Time Dilation

[chinglu]

This is not what he calculated, you do not understand what he's doing.
He is claiming that :

$$t'=\gamma (t-\frac{vx}{c^2})$$

accepts a solution of the form

$$t'=t$$

(which is true but irrelevant) thus disproving time dilation (which is obviously false).


As an alternative, he uses the other Lorentz transform:

$$x'=\gamma(x-vt)$$ and , by making $$x'=-x$$ he concludes, just as incorrectly, that he has disproven time dilation.
Chinglu does not even begin to understand the notion of $$\Delta$$ as in $$\Delta x$$ or $$\Delta t$$. He has lied about having taken calculus.

I take this one step at time.

I am not doing that yet.

As a recovering Δx = 0 addict, should you first come clean that you are wrong that this describes the motion of x' moving to the origin of the rest frame?

 

[chinglu]

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.

You are wrong.

You cannot prove time dilation without the corresponding motion that describes the event based on the relativity postulate.

Each frame is watching some event of motion and each frame is recording the time required for that motion to beginning to end.

That is how time dilation is calculated. It is the ratio of these time intervals for two agreed upon events of beginning to end.

 

[chinglu]

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.

Very good.

Now, accept what you have figured out. I reversed the normal time dilation logic.

I took the primed frame as the origin and stationary with x' = -vt'. This proves t'/γ = t. So, the primed origin claims time dilation as expected.

This is where you are.

Now must continue with all thoughts.

In the view of the unprimed frame, x' moved to the unprimed origin. It was length contracted.

So t' = tγ. When you now take the unprimed frame as stationary, the unprimed origin concludes t' is time expanded. That is the problem.

The moving x' beat faster than the unprimed origin clock.

 

[James R]

chinglu:

Because the time interval from start to end is longer for the primed frame interval for the motion over the unprimed.

I have clearly shown in post #117 why this is not a problem for relativity.

Only 2 events. (2.) is a consequence of (1.) It is not an event. The rest of post is error.

No. My analysis is correct. A spacetime event is anything that takes place at a specific location in space at a specific time. Your scenario involves three separate events, as I have listed in post #117.

Very good.

Now, accept what you have figured out. I reversed the normal time dilation logic.

I took the primed frame as the origin and stationary with x' = -vt'. This proves t'/γ = t. So, the primed origin claims time dilation as expected.

This is where you are.

Now must continue with all thoughts.

In the view of the unprimed frame, x' moved to the unprimed origin. It was length contracted.

So t' = tγ. When you now take the unprimed frame as stationary, the unprimed origin concludes t' is time expanded. That is the problem.

The moving x' beat faster than the unprimed origin clock.

I have pointed out your error with the time intervals in post #117, above.

Here, you are making the same kind of error with length contraction. Note that a proper length is a length measured between two events that occur at the same time in a particular reference frame.

For events 1, 2 and 3, can you work out which pairs produce a proper length in the two reference frames?

 

[chinglu]

This is not what he calculated, you do not understand what he's doing.
He is claiming that :

$$t'=\gamma (t-\frac{vx}{c^2})$$

accepts a solution of the form

$$t'=t$$

(which is true but irrelevant) thus disproving time dilation (which is obviously false).


As an alternative, he uses the other Lorentz transform:

$$x'=\gamma(x-vt)$$ and , by making $$x'=-x$$ he concludes, just as incorrectly, that he has disproven time dilation.
Chinglu does not even begin to understand the notion of $$\Delta$$ as in $$\Delta x$$ or $$\Delta t$$. He has lied about having taken calculus.

What do you think $$t'=t$$ mean?

I would not answer, but you think you smart.

 

[Tach]

What do you think $$t'=t$$ mean?

I would not answer, but you think you smart.

It is your idea of refuting time dilation. Perfectly incorrect.

 

[chinglu]

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.

This is your error.
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$

You have put a negative elapsed time on clock at x' from view of unprimed.

So, unprimed frame views origin of primed frame as elapsed 0 and x' as elapsed $$-\frac{\gamma k v}{c^2}$$

These are the times on the clocks so to speak. Who cares. We are not concerned with the times on the clock but the elapsed times since the origins were same.

So what is clock had x. Who cares. What if clock had y. Who cares. You get this yet?

Once the start of events has commenced, times on clock at that instant is irrelevant. It is the elapsed times on clocks after that instant. This is a common error.

So your step 2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$ has nothing to do with this problem.

 

[chinglu]

It is your idea of refuting time dilation. Perfectly incorrect.

So explain the meaning.

 

[James R]

This is your error.
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$

You have put a negative elapsed time on clock at x' from view of unprimed.

Two spatially-separated events in spacetime cannot be simultaneous in more than one reference frame. Any two spatially-separated events that are simultaneous in one reference frame will not be simultaneous in the other.

In this case, you have set up a problem where the events I have labelled 1 and 2 occur simultaneously in the primed frame. Therefore, they do not occur simultaneously in the unprimed frame. In fact, in the unprimed frame, event 2 occurs before event 1.

So, unprimed frame views origin of primed frame as elapsed 0 and x' as elapsed $$-\frac{\gamma k v}{c^2}$$

Correct.

These are the times on the clocks so to speak. Who cares. We are not concerned with the times on the clock but the elapsed times since the origins were same.

These are not times on a clock. They are the actual times at which those events occur in the two reference frames. If you like, you could put a clock at the locations of the two events. One of those clocks would be moving with the primed frame, and the other with the unprimed frame. They would read the values I have shown in post #117, provided they were synchronised with clocks at x=0 and x'=0 in their respective rest frames.

So what is clock had x. Who cares. What if clock had y. Who cares. You get this yet?

Sure I get it. Do you? I've explained it very thoroughly to you.

Once the start of events has commenced, times on clock at that instant is irrelevant. It is the elapsed times on clocks after that instant. This is a common error.

I have calculated the elapsed times by subtracting the final time from the initial time in each case. So, you're right - the particular reading on the clock at the start of the interval is irrelevant. Only the elapsed time will matter in the end. I have calculated the elapsed time in each frame assuming that the initial values at x=0 and x'=0 were set to zero at t=t'=0. If you want to set those initial values to some other constant, go right ahead. The elapsed time calculation still gives the same answer that I calculated.

You get this yet?

So your step 2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$ has nothing to do with this problem.

It's a simple application of the Lorentz transformation, as I wrote in post #117.

Do you understand the Lorentz tranformation?
Do you believe in the Lorentz transformation?

If not, then what is your argument?

Are you claiming that relativity is wrong, or that it is inconsistent?

I have shown that it is not inconsistent. You get this yet?

 

[chinglu]

Two spatially-separated events in spacetime cannot be simultaneous in more than one reference frame. Any two spatially-separated events that are simultaneous in one reference frame will not be simultaneous in the other.

This has nothing to do with problem. You are quoting typical quotes.

None has claimed two spatially-separated events in spacetime are simultaneous. Let's agree the primed clocks are not synchronized. So, one has time t'1 and the other t'2 on their clocks.

That is where you get error.

We are concerned with elapsed times not initial times. You are claiming initial times on clocks affects the outcomes of events of the universe.

If that is true, then if I am late for a train, I simply adjust the clock at the station to whatever I want and I am never late. See how initial clock values do not affect elapsed times in the universe? I am thinking this will take several pages.

In short, to prove your theory, you are adjusting the clock at the train station to whatever you like and then claim you are not late for the train.

Special relativity is about time intervals and elapsed times, not intial times.

That is your error.

 

[Tach]

So explain the meaning.

To you - not possible.

 

[James R]

chinglu:

This has nothing to do with problem. You are quoting typical quotes.

I'm telling you what relativity says. Do you think I've made a mistake in what relativity says, or do you think relativity is wrong?

None has claimed two spatially-separated events in spacetime are simultaneous.

Good. So you agree that spatially-separated clocks, such as at position -x and 0 in the primed frame, if they are synchronised in the primed frame they cannot be synchronised in the unprimed frame. Correct?

Let's agree the primed clocks are not synchronized. So, one has time t'1 and the other t'2 on their clocks.

No, we can't assume that. The definition of a reference frame includes the fact that all clocks in that frame are synchronised with each other.

If you have two clocks in a frame that are not synchronised in that frame, then either (a) you have a faulty clock or (b) you're not using the standard definition of a reference frame.

We are concerned with elapsed times not initial times. You are claiming initial times on clocks affects the outcomes of events of the universe.

No. Read my previous post where I explained this exact point to you. Here's the relevant part again:

I have calculated the elapsed times by subtracting the final time from the initial time in each case. So, you're right - the particular reading on the clock at the start of the interval is irrelevant. Only the elapsed time will matter in the end. I have calculated the elapsed time in each frame assuming that the initial values at x=0 and x'=0 were set to zero at t=t'=0. If you want to set those initial values to some other constant, go right ahead. The elapsed time calculation still gives the same answer that I calculated.

You get this yet?

See?

If that is true, then if I am late for a train, I simply adjust the clock at the station to whatever I want and I am never late. See how initial clock values do not affect elapsed times in the universe? I am thinking this will take several pages.

You're right. So, we agree.

In short, to prove your theory, you are adjusting the clock at the train station to whatever you like and then claim you are not late for the train.

No.

Re-read post #117 and tell me exactly where you think I made a mistake. Go through it line by line if you have to.

Because it seems to me that either you didn't read that post, or you didn't understand it.

---

On a side-note, I must comment that you are very ill-mannered.

I have spent a considerable amount of time showing you exactly where you went wrong. And either you haven't read posts #117 and #118 or you are ignoring what I showed you there.

You can't pretend that I haven't shown you your mistake. It is quite clearly set out in post #118.

If you wish to dispute this matter further, you will need to go through posts #117 and #118 and show where I have made a mistake.

If you will not argue your case in good faith and with honesty and integrity, as any scholar would, then I don't think I will waste any more time on you.

Are you willing to look at posts #117 and #118 honestly? We'll see.

You should be thanking me for helping you, not telling me that I have made mistakes. I am educating you, and for free!

On the other hand, if you can't understand posts #117 and #118, please ask me questions and I will do my best to answer them for you. I understand that relativity can be difficult for beginners such as yourself.

 

[chinglu]

chinglu:



I'm telling you what relativity says. Do you think I've made a mistake in what relativity says, or do you think relativity is wrong?



Good. So you agree that spatially-separated clocks, such as at position -x and 0 in the primed frame, if they are synchronised in the primed frame they cannot be synchronised in the unprimed frame. Correct?



No, we can't assume that. The definition of a reference frame includes the fact that all clocks in that frame are synchronised with each other.

If you have two clocks in a frame that are not synchronised in that frame, then either (a) you have a faulty clock or (b) you're not using the standard definition of a reference frame.



No. Read my previous post where I explained this exact point to you. Here's the relevant part again:



See?



You're right. So, we agree.



No.

Re-read post #117 and tell me exactly where you think I made a mistake. Go through it line by line if you have to.

Because it seems to me that either you didn't read that post, or you didn't understand it.

---

On a side-note, I must comment that you are very ill-mannered.

I have spent a considerable amount of time showing you exactly where you went wrong. And either you haven't read posts #117 and #118 or you are ignoring what I showed you there.

You can't pretend that I haven't shown you your mistake. It is quite clearly set out in post #118.

If you wish to dispute this matter further, you will need to go through posts #117 and #118 and show where I have made a mistake.

If you will not argue your case in good faith and with honesty and integrity, as any scholar would, then I don't think I will waste any more time on you.

Are you willing to look at posts #117 and #118 honestly? We'll see.

You should be thanking me for helping you, not telling me that I have made mistakes. I am educating you, and for free!

On the other hand, if you can't understand posts #117 and #118, please ask me questions and I will do my best to answer them for you. I understand that relativity can be difficult for beginners such as yourself.

I understand that relativity can be difficult for beginners such as yourself

Yea.

You are doing way to much typing.

I told you what I am doing an you just keep at it because it went over you.

I will now reverse it into a form you know.

O.
O'
v--->---------------------x=vt

Have you seen this before. It is the proof from Einstein for time dilation.

In the O frame, you can plug in x=vt and calculate t' = t/γ. Everyone know this. So, in absolute sense one conclude moving clock time dilated.

But, what does O' calculate? You are trying to claim an adjustment factor must be used for the x coordinate same way you think falsely it must be used for x' coordinate.

There is no correction to the clock. It is only elapsed times since origins same.

So, what does O' calculate?

t' = x/(vγ). Since t = x/v. As can see t' = t/γ. So, O' frame as stationary thinks moving x clock beats slower exactly as I have typing.

Look how both frame agree t' = t/γ. That is the way it works under relativity.

Here is something to help you understand relativity better. When you calculate t in frame and use that to calculate t', if you did your calculations correctly, then using t' in Lorentz Transforms will result in t.

If not, then you failed. You can check your work on your own and find your own failure.

 

[James R]

chinglu:

I think we're in agreement. Both frames see clocks in the other frame as running slow. Correct?

Ok, are we done?

I assume we are finished, since you agree with me and you agree with what I wrote in posts #117 and #118.

Thankyou for the conversation. I have had fun educating you.

 

[chinglu]

chinglu:

I think we're in agreement. Both frames see clocks in the other frame as running slow. Correct?

Ok, are we done?

I assume we are finished, since you agree with me and you agree with what I wrote in posts #117 and #118.

Thankyou for the conversation. I have had fun educating you.

No, typing mistake on my part.

t' = x/(vγ). Since t = x/v. As can see t' = t/γ. So, O' frame as stationary thinks moving x clock beats faster exactly as I have typing.

As we can see, in the primed frame for x moving, t' = t/γ. Note how t is faster for the moving x.

 

[chinglu]

chinglu:

I think we're in agreement. Both frames see clocks in the other frame as running slow. Correct?

Ok, are we done?

I assume we are finished, since you agree with me and you agree with what I wrote in posts #117 and #118.

Thankyou for the conversation. I have had fun educating you.

Also,
Lets assume you correct.

$$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$

Let's see you translation using Lorentz assuming light emits from origins when same.


Make sure you use your theory above.

 

[James R]

chinglu:

t' = x/(vγ). Since t = x/v. As can see t' = t/γ. So, O' frame as stationary thinks moving x clock beats faster exactly as I have typing.

As we can see, in the primed frame for x moving, t' = t/γ. Note how t is faster for the moving x.

Yes. I explained this observation in post #117. It is important to correctly identify which time interval is a proper time and which is not.

See?

You still haven't pointed out any error in posts #117 or #118.

Also,
Lets assume you correct.

$$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$

Let's see you translation using Lorentz assuming light emits from origins when same.

Make sure you use your theory above.

I'm happy to do this, but I don't understand exactly what it is that you're asking me to do. Can you please be specific?

Light emits from the origin when? Towards what? Are you asking me to calculate the flight time for a light pulse in each frame?

Once you say what you want me to do, I'll do it.

You really should point out any mistake in posts #117 and #118 first, though.

 

[rpenner]

$$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$.

$$x' = \gamma (x - vt) = \gamma \left( -\gamma k - v (- \frac{\gamma k v}{c^2}) \right) = - \gamma^2 k \left( 1 - \frac{v^2}{c^2} \right) = -k$$
$$t' = \gamma (t - \frac{v}{c^2} x) = \gamma \left( -\frac{\gamma k v}{c^2} - \frac{v}{c^2} (-\gamma k) \right) = - \gamma^2 k \left( \frac{v}{c^2} - \frac{v}{c^2} \right) = 0$$

So saying $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$ is the same thing as saying $$(x', t') = ( -k, 0)$$ -- (x,t) and (x', t') are two different names for the same point in space and time, just like (r,θ) and (x,y) are different names for the same point on the Euclidean plane.
$$x = r \cos \theta\\y = r \sin \theta$$

This is JamesR's event 2 from post 117.
When k = 5/6 light-seconds this is event $$P_{t'=0}$$ in post 18.

 

[chinglu]

$$x' = \gamma (x - vt) = \gamma \left( -\gamma k - v (- \frac{\gamma k v}{c^2}) \right) = - \gamma^2 k \left( 1 - \frac{v^2}{c^2} \right) = -k$$
$$t' = \gamma (t - \frac{v}{c^2} x) = \gamma \left( -\frac{\gamma k v}{c^2} - \frac{v}{c^2} (-\gamma k) \right) = - \gamma^2 k \left( \frac{v}{c^2} - \frac{v}{c^2} \right) = 0$$

So saying $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$ is the same thing as saying $$(x', t') = ( -k, 0)$$ -- (x,t) and (x', t') are two different names for the same point in space and time, just like (r,θ) and (x,y) are different names for the same point on the Euclidean plane.
$$x = r \cos \theta\\y = r \sin \theta$$

This is JamesR's event 2 from post 117.
When k = 5/6 light-seconds this is event $$P_{t'=0}$$ in post 18.

I thought you were resting peacefully in old folks home.

Apply this argument to Einstein's proof of time dilation.

 

[chinglu]

chinglu:



Yes. I explained this observation in post #117. It is important to correctly identify which time interval is a proper time and which is not.

See?

You still haven't pointed out any error in posts #117 or #118.



I'm happy to do this, but I don't understand exactly what it is that you're asking me to do. Can you please be specific?

Light emits from the origin when? Towards what? Are you asking me to calculate the flight time for a light pulse in each frame?

Once you say what you want me to do, I'll do it.

You really should point out any mistake in posts #117 and #118 first, though.

First, you need to apply your argument to Einstein's proof of time dilation


You will then find the coordinate moving toward the origin is time expanded.

Next, assume a light pulse is emitted from origins. That is a timing device.

Prove time dilation negative x-axis point and positive x-axis point.

Let me know when you have failed.

Remember, you told the readers you are the teacher.