Time Dilation

[chinglu]

Oh, so $$\Delta x = v\Delta t$$ is false for you?

It is false for anyone who understands physics. You are excluded . Forever.

Here is Einstein.

Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt and
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Since time start when origins same, x coordinate is $$\Delta x = x - 0$$

Since t = 0 when origins same, $$\Delta t = t - 0$$

This means Einstein wrote $$\Delta x = v\Delta t$$.

You are not very good at all.

 

[James R]

Ok. Let's try this again.

Assume clock is located at x' < 0 in the primed coordinates.

Assume standard configuration and time interval begins for both frame when origins same.

Assume time interval ends when x' and unprimed origin same.

What is this time interval in each frame.

Let's assume the clock is initially located at x'=-k, with k a positive constant.

The primed frame sees the x=0 origin moving in the negative direction with speed v. For x=0 to coincide with x'=-k, the x=0 origin must move distance k at speed v, which in the primed frame takes time

$$\Delta t' = \frac{k}{v}$$

In the unprimed frame, at t=0 the distance to the point x'=-k is contracted by the Lorentz factor. Therefore, in the unprimed frame:

$$\Delta t = \frac{k}{\gamma v}$$

I think this is what chinglu originally calculated. His answers are correct for this particular problem.

So, chinglu, what I don't understand is why you believe this is a problem for relativity?

 

[Tach]

Since time start when origins same, x coordinate is $$\Delta x = x - 0$$

Wrong. You claimed you learned calculus, you lied.

Since t = 0 when origins same, $$\Delta t = t - 0$$

Wrong. You claimed you learned calculus, you lied.

This means Einstein wrote $$\Delta x = v\Delta t$$.

You are not very good at all.

no , this only means you don't have a clue. You never will.

 

[Tach]

Ok. Let's try this again.



Let's assume the clock is initially located at x'=-k, with k a positive constant.

The primed frame sees the x=0 origin moving in the negative direction with speed v. For x=0 to coincide with x'=-k, the x=0 origin must move distance k at speed v, which in the primed frame takes time

$$\Delta t' = \frac{k}{v}$$

In the unprimed frame, at t=0 the distance to the point x'=-k is contracted by the Lorentz factor. Therefore, in the unprimed frame:

$$\Delta t = \frac{k}{\gamma v}$$

I think this is what chinglu originally calculated. His answers are correct for this particular problem.

So, chinglu, what I don't understand is why you believe this is a problem for relativity?

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.

 

[James R]

Just to add a little to my previous post, 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.
2. The clock is 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)

Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame:

1. (x,t) = (0,0)
2. $$(x,t)=(-\gamma k, \frac{\gamma k v}{c^2})$$
3. $$(x,t)=(0,\frac{k}{\gamma v})$$

Note that using events 1 and 3, the time intervals are as calculated in my previous post, which are also in agreement with chinglu's original post.

But what is interesting is event 2. Note that events 1 and 2 are simultaneous in the primed frame, but NOT in the unprimed frame.

The time interval between events 2 and 3 is k/v in the primed frame, but in the unprimed frame the time interval between those two events is:

$$\Delta t = \frac{\gamma k}{v}\left(1-\frac{2v^2}{c^2}\right)$$

 

[James R]

This is not what he calculated, you do not understand what he's doing.

I have reproduced the relevant parts of his post that I concentrated on in my posts #82 and #85.

Do you assert that I have not applied his prescription correctly?

I have no interest in what he has said or done since that post. I have merely concentrated on what he initially said.

I do not particularly appreciate your telling me that I don't understand. If you wish to make such an assertion, then I suggest you point out exactly where I have gone wrong in posts #82 and #85.

 

[Tach]

I have reproduced the relevant parts of his post that I concentrated on in my posts #82 and #85.

Do you assert that I have not applied his prescription correctly?

yes, you didn't

I do not particularly appreciate your telling me that I don't understand. If you wish to make such an assertion, then I suggest you point out exactly where I have gone wrong in posts #82 and #85.

i explained to you what is the mechanism of his idea. it is different from what you understand it to be.

 

[Tach]

Just to add a little to my previous post, 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.
2. The clock is 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)

you mean 3. (x',t') = (-k, -k/v). otherwise , you you couldn't possibly have

3. $$(x,t)=(0,\frac{k}{\gamma v})$$

The time interval between events 2 and 3 is k/v in the primed frame, but in the unprimed frame the time interval between those two events is:

$$\Delta t = \frac{\gamma k}{v}\left(1-\frac{2v^2}{c^2}\right)$$

This appears to be incorrect, you may want to check your calculations.

 

[James R]

Tach:

I have reproduced the relevant parts of his post that I concentrated on in my posts #82 and #85.

Do you assert that I have not applied his prescription correctly?

yes, you didn't

Then go back to posts #82 and #85 and point out exactly where I went wrong.

i explained to you what is the mechanism of his idea. it is different from what you understand it to be.

I'll wait for you to point out the errors in posts #82 and #85 before I discuss what his "mechanism" might be.

you mean 3. (x',t') = (-k, -k/v). otherwise , you you couldn't possibly have

3. $$(x,t)=(0,\frac{k}{\gamma v})$$

No. I mean what I wrote. Why don't you apply the Lorentz transformation to the (x,t) coordinates given and tell me what you get for (x',t')?

This appears to be incorrect, you may want to check your calculations.

I think you need to check yours. Mine are correct, as far as I can tell.

 

[Tach]

Tach:



Then go back to posts #82 and #85 and point out exactly where I went wrong.



I'll wait for you to point out the errors in posts #82 and #85 before I discuss what his "mechanism" might be.



No. I mean what I wrote. Why don't you apply the Lorentz transformation to the (x,t) coordinates given and tell me what you get for (x',t')?

You don't write it but it is obvious that you are trying to use:

$$t=\gamma(t'-vx'/c^2)$$

In order to get what you wish to get you must have $$t'=-k/v$$ which is consistent with your result 3.

I think you need to check yours. Mine are correct, as far as I can tell.

Well, you have mistakes.

 

[James R]

Tach:

Yeah. You already made that claim.

Assertions without evidence are worthless.

 

[Tach]

Tach:

Yeah. You already made that claim.

Assertions without evidence are worthless.

I added the math. The second mistake is more serious, try showing your steps and you'll find your error. Events 2 and 3 have both a temporal and a spatial separation, you calculated only the temporal separation.

 

[James R]

Tach:

You've made the error, not me.

The correct Lorentz transformation is:

$$t = \gamma (t' + vx'/c^2)$$

You had the sign wrong.

There's no error in the other result, either.

Will you admit your mistake now, or will you keep insisting you are right, as I have observed you do on many occasions when you have been proved wrong? Let's see...

 

[Tach]

Tach:

You've made the error, not me.

The correct Lorentz transformation is:

$$t = \gamma (t' + vx'/c^2)$$

You had the sign wrong.

If that is the case, then you got

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

wrong, since according to you 2. (x',t') = (-k,0).

It is difficult to know what you did since you don't show the steps. Either 2 or 3 is wrong. You decide.

 

[James R]

Tach:

You're right in post #94. I did make a mistake.

As for following the steps, all I did was to apply the Lorentz transformations. I didn't think it was necessary to include the algebra, since I assume everybody following this conversation knows the transformations and can do the algebra themselves.

Are you willing to concede that you, like me, made a mistake with the sign in the transformation?

 

[James R]

Here's a corrected repost of my post #82...

--------

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. The clock is 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)

Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame:

1. (x,t) = (0,0)
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$
3. $$(x,t)=(0,\frac{k}{\gamma v})$$

Note that using events 1 and 3, the time intervals are as calculated in my previous post, which are also in agreement with chinglu's original post.

But what is interesting is event 2. Note that events 1 and 2 are simultaneous in the primed frame, but NOT in the unprimed frame.

The time interval between events 2 and 3 is k/v in the primed frame, but in the unprimed frame the time interval between those two events is:

$$\Delta t = \frac{\gamma k}{v}$$

Note that this time interval between events 1 and 3 is SHORTER in the unprimed frame than it is in the primed frame, whereas the time interval between events 2 and 3 is LONGER in the unprimed frame than in the primed frame.

 

[Tach]

Tach:

You're right in post #94. I did make a mistake.

As for following the steps, all I did was to apply the Lorentz transformations. I didn't think it was necessary to include the algebra, since I assume everybody following this conversation knows the transformations and can do the algebra themselves.

Are you willing to concede that you, like me, made a mistake with the sign in the transformation?

I told you that you have EITHER event 2 or event 3 computed incorrectly. You need to learn how to own when you make a mistake and not try to shift the burden on the other person. I had no idea what transform you used , based on 2 I assumed that you used the minus sign. You claim that you used the plus sign, that makes your event 3 correct but your event 2 incorrect. Either way, one was incorrect.

You have not acknowledged a much more serious mistake, in calculating $$\Delta t$$. Do you understand your mistake?

 

[Tach]

.

The time interval between events 2 and 3 is k/v in the primed frame,

Yes.

but in the unprimed frame the time interval between those two events is:

$$\Delta t = \frac{\gamma k}{v}$$

How did you get that?

 

[James R]

I told you that you have EITHER 2 or 3 computed incorrectly.

No you didn't.

And please don't keep going back and editing your posts. You start to look dishonest when you do that.

You need to learn how to own when you make a mistake and not try to shift the burden on the other person.

I owned my mistake as soon as we had together worked out what it was.

You have yet to own yours. In fact, I can't recall you ever owning any mistake you've made since you joined this forum.

I had no idea what transform you used , based on 2 I assumed that you used the minus sign. You claim that you used the plus sign, that makes your 3 correct but your 2 incorrect.

In post #82 I explicitly stated that I used the Lorentz transformations. What more is needed? Note: I was also explicit about which was the primed frame and which was the unprimed frame.

You have not acknowledged a much more serious mistake, in calculating $$\Delta t$$. Do you understand your mistake?

What mistake? Are you saying I've made another mistake in my corrected post above?

 

[James R]

How did you get that?

I subtracted the time of event 2 from the time of event 3.