You have to do better than the above. Saying 1=1 or two events are infact the same event measured in proper time is not an argument for anything.
Time Dilation
- Thread starter chinglu
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You have to do better than the above. Saying 1=1 or two events are infact the same event measured in proper time is not an argument for anything.
I can not know what you mean.
It is one event when x' and x are at same place time starts when origins same place.
There is a space interval between them.
Here is picture what unprime frame sees when origins same. x'/γ moves toward x coordinate.
x'/γ-------------O-------------------x
Here is picture what unprime frame sees when origins same. x/γ moves toward x' coordinate.
x'----------------O'---------------x/γ
Then use an analogy. Or make a thought experiment.
I can not simplify it any further.
It is one event when x' and x are at same place time starts when origins same place.
There is a space interval between them.
Here is picture what unprime frame sees when origins same. x'/γ moves toward x coordinate.
x'/γ-------------O-------------------x
Here is picture what unprime frame sees when origins same. x/γ moves toward x' coordinate.
x'----------------O'---------------x/γ
Why would you assume the above? Physics is not numerology.
What you have to assume is that you are measuring the time interval on a clock in one of the frames. The spatial separation is 0:
$$\Delta x=0$$
So:
$$(c \Delta t')^2 - (\Delta x')^2 = (c \Delta t)^2 -0$$
This means:
$$\Delta t=\Delta t' \sqrt{1-(v'/c)^2}$$
where
$$v'=\frac{\Delta x'}{\Delta t'}$$
I bet that you've been shown this before. I also bet that you will never understand it and that you will continue arguing to death.
No. I think you just don't want too for some reason.
You are claiming the spatial separation between x' and x in the view of the unprimed frame is 0? How did you arrive at this conclusion? This is about an observer at x having the primed coordinate x' come toward it. The spatial separation between the two is x - x'/ γ when origins same. The start of the event is the origins being same. The end of the event is when x' and x are the same place. That is not a zero spatial separation.
Why would you not then claim $$\Delta x'=0$$ in the view of the primed frame.
Otherwise you make absolute claims.
$$\Delta x=0$$
$$v\Delta t=\Delta x=0$$
So
$$v=0$$
or
$$\Delta t\ = 0$$
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What?
No, this is not what I am claiming. You don't know the meaning of $$\Delta x$$. One day, in a few years, it will come to you. Give it time, you aren't ready yet.
Oh, I guess it is some deep thought.
You have
$$\Delta x=0$$.
But, you also have $$\Delta x'<>0$$.
How do you work this out given the relativity postulate?
Also, $$\Delta x=0$$ mean no spatial separation between x' and x. How do you make this true? I would sure like to understand how the observer at x' moved to x in the unprimed frame and yet no distance was between them but time was between them.
Simple, it has nothing to do with PoR. You can easily have $$\Delta x=0$$ while $$\Delta x'<>0$$.
See? You are starting to get it. It means that in the unprimed frame we have two events with pure temporal separation, like two clock ticks.
Err, you are wrong again, $$\Delta x$$ has NOTHING to do with the separation between x' and x. You must think some more, one day you'll get it.
It didn't.
This not what lorentz transforms say.
x' = ( x- vt )γ
x' /γ = x - vt
x' /γ - x = vt = Δx
Sorry your thoughts are not the same as lorentz transforms.
See? You are starting to get it. It means that in the unprimed frame we have two events with pure temporal separation, like two clock ticks.
I must apply relativity postulate making your argument false.
if Δx = 0 then Δx' = 0.
It is not my problem that you confuse physics with numerology.
x' = ( x- vt )γ
implies
$$\Delta x' =\gamma ( \Delta x- v \Delta t ) $$
I am quite sure others tried to teach you tha. Don't worry, in a few years, you'll get it.
One day you'll get it. It will take you many years but you are still young.
numerology
Yes I am seeing your point.
I have coordinate at x' and one at x.
Can you explain $$\Delta x$$ in unprimed frame and $$\Delta x'$$ in primed frame.
Many apologies to interrupt your view of self but x' and x are coordinates not deltas.
Because this is what basic calculus tells you. Do they teach calculus in China?
You are welcome. See what classes they teach in your province. Come back when you passed them.
Many sorries but I passed them. They are coordinate mappings not deltas.
We learn to admit when we are wrong. That way we do not make same simple error over and over.
Do you have this course? If so, you failed.
Lorentz transforms map coordinates not deltas.
Then, you must have bribed your teachers. You are utterly hopeless.
Too bad that you are not applying this, you are making a fool of yourself everywhere on the physics forums.
By basic differentiation, they also map coordinate differences.
You say that you passed calculus? You lie.
It seem you finally understand Lorentz map coordinates. Now you should admit you are wrong so as to not make that simple error again. You can do this with self as we are taught.
And since we are dealing with coordintes, what is derivative of a coordinate?
For example, x=0 is contant, origin.
t' = ( t - xv/c² )γ = tγ.
So moving clock at x' beat time expanded in view of rest frame.
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?
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