Re: TIME DILATION


"Stamenin" <tasko.s@xxxxxxxxxxx> wrote in message news:1155510752.965542.178220@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
"Stamenin" <tasko.s@xxxxxxxxxxx> wrote in message news:1155445599.480157.190190@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
"Stamenin" <tasko.s@xxxxxxxxxxx> wrote in message news:1155362119.887490.190620@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

[snip]

The time dilatation or the relativity of the time is a joke of the
mathematics. The relation
t'=t(1-v^2/c^)^1/2 Is mistaken. If you mean that t is the time at the
rails and t' at a train than the relation should be:
t=t'/R where R=(1-v^2/c^2)^1/2, for the determination of the time at
the railwayand,
t'=t/R for the determination of the time at the train. If you respect
the Lorentz transformation you can't invert these relations.

It seems that you don't understand the meaning of the variables
t and t' in these relations.
You see, t is not just like you say "the time at the rails", and t' is
not "the time at the train".
If t and t' were like you think they are, then indeed there is
something wrong.
This is a *very* common mistake and misunderstanding.

The variable t is "the time Of Some Event according to a clock
on the rails", whereas the variable t' is "the time Of Some Event
according to a clock on the train.
The specification "Of Some Event" is the most important part
in these definitions. Ignoring it guarantees immediate failure.

The relation
t' = t sqrt(1-v^2/c^2)
says something about an event that happens on the train.
whereas the relation
t = t' sqrt(1-v^2/c^2)
says something about an event that happens on the rails.

These relations say something about very different situations.
Do you understand this?

Dirk Vdm

The time is time and is masured in seconds.

hm, so you don't understand.

If you mean that there the
t and t' are differences of time relative to a event then have to
represent them as Dt and Dt'.

No, actually that is not necessary. The t and t' can mean
time differences with respect to events with t = t' = 0.
In that case t = Dt and t' = Dt'.
But that is not the point.

In this case you again, do not obtain a
beter situation. Try to repeat the calculation and will get the same
result that the intevals will grow as meny times as times you repeat
the calculation. So what is the difference. In fact by making the
difference of two moments wetween which apears that event for examle
from:
t=1/R(t'+vx'^2/c^2)
You get rid of the term (vx'^2/c^2) and would have the relations:
Dt=Dt'/R and Dt'=Dt/R. The bad thing is in these caces that the
condition x'=cons. and
x=cons.
These conditions put the question, what kind of events are they
if the material body M hose coordinates are x and x' is in a state of
staying. And whot happens with the time when x and x' are not constant?

Hm, so indeed you really have no idea what the variables represent.
Let's add the D-symbols then and talk about explicit differences.

The relation
Dt' = Dt sqrt(1-v^2/c^2)
says something about two events that happen on the train
at the same place (meaning Dx' = 0), whereas the relation
Dt = Dt' sqrt(1-v^2/c^2)
says something about two events that happen on the rails
at the same place (meaning Dx = 0).

Now, as an exercise for you, if you take the equations
together and write
Dt' = Dt sqrt(1-v^2/c^2)
Dt = Dt' sqrt(1-v^2/c^2) ,
about which events are you talking?
And how is this reflected by the equations?

Dirk Vdm

These relations can't be taken in consideration because they can't be
dedused from the
Lorentz transformation.

Of course they can be deduced from the transformation.
Look, this is the transformation:
{ x' = ( x - v t ) / sqrt(1-v^2/c^2)
{ t' = ( t - v x /c^2) / sqrt(1-v^2/c^2)
from which you get
{ Dx' = ( Dx - v Dt ) / sqrt(1-v^2/c^2)
{ Dt' = ( Dt - v Dx /c^2) / sqrt(1-v^2/c^2) .
For events satisfying Dx = 0 the last equation gives
Dt = Dt' sqrt(1-v^2/c^2)

Solving the system for Dx and Dt gives
{ Dx = ( Dx' + v Dt' ) / sqrt(1-v^2/c^2)
{ Dt = ( Dt' + v Dx' /c^2) / sqrt(1-v^2/c^2)
For events satisfying Dx' = 0 the last equation gives
Dt' = Dt sqrt(1-v^2/c^2)

So you see that the equations can be deduced from the
transformation.

Now you can try to make the exercise again:
If you take the equations together and write
{ Dt' = Dt sqrt(1-v^2/c^2)
{ Dt = Dt' sqrt(1-v^2/c^2) ,
about which events are you talking and what
can they represent physically?
And how is this reflected purely by the equations?

Dirk Vdm


.

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