Re: time intervals
- From: Darwin123 <drosen0000@xxxxxxxxx>
- Date: Sat, 5 Apr 2008 15:39:22 -0700 (PDT)
On Mar 29, 6:35 pm, rbwinn <rbwi...@xxxxxxxx> wrote:
On Mar 29, 1:19�pm, "Pmb" <some...@xxxxxxxxxxxxx> wrote:If that is what you saw, it is probably a typographical mistake. The
"rbwinn" <rbwi...@xxxxxxxx> wrote in message
news:fed96972-3a50-4c36-ba36-02fb7830d2d4@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
One scientist has brought it to my attention that scientists are using
a different equation for t' than t'=(t-vx/c^2)gamma in calculating
time intervals, and that this is taught in physics textbooks, which I
also recall as being true. �He said that the time interval in S' is
measured by t'=t gamma.
I don't recall saying that Robert. I tried to explain the meaning of the
time dilation expression, what
t' = gamma t
means. It relates the proper time, t (which is measured in S) and the
coordinate time t' (as measured in S'). By definition of "proper time" the
events take place at the same spatial location. This must be because we're
considering the rate of a clock which is *stationary* in S as compared to
the same clock which is moving in S'. Thus the spatial coordinates must be
the same. � The expression t'=(t-vx/c^2)gamma gives the temporal coordinate
of an event. Time intverval between two events does require keeping the x
variable since the events don't generally take place at the same spatial
location. I appologize if I didn't make that clear.
There is still something haywire about what you are doing. Go back to
the original problem as Einstein set it up. Light is beamed in the
direction of motion of S' when the origins of S and S' coincide.
After a time of t, S' has moved a distance of vt. The time in both
frames of reference was t=t'=0 when the light was turned on. So a
photon traveled from the origin of S to x in a time of t, and the same
photon traveled from the origin of S' to x' in a time of t'. Figuring
this with the Lorentz equations and your previous explanation, we have
t2=t, t1=0. t2-t1=t t is what a clock in S will say.
Now you want to figure the same interval of time in S'. The photon
starts at the origin of S' and ends up at x'. t'=x'/c t'1=x'1/c=0/
c=0
t'2=x'2/c=[(x-vt)/c]gamma = (t-vx/c^2)gamma = t'2-t'1=t'
So where is the S clock in your calculations, and where is the S'
clock?
�I told him that t' cannot equal t gamma
because t' is already defined to be t'=(t-vx/c^2)gamma.
You have to be very careful when using these transformations because you
have to keep in mind the *precise* interpretation of what the variables
represent. Transfroming from (ct, x, y, z) to (ct', x', y', z') requires
that you understand the physical meaning of the variables. t' = gamma t
compares the rates between two clocks in relative motion whie
t'=(t-vx/c^2)gamma represents the coordinate of an event.
Well, if you are going to be precise, you do not use the same variable
for two different meanings. I know where I saw this equations now,
t'=t(sqrt (1-v^2/c^2) It was in Einstein's book. Einstein himself
came up with what you are trying to say, but so far I can't tell what
it is. t' in the Lorentz equations comes up with a smaller time than
t. I cannot see where this t'=t(1-v^2/c^2) comes from unless it is a
mistake.
Lorentz transform for the S' coordinate traveling in the direction of
the x-axis is:
(gamma)=(1-v^2/c^2)^-0.5
x'=(gamma)(x-vt)
y'=y
z'=z
t'=(gamma)t(1+xv/c^2)
The (gamma) is a factor in the time term, and (1+xv/c^2) is
another term. The Lorentz time dilation factor corresponds to x=0.
BTW, the xv/c^2 factor is the term that resolves the so-called
"twin paradox." This is the term that makes "acceleration" and "force"
real physical quantities in SR.
The Lorentz transform as stated above is exact only if the
observers in both S and S' are inertial. However, let us assume that
S' is not inertial. The observer of mass m is subject to a force F in
the x direction. Therefore, S' has a net acceleration,
a=F/m.
Okay. So now "v" isn't constant. In fact,
dv/dt=a.
Okay, NOW calculate
dt'/dt,
being careful to use all those calculus rules of differentiation. Not
that a term in dt'/dt is generated proportional to "a." That term
solves all the "twin paradox" problems.
Note that S' is not an inertial frame anymore. However, t' is
still well determined. And it is still SR, not GR.
.
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