Time Dilation
- Thread starter chinglu
- Start date
chinglu:
To know how long something takes to move from one place to another, you need to know the time it started moving and the time it finished moving. Those two times are two separate events in spacetime.
Well, let's see. You're saying that when $$(x,t)=(0,0)$$ and $$(x',t')=(0,0)$$ (so that the origins are the same) then another event occurs at $$(x,t)=(k/\gamma,0)$$. That's in the unprimed frame.
But that other event does not occur at $$t'=0$$!
Check it for yourself by calculating the Lorentz transformed coordinates of the event. Do you know how to do that? If not, see posts #117 and #118 where I explicitly gave you all the maths you need.
But ultimately you want to compare the time taken in the primed frame and in the unprimed frame for the same events, don't you? If you don't use the same events, but think you are using the same events, then you'll make a mistake and reach an incorrect conclusion, just like you did before. I explained where and why you made the error in posts #117 and #118, as you will recall.
To know how long something takes to move from one place to another, you need to know the time it started moving and the time it finished moving. Those two times are two separate events in spacetime.
Well, let's see. You're saying that when $$(x,t)=(0,0)$$ and $$(x',t')=(0,0)$$ (so that the origins are the same) then another event occurs at $$(x,t)=(k/\gamma,0)$$. That's in the unprimed frame.
But that other event does not occur at $$t'=0$$!
Check it for yourself by calculating the Lorentz transformed coordinates of the event. Do you know how to do that? If not, see posts #117 and #118 where I explicitly gave you all the maths you need.
But ultimately you want to compare the time taken in the primed frame and in the unprimed frame for the same events, don't you? If you don't use the same events, but think you are using the same events, then you'll make a mistake and reach an incorrect conclusion, just like you did before. I explained where and why you made the error in posts #117 and #118, as you will recall.
James R
All I am trying to get you to do here is tell the the distance -k in the moving coordinates is from the origin when the origins are the same. We are not trying to determine the anything about the clock time at that coordinate in the moving system.
We are only calculating the distance to the origin in the view of unprimed frame when origins are same.
So, what is answer? I calculate the distance is k/γ based on length contraction. What do you calculate for the distance?
chinglu:
In the moving frame, you're talking about two events here. The moving frame origin at $$(x',t')=(0,0)$$ and the event $$(x',t')=(-k,0)$$.
Now, if the primed origin is at $$x=0$$ at time $$t=0$$ in the unprimed coordinates, then the event $$(x',t')=(-k,0)$$ is at $$(x,t)=(-\gamma k, -\frac{\gamma kv}{c^2})$$.
Note that this event occurs at a time in the unprimed frame when the origins are not co-located, even though the same event occurs simultaenously with the origins being co-located in the primed frame.
This is because events that spatially separated and simultaneous in one frame cannot be simultaenous in any other frame.
I explained this to you in detail in posts #117 and #118, above.
Now you're talking about the event $$(x,t)=(-\frac{k}{\gamma},0)$$, which as you can see from above, is NOT the same event as the one at $$(x',t')=(-k,0)$$.
The event $$(x,t)=(-\frac{k}{\gamma},0)$$ is the same as $$(x',t')=(-k,\frac{kv}{c^2})$$
Your mistake is that you are confusing two different events and thinking they are the same.
To summarise, here are two events, with their spacetime coordinates in both frames:
Event 2: $$(x',t')=(-k,0); (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2})$$.
Event 3: $$(x',t')=(-k,\frac{kv}{c^2}); (x,t)=(-\frac{k}{\gamma},0)$$
Note that event 2 occurs when the origins are co-located in the primed frame, and event 3 occurs when the origins are co-located in the unprimed frame.
In the moving frame, you're talking about two events here. The moving frame origin at $$(x',t')=(0,0)$$ and the event $$(x',t')=(-k,0)$$.
Now, if the primed origin is at $$x=0$$ at time $$t=0$$ in the unprimed coordinates, then the event $$(x',t')=(-k,0)$$ is at $$(x,t)=(-\gamma k, -\frac{\gamma kv}{c^2})$$.
Note that this event occurs at a time in the unprimed frame when the origins are not co-located, even though the same event occurs simultaenously with the origins being co-located in the primed frame.
This is because events that spatially separated and simultaneous in one frame cannot be simultaenous in any other frame.
I explained this to you in detail in posts #117 and #118, above.
Now you're talking about the event $$(x,t)=(-\frac{k}{\gamma},0)$$, which as you can see from above, is NOT the same event as the one at $$(x',t')=(-k,0)$$.
The event $$(x,t)=(-\frac{k}{\gamma},0)$$ is the same as $$(x',t')=(-k,\frac{kv}{c^2})$$
Your mistake is that you are confusing two different events and thinking they are the same.
To summarise, here are two events, with their spacetime coordinates in both frames:
Event 2: $$(x',t')=(-k,0); (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2})$$.
Event 3: $$(x',t')=(-k,\frac{kv}{c^2}); (x,t)=(-\frac{k}{\gamma},0)$$
Note that event 2 occurs when the origins are co-located in the primed frame, and event 3 occurs when the origins are co-located in the unprimed frame.
We are not writing about the moving frame.
Are you going to give the distance of moving k in the unprimed frame?
If not, you will have to conceed to my calculations.
In the unprimed frame, that distance is k/γ when the origins are same.
You can not disagree since that mean you refute length contraction.
I guess you can not know this value. It might be too hard.
Yes we are. You are assuming the primed frame moves at speed v in the unprimed frame. Your original question asked when the x'=-k coordinate was in the same place as the x=0 coordinate, assuming the x'=0 coordinate coincides with x=0 at t=t'=0.
I already did that in post #117, and repeated myself in my post immediately before this one.
Yes, when the origins are the same in the unprimed frame. Of course, at that particular time the origins are NOT the same in the primed frame.
I don't refute length contraction. As I explained to you in post #118, you have to be careful to apply length contraction the "right way" round. You need to identify which length is the proper length and which is not. Your mistake is that you got it wrong. You assumed something is a proper length when it is not.
Your job is to refute posts #117 and #118, if you can. Your insults won't work.
Maybe we can make this simpler for both of us.
If the moving frame has a rod of length k, how long is that rod in the view of the rest frame?
Does its position in the moving frame change its length in the view of the rest frame?
Now, what if I place one end of the rod at the origin of the moving frame and the otther along the negative x-axis, does this matter?
Can you answer these questions?
If you are unable to answer these simple questions, let's just go our separate ways.
chinglu:
The rod has length $$\gamma k$$ in the rest frame of the rod if the frame that sees the rod moving measures length k. This result is derived by considering spacetime events at the two ends of the rod in each frame. Notice that, once again, three events are involved, not two, because events at opposite ends of the rod that are simultaneous in one frame are NOT simultaneous in the other, and to measure a length we need two simultaneous ends of the rod.
No.
No, it doesn't matter.
I've answered all your questions. I showed you exactly where you went wrong, first in posts #117 and #118, and then again a few posts above this one.
Are you unable to comprehend the explanations I have given to you? If you are unable to understand how to measure lengths, then let's just go our separate ways.
The rod has length $$\gamma k$$ in the rest frame of the rod if the frame that sees the rod moving measures length k. This result is derived by considering spacetime events at the two ends of the rod in each frame. Notice that, once again, three events are involved, not two, because events at opposite ends of the rod that are simultaneous in one frame are NOT simultaneous in the other, and to measure a length we need two simultaneous ends of the rod.
No.
No, it doesn't matter.
I've answered all your questions. I showed you exactly where you went wrong, first in posts #117 and #118, and then again a few posts above this one.
Are you unable to comprehend the explanations I have given to you? If you are unable to understand how to measure lengths, then let's just go our separate ways.
James R:
Yes, I do understand your posts. That is not what I am doing.
We have not resolved the simple matter of length contraction yet.
If the coordinate is -k in the moving system, then it is length contracted in the view of the rest coordinates, hence its length is k/γ in the view of the rest system.
You wrote k γ is the length in the rest system. Did you mean this or what it a typing mistake? This is not length contraction of a moving rod in the view of the rest frame.,
chinglu:
Coordinates are not length contracted. Spatial intervals are length contracted.
Every length has two ends to it. Each end is one event in spacetime. A length is the spatial interval between two events. A proper length is a spatial interval between two events that occur at the same time. Therefore, a proper length in one reference frame CANNOT be a proper length in any other frame due to the relativity of simultaneity.
I was very specific. I used the spacetime events that you specified. The length k that you gave was not a proper length, but a contracted length. Changing frames in that case gives the proper length, which is longer by a factor of $$\gamma$$.
Your mistake, as I pointed out in post #117 was to confuse which length is a proper length and which is not. You're making the same mistake here.
Coordinates are not length contracted. Spatial intervals are length contracted.
Every length has two ends to it. Each end is one event in spacetime. A length is the spatial interval between two events. A proper length is a spatial interval between two events that occur at the same time. Therefore, a proper length in one reference frame CANNOT be a proper length in any other frame due to the relativity of simultaneity.
I was very specific. I used the spacetime events that you specified. The length k that you gave was not a proper length, but a contracted length. Changing frames in that case gives the proper length, which is longer by a factor of $$\gamma$$.
Your mistake, as I pointed out in post #117 was to confuse which length is a proper length and which is not. You're making the same mistake here.
You can read here if L0 = x2' - x1' is measured as a distance in the moving system, then L0/γ is the measured distance in the stationary frame.
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html
Since we have L0 = 0 - (-k) = k as the diatance in the moving system, then k/γ is the distance in the stationary system.
Here is another link
http://physics.tamuk.edu/~hewett/Mo...ativity/RelativeView/Space/Length/Length.html
chinglu:
The problem is that you need to decide which frame is "stationary".
For example, suppose I watch a plane flying past as I stand on the ground. As far as I'm concerned, I am stationary and the plane is length contracted. But a person on the plane thinks he is stationary and I am length contracted.
The thing is: any frame can be regarded as stationary, so working out what is a contracted length and what is an uncontracted length isn't always obvious.
That's why you need to work out who is measuring the proper length or rest length, and who is not, whenever you deal with a length.
The problem is that you need to decide which frame is "stationary".
For example, suppose I watch a plane flying past as I stand on the ground. As far as I'm concerned, I am stationary and the plane is length contracted. But a person on the plane thinks he is stationary and I am length contracted.
The thing is: any frame can be regarded as stationary, so working out what is a contracted length and what is an uncontracted length isn't always obvious.
That's why you need to work out who is measuring the proper length or rest length, and who is not, whenever you deal with a length.
James R:
I thought we already decided which frame we are taking as stationary for right now, it is the unprimed frame.
So, k < 0 is in the measurements of the primed frame and it is moving.
Now that we have resolved this issue, then I must assume you agree
Δt1 = (k/γ)/v is the time it takes for k to move to the unprimed origin based on simple length contraction.
If an object starts at $$x=-\frac{k}{\gamma}$$ at time $$t=0$$, then I agree with you that if it travels in the +x direction at speed v then it reaches $$x=0$$ at time $$t=\frac{k}{\gamma v}$$.
This is different from your initial scenario, where the coordinate was not at $$x=-\frac{k}{\gamma}$$ at t=0.
If you're not sure about that, please read posts #117 and #118 again, where I explained it all to you in detail.
Alternatively, look at post #165:
James R said:
Now, that we have that straight that the unprimed frame will conclude
Δt = (k/γ)/v
let's calculate the view of the primed frame now.
It sees the origin frame move to the coordinate -k.
It calculates
Δt' = k/v.
Unprimed Frame
So, these from the time the origins are same to the time k reaches the unprimed origin, Δt = (k/γ)/v
Primed Frame
So, these from the time the origins are same to the time unprimed origin reaches k, Δt' = k/v
There is agreement on the stop and start of these two events between the frames.
We can understand Δt = Δt'/γ.
So, in view of unprimed frame at rest, Δt = Δt'/γ for the coordinate k in the moving system to reach the unprimed origin. Thus, the moving clock beats time expanded.
Let's use Lorentz Transforms to check out work.
We have x=0 because at origin in unprimed frame.
t' = ( t - vx/c² )γ = t γ, so t = t'/γ, just as we calculated.
No time dilation for clock moving to the rest origin when in coordinates of moving frame.
We just proved it.
chinglu:
Ok. But first let's make sure we know which events we're dealing with. The time interval you give here starts at t=0, when the origins of the primed and unprimed frames are in the same place. And at t=0 you're interested in the other point located at $$(x,t)=(-\frac{k}{\gamma},0)$$.
Now, let's set the clock at the origin in the primed frame to read t'=0 when the origins of the two frames are in the same place. Now, we calculate the coordinates of the other event:
$$(x,t)=(-\frac{k}{\gamma},0); (x',t') = (-k, \frac{kv}{c^2})$$
Notice that the event at $$x=-k/\gamma$$ does NOT occur at t'=0 in the primed frame.
Yes.
Yes. This is the time interval for the origin of the unprimed frame to move to the primed coordinate x'=-k.
Now you're talking about different events.
If you want to know where the primed coordinate x'=-k is at t=0 in the unprimed frame you need to calculate it:
$$x=\gamma (x' + vt') = \gamma x' =-\gamma k$$
No. In the primed frame, the distance travelled by unprimed origin is $$\gamma k$$, which takes time
$$\Delta t' = \frac{\gamma k}{v}$$
Which two events? You've used three events here.
Which is the "moving" clock?
It's better not to make assumptions about who is moving and who is not. Better to make sure you clearly specify which spacetime events you're talking about.
So the event you're transforming here is located at x=0 at some time t=t. Note that you're not transforming a time interval here, but a coordinate.
It's not clear what you proved. You mixed up two different events, for a start.
Ok. But first let's make sure we know which events we're dealing with. The time interval you give here starts at t=0, when the origins of the primed and unprimed frames are in the same place. And at t=0 you're interested in the other point located at $$(x,t)=(-\frac{k}{\gamma},0)$$.
Now, let's set the clock at the origin in the primed frame to read t'=0 when the origins of the two frames are in the same place. Now, we calculate the coordinates of the other event:
$$(x,t)=(-\frac{k}{\gamma},0); (x',t') = (-k, \frac{kv}{c^2})$$
Notice that the event at $$x=-k/\gamma$$ does NOT occur at t'=0 in the primed frame.
chinglu said:
Yes.
Yes. This is the time interval for the origin of the unprimed frame to move to the primed coordinate x'=-k.
Now you're talking about different events.
If you want to know where the primed coordinate x'=-k is at t=0 in the unprimed frame you need to calculate it:
$$x=\gamma (x' + vt') = \gamma x' =-\gamma k$$
No. In the primed frame, the distance travelled by unprimed origin is $$\gamma k$$, which takes time
$$\Delta t' = \frac{\gamma k}{v}$$
Which two events? You've used three events here.
Which is the "moving" clock?
It's better not to make assumptions about who is moving and who is not. Better to make sure you clearly specify which spacetime events you're talking about.
So the event you're transforming here is located at x=0 at some time t=t. Note that you're not transforming a time interval here, but a coordinate.
It's not clear what you proved. You mixed up two different events, for a start.
No, I did not mix up anything.
Let's keep it simple.
Note that you're not transforming a time interval here, but a coordinate.
Are you talking about a time coordinate? Be specific is it a GMT time or an elapsed time?
What do you think time means under Lorentz transforms?
Unprimed Frame
So, these from the time the origins are same to the time k reaches the unprimed origin, Δt = (k/γ)/v
Primed Frame
So, these from the time the origins are same to the time unprimed origin reaches k, Δt' = k/v
I wrote above. True or false.
chinglu:
Yes you did. I have now explained it to you in posts #117, #118, #165 and #177. It seems you're a slow learner, because you keep repeating the same mistake over and over again.
Let's not dumb it down so much that we start to make mistakes.
Every event in spacetime has coordinates. To specify the coordinates of an event, you need a location and the time the event occurs. For example, (x,t) or (x',t'). Coordinates for the same event are generally different in different frames.
An interval is the difference between two coordinates. For example, a time interval is $$\Delta t = t_1 - t_2$$ where $$t_1, t_2$$ are the times of two different events. A length (space interval) is $$\Delta x = x_1 - x_2$$ where $$x_1, x_2$$ are the two locations in space where two different events occur. You can also have a spacetime interval, which takes both the x and t coordinates into account.
Time is a coordinate. The Lorentz transformations tell us how to transform a coordinate from one frame to another.
Read post #177! I went over it all in detail for you.
Yes you did. I have now explained it to you in posts #117, #118, #165 and #177. It seems you're a slow learner, because you keep repeating the same mistake over and over again.
Let's not dumb it down so much that we start to make mistakes.
Every event in spacetime has coordinates. To specify the coordinates of an event, you need a location and the time the event occurs. For example, (x,t) or (x',t'). Coordinates for the same event are generally different in different frames.
An interval is the difference between two coordinates. For example, a time interval is $$\Delta t = t_1 - t_2$$ where $$t_1, t_2$$ are the times of two different events. A length (space interval) is $$\Delta x = x_1 - x_2$$ where $$x_1, x_2$$ are the two locations in space where two different events occur. You can also have a spacetime interval, which takes both the x and t coordinates into account.
Time is a coordinate. The Lorentz transformations tell us how to transform a coordinate from one frame to another.
Read post #177! I went over it all in detail for you.
Let's try to remember what you said.
You wrote here in the primed frame
∆t' = k/v.
http://www.sciforums.com/showpost.php?p=2668002&postcount=158
You wrote here in the unprimed frame
∆t = (k/γ)/v.
http://www.sciforums.com/showpost.php?p=2669157&postcount=175
I assume the frames agree on the start event when the frames are the same.
We also have your previous agreements above.
Now, there is only the question as to whether when k and origin of unprimed are same in primed frame and when k and unprimed origin are same in unprimed frame.
If you disagree, perhaps you can explain why these events are not the same. If you try to use time, then you are confused since time does not work between the frames.
Note that your calculations try to cross frames and predict some time on the clock in the primed frame when calculating in the unprimed frame. We do not need this. We simply look at each frame's calculations individually.
This method you use above is called frame mixing.