Time Dilation

[chinglu]

You claimed that you took calculus. So, you lied.



I am sure that you've been shown this before. Since you lied about taking calculus, I do not expect you to understand what it means. Here:

$$ \Delta t'=\gamma (\Delta t - \frac{v}{c^2} \Delta x)$$

What happens if a clock at rest in the unprimed frame ($$\Delta x=0$$) is viewed from the primed frame?

You do not seem to understand, Lorentz transforms do not map intervals, they map coordinates. The interval is built in.

$$ x'=\gamma ( x - vt)$$

x' and x are coordinates not deltas. you can call t a delta and that part is OK.
Here is a basic fact.

$$\Delta x = v\Delta t$$

Now if have moving observer at x' and want to know what happens for it to move to x when all events begin when origins same, you calculate the distance between the two applying length contraction to x' since it is measured in the moving frame.

$$\Delta x = x - x'/\gamma$$

Substitute this result into

$$\Delta x = v\Delta t$$

$$ x - x'/\gamma= v\Delta t$$

$$ x'/\gamma= x - v\Delta t$$

$$ x'= \gamma(x - v\Delta t)$$

This is Lorentz transform. As you can understand the interval is built in from the coordinates x' and x.

Now the point of this thread is the comparison of the time intervals between the frames for a moving clock at x'<0 to move to the stationary origin.

All math thus far have proved the moving clock beats faster than the clock at the stationary origin.

 

[Tach]

You do not seem to understand, Lorentz transforms do not map intervals, they map coordinates.

Find a different hobby, your trolling is getting boring.

 

[chinglu]

Find a different hobby, your trolling is getting boring.

You have the maths I typed. You did not address maths so I must conclude you give up.

I can understand after reading your posts.

 

[Tach]

You do not seem to understand, Lorentz transforms do not map intervals, they map coordinates. The interval is built in.

$$ x'=\gamma ( x - vt)$$

x' and x are coordinates not deltas. you can call t a delta and that part is OK.
Here is a basic fact

A basic fact is that you lied when you said you learned calculus. See the proof:

$$ x'=\gamma ( x - vt)$$

means:

$$ x_1'=\gamma ( x_1 - vt_1)$$

$$ x_2'=\gamma ( x_2 - vt_2)$$

Subtract and you get:

$$ x_2'-x_1'=\gamma ( (x_2 -x_1)- v(t_2-t_1))$$


You never took calculus.

 

[chinglu]

A basic fact is that you lied when you said you learned calculus. See the proof:

$$ x'=\gamma ( x - vt)$$

means:

$$ x_1'=\gamma ( x_1 - vt_1)$$

$$ x_2'=\gamma ( x_2 - vt_2)$$

Subtract and you get:

$$ x_2'-x_1'=\gamma ( (x_2 -x_1)- v(t_2-t_1))$$


You never took calculus.

You maths correct except should look like

$$ x_2'-x_1'=\gamma ( (x_2 -x_1)- v( \Delta t_2 - \Delta t_1 ))$$

This is not Lorentz transform. Also, why not continue forward with this and prove unrestricted time dilation is false.

Also, you set $$\Delta x' = 0$$ in prior post.

Can you now understand how false that is. It implies exact motion forward and back, no motion at all or no time. What good is that?

Next, you thought $$\Delta x'$$ had something to do with coordinate transformations as this thread is doing. Can you now see all your posts were false? Sometimes it is good to write things down to see exactly what you mean and where you are wrong.

So can you refute the conclusions of the OP with maths? That is point of this thread.

 

[Tach]

you maths correct except should look like

$$ x_2'-x_1'=\gamma ( (x_2 -x_1)- v( \Delta t_2 - \Delta t_1 ))$$

lol

this is not lorentz transform.

lol.

So can you refute the conclusions of the OP with maths? That is point of this thread.

Yes, I already did. The problem is that you understand neither the "maths" nor the physics. And I can't fix that.

 

[James R]

$$(c \Delta t')^2 - (\Delta x')^2 = \left( \frac{c^4}{v^2 (c^2 - v^2)} \right) c^2 \gamma^2$$ but
$$(c \Delta t)^2 - (\Delta x)^2 = \left(\frac{c^2}{v^2} - 1 \right) c^2 \gamma^2$$
So it looks like you are not talking about the same space-time interval throughout.
That's because at x'=-c, and t'=0, t is not also 0.

Yeah. I was a bit hasty with that post. It's hard to work out what chinglu's scenario actually is. I might try again later if I have some spare time.

 

[chinglu]

Yes, I already did. The problem is that you understand neither the "maths" nor the physics. And I can't fix that.

No, you set delta x = 0. I already proved this is no good and does nothing.

Are you still wanting this?

Otherwise you proved nothing as usual.

 

[chinglu]

lol

Can you show your equation in Einstein's paper?

I can not find it.

You claimed your equation is Lorentz transform.

 

[Tach]

No, you set delta x = 0.

Yes, I can see that you still have trouble understanding this. Don't worry, you will NEVER understand it.

 

[Tach]

You claimed your equation is Lorentz transform.

It is. It is also your problem that you can't recognize it.

 

[chinglu]

Yes, I can see that you still have trouble understanding this. Don't worry, you will NEVER understand it.

Oh, so when delta x' = 0 and not motion resulted you have found some great idea?

Why not give direct explanations instead of evasive.

Readers know this mean you can not answer because you are wrong.

 

[chinglu]

It is. It is also your problem that you can't recognize it.

If your equation is lorentz transform readers would like to see the link in Einstein's paper as you can find mine.

That gives you no credability if you can not show that in Einstein's paper. Readers will think you crackpot without this link.

 

[Tach]

Oh, so when delta x' = 0 and not motion resulted you have found some great idea?

$$\Delta x =0$$ means that we are measuring the tempotal interval on a clock at rest in the unprimed frame. You still don't understand this?

Readers know this mean you can not answer because you are wrong.

No, readers have figured out that you are retarded.

 

[chinglu]

$$\Delta x =0$$ means that we are measuring the tempotal interval on a clock at rest in the unprimed frame. You still don't understand this?

Given the relativity postulate, the lorentz transforms indicate the motion as seen by the stationary coordinates. There may be a temporal interval, but that must be associated with some $$\Delta x >0$$ motion. This elementary.

Now, children know this.

$$\Delta x = v\Delta t$$

Since you have $$\Delta t>0$$, obviously you must have $$\Delta x>0$$ or $$\Delta x = v\Delta t=0$$.

Simple mind can understand this.

 

[Tach]

Given the relativity postulate, the lorentz transforms indicate the motion as seen by the stationary coordinates. There may be a temporal interval, but that must be associated with some $$\Delta x >0$$ motion.

No, it isn't. You are making up your own demented rules.

Now, children know this.

$$\Delta x = v\Delta t$$

Nope, v is the relative speed between frames, it has nothing to do with $$\Delta x$$. You need to start studying and to stop spewing stupidities.

 

[chinglu]

No, it isn't. You are making up your own demented rules.

Nope, v is the relative speed between frames, it has nothing to do with $$\Delta x$$. You need to start studying and to stop spewing stupidities.

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

What was your reason again this false?

v is the relative speed between frames, it has nothing to do with $$\Delta x$$

 

[Tach]

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

In the Lorentz transforms , v represents the speed between frames. You need to start learning and to stop spewing stupidities.

 

[chinglu]

In the Lorentz transforms , v represents the speed between frames. You need to start learning and to stop spewing stupidities.

You did not answer question.

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

 

[Tach]

You did not answer question.

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

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