Re: Challenging exercises in relativity theory

Spoonfed wrote:
Spoonfed wrote:
Spoonfed wrote:
Ben Rudiak-Gould wrote:
Spoonfed wrote:
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)

I don't see why it would suggest anything at all. This is a very strange
calculation you're asking people to perform.

It's challenging, but it has a definite answer.

It reminds me of Feynman's
anecdote about adding the temperatures of stars. Why would there be any
significance to angles obtained in this way? For one thing, it's not clear
why angles defined by projections of light beams (null lines) onto a 3-plane
would have any significance.

Indeed it is not clear at all. However, I have read five or six books
where the journey from the equator to the north pole, back to the
equator and back to the original position is described as a triangle
with three 90 degree angles in it. This is used to explain the
ephemeral concept of positively curved space. However, I have never
seen a single book containing the calculation I am suggesting.

If you prefer, you can use any object traveling fast enough to catch up
with the three rockets. Or you could just calculate the angle between
the two rockets as viewed from earth + the angle between the earth and
rocket from each of the two rockets.

After you do this calculation, we can consider the question of whether
this has more or less validity than adding the temperature of stars.

For another, the light doesn't care about the
velocity of the rocketships which absorb and re-emit it, yet you're asking
us to calculate angles which depend on those velocities.


No, but what is the appropriate place to measure the angle? Who would
be in the position to measure that angle? The people in the
rocketships, and the people on earth. Just like it is the ants on a
ball that must measure the angles on the surface of a sphere, it is
people with protractors that measure angles in space.

Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.

If you want a Lorentz-invariant notion of "evenly spaced velocities", this
isn't the way to do it. Exercise: calculate the momenta of particles k-1 and
k+1 with respect to the rest frame of particle k. (They won't be negatives
of each other.) Exercise 2: figure out the Lorentz-invariant distribution.


You're way ahead of me here. I had done this a different way,
calculating the density from the assumptions and arrived at a perfect
hyperbolic sine curve, which is Lorentz invariant as far as I know.
But I shall have to try your way, and see if I can resolve the
difference in our results.


Actually the last time I worked on this, the value I got for the linear
density was
(c^3*m)/(h*t*(c^2 - x^2/t^2)^(3/2))
and I was too busy worrying about the fact that the units came out to
be meter^-2 instead of meter^-1. This I have more-or-less resolved by
realizing h in our equation is not exactly Planck's constant, though I
need to work it out in more detail.

I had forgotten about the suggestion that the solution should be
Lorentz invariant until you just mentioned it, but it's gone through my
head a few times that linear momentum, unlike velocity, adds simply in
relativity. For instance, the frames in this animation
http://en.wikipedia.org/wiki/Image:Animated_Lorentz_Transformation.gif
were created by adding a constant value of momentum to the boost in
each consecutive frame.

Oh b{r}other. The value to which a constant was added was not momentum.
It was v/sqrt(1-(v/c)^2). This quantity only becomes momentum when it
is multiplied by the mass of an object.

SR is simply 3d +1t Pythagoras. Rather hard to disprove.


Unfortunately, though this is
probably the most important quantity in Special Relativity, I have not
found a name for it in any of the literature. If anybody knows an
official term for it, I would sure appreciate it.

Anyway, through the rest of this discussion I was thinking of the
quantity v/sqrt(1-(v/c)^2); sort of a mass-independent quantity of
momentum.

No mass, no physics. It is an important mathematical relation.
It has been since the days of Pythagoras.


I just thought I'd try to clear this up. In this problem with
particles all of equal mass, it hardly matters whether we talk about
mv/sqrt(1-(v/c)^2) or v/sqrt(1-(v/c)^2), although it might be confusing
if you thought I meant that changing the momentum of a 70 kg observer
by 1 m/s would cause him to see 1 g objects previously stopped,
suddenly going 70 km/s. I wanted to clarify because I didn't want you
to think this is what I meant. Even though I guess it was what I said.
But strike what I said. Replace momentum with v/sqrt(1-(v/c)^2).
Thanks.

SR takes exception to transactions with mass and energy
which is why you can't prove or disprove it by experiment.



I will continue to work on the problem, but the trouble is the
simplicity of my construction of the problem is at issue. I would
perform the Lorentz transformation at time t=0 and position x=0 by
changing the momentum of the observer. Then the momentum of each
particle is changed by an equal amount. Since every momentum was
represented in the original problem, they would still be represented in
the new construction of the problem, i.e. it is the same problem, and
therefore it is Lorentz Invariant.

You ***Really Need*** to look at ar REAL application.
The time to accelerate a charge from a distance must be allowed
for. The Lorentz transform predicts this.

"Retarded potential"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html


I think if I can show that linear momentum is added simply, then the
rest follows. That is, if changing the momentum of the observer by
DeltaP has the effect of changing the relative velocity of all objects
he is observing by DeltaP, then the construction I gave you is Lorentz
invariant. I will continue to look into this.

If he is observing then something must accelerate the charges in his
eyeball.
"Retarded potential"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html

Anyway, I thank you for taking the time to look carefully at problems 2
and 3. I feel I may have to split problem 1 into parts, as no one has
yet attempted it.

Get physical! :o)
"Retarded potential"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html

Sue...

.

Relevant Pages

  • Re: Challenging exercises in relativity theory
    ... Two rocketships leave earth simultaneously at .866c, ... message is beamed from earth to the first rocket, to the second rocket, ... Calculate the sum of the angles of this triangle by ... head a few times that linear momentum, unlike velocity, adds simply in ...
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  • Re: Challenging exercises in relativity theory
    ... Two rocketships leave earth simultaneously at .866c, ... message is beamed from earth to the first rocket, to the second rocket, ... Calculate the sum of the angles of this triangle by ... head a few times that linear momentum, unlike velocity, adds simply in ...
    (sci.physics.relativity)
  • Re: Challenging exercises in relativity theory
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