Understanding Special Relativity
Questions and Answers
In the explanation about the twins paradox; During
the U-turn the difference between Traal's and Earth's clocks is changing
from -7.5 to 7.5 in a short time relative to Bob. Does that mean that
activity on Earth goes at tremendous speed during this period? And if it
does, isn't one of the consequences of it that bodies on Earth are moving faster than light
relative to Bob?
Relativity states that the relative velocity between two
bodies cannot exceed the light speed. It doesn't mean that an observer on a
third frame cannot see the relative speed between them beeing grater than
light speed. If, for example, two spaceships are going away from an observer
in opposite directions at a speed of 0.8c each, then the observer measures a
relative speed of 1.6c between them, though, due to the law of velocities
composition, each of the spaceships sees the other going away from it slower
than the light speed. Thus, there is no problem in saying that bodies on
Earth are moving at a speed greater than the light speed relative to each
other in Bob's frame.
The question is at what speed each body is moving relative to Bob. Special
Relativity deals with inertial frames, and its rules applies to such frames.
During an acceleration things are more complicated. Though at any point
on the acceleration line in time-space all the rules are preserved, when we
calculate the integral along a range we get strange things. One can also ask
the following question: When Bob leaves Earth Traal is 8.66 light years away
from him. A short time later, when he completes his acceleration Traal
is a few light years closer. Does Traal approach Bob faster then light? The
question is if we can call this approach a motion. If you check the relative
speed between Bob and the frame of Earth and Traal at any point during the
acceleration it will be lower then c (the slope of the tangent of the curve
in the time-space diagram is greater then 45 degrees). Bob is continuously
changing reference frames during the acceleration. Traal is getting closer
to Bob due to change in the length contraction and not because it moves in
space relative to any of these frames.
There is one thing, however, that confuses me, not in
your paper but in general. Upon the explosion of a supernova, parts of it
travel in all directions at high relativistic speeds. The light emitted
from these parts is red shifted or blue shifted depending on the part
velocity. It is measured that the red shifted light arrives to the earth
just a few days after the blue shifted light. The Supernova to Earth
distance is measured in millions of light years. It would take centuries
to get blue shifted light after the red shifted light because according
to SRT (and your explanation) the departure times (as seen by the
observer on Earth) are different, they are proportional to gamma * vx/c^2
(where v is varying say from -0.9c to +0.9c). As the arrival times are
virtually the same (days instead of centuries), why does Einstein state that
the observer on the moving train receives the 2 flashes at different
times? I see no substantial difference between the supernova and the
relativistic train experiment and I suppose a real experiment should be
more relevant than a thought experiment.
As you know a supernova is a huge phenomenon. Matter
moves towards Earth at the part of the supernova that is closer to Earth.
Matter moves away from Earth at the part that is further from Earth. The
diameter of the supernova is negligible relative to the distance between
Earth and the supernova, but it is still a significant distance.
The time difference between the light coming from matter that moves towards
Earth (blue-shifted) and matter that moves away from Earth (red-shifted) is
affected by two things: The longer way that the red shifted light pass and
(as you correctly pointed) the time difference between the emission of the
light relative to Earth due to the relativity of simultaneity.
Your confusion comes from choosing the wrong distance and velocity. As
written in the article, simultaneity difference is proportional to the
distance between the events and the relative velocity between the frames of
reference. The simultaneity difference is not affected by the distance
between the observer and the events. The woman in the train would measure
the same time difference between the lightning bolts if the train was
hundred miles further away on the track. Now, what is the reference frame on
which the two events are simultaneous? This is not the reference frame of
the moving matter, but the one of the center of the supernova, relative to
which the pieces of matter are moving at the same (high) velocity in
opposite directions. This reference frame is moving away from Earth due to
the universe expansion.
So the distance you should use is the diameter of the supernova and the
velocity is the velocity of the motion of the whole supernova away from
Earth as it was when the supernova happened. I guess when you substitute
these values in the equations you get time difference of days.
For the sake of simplicity, I will not take the
expansion of the universe into account and I will assume that the
relative speed between the Earth and the supernova is 0. Also I will
assume the explosion of the supernova creates two parts moving at
velocities +v and -v relative to the earth (and the center of the
supernova). The first part (approaching earth with +v speed) creates
blue shift and the second part (moving away from the earth with -v
speed) creates red shift.
I know the events
are (nearly) simultaneous relative to the supernova. It appears to me
they are not simultaneous for the observer on earth, even if the earth
is at rest relative to supernova. This is also true in the case of the
twin paradox: When a twin passes near a distant star, he will see
different times on the star and on the earth, even if the the star is at
rest relative to the earth.
The velocity you put
in Lorenz transformation is a relative velocity between two frames of
reference, not a velocity of something that is moving relative to the
reference frame. You have to decide what your reference frame is. If two
events happen simultaneously for one point on a reference frame they are
simultaneous everywhere on this frame. It doesn't matter if they happen
on stationary or moving parts. They can only have time difference when
measured from a different frame. If two pulses of light are emitted
simultaneously relative to the supernova center and Earth is not moving
relative to this center, then they are emitted simultaneously relative
to Earth too. If you like, the relative velocity between the frames is
0, beta = 0 and gamma = 1.
Also, if you say
that the two events happen at approximately the same point, they will be
simultaneous relative to any reference frame, no matter how fast it
moves. So the only thing that makes difference between the times they
arrive to Earth is the diameter of the supernova.
During the U-turn in the twins paradox, 15 years pass at
Earth from Bob's point of view. What happens if Bob changes his mind,
continues his turn and return to his original course? Is the time on Earth
goes backwards for him? Does his brother Al earn 15 years back, or perhaps
this time cannot be returned and time continues from where it reached?
When Bob gets to Traal as described in the twins paradox,
there are two reference frames: This of the spaceship and this of the plants
Traal and Earth. In the frame of the planets the distance to Earth is 8.66
light years and the time on Earth is 10 years after the launch. In the frame
of the spaceship the distance to Earth is 4.33 light years and the time on
it is 2.5 light years from launch. If Bob suddenly stops (makes a half of
the U-turn) he moves to the reference frame of the planets, and thus time on
Ears turns to be 10 years from launch for him, that is 7.5 years later. If
he then immediately accelerate and return to his original speed, he is back
in his old frame again (actually he is a little behind in the time axis
because it took him time to stop and accelerate, but this difference is
negligible), and thus time on Earth is again only 2.5 years after launch.
Apparently his brother gained 7.5 years back. Actually, these 7.5 years do
not really pass, not forward nor backward. What changes is only what Bob
calls "now". It is a different thing in the two frames.
Imagine a man standing in a wood. He defines a coordinate
system where the line left-right is the x axis and the line backward-forward
is the y axis. All the trees that are exactly to his left or exactly to his
right has a value y=0. Trees that are behind this line relative to his
position have negative value of y and trees that are in front of it have
positive y value. Now the man turns a little to his left and set a new
coordinate system relative to his new position. All the trees to his left
have now grater y values and the trees to his right have smaller y values.
Clearly neither the trees nor the man have moved in any direction. The new y
values are only due to the change in the coordinate system.
In relativity time is also an axis in time-space. When
Bob moves from one frame to another the line he calls "now" changes, just as
line y=0 have changed when the man changed his coordinate system. The line
"now" in the new frame goes through different set of events in time-space.
Al don't gain years just as the trees in the wood will not move back if the
man will return to his original position.
The laws of nature also make sure that Bob will not be
able to see this change. Signals that are sent from Earth in light speed
will always get to Bob in the order they left. If he has a strong telescope,
he will see life on Earth moving always forward. He might see them in
slow-motion or in fast forward, but never in rewind.
I am trying to understand what happens when a very
long body accelerates. In the explanation about the twin paradox we
assume that after the acceleration the spaceship remains approximately
at the same point and the same time as it was before, relative to Earth.
This was good enough for a point mass. But in the case of a very long
solid spaceship that accelerates, we cannot assume the same about both
ends of it. After the end of the acceleration, if we look from the
original frame, there is a time difference between the ends of the
spaceship and its length is smaller than it was before the acceleration.
If we place clocks at fixed distances along the spaceship and in front
of every clock we place another clock on the inertial space station on
which it anchors, after the acceleration the distance between the clocks
on the spaceship as observed from the space station will be shorter than
the distance between the clocks on the station. If the two bodies are
long enough, one of the clocks must be approximately in front of the
same clock as it was before the acceleration. How can we determine which
clock it will be? What is the difference between the points? Also, the
time difference between the clocks changes from point to point. Again,
how can we determine what is the time difference at each point?
It seems to me that there are two degrees of freedom
here while in reality the state of the spaceship after the acceleration
must be well defined.
According to relativity strange things happen during
acceleration. During the acceleration, time doesn’t pass at the same
rate at all the points of the spaceship. Clocks that are closer to the
front of the spaceship will move faster than clocks closer to the back
of it. When the spaceship stops accelerating and becomes inertial again,
the clocks are synchronized in the new reference frame. Clocks at the
rear will show earlier time than clocks at the front. We can't speak
about the time that passed in the spaceship during the acceleration.
Different time passed at every point of the spaceship. If we want to
resynchronize the clocks we can freely choose which clock shows the
correct time. Apparently there is a degree of freedom here, but it is
all about the origin of one reference frame relative to the other.
And what about distances? As observed from the space
station, all the points along the spaceship start accelerating
synchronously. But in this reference frame the acceleration is not
constant. Due to the non linear composition of velocities the
acceleration is reduced with time. In addition, in the reference frame
of the spaceship all the points of it stop accelerating simultaneously.
But this frame is moving relative to the frame of the station so in this
frame it will not happen simultaneously. Rear points will stop
accelerating before frontal points. In the frame of the station the
acceleration ends when the front end of the spaceship stops
accelerating. Since all the points end up at the same speed, the rear
points had grater acceleration (they reached the final speed faster) and
thus moved longer distance then the front points. At the end of the
acceleration all the points on the spaceship will be shifted from the
points on the station in the direction of the motion of the spaceship.
The front end of it will be closest to its original position. As we get
closer to the back of the spaceship, there will be greater shift of the
point. The distance between two points on the spaceship will be smaller
then the distance of the matching points on the spaceship (as observed
from the station). This is the expected length contraction.
I am trying to find a formula for
elastic collision. I know that the total energy and the total momentum
are preserved in elastic collision, but I can't solve the equations for
the velocities after the collision as we do in classic mechanics.
The answer is too
long for this page. I have put it here.