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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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 Posted: Tue Jun 27, 2006 5:10 am Post subject:
Re: Challenging exercises in relativity theory
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Ben Rudiak-Gould wrote:
| Quote: | Spoonfed wrote:
My concept is that the original angle is 90 degrees because that is the
way we set up the problem. I chose .866c because the Lorentz Factor is
2, then I thought of the plane that contains earth and RocketShipB as a
receding flat perpendicular surface, on which, due to time dilation,
everything goes at half the speed. So the plane is receding at .866c
and the rocketship is moving along the plane at .433c from
RocketShipA's perspective. This forms a 30 degree angle, and if
correct, it would be another 30 degrees at RocketShipB's perspective.
You'd need to use the redshift formula rather than the time dilation formula
for that. I don't recommend this approach. I think it would be hard to get
right.
|
I think I feel comfortable enough with this approach, at least, to have
it as an independent method. Best to have two or three ways to
understand something if you want to be sure.
| Quote: | But in fact, I've never had it as my goal to solve Einstein's Field
Equation.
We were talking Newton, not Einstein.
I tried for some time to find out what the question was, but
unfortunately, most of what I've read has either not explained it, or
gone into some preposterous physical example or incomprehensible code
of undeveloped 10 X 10 matrices before it got to the point. I can't
read a book where I have to guess what all of the matrices look like,
and so far, I've never seen one which simply showed the matrices,
instead of hiding the information in subscripts and superscripts.
I share your frustration with textbooks that are heavy on formalism and
light on examples. Unfortunately, as you get into the more advanced physics,
there are more and more textbooks like that. I took general relativity at a
university, and we worked through some explicit calculations in gory detail
on the blackboard and in homework problems. If you want to self-study, I can
assign you some homework along those lines. If you want me to write out some
matrices explicitly, I can do that (but you'll have to be more specific).
You'll definitely need to know linear algebra before you tackle general
relativity. If you're not comfortable with dual spaces and similarity
transformations and whatnot, you should work through a linear algebra
textbook first.
|
Well, I've got two full semesters of grad school in math coming up, and
hopefully during that time I'll get comfortable. As luck would have
it, there is a considerable amount of material on dual spaces on my
test this Thursday.
| Quote: | Anyway, I've pretty much given up on that route. It seems
like everybody going that way is stuck at a dead end anyway, with all
the mystery and confusion you hear in the media.
Reporters always get everything wrong. This doesn't reflect on the state of
understanding among scientists.
|
Neither does sci.physics.relativity. :-)
| Quote: | Alright. Let Q and R be points in space not equal to P. Now we wish to
calculate the force at point P by calculating the force from particles
within sphrical shells around a given origin. If the number of
particles is finite, then we will get the same answer regardless of
which point we choose. If the number of particles is infinite, we will
get different answers for each point.
First, what matters is whether the matter distribution is bounded,
|
Good point. i.e. a bounded distribution of matter should result in the
same force no matter what order you used to perform the volume
integral. I think (wonder if) there might be differences if you choose
a different reference frame.
| Quote: | not
whether the number of particles is finite. We're not dealing with particles
here but with differential bits of mass, of which there are always an
uncountably infinite number. Second, even if the matter distribution is
bounded, there are still infinitely many solutions to the field equation.
The difference is that you can impose the boundary condition that the field
is zero at infinity, and that always leaves you with a unique solution (I
think). Third, adding contributions from spherical shells only works if you
can decompose the matter distribution into spherical shells. In the bounded
case, your distribution has to be spherically symmetric and P has to be the
center of mass. In the unbounded case, I'm a bit leery of the whole thing.
That's why I prefer the differential rather than the integral equation in
that case.
You are saying that it is valid to calculate the
force from point Q, which you will select over P or point Q, presumably
for some *reason*. From here, you will discover that the force on
point P is toward point Q.
You've consistently been talking about procedures for *finding* a solution,
whereas I've consistently been talking about what the solutions *are*. When
I said that F(r) = -4/3 pi G rho (r-r0) was a family of solutions to the
field equation, I meant only that. My procedure for finding them was to
remember that the answer had the form F(r) = k (r-r0), plug that into the
field equation, and solve for k. I never chose r0 as the center of anything.
Anyway, the whole argument seems to lie on the homogeneous static
universe idea, "dust" as they sometimes call it, which seems to me a
rather obviously flawed construction considering Hubble's Law.
It's homogeneous and isotropic, but not static. It's expanding. You're going
to have to explain what you think is the conflict between this and Hubble's Law.
"Dust" refers to a particular idealized kind of matter, which happens to be
homogeneously distributed in big bang cosmology but needn't be in general.
There's GOT to be some relationship between the effect of the
non-Lorentz-Invariant p[n]=h*n distribution and the Pauli Principle.
You mean the Pauli exclusion principle? I doubt it. This is the second time
you've implied that your h has something to do with quantum mechanics.
Planck's constant doesn't have units of momentum.
-- Ben
|
My problem is an impossible situation because it has precisely defined
momentum in a precisely defined space... Or at least, as far as I know
that's impossible due to the Heisenberg Uncertainty Principle. I use h
badly, I know. How about p[n]=h*n/R[n] where R[n] is the scale factor
of the universe between the particle with zero momentum and the one
with momentum n. Only I have no idea what precisely I mean by scale
factor of the universe, so I just say h. Then it would be neat if
calculating the density and insisting on Lorentz Invariance rendered a
nice neat solution. Then we could decide whether R[n] was perfectly
accounted for by gravity or by some quantum mechanical effect.
I'm pretty sleepy right now, so you can tell me whether I'm making
sense later. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Tue Jun 27, 2006 5:22 am Post subject:
Re: Challenging exercises in relativity theory
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Ben Rudiak-Gould wrote:
| Quote: | I wrote:
Spoonfed wrote:
Anyway, the whole argument seems to lie on the homogeneous static
universe idea, "dust" as they sometimes call it, which seems to me a
rather obviously flawed construction considering Hubble's Law.
It's homogeneous and isotropic, but not static. It's expanding. You're
going to have to explain what you think is the conflict between this and
Hubble's Law.
I just realized you're not talking about big bang cosmology. Pretend I
didn't write this.
-- Ben
|
Right. In my view a homogeneous isotropic expanding group of particles
would follow Hubble's Law, and would not be static. The problem with
this mass distribution is, if it were infinite in expanse, it would
have members that were moving faster than the speed of light from one
another, and besides that, it would not be Lorentz Invariant. |
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Phineas T Puddleduck
science forum Guru
Joined: 01 Jun 2006
Posts: 759
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| Posted: Tue Jun 27, 2006 7:30 am Post subject:
Re: Challenging exercises in relativity theory
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In article <1151385779.048249.80210@m73g2000cwd.googlegroups.com>,
Spoonfed <good4usoul@yahoo.com> wrote:
| Quote: | Right. In my view a homogeneous isotropic expanding group of particles
would follow Hubble's Law, and would not be static. The problem with
this mass distribution is, if it were infinite in expanse, it would
have members that were moving faster than the speed of light from one
another, and besides that, it would not be Lorentz Invariant.
|
Since it is sace itself that is expanding, there is no issue with
superluminal expansion of space.
--
The greatest enemy of science is pseudoscience.
Jaffa cakes. Sweet delicious orangey jaffa goodness, and an abject lesson why
parroting information from the web will not teach you cosmology.
Official emperor of sci.physics, head mumbler of the "Cult of INSANE SCIENCE".
Please pay no attention to my butt poking forward, it is expanding.
Relf's Law?
"Bullshit repeated to the limit of infinity asymptotically approaches
the odour of roses." |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Tue Jun 27, 2006 3:55 pm Post subject:
Re: Challenging exercises in relativity theory
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Phineas T Puddleduck wrote:
| Quote: | In article <1151385779.048249.80210@m73g2000cwd.googlegroups.com>,
Spoonfed <good4usoul@yahoo.com> wrote:
Right. In my view a homogeneous isotropic expanding group of particles
would follow Hubble's Law, and would not be static. The problem with
this mass distribution is, if it were infinite in expanse, it would
have members that were moving faster than the speed of light from one
another, and besides that, it would not be Lorentz Invariant.
Since it is sace itself that is expanding, there is no issue with
superluminal expansion of space.
|
I cannot accept that answer. You start with assumptions about the
distribution of matter in the universe that I disagree with, and arrive
at a simple, glib, and completely ambiguous explanation to how it is
possible to have this matter distribution.
Anyway, if space is expanding in such a way as to create particles
which are traveling faster than the speed of light from one another,
*the most distant particle moving slower than the speed of light cannot
ever interact with the adjacent slowest particle moving faster than the
speed of light*. For all intents and purposes these two particles are
from different universes.
*proof* if a particle is moving faster than the speed of light away
from us, then no beam of light from our position could reach that
particle. By symmetry no light could ever reach us from that particle.
If the most distant particle with v<c adjacent to the nearest to
particle with v>c were to interact, then we could send messages through
that interaction. These messages can't travel faster than light.
I am trying to approach the question from a starting point that allows
me to ask more questions and to move forward rather than accepting an
ambiguous answer, which seems to be a dead end.
| Quote: | --
The greatest enemy of science is pseudoscience.
Jaffa cakes. Sweet delicious orangey jaffa goodness, and an abject lesson why
parroting information from the web will not teach you cosmology.
Official emperor of sci.physics, head mumbler of the "Cult of INSANE SCIENCE".
Please pay no attention to my butt poking forward, it is expanding.
Relf's Law?
"Bullshit repeated to the limit of infinity asymptotically approaches
the odour of roses." |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Tue Jun 27, 2006 4:09 pm Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote:
| Quote: | Ben Rudiak-Gould wrote:
Spoonfed wrote:
My concept is that the original angle is 90 degrees because that is the
way we set up the problem. I chose .866c because the Lorentz Factor is
2, then I thought of the plane that contains earth and RocketShipB as a
receding flat perpendicular surface, on which, due to time dilation,
everything goes at half the speed. So the plane is receding at .866c
and the rocketship is moving along the plane at .433c from
RocketShipA's perspective. This forms a 30 degree angle, and if
correct, it would be another 30 degrees at RocketShipB's perspective.
You'd need to use the redshift formula rather than the time dilation formula
for that. I don't recommend this approach. I think it would be hard to get
right.
I think I feel comfortable enough with this approach, at least, to have
it as an independent method. Best to have two or three ways to
understand something if you want to be sure.
But in fact, I've never had it as my goal to solve Einstein's Field
Equation.
We were talking Newton, not Einstein.
I tried for some time to find out what the question was, but
unfortunately, most of what I've read has either not explained it, or
gone into some preposterous physical example or incomprehensible code
of undeveloped 10 X 10 matrices before it got to the point. I can't
read a book where I have to guess what all of the matrices look like,
and so far, I've never seen one which simply showed the matrices,
instead of hiding the information in subscripts and superscripts.
I share your frustration with textbooks that are heavy on formalism and
light on examples. Unfortunately, as you get into the more advanced physics,
there are more and more textbooks like that. I took general relativity at a
university, and we worked through some explicit calculations in gory detail
on the blackboard and in homework problems. If you want to self-study, I can
assign you some homework along those lines. If you want me to write out some
matrices explicitly, I can do that (but you'll have to be more specific).
You'll definitely need to know linear algebra before you tackle general
relativity. If you're not comfortable with dual spaces and similarity
transformations and whatnot, you should work through a linear algebra
textbook first.
Well, I've got two full semesters of grad school in math coming up, and
hopefully during that time I'll get comfortable. As luck would have
it, there is a considerable amount of material on dual spaces on my
test this Thursday.
Anyway, I've pretty much given up on that route. It seems
like everybody going that way is stuck at a dead end anyway, with all
the mystery and confusion you hear in the media.
Reporters always get everything wrong. This doesn't reflect on the state of
understanding among scientists.
Neither does sci.physics.relativity. :-)
Alright. Let Q and R be points in space not equal to P. Now we wish to
calculate the force at point P by calculating the force from particles
within sphrical shells around a given origin. If the number of
particles is finite, then we will get the same answer regardless of
which point we choose. If the number of particles is infinite, we will
get different answers for each point.
First, what matters is whether the matter distribution is bounded,
Good point. i.e. a bounded distribution of matter should result in the
same force no matter what order you used to perform the volume
integral. I think (wonder if) there might be differences if you choose
a different reference frame.
not
whether the number of particles is finite. We're not dealing with particles
here but with differential bits of mass, of which there are always an
uncountably infinite number. Second, even if the matter distribution is
bounded, there are still infinitely many solutions to the field equation.
The difference is that you can impose the boundary condition that the field
is zero at infinity, and that always leaves you with a unique solution (I
think). Third, adding contributions from spherical shells only works if you
can decompose the matter distribution into spherical shells. In the bounded
case, your distribution has to be spherically symmetric and P has to be the
center of mass. In the unbounded case, I'm a bit leery of the whole thing.
That's why I prefer the differential rather than the integral equation in
that case.
You are saying that it is valid to calculate the
force from point Q, which you will select over P or point Q, presumably
for some *reason*. From here, you will discover that the force on
point P is toward point Q.
You've consistently been talking about procedures for *finding* a solution,
whereas I've consistently been talking about what the solutions *are*. When
I said that F(r) = -4/3 pi G rho (r-r0) was a family of solutions to the
field equation, I meant only that. My procedure for finding them was to
remember that the answer had the form F(r) = k (r-r0), plug that into the
field equation, and solve for k. I never chose r0 as the center of anything.
Anyway, the whole argument seems to lie on the homogeneous static
universe idea, "dust" as they sometimes call it, which seems to me a
rather obviously flawed construction considering Hubble's Law.
It's homogeneous and isotropic, but not static. It's expanding. You're going
to have to explain what you think is the conflict between this and Hubble's Law.
"Dust" refers to a particular idealized kind of matter, which happens to be
homogeneously distributed in big bang cosmology but needn't be in general.
There's GOT to be some relationship between the effect of the
non-Lorentz-Invariant p[n]=h*n distribution and the Pauli Principle.
You mean the Pauli exclusion principle? I doubt it. This is the second time
you've implied that your h has something to do with quantum mechanics.
Planck's constant doesn't have units of momentum.
-- Ben
My problem is an impossible situation because it has precisely defined
momentum in a precisely defined space... Or at least, as far as I know
that's impossible due to the Heisenberg Uncertainty Principle. I use h
badly, I know. How about p[n]=h*n/R[n] where R[n] is the scale factor
of the universe between the particle with zero momentum and the one
with momentum n. Only I have no idea what precisely I mean by scale
factor of the universe, so I just say h. Then it would be neat if
calculating the density and insisting on Lorentz Invariance rendered a
nice neat solution. Then we could decide whether R[n] was perfectly
accounted for by gravity or by some quantum mechanical effect.
I'm pretty sleepy right now, so you can tell me whether I'm making
sense later.
|
Well, for one, by making R[n] the scale factor instead of the distance
between the two particles, I didn't correct the unit problem. (The
scale factor is unitless.) Momentum is kg m^2/s, so divide this by the
*distance between the two particles* (rather than the scale factor),
and then use Planck's constant for h and the units come out right.
Of course, having this distance in here makes the momentum
undeterminable when the size of the universe is a point. As R[1] -->
0, p[1] must approach infinity. That is, the first two adjacent
particles out of the singularity had to have infinite momentum with
respect to one another; either {v>0, mass=infinity} or {velocity=c,
mass>0}.
Either that or R[1] was never quite zero. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Wed Jun 28, 2006 8:44 am Post subject:
Re: Challenging exercises in relativity theory
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Ben Rudiak-Gould wrote:
| Quote: | Spoonfed wrote:
My concept is that the original angle is 90 degrees because that is the
way we set up the problem. I chose .866c because the Lorentz Factor is
2, then I thought of the plane that contains earth and RocketShipB as a
receding flat perpendicular surface, on which, due to time dilation,
everything goes at half the speed. So the plane is receding at .866c
and the rocketship is moving along the plane at .433c from
RocketShipA's perspective. This forms a 30 degree angle, and if
correct, it would be another 30 degrees at RocketShipB's perspective.
You'd need to use the redshift formula rather than the time dilation formula
for that. I don't recommend this approach. I think it would be hard to get
right.
|
I'm now quite certain it is 150 degrees. Consider the idea of a larger
spacecraft going by at .866c. On this spacecraft, there is a design
that appears to have a line oriented straight toward you, and another
oriented forty-five degrees. As the spacecraft goes by, I can imagine
the spacecraft grabbing one of the rocketships, and the other
rocketship moving straight along the forty-five degree stripe. The
rocket that was grabbed would now see the ship in all of its
UN-contracted form. What was a forty-five degree angle becomes a 30
degree angle. 90+30+30=150.
I'll have to work my way through what you did and see any place where
our assumptions differ.
| Quote: | But in fact, I've never had it as my goal to solve Einstein's Field
Equation.
We were talking Newton, not Einstein.
I tried for some time to find out what the question was, but
unfortunately, most of what I've read has either not explained it, or
gone into some preposterous physical example or incomprehensible code
of undeveloped 10 X 10 matrices before it got to the point. I can't
read a book where I have to guess what all of the matrices look like,
and so far, I've never seen one which simply showed the matrices,
instead of hiding the information in subscripts and superscripts.
I share your frustration with textbooks that are heavy on formalism and
light on examples. Unfortunately, as you get into the more advanced physics,
there are more and more textbooks like that. I took general relativity at a
university, and we worked through some explicit calculations in gory detail
on the blackboard and in homework problems. If you want to self-study, I can
assign you some homework along those lines. If you want me to write out some
matrices explicitly, I can do that (but you'll have to be more specific).
You'll definitely need to know linear algebra before you tackle general
relativity. If you're not comfortable with dual spaces and similarity
transformations and whatnot, you should work through a linear algebra
textbook first.
|
I'm taking another year of math right now, so hopefully I'll soon be
able to stop just complaining. I appreciate your offer, and might take
you up on it sometime. It's been nearly a year since I gave up
learning Tensors and GR from self-study, though. I hope maybe tensors
are not so hard if you don't try to learn GR at the same time.
| Quote: | Anyway, I've pretty much given up on that route. It seems
like everybody going that way is stuck at a dead end anyway, with all
the mystery and confusion you hear in the media.
Reporters always get everything wrong. This doesn't reflect on the state of
understanding among scientists.
Alright. Let Q and R be points in space not equal to P. Now we wish to
calculate the force at point P by calculating the force from particles
within sphrical shells around a given origin. If the number of
particles is finite, then we will get the same answer regardless of
which point we choose. If the number of particles is infinite, we will
get different answers for each point.
First, what matters is whether the matter distribution is bounded, not
whether the number of particles is finite. We're not dealing with particles
here but with differential bits of mass, of which there are always an
uncountably infinite number. Second, even if the matter distribution is
bounded, there are still infinitely many solutions to the field equation.
The difference is that you can impose the boundary condition that the field
is zero at infinity, and that always leaves you with a unique solution (I
think). Third, adding contributions from spherical shells only works if you
can decompose the matter distribution into spherical shells. In the bounded
case, your distribution has to be spherically symmetric and P has to be the
center of mass. In the unbounded case, I'm a bit leery of the whole thing.
That's why I prefer the differential rather than the integral equation in
that case.
|
How does the differential form differ? Would it, by any chance,
calculate the force at point P at point P?
All the spherical shells do is determine what order you do the
calculation in. I think you actually taught me that several months
ago. Hey, I just had a thought. Since you have a finite number of
shells around point P-r0, with radius < r0 and an infinite number of
shells with radius greater than r0, how about if we do the infinite
number of shells BEFORE we get to the finite number. Then we'd never
finish, but our sum would always be zero!
| Quote: | You are saying that it is valid to calculate the
force from point Q, which you will select over P or point Q, presumably
for some *reason*. From here, you will discover that the force on
point P is toward point Q.
You've consistently been talking about procedures for *finding* a solution,
whereas I've consistently been talking about what the solutions *are*. When
I said that F(r) = -4/3 pi G rho (r-r0) was a family of solutions to the
field equation, I meant only that. My procedure for finding them was to
remember that the answer had the form F(r) = k (r-r0), plug that into the
field equation, and solve for k. I never chose r0 as the center of anything.
|
I don't get it. Your answer is the answer to a question about the
force on an object standing on the surface of the sphere of constant
density. Why would you apply that to a question of the force on an
object in an infinite uniform density region?
| Quote: | Anyway, the whole argument seems to lie on the homogeneous static
universe idea, "dust" as they sometimes call it, which seems to me a
rather obviously flawed construction considering Hubble's Law.
It's homogeneous and isotropic, but not static. It's expanding. You're going
to have to explain what you think is the conflict between this and Hubble's Law.
"Dust" refers to a particular idealized kind of matter, which happens to be
homogeneously distributed in big bang cosmology but needn't be in general.
There's GOT to be some relationship between the effect of the
non-Lorentz-Invariant p[n]=h*n distribution and the Pauli Principle.
You mean the Pauli exclusion principle? I doubt it. This is the second time
you've implied that your h has something to do with quantum mechanics.
Planck's constant doesn't have units of momentum.
-- Ben
|
Corrected as p[n]=h*n/R. Solving for R to get a Lorentz invariant
solution (or something like that) might become a candidate for my next
set of challenging problems. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Wed Jun 28, 2006 11:48 pm Post subject:
Re: Challenging exercises in relativity theory
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Ben Rudiak-Gould wrote:
| Quote: | Spoonfed wrote:
Ben Rudiak-Gould wrote:
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.
Well, so does Feynman's problem, but that doesn't save it from physical
irrelevance. But I've thought about it and realized that I was wrong to
criticize your problem. My intuition was that the observed earth-ship angles
would change with time. I think your rephrasing is better, since it makes it
clearer that they don't:
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.
I can tell you immediately that the sum will be less than 180, but I'll work
through the problem anyway. I'll solve a more general problem where the
angle between the ships from the earth is theta, and the speed of each ship
is beta. I'll let one ship move along the +x axis, and the other will have a
four-velocity V given by
V_t = gamma
V_x = gamma beta cos theta
V_y = gamma beta sin theta
Boosting into a primed coordinate system where the first ship is at rest, we get
V_t' = gamma (V_t - beta V_x) = gamma^2 (1 - beta^2 cos theta)
V_x' = gamma (V_x - beta V_t) = gamma^2 beta (cos theta - 1)
V_y' = V_y = gamma beta sin theta
The earth is moving along the -x axis, so the apparent angle between the
earth and the second ship is
phi = tan^-1 (-V_y' / V_x')
= tan^-1 (sin theta / (gamma (1 - cos theta)))
Hmm, not very reasonable-looking, so maybe I made a mistake. For the special
case of theta = 90 we get phi = tan^-1 (1/gamma), which does look
reasonable. It always gives a sum of less than 180 degrees for gamma > 1.
For gamma=2 I get 143 degrees. Is that what you got?
|
Oh, I made a trig mistake. I was calculating ArcCos[1/2] when I should
have been calculating ArcTan[1/2]. Anyway, after looking at your
method, I re-did the problem cosidering sets of events.
For instance, events on earth
E0[n]={n,0,0}
Events where rocket 1 passes distances n along the x axis:
E1[n]={n,ß*n,0}
and events where rocket 2 passes distances n along its path
E2[n]={n,ß*n*cos[θ],beta*n*sin[θ]}
I wanted to use sets of events as a check to make sure all the events
remained in a line at the end.
E0' = {n/Sqrt[1 - ß^2], -((n*ß)/Sqrt[1 - ß^2]), 0}; Appropriate,
because all these events are in the -x direction.
E1' = {n/Sqrt[1 - ß^2] - (n*ß^2)/Sqrt[1 - ß^2], 0, 0}; Appropriate,
because all these events are at x'=y'=0.
E2'={n/Sqrt[1 - ß^2] - (n*ß^2*Cos[θ])/Sqrt[1 - ß^2],
-((n*ß)/Sqrt[1 - ß^2]) + (n*ß*Cos[θ])/Sqrt[1 - ß^2], n*ß*Sin[θ]}
Plugging in beta=.866 and theta=Pi/2,
E2[n]'= {1.99982 n, -1.73185 n, 0.866 n}
So the angle towards E0' is 180 degrees and the angle toward E2' is
153.433
2*26.5670 + 90=143.134
and so yes, you had the correct answer. Sorry about that. Again, I
just made a silly mistake. Taking ArcCos instead of ArcTan. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Thu Jun 29, 2006 12:01 am Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote:
| Quote: | Ben Rudiak-Gould wrote:
Spoonfed wrote:
My concept is that the original angle is 90 degrees because that is the
way we set up the problem. I chose .866c because the Lorentz Factor is
2, then I thought of the plane that contains earth and RocketShipB as a
receding flat perpendicular surface, on which, due to time dilation,
everything goes at half the speed. So the plane is receding at .866c
and the rocketship is moving along the plane at .433c from
RocketShipA's perspective. This forms a 30 degree angle, and if
correct, it would be another 30 degrees at RocketShipB's perspective.
You'd need to use the redshift formula rather than the time dilation formula
for that. I don't recommend this approach. I think it would be hard to get
right.
I'm now quite certain it is 150 degrees.
|
If you're going to make a mistake, make it loud. That's what my
marching band director used to say.
| Quote: | Consider the idea of a larger
spacecraft going by at .866c. On this spacecraft, there is a design
that appears to have a line oriented straight toward you, and another
oriented forty-five degrees. As the spacecraft goes by, I can imagine
the spacecraft grabbing one of the rocketships, and the other
rocketship moving straight along the forty-five degree stripe. The
rocket that was grabbed would now see the ship in all of its
UN-contracted form. What was a forty-five degree angle becomes a 30
degree angle. 90+30+30=150.
|
When the UN-contracted form comes out, you have the y-coordinate stay
the same, while the x-coordinate doubles. So you go from an angle
whereTan(theta)= 1 to where Tan[theta]=2. I was calculating
ArcCos[1/2] instead of ArcTan[1/2].
| Quote: | I'll have to work my way through what you did and see any place where
our assumptions differ.
|
Nope, no difference in assumptions.
Good job, Ben. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Wed Jul 05, 2006 4:36 pm Post subject:
Re: Challenging exercises in relativity theory
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|
|
Ben Rudiak-Gould wrote:
| Quote: | 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 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. 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.
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.
|
I've been working on this all week and have found the solution to
Exercise 2.
The Lorentz Invariant linear density function is rho[b]=1/(1-b^2) where
b=v/c and rho[b] is the velocity density.
Integrate this and you have a partition function Log[sqrt[(b+1)/(b-1)]]
+ constant
Add the constant, sqrt(-1) and you have a partition function
f=Log[sqrt[(1+b)/(1-b)]] which is equal to the rapidity. (i.e. for
each available rapidity there should be 1 particle.)
Assuming available rapidities are distributed equally by 1 particle per
h rapidity, we can set h*n=f[b]
setting h*n=f[b] and solving for b, you get
b=tanh[h*n]
By performing Lorentz Transformation, you'll find that the velocity
between particle n and particle n+1 is always Tanh[h].
Thus, this velocity distribution is Lorentz invariant, and so to first
order, I would guess that the linear velocity density of the universe
is 1/(1-(v/c)^2), though further analysis is in order. For instance, I
still need to find out or prove whether this is the ONLY such Lorentz
Invariant distribution.
Ben, thank you again for pointing out that equipartion of linear
momentum did NOT lead to a Lorentz Invariant distribution. Your method
of verification was a huge help. |
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Sorcerer
science forum Guru
Joined: 09 Jun 2006
Posts: 410
|
| Posted: Wed Jul 05, 2006 8:32 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
"Spoonfed" <good4usoul@yahoo.com> wrote in message
news:1152117383.684539.32810@l70g2000cwa.googlegroups.com...
|
| 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 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. 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.
| >
| > > 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.
| >
|
| I've been working on this all week and have found the solution to
| Exercise 2.
|
| The Lorentz Invariant linear density function is rho[b]=1/(1-b^2) where
| b=v/c and rho[b] is the velocity density.
|
| Integrate this and you have a partition function Log[sqrt[(b+1)/(b-1)]]
| + constant
|
| Add the constant, sqrt(-1) and you have a partition function
| f=Log[sqrt[(1+b)/(1-b)]] which is equal to the rapidity. (i.e. for
| each available rapidity there should be 1 particle.)
|
| Assuming available rapidities are distributed equally by 1 particle per
| h rapidity, we can set h*n=f[b]
|
| setting h*n=f[b] and solving for b, you get
|
| b=tanh[h*n]
|
| By performing Lorentz Transformation, you'll find that the velocity
| between particle n and particle n+1 is always Tanh[h].
|
| Thus, this velocity distribution is Lorentz invariant, and so to first
| order, I would guess that the linear velocity density of the universe
| is 1/(1-(v/c)^2), though further analysis is in order. For instance, I
| still need to find out or prove whether this is the ONLY such Lorentz
| Invariant distribution.
|
| Ben, thank you again for pointing out that equipartion of linear
| momentum did NOT lead to a Lorentz Invariant distribution. Your method
| of verification was a huge help.
"Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message
news:P4Hqg.60105$Lm5.3167@newssvr12.news.prodigy.com...
| This is PHYSICS, not math or logic, and "proof" is completely irrelevant.
No sense in arguing with a bigot.
'nuff said. <shrug>
Androcles. |
|
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|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Wed Jul 05, 2006 10:50 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Sorcerer wrote:
| Quote: | "Spoonfed" <good4usoul@yahoo.com> wrote in message
news:1152117383.684539.32810@l70g2000cwa.googlegroups.com...
|
| 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 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. 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.
|
| > > 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.
|
|
| I've been working on this all week and have found the solution to
| Exercise 2.
|
| The Lorentz Invariant linear density function is rho[b]=1/(1-b^2) where
| b=v/c and rho[b] is the velocity density.
|
| Integrate this and you have a partition function Log[sqrt[(b+1)/(b-1)]]
| + constant
|
| Add the constant, sqrt(-1) and you have a partition function
| f=Log[sqrt[(1+b)/(1-b)]] which is equal to the rapidity. (i.e. for
| each available rapidity there should be 1 particle.)
|
| Assuming available rapidities are distributed equally by 1 particle per
| h rapidity, we can set h*n=f[b]
|
| setting h*n=f[b] and solving for b, you get
|
| b=tanh[h*n]
|
| By performing Lorentz Transformation, you'll find that the velocity
| between particle n and particle n+1 is always Tanh[h].
|
| Thus, this velocity distribution is Lorentz invariant, and so to first
| order, I would guess that the linear velocity density of the universe
| is 1/(1-(v/c)^2), though further analysis is in order. For instance, I
| still need to find out or prove whether this is the ONLY such Lorentz
| Invariant distribution.
|
| Ben, thank you again for pointing out that equipartion of linear
| momentum did NOT lead to a Lorentz Invariant distribution. Your method
| of verification was a huge help.
"Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message
news:P4Hqg.60105$Lm5.3167@newssvr12.news.prodigy.com...
| This is PHYSICS, not math or logic, and "proof" is completely irrelevant.
No sense in arguing with a bigot.
'nuff said. <shrug
Androcles.
|
Tom Roberts hasn't participated in this conversation. |
|
| Back to top |
|
|
Sorcerer
science forum Guru
Joined: 09 Jun 2006
Posts: 410
|
| Posted: Thu Jul 06, 2006 5:22 am Post subject:
Re: Challenging exercises in relativity theory
|
|
|
"Spoonfed" <good4usoul@yahoo.com> wrote in message
news:1152139813.612600.295000@l70g2000cwa.googlegroups.com...
|
| Sorcerer wrote:
| > "Spoonfed" <good4usoul@yahoo.com> wrote in message
| > news:1152117383.684539.32810@l70g2000cwa.googlegroups.com...
| > |
| > | 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 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. 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.
| > | >
| > | > > 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.
| > | >
| > |
| > | I've been working on this all week and have found the solution to
| > | Exercise 2.
| > |
| > | The Lorentz Invariant linear density function is rho[b]=1/(1-b^2)
where
| > | b=v/c and rho[b] is the velocity density.
| > |
| > | Integrate this and you have a partition function
Log[sqrt[(b+1)/(b-1)]]
| > | + constant
| > |
| > | Add the constant, sqrt(-1) and you have a partition function
| > | f=Log[sqrt[(1+b)/(1-b)]] which is equal to the rapidity. (i.e. for
| > | each available rapidity there should be 1 particle.)
| > |
| > | Assuming available rapidities are distributed equally by 1 particle
per
| > | h rapidity, we can set h*n=f[b]
| > |
| > | setting h*n=f[b] and solving for b, you get
| > |
| > | b=tanh[h*n]
| > |
| > | By performing Lorentz Transformation, you'll find that the velocity
| > | between particle n and particle n+1 is always Tanh[h].
| > |
| > | Thus, this velocity distribution is Lorentz invariant, and so to first
| > | order, I would guess that the linear velocity density of the universe
| > | is 1/(1-(v/c)^2), though further analysis is in order. For instance,
I
| > | still need to find out or prove whether this is the ONLY such Lorentz
| > | Invariant distribution.
| > |
| > | Ben, thank you again for pointing out that equipartion of linear
| > | momentum did NOT lead to a Lorentz Invariant distribution. Your
method
| > | of verification was a huge help.
| >
| > "Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message
| > news:P4Hqg.60105$Lm5.3167@newssvr12.news.prodigy.com...
| >
| > | This is PHYSICS, not math or logic, and "proof" is completely
irrelevant.
| >
| > No sense in arguing with a bigot.
| > 'nuff said. <shrug>
| > Androcles.
|
| Tom Roberts hasn't participated in this conversation.
Humpty Roberts is part author to the Relativity FAQs (endorsed
by Professor Baez of USC) and Roberts' testimony goes to credibility.
You may as well burn the FAQs now, the truth is out.
Say all you want, the relativity bus is for cranks like you to ride straight
to the nuthouse, no proof required.
Androcles.
| |
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