(x,t)=(0,-k/\gamma v)$$
Here's the picture in the unprimed frame:
Code:
At t = gkv/c^2:
----------k------0------>x'
-------gk---------0------>x t=gkv/c^2
E2
At t=0:
----------k------0------>x'
-------gk--k/g----0------>x t=0
E1
At t=-k/gv:
----------k------0------>x'
-------gk--k/g----0------>x t= -k/gv
E3Things to note:
- Events E1 and E2 are not simultaneous in the unprimed frame. E1 occurs after E2.
- In the top diagram, we see that the distance between the clock at x'=k and the primed origin is shorter than the distance between that clock and the unprimed origin. The unprimed frame sees distances in the primed frame as being contracted, since the primed frame is moving according to the unprimed frame.
- chinglu's start event E1 occurs in the second picture at t=0. In this picture the clock at x'=k is now located at $$x=k/\gamma$$. The contraction of the distance from x'=k to x'=0 remains the same as in the first picture.
- chingu's end event E3 occurs at $$t=-k/\gamma v$$
You are changing simple to complex based on your lack of understanding of SR.
You have already agreed both frames claim the origin of the unprimed frame will elapse -k/(vγ).
Now, see if you are able to be an observer at k. At t'=0, the unprimed frame is a distance -k from you and moving toward you.
If that origin is a distance -k and moving at v, then your clock will elapse -k/v. This is very simple stuff.
Let me know when you have reached this simple point.
You wrote.
For events E1 and E3 the time intervals are:
$$\Delta t' = -k/v, \Delta t = -k/\gamma v$$
This is exactly what I was trying to teach you and you finally got it correct. Can you explain this to Pete?
Anyway, both frames agree, even with the unprimed origin at rest, the moving clock at k elapses more time on the interval than does the clock at the unprimed origin.
Hence, time dilation is false.
You have graduated to your first step in understanding SR is false.
chinlgu:
Your conclusion that time dilation is false does not follow from your thought experiment. I have clearly shown you this over and over again, including in the post you quoted in your latest reply. I even drew some nice diagrams to help you. What's your problem? Are you really too thick to understand, or are you just a troll?
Why haven't you responded to my challenges:
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
I know why you haven't responded. Because you cannot respond to question 1 without either being blatantly wrong or conceding that this entire thread is based on a misunderstanding of yours.
I'll tell you what, chinglu. I'll give you one chance to show some honesty in responding to Pete's latest posts and my latest posts. I eagerly await your response.
Pete and I have been endlessly patient with your trolling behaviour. In the past day alone I posted two large posts explaining exactly where you went wrong (again), and showing you the right answer. You have ignored those to respond with a couple of lines of arrogant nonsense. Didn't your mother teach you any manners?
Your conclusion that time dilation is false does not follow from your thought experiment. I have clearly shown you this over and over again, including in the post you quoted in your latest reply. I even drew some nice diagrams to help you. What's your problem? Are you really too thick to understand, or are you just a troll?
Why haven't you responded to my challenges:
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
I know why you haven't responded. Because you cannot respond to question 1 without either being blatantly wrong or conceding that this entire thread is based on a misunderstanding of yours.
I'll tell you what, chinglu. I'll give you one chance to show some honesty in responding to Pete's latest posts and my latest posts. I eagerly await your response.
Pete and I have been endlessly patient with your trolling behaviour. In the past day alone I posted two large posts explaining exactly where you went wrong (again), and showing you the right answer. You have ignored those to respond with a couple of lines of arrogant nonsense. Didn't your mother teach you any manners?
Didn't your mother teach you intelligence?
You gave me a warning.
Here is the OP.
Have coordinate at x' = -5/6ls.
What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0.
t' = ( t - vx/c² )γ
t' = ( t - 0 )γ = tγ.
So, moving clock coming toward rest origin beat not time dilated.
Is this wrong?
So, can you explain exactly why my OP as I said it above is inconsistent with the post you warned me on.
When I prove you are wrong, I expect the warning to be removed.
Moderator note: chinglu has been banned from sciforums for 3 days for trolling.
------------
When you get back, chinlgu, you will need to address at least one of the following points:
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
3. Go through Pete's post #332 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in that post.
4. Address post #334, which currently remains unexamined by you.
5. Go through post #333 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in that post.
See you in three days time, chinglu. Let's hope we start getting some honesty from you then.
------------
When you get back, chinlgu, you will need to address at least one of the following points:
1. Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events? Yes or no?
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in those posts.
3. Go through Pete's post #332 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in that post.
4. Address post #334, which currently remains unexamined by you.
5. Go through post #333 line-by-line, saying clearly whether you agree or disagree with each step and explaining clearly any errors you can find in that post.
See you in three days time, chinglu. Let's hope we start getting some honesty from you then.
So you agree that the starting event is at x'=k, t'=0?
Yes, this has been said several times.
James said it as early as [post=2665420]post 117[/post], and as recently as a couple of posts ago.
I spelled it out several times, eg in [post=2693649]post 307[/post]:
Pete said:
Enjoy your break. You might like to take the time to think carefully about exactly what events define the beginning and end of the elapsed time on the clock.
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