Yes, you still have it wrong.
Time Dilation
- Thread starter chinglu
- Start date
Yes, you still have it wrong.
Events 2 and 3 have both a temporal and a spatial separation. You can't use such events in order to demonstrate time dilation. In a few other posts, I showed how you demonstrate time dilation correctly.
In the unprimed frame, events 1 and 3 occur at the same spatial location. In the primed frame events 2 and 3 occur at the same spatial location.
Tell that to chinglu. I haven't attempted to use these events to demonstrate time dilation.
Well, thanks for your efforts, but I already know all about that.
Then, the whole exercise is pointless.
Well, I can show you how to do this correctly. It takes much fewer steps than your approach , making it immune to errors. Do you want to see?
In fact, if we look at my post #96, we see a rather nice demonstration of time dilation.
Consider the primed frame to be "stationary" and the unprimed frame to be "moving". (Note: this is the opposite of what is usual done in this kind of demonstration.)
Now look at events 1 and 3.
1. The origins of the primed and unprimed frames coincide at t=t'=0.
3. The coordinate x'=-k coincides with the coordinate x=0 at some later time.
From the point of view of an observer in the primed frame, the unprimed (x) axis moves in the negative x direction with speed v. The time it takes to move distance k is k/v as measured by clocks in the primed frame.
As post #96 shows, the time interval between these two events in the unprimed frame is SHORTER by the Lorentz factor. But in the unprimed frame these two events occur at the same spatial location. Hence, it is the unprimed frame here that measures the proper time. And we see that the proper time is shorter than the time in the other frame.
In other words, this confirms that moving clocks run slow.
Neat, eh?
Consider the primed frame to be "stationary" and the unprimed frame to be "moving". (Note: this is the opposite of what is usual done in this kind of demonstration.)
Now look at events 1 and 3.
1. The origins of the primed and unprimed frames coincide at t=t'=0.
3. The coordinate x'=-k coincides with the coordinate x=0 at some later time.
From the point of view of an observer in the primed frame, the unprimed (x) axis moves in the negative x direction with speed v. The time it takes to move distance k is k/v as measured by clocks in the primed frame.
As post #96 shows, the time interval between these two events in the unprimed frame is SHORTER by the Lorentz factor. But in the unprimed frame these two events occur at the same spatial location. Hence, it is the unprimed frame here that measures the proper time. And we see that the proper time is shorter than the time in the other frame.
In other words, this confirms that moving clocks run slow.
Neat, eh?
It's not my fault that you can't see the point. I think I've rather neatly shown where chinglu went wrong - something you failed to do over the course of 80 posts or so.
Sure. It would make a nice change for you to actually do something, rather than simply claim that other people are making mistakes.
You can't prove chinglu wrong. No one can. You will not be able to do that, just wait for his answers.
Well, you made quite a few mistakes in your post 85 and I showed you how to correct them. I think that this is pretty constructive.
Tach:
Let's see.
I made a mistake in post #85.
I admitted the mistake in post #95 and posted a revised version of post #85 in post #96.
Between posts #86 and #94, you repeatedly told me I have made a mistake, although you were initially unsure where I had gone wrong. In fact, in post #90, you made your own mistake by getting the sign wrong in the Lorentz transformation.
Nowhere between #86 and #94 do I see you "showing me" how to correct my mistake. All you did was to assert repeatedly that there was a mistake there somewhere. I found what it was myself.
You still haven't admitted your own mistake in #90. Will you now do that?
Let's see.
I made a mistake in post #85.
I admitted the mistake in post #95 and posted a revised version of post #85 in post #96.
Between posts #86 and #94, you repeatedly told me I have made a mistake, although you were initially unsure where I had gone wrong. In fact, in post #90, you made your own mistake by getting the sign wrong in the Lorentz transformation.
Nowhere between #86 and #94 do I see you "showing me" how to correct my mistake. All you did was to assert repeatedly that there was a mistake there somewhere. I found what it was myself.
You still haven't admitted your own mistake in #90. Will you now do that?
You made two mistakes, you got either 2 or 3 wrong and you also got $$\Delta t$$ wrong.
I guessed you inverse Lorentz transform wrong. If you use one sign, then your 2 is wrong, if you use the other sign, then your 3 is wrong. Eiither way, one of the events is wrong, you take your pick. So is your $$\Delta t$$. Next time I will not try to guess your mistakes, I will ask you outright to show your steps.
Also, Tach, it turns out you were also wrong in post #103, where you claimed that events 2 and 3 cannot be used to demonstrate time dilation. Since those two events both occur at the same spatial location in the primed frame, the time interval between them is a proper time in the primed frame. Therefore, they can be used to demonstrate time dilation.
Do you agree?
For comparison, see post #106.
I'd say I've given quite a nice demonstration of time dilation here. Wouldn't you?
Do you agree?
For comparison, see post #106.
I'd say I've given quite a nice demonstration of time dilation here. Wouldn't you?
Well, a single sign error in the time of event 2 led to a consequential error in the calculation of $$\Delta t = t_3 - t_2$$.
You can call that 2 mistakes if you like, but if I was marking an exam and a student did that I wouldn't be penalising them twice.
Well, considering that you are a moderator that should be held to higher ethical and scientific standards and that you were so aggressive towards me in denying any wrongs, I would .
This is a bad start, there is no such thing as "stationary" frame in SR.
This is hard to follow, it needs the math that goes with it.
While correct, this is a bad conclusion, you want to demonstrate time dilation , not time contraction. You don't want to start justifying why the contraction is actually a dilation.
Not very, try putting all this in a mathematical formulation, let's see how it looks.
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Ok. Here we go...
We have a primed frame with event coordinates (x',t'), and an unprimed frame with coordinates (x,t). In the unprimed coordinates, the primed frame moves in the positive x direction with speed v. (Note: in what follows, v is always taken to be a positive constant.) In the primed coordinates, the unprimed frame moves in the negative x' direction with speed v.
Let's look at the Lorentz transformations. Note that we are considering THREE events here:
1. The origins of the primed and unprimed frames coincide at t=t'=0.
2. Some irrelevant object (e.g. a clock) is initially located at position x'=-k in the primed frame at t'=0.
3. The coordinate x'=-k coincides with the coordinate x=0 at some later time.
In the primed frame, the spacetime coordinates of these three events are:
1. (x',t') = (0,0)
2. (x',t') = (-k,0)
3. (x',t') = (-k, k/v)
The coordinates of event 3 follow from the fact that in the primed frame the unprimed x axis moves a distance k at speed v in time k/v, where the unprimed axis moves in the negative x' direction in the primed frame.
Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame as follows:
$$x = \gamma (x' + vt'); t = \gamma (t' + vx'/c^2); \gamma = \frac{1}{\sqrt{1-(v/c)^2}$$
1. (x,t) = (0,0)
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$
3. $$(x,t)=(0,\frac{k}{\gamma v})$$
In the unprimed frame the time intervals between pairs of events (calculated by subtracting the time coordinate of one event from the other) are:
1 and 3: $$\Delta t = \frac{k}{\gamma v}$$
2 and 3: $$\Delta t = \frac{\gamma k}{v}$$
In the primed frame, the corresponding intervals are:
1 and 3: $$\Delta t' = \frac{k}{v}$$
2 and 3: $$\Delta t' = \frac{k}{v}$$
Now, in the unprimed frame, the spatial coordinates of events 1 and 3 are the same, whereas in the primed frame they are different. So, the time measured between 1 and 3 in the unprimed frame is a proper time. Note that the time interval between events 1 and 3 in the primed frame is LONGER than in the unprimed frame by a factor of $$\gamma$$.
Relativity tells us that the proper time is always the SHORTEST time interval between two events, so this is the result we expect.
Similarly, in the primed frame, the spatial coordinates of events 2 and 3 are the same, so the time interval between those events in the primed frame is a proper time. Comparing the time interval between events 2 and 3 in the unprimed frame, we again see that it is LONGER by a factor of $$\gamma$$. So, again, we see that the proper time is the shorter time of the two.
All of this is consistent with the relativistic result often expressed sloppily as "moving clocks run slow".
Considering events 1 and 3, a clock has to move in the primed frame to record both events at different locations, so for those events the primed frame is "moving", while the unprimed frame is "stationary".
Considering events 2 and 3, a clock has to move in the unprimed frame to record both events at different locations, so for those events the unprimed frame is "moving", while the primed frame is "stationary".
The mathematics shows that in both cases the "moving" clocks run slow - just as Einstein said.
We have a primed frame with event coordinates (x',t'), and an unprimed frame with coordinates (x,t). In the unprimed coordinates, the primed frame moves in the positive x direction with speed v. (Note: in what follows, v is always taken to be a positive constant.) In the primed coordinates, the unprimed frame moves in the negative x' direction with speed v.
Let's look at the Lorentz transformations. Note that we are considering THREE events here:
1. The origins of the primed and unprimed frames coincide at t=t'=0.
2. Some irrelevant object (e.g. a clock) is initially located at position x'=-k in the primed frame at t'=0.
3. The coordinate x'=-k coincides with the coordinate x=0 at some later time.
In the primed frame, the spacetime coordinates of these three events are:
1. (x',t') = (0,0)
2. (x',t') = (-k,0)
3. (x',t') = (-k, k/v)
The coordinates of event 3 follow from the fact that in the primed frame the unprimed x axis moves a distance k at speed v in time k/v, where the unprimed axis moves in the negative x' direction in the primed frame.
Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame as follows:
$$x = \gamma (x' + vt'); t = \gamma (t' + vx'/c^2); \gamma = \frac{1}{\sqrt{1-(v/c)^2}$$
1. (x,t) = (0,0)
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$
3. $$(x,t)=(0,\frac{k}{\gamma v})$$
In the unprimed frame the time intervals between pairs of events (calculated by subtracting the time coordinate of one event from the other) are:
1 and 3: $$\Delta t = \frac{k}{\gamma v}$$
2 and 3: $$\Delta t = \frac{\gamma k}{v}$$
In the primed frame, the corresponding intervals are:
1 and 3: $$\Delta t' = \frac{k}{v}$$
2 and 3: $$\Delta t' = \frac{k}{v}$$
Now, in the unprimed frame, the spatial coordinates of events 1 and 3 are the same, whereas in the primed frame they are different. So, the time measured between 1 and 3 in the unprimed frame is a proper time. Note that the time interval between events 1 and 3 in the primed frame is LONGER than in the unprimed frame by a factor of $$\gamma$$.
Relativity tells us that the proper time is always the SHORTEST time interval between two events, so this is the result we expect.
Similarly, in the primed frame, the spatial coordinates of events 2 and 3 are the same, so the time interval between those events in the primed frame is a proper time. Comparing the time interval between events 2 and 3 in the unprimed frame, we again see that it is LONGER by a factor of $$\gamma$$. So, again, we see that the proper time is the shorter time of the two.
All of this is consistent with the relativistic result often expressed sloppily as "moving clocks run slow".
Considering events 1 and 3, a clock has to move in the primed frame to record both events at different locations, so for those events the primed frame is "moving", while the unprimed frame is "stationary".
Considering events 2 and 3, a clock has to move in the unprimed frame to record both events at different locations, so for those events the unprimed frame is "moving", while the primed frame is "stationary".
The mathematics shows that in both cases the "moving" clocks run slow - just as Einstein said.
I should also add that the above post shows what's wrong with chinglu's picture from post #6 of this thread.
In this post, chinglu is comparing what I have called events 1 and 3. He concludes, correctly, that the time interval between events 1 and 3 in the primed frame is longer than between those two events in the unprimed frame.
Since these kinds of problems mostly start with a proper time in the primed frame, the usual conclusion is that the primed clocks ran "slow" compared to the unprimed clocks.
However, chinglu has managed to choose events such that the proper time is actually measured in the unprimed frame, in which case we expect the unprimed frame to run "slow".
As you can see clearly from my calculations, there is no problem for relativity here. The results are entirely consistent with the special theory of relativity.
The lesson chinglu needs to take away from this is to be careful about what is and is not a proper time interval.
I'll be happy to accept the thanks of chinglu and Tach for producing such a comprehensive and clear explanation and solving the question of the thread. Let's see if either of those two posters have any sense of manners and good grace.
In this post, chinglu is comparing what I have called events 1 and 3. He concludes, correctly, that the time interval between events 1 and 3 in the primed frame is longer than between those two events in the unprimed frame.
Since these kinds of problems mostly start with a proper time in the primed frame, the usual conclusion is that the primed clocks ran "slow" compared to the unprimed clocks.
However, chinglu has managed to choose events such that the proper time is actually measured in the unprimed frame, in which case we expect the unprimed frame to run "slow".
As you can see clearly from my calculations, there is no problem for relativity here. The results are entirely consistent with the special theory of relativity.
The lesson chinglu needs to take away from this is to be careful about what is and is not a proper time interval.
I'll be happy to accept the thanks of chinglu and Tach for producing such a comprehensive and clear explanation and solving the question of the thread. Let's see if either of those two posters have any sense of manners and good grace.
Because the time interval from start to end is longer for the primed frame interval for the motion over the unprimed.
This mean moving clock beat time expanded.
Only 2 events. (2.) is a consequence of (1.) It is not an event. The rest of post is error.