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07:16
@JohnRennie Hi !
@KavinIshwaran Hi :-)
I was reading your answer in this Q and A
85
A: What is time dilation really?

John RennieIntroduction This answer will use ideas discussed in the answers to What is time, does it flow, and if so what defines its direction?, so you really need to read the answers to that question before tackling this one. The key concept you need in order to understand time dilation is that a clock ...

OK ... ?
In this You have given two graphs
saying that proper time is invariant
07:20
In that graph we can clearly see that the two red arrows (a straight and curved one) are different in length, but they are actually the same. how they can be invariant
I suppose it is the way our universe built on ?
and how do you actually define proper time
In regular geometry we can measure the length of the line using Pythagoras' theorem. If we have a line from (0,0) to (x,y) then the length of that line is s² = x² + y². Yes?
And this extends to three dimensions s² = x² + y² + z², and in principle to any number of dimensions.
I am fully clear on the Euclidean metric part
you were explaining this a while ago :-) but, I cant understand the invariant thing
Well this is the weird thing about SR. We have four dimensions (t, x, y, z) but the distances are not given by the Euclidean metric but rather by: s² = -c²t² + x² + y² + z²
@KavinIshwaran Didn't we talk about this? I remember explaining using the example of an arrow. If you rotate or move the arrow it doesn't change length.
07:25
yes but what about a curved path ?
A curved path is just a collection of infinitesimally small straight paths.
So you can treat the curved path as a sequence of infinitesimally small arrows arranged head to tail, and as we discussed the length of those arrows doesn't change just because we changed our coordinate system, and hence the total length of the curve doesn't change.
lets take the same example
you are being stationary
and I am moving at velocity v and both space time graph have the common point 0,0
along x axis*
OK. So we zero our clocks as we pass each other so we both started from the same point (0, 0)
07:37
yes
And you move until you reach some point that we have labelled P. e.g. we could literally stick a flag in the ground with "P" written on it to mark that point.
And we both record the time shown by our clocks when you reach that point P.
Yes ?
yes, so when I reach point p in time t and after some time t you have moved a distance ct in space time and distance between 0 and point p is root (-c2t2 + v2t2) yes ?
@JohnRennie yes
In my rest frame I am stationary at the origin, and you are moving with velocity v. You travel for a time on my clock t, and because you are moving at velocity v the distance you move (as observed by me) is x = vt. Yes?
So in my frame the proper time for your motion is s² = -c²t² + v²t²
Which is what you said above :-)
I just wanted to be clear exactly what we were calculating.
So, is this all OK so far?
07:44
yes and
this is where I am having a doubt
how do we define proper time
You mean how do we relate times to this quantity "s²" ?
OK, suppose I am at rest in my frame and I am holding a clock - so the clock is also at rest.
After some time T I have reached the point (T, 0) i.e. I moved T in time and 0 in space.
OK so far?
So if we substitute this into the metric we get: s² = -c²T²
where T is the time shown on my clock.
And c is a constant, so s² ∝ T² i.e. the "distance" s and the time T are related by a constant.
We call s the proper distance and T the proper time. They are basically the same thing because they differ only by a constant.
In fact there are two ways of writing the metric:
s² = -c²t² + x²
c²T² = c²t² - x²
07:51
@JohnRennie and that constant is lorentz factor ?
No, s² = -c²T² and that constant c is the speed of light.
ok, But at the end we see that t and T differ by a factor of the gamma (?)
Hang on. You asked me what the proper time means, and that' what I'm telling you. The proper time for an observer travelling between two points is the time recorded by that observer's clock.
The reason t and T differ in our example is because t is not a proper time while T is a proper time.
@JohnRennie yes
then that is why we say that the length s is invariant ?
@KavinIshwaran I'm not sure what you are asking ...
The proper length, and the proper time since the two are proportional, is the length of a four-vector i.e. the length of a 4D arrow.
And it's constant because arrows don't change length.
07:58
Ok.. Got it !
I need to drop out for a while, but I will be around later.
 
2 hours later…
10:29
@JohnRennie, Hi John, do you have a few minutes? :)
@user157860 I'm busy answering a question in another room at the moment, but I shouldn't be long.
ok, I'll wait
10:41
@user157860 Hi, I'm free now :-)
Hi, recently you told me that the electron in H atom has no orbit but it's spread in a cloud all around the nucleus, can you explain how that is compatible with classical laws that are still valid? Coulomb's law needs a well-defined distance and and so does centrifugal law: v^2/r besides a velocity. Or such laws are not valid?
The classical laws do not apply to quantum particles like electrons in atoms.
so the electron is not attracted to proton?
The electron is attracted to the proton, but the attractive force is more complicated than Coulomb's law.
OK, I see, thanks :)
11:39
@JohnRennie Hi :)
Please tell me when you are free
Hi @JohnRennie.
I just have a simple question. If two masses one 5kg and 1kg are hanging in a Atwood machine and the pulley is hanging from a spring balance, then as the masses let go how much the spring balance meansure during acceleration?
On second thought let’s say there are two weights 5kg-f and 1kg-f.

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