Re: time dilation

Am Thu, 10 Apr 2008 16:40:17 -0700 schrieb rbwinn <rbwinn3@xxxxxxxx> in
dd0d3b7a-58e7-48e4-b9ed-c269b00d1a88@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

[...]
This brings us to another question about falling objects which
arises from the idea of dropping an object in a moving train car, which
writers of textbooks about relativity often use to show how the Lorentz
equations work. If a weight is dropped from the top of a train car to
the floor, it falls a distance of y'. In any transformation equations
this is always expressed as y'=y. The object travels the same distance
vertically in S' as it does in S. In Galileo's equations, it takes the
same amount of time for the object to travel from the roof of the train
car to the floor in either frame of reference. t'=t.
In the Lorentz equations, a clock in S', the frame of reference
of the train car, is slower than a clock in S, the frame of reference of
the train tracks.

That statement is true (in Relativity) only for colocal events. That is
why the Lorentz transformation is used if you can't assure colocality.

t'=(t-vx/c^)/sqrt(1-v2/c^2). According to this equation, it takes less
time for the object to fall from the roof of the train car to the floor
in S' than it does in S. So how are the laws of physics the same in
both frames of reference?

Each event in each of the two frames has coordinates of the form
[t;x;y;z] and the Lorentz transformation, which only applies to frames
that are moving inertially, relates the coordinates from one frame to
another. So if in the moving frame we have [t';x';y';z'] and in the
stationary frame we have [t;x;y;z] then

t' = gamma (t - v*x/c^2)
x' = gamma (x - v*t) (direction of motion)
y' = y
z' = z
gamma = 1/sqrt[1 - v^2/c^2]

What I usually do in solving problems like this, is to make a table of
events like this:

S':[t';x';y'] S:[t;x;y]
start drop [0;0;0] [0;0;0]
end of drop [(2y'/g)^(1/2);0;y'] [?;?;?]

In this I am assuming a weak gravitational field, so that y'=1/2*g*t'^2
inside the train. Now we just apply the Lorentz transformation and get

[gamma*(2y/g)^(1/2);gamma*v*(2y/g)^(1/2);y]

with y=y' as the coordinates for the event of the object hitting the
floor as seen in the rest frame.

The Galilean/Newtonian condition is recovered as v->0 where t->t', x->x'
and y=y'. However, clearly that is true only when the object is slow.
Notice that the object will never hit the floor in finite time when v->c,
ie x->infinity.

Now, about the laws of nature. Basically in this case we are talking
about Newton's three laws: specifically, that the momentum will change
when a force like gravity is applied. In math it is written:

F=dp'/dt'= d/dt' (m dy'/dt') in the moving frame and
F=dp/dt= d/dt (gamma m dy/dt) in the stationary frame.

The force of gravity, F=m*g, is the same in both frames. Now the law in
the moving frame reduces to just

m*g=m*d^2y'/dt'^2 or
g = d^2y'/dt'^2

meaning that acceleration is constant in the moving frame.

In the stationary frame

m*g= d/dt (gamma m dy/dt)
= m dgamma/dt dy/dt + m gamma d^2y/dt^2

Since the train is moving inertially, v is constant and so gamma is
constant resulting in

g = gamma d^2y/dt^2

At this point we turn to the Lorentz transformation and use

y' = y
t' = gamma (t - v*x/c^2)
= gamma (t - v^2*t/c^2) because the train is moving x=v*t
= gamma t (1 - v^2/c^2)
= gamma t / gamma^2
= t/gamma

Taking derivatives we get

dy' = dy
dt' = dt/gamma - t/gamma^2*dgamma

However since the train is moving inertially dgamma=0 so

dt' = dt/gamma

Dividing,

dy'/dt' = gamma dy/dt

and since dgamma=0

d^2y'/dt'^2 = gamma d^2y/dt^2

ie g=g is constant acceleration in both frames.

Thus Newton's second law being valid in both frames leads to the fact
that the acceleration due to gravity is constant in both frames even
though the way acceleration is calculated in different frames is
different.

The point to Relativity is that the physics doesn't change just because
your viewpoint changes, even if you choose non-priviledged coordinate
systems.

[...]

Great question. Thanks for the post.
.

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