Re: time dilation
- From: Darwin123 <drosen0000@xxxxxxxxx>
- Date: Thu, 10 Apr 2008 18:39:44 -0700 (PDT)
On Apr 10, 7:40 pm, rbwinn <rbwi...@xxxxxxxx> wrote:
The work of famous scientist Galileo Galilei provides us with aWell, the law of physics is the same in both frames.
question about time dilation and Dr. Albert Einstein's statement that
the laws of physics must remain the same in all frames of reference.
Galileo carried two lead weights of unequal sizes to the top of the
leaning tower of Pisa and dropped them at the same time, disproving
the idea of scientists of his time that the heavier of the two weights
would strike the ground first. Of course, it took some time before
scientists accepted the results of his experiment. They did not all
believe in the principle of equivalence the moment the two lead
weights hit the ground.
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.
t'=(t-vx/c^)/sqrt(1-v^2/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?
In your case, the initial conditions are different.
In S,
2at=v^2-v_0^2
In S',
2a't'=v'^2-v_0'^2
In your example, the boundary conditions are the same. In other
words, the distance from roof to floor is the same in both frames.
Obviously v is not v', v_0 is not v_0', and a is not a'. One has
to perform a Lorentz transformation on each quantity. However, the
equation (i.e., the law of physics) is exactly the same.
The laws of physics are the same in the two frames. The initial
and boundary conditions do not have to be the same in the two
frames.
If you want to include force laws, you have to Lorentz transform
the forces. In SR, the transformation of the gravitational field is
exactly the same as the transformation of an electromagnetic field.
Therefore, when you transform the gravitational field you will get a
gravitomagnetic field, which is analogous to the magnetic field. The
gravitomagnetic field is what transforms turns a to a'. However, both
S and S' describe the drop with the same equations. The S' frame may
use a zero gravitomagnetic field, while the S frame has a nonzero
gravitomagnetic field.
I know, some purists will say it is a GR problem and to some
extent I agree. I just wish to point out that SR is a valid
approximation in this case.
The accurate laws of physics are the same in S and S', because the
most accurate laws are Lorentz invariant. If you use a law that isn't
Lorentz invariant (like a=a'), you are using a law that isn't
accurate. What you are doing is assuming an inaccurate law, a=a'. This
law isn't true for relative velocities close to c.
.
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