|
|
| Author |
Message |
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
 Posted: Sat Jun 24, 2006 5:01 am Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | 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 bother. 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. 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.
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.
| Quote: | 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.
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.
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. |
|
|
| Back to top |
|
 |
Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
|
| Posted: Sat Jun 24, 2006 9:55 am Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | 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.
| Quote: | 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.
| Quote: |
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.
| Quote: |
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
| Quote: |
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
| Quote: |
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!  )
"Retarded potential"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html
Sue... |
|
| Back to top |
|
|
Dirk Van de moortel
science forum Guru
Joined: 01 May 2005
Posts: 3019
|
| Posted: Sat Jun 24, 2006 11:14 am Post subject:
Re: Challenging exercises in relativity theory
|
|
|
"Spoonfed" <good4usoul@yahoo.com> wrote in message news:1151125310.803299.266050@b68g2000cwa.googlegroups.com...
[snip - unread]
| Quote: |
Oh bother. 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. 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.
|
With x and t coordinate distance and time of a particle, and
T proper time (tau) of the particle, from the chain rule you get
v gamma = dx/dt dt/dT = dx/dT ,
so this quantitiy
v / sqrt(1-(v/c)^2)
is the derivative of coordinate distance x w.r.t. proper time T,
so you can call it the proper velocity.
| Quote: |
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.
|
yes, momentum per unit of mass.
Dirk Vdm |
|
| Back to top |
|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Sat Jun 24, 2006 1:40 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Sue... wrote:
| Quote: | 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.
|
Are you trying to keep me from acting like Winnie the Pooh?
| Quote: | 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.
|
Now that's a vague statement if I ever heard one. Yes it is hard to
disprove something so completely devoid of definition or
quantification.
| Quote: |
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.
|
But in your last statement you seemed to be placing SR in the realm of
geometry, not physics.
| Quote: |
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.
|
You ***Really Need*** to stop trying to change the subject. I would
encourage you to try to attempt some of the geometry problems I have
given you. |
|
| Back to top |
|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Sat Jun 24, 2006 1:40 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Sue... wrote:
| Quote: | 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.
|
Are you trying to keep me from acting like Winnie the Pooh?
| Quote: | 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.
|
Now that's a vague statement if I ever heard one. Yes it is hard to
disprove something so completely devoid of definition or
quantification.
| Quote: |
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.
|
But in your last statement you seemed to be placing SR in the realm of
geometry, not physics.
| Quote: |
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.
|
You ***Really Need*** to stop trying to change the subject. I would
encourage you to try to attempt some of the geometry problems I have
given you. |
|
| Back to top |
|
|
Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
|
| Posted: Sat Jun 24, 2006 2:08 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
snip
| Quote: |
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.
Are you trying to keep me from acting like Winnie the Pooh?
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.
Now that's a vague statement if I ever heard one. Yes it is hard to
disprove something so completely devoid of definition or
quantification.
|
A Cartesian coordinate system is a human construct.
Tho' orthogonal forces act uniquely so it has a natural basis.
If you can find a fire whose heat does not diminish by 1/r^2
we could say a polar system is not a natural construct.
| Quote: |
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.
But in your last statement you seemed to be placing SR in the realm of
geometry, not physics.
|
Indeed... I was.
The authors ambiguity about observation, and observation where
a charge requires a force to move leaves us no choice but to
accept only its geometrical interpretation.
| Quote: |
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.
You ***Really Need*** to stop trying to change the subject. I would
encourage you to try to attempt some of the geometry problems I have
given you.
|
Geometry is maths.. I study physics. )
Heft a few Coulombs and you notice the difference right away.
http://230nsc1.phy-astr.gsu.edu/hbase/electric/elefor.html
Sue... |
|
| Back to top |
|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Sat Jun 24, 2006 3:16 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | Ben Rudiak-Gould wrote:
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.)
|
This is a really tricky problem. I worked on it for about two and a
half hours this morning, trying to do it for Lorentz Transformations
around point (ct, 0). Hopefully I'll be able to give you a general
form solution before the end of the weekend. I think you are right
that if they are not negatives of each other, I will have to admit that
this is not a natural matter distribution. But even so, it is not hard
to define and develop, so it is certainly as good a distribution to
consider as the "dust" distribution typically found in the literature. |
|
| Back to top |
|
|
Sorcerer
science forum Guru
Joined: 09 Jun 2006
Posts: 410
|
| Posted: Sat Jun 24, 2006 5:04 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:q89ng.500122$j61.12591229@phobos.telenet-ops.be...
|
| "Spoonfed" <good4usoul@yahoo.com> wrote in message
news:1151125310.803299.266050@b68g2000cwa.googlegroups.com...
|
| [snip - unread]
And my turn...
Androcles |
|
| Back to top |
|
|
Ben Rudiak-Gould
science forum Guru
Joined: 04 May 2005
Posts: 382
|
| Posted: Sun Jun 25, 2006 9:27 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | 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:
| Quote: | 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?
How did I know the answer would be less than 180? Because the geometry of
special relativity is the same as the FRW geometry with Omega=Lambda=0, and
you problem is equivalent to adding the vertices of a triangle in that
geometry, and I know that it's hyperbolic for Omega < Omega_c.
| Quote: | Don't you want some units on that?
Good enough.
|
:-)
| Quote: | what point should you use to determine the center of
that coordinate system?
Er, it had better not matter, else your theory is inconsistent.
Shouldn't you use the point P to calculate the force at point P?
|
The choice of origin is dictated by convenience. If I'm solving the
gravitational two-body problem, the most convenient origin is the center of
mass of the system, even though neither body is ever located there. I'm free
to choose the center of mass because the answer is the same no matter what
origin I choose. You can't claim that someone's calculation is wrong just
because the point they picked isn't the one you would have picked.
| Quote: | F(r) = -4/3 pi G rho (r - r0)
where r0 is an arbitrary point in space. r0 is the center of a radial
gravitational field which increases without bound as you move farther from
the center.
Correct. This is essentially what Einstein calculated. To get
infinite force at point P, then, he used r=infinity.
|
Maybe you're right. I don't really understand what he was arguing at the end
of the paragraph.
| Quote: | So what's the significance of this? You could argue that the gravitational
field is not well-defined, since it depends on this arbitrary point, and you
could use this to make the case for a modification of Newton's theory
(perhaps an added exponential falloff).
... which is essentially what Einstein did?
|
Maybe.
| Quote: | But there's another way out. The
difference between any two of these fields (with different values of r0) is
a constant field, i.e. you can move r0 around by adding a constant field.
What's the added effect of the constant field? Nothing!
That is similar to my idea of taking the average field which I
mentioned to Sue.
|
I assume you're thinking of something like the following: to calculate the
field at P, it makes sense to take r0=P; so the field everywhere works out
to zero. Or, it makes sense to take an average over all the solutions for
all values of r0, and that average is zero everywhere.
These arguments look reasonable, but they must be wrong, for the simple
reason that F(r)=0 does not, in fact, satisfy the field equation that it was
our original goal to solve! You can concoct a solution to a differential
equation by any method you like, no matter how wacky (think image charges),
but only if it's actually a solution. If you're interested I can explain the
flaw in each argument, but I only have to look at the conclusion to know
that there is a flaw.
-- Ben |
|
| Back to top |
|
|
Ben Rudiak-Gould
science forum Guru
Joined: 04 May 2005
Posts: 382
|
| Posted: Sun Jun 25, 2006 10:05 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | Ben Rudiak-Gould wrote:
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.)
This is a really tricky problem. I worked on it for about two and a
half hours this morning, trying to do it for Lorentz Transformations
around point (ct, 0).
|
I feel guilty for asking you to work this out, since I don't think the
answer is very clean. Anyway, here's one approach. I'll use the fact that
the inner product of two four-velocities is the gamma factor of either
object in the rest frame of the other.
The momentum of particle k is hk (in the +x direction, say), so its
four-momentum p(k) is given by
p(k)_x = hk
p(k)_t = sqrt(m^2 + (hk)^2)
to get the four-velocity you divide by m:
v(k)_x = (h/m) k
v(k)_t = sqrt(1 + (h/m)^2 k^2)
The inner product of the velocities of particles k and k+1 is
gamma(k) = v(k).v(k+1)
= sqrt(1 + (h/m)^2 k^2) sqrt(1 + (h/m)^2 (k+1)^2) - (h/m)^2 k (k+1)
which depends monotonically on k, though it's not immediately obvious that
it does (like I said, ugly). At this point we're done, since if the
distribution were uniform then gamma(k) would be independent of k. But since
I asked you to calculate the momentum, I'll do that too, using another
trick. The reason the dot product above gets you gamma is that it
effectively selects out the t component of the other four-velocity, which is
equal to gamma. The x component is gamma v, which is just a factor of m away
from the momentum. In 1+1 dimensions we can get the x component with a sort
of cross product:
m v(k) cross v(k+1)
= m [v(k)_t v(k+1)_x - v(k)_x v(k+1)_t]
= m [(h/m)(k+1)sqrt(1+(h/m)^2 k^2) - (h/m) k sqrt(1+(h/m)^2(k+1)^2)]
= h [(k+1) sqrt(1 + (h/m)^2 k^2) - k sqrt(1 + (h/m)^2 (k+1)^2)]
Yuck.
-- Ben |
|
| Back to top |
|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Sun Jun 25, 2006 11:35 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Ben Rudiak-Gould wrote:
| Quote: | Spoonfed wrote:
Ben Rudiak-Gould wrote:
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.)
This is a really tricky problem. I worked on it for about two and a
half hours this morning, trying to do it for Lorentz Transformations
around point (ct, 0).
I feel guilty for asking you to work this out, since I don't think the
answer is very clean. Anyway, here's one approach. I'll use the fact that
the inner product of two four-velocities is the gamma factor of either
object in the rest frame of the other.
|
It's quite alright, because this question has flicked in and out of my
head before, but I usually shrugged it off, thinking, "of course it is
invariant; how could it not be." I worked it out with Mathematica I'm
not ready to throw the problem away just yet, but it appears you were
right--it is unbelievably close to being Lorentz Invariant, but it's
just not quite.
I made 5 events:
E0={-ct,0} is the event where everything came from.
E1={0,0} is here and now. The place where we're performing this
transform from.
E2={0,v[n-1] t}
E3={0,v[n] t}
E4={0,v[n+1] t}
E2, E3, and E4 are the events which determine the current time and
current positions of particles n-1, n, and n+1.
Then I do the Lorentz Transform on each event, where the velocity is
v[n] = (c*h*n)/Sqrt[c^2*m^2 + h^2*n^2], to get the ordered pairs
{c t0,x0},{c t1,x1},...,{c t4,x4}
And find the final velocities by taking
(x2-x0)/(t2-t0) and (x4-x0)/(t4-t0)
Adding the two together, you get
((c*h*(-1 + n)*t)/(Sqrt[c^2*m^2 + h^2*(-1 + n)^2]*Sqrt[1 -
(h^2*n^2)/(c^2*m^2 + h^2*n^2)]) -
(c*h*n*t)/(Sqrt[c^2*m^2 + h^2*n^2]*Sqrt[1 - (h^2*n^2)/(c^2*m^2 +
h^2*n^2)]))/
(t/Sqrt[1 - (h^2*n^2)/(c^2*m^2 + h^2*n^2)] - (h^2*(-1 +
n)*n*t)/(Sqrt[c^2*m^2 + h^2*(-1 + n)^2]*Sqrt[c^2*m^2 + h^2*n^2]*
Sqrt[1 - (h^2*n^2)/(c^2*m^2 + h^2*n^2)])) +
(-((c*h*n*t)/(Sqrt[c^2*m^2 + h^2*n^2]*Sqrt[1 - (h^2*n^2)/(c^2*m^2 +
h^2*n^2)])) +
(c*h*(1 + n)*t)/(Sqrt[c^2*m^2 + h^2*(1 + n)^2]*Sqrt[1 -
(h^2*n^2)/(c^2*m^2 + h^2*n^2)]))/
(t/Sqrt[1 - (h^2*n^2)/(c^2*m^2 + h^2*n^2)] - (h^2*n*(1 +
n)*t)/(Sqrt[c^2*m^2 + h^2*n^2]*Sqrt[c^2*m^2 + h^2*(1 + n)^2]*
Sqrt[1 - (h^2*n^2)/(c^2*m^2 + h^2*n^2)]))
.... Which looks even less clean than your answer!
After a considerable amount of procrastination, plugging in a few
numbers, I found that it was not zero. Eerily close to zero, though.
I'm still a bit troubled by it being nonzero though, so I'm not quite
ready to dismiss this as just another uninteresting problem.
| Quote: | The momentum of particle k is hk (in the +x direction, say), so its
four-momentum p(k) is given by
p(k)_x = hk
p(k)_t = sqrt(m^2 + (hk)^2)
to get the four-velocity you divide by m:
v(k)_x = (h/m) k
v(k)_t = sqrt(1 + (h/m)^2 k^2)
The inner product of the velocities of particles k and k+1 is
gamma(k) = v(k).v(k+1)
= sqrt(1 + (h/m)^2 k^2) sqrt(1 + (h/m)^2 (k+1)^2) - (h/m)^2 k (k+1)
which depends monotonically on k, though it's not immediately obvious that
it does (like I said, ugly). At this point we're done, since if the
distribution were uniform then gamma(k) would be independent of k. But since
I asked you to calculate the momentum, I'll do that too, using another
trick. The reason the dot product above gets you gamma is that it
effectively selects out the t component of the other four-velocity, which is
equal to gamma. The x component is gamma v, which is just a factor of m away
from the momentum. In 1+1 dimensions we can get the x component with a sort
of cross product:
m v(k) cross v(k+1)
= m [v(k)_t v(k+1)_x - v(k)_x v(k+1)_t]
= m [(h/m)(k+1)sqrt(1+(h/m)^2 k^2) - (h/m) k sqrt(1+(h/m)^2(k+1)^2)]
= h [(k+1) sqrt(1 + (h/m)^2 k^2) - k sqrt(1 + (h/m)^2 (k+1)^2)]
Yuck.
-- Ben
|
You also asked me to find a Lorentz Invariant mass distribution. Was
that also a problem that has a solution that is not very "clean?" |
|
| Back to top |
|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Mon Jun 26, 2006 5:15 am Post subject:
Re: Challenging exercises in relativity theory
|
|
|
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:
|
It is kind of odd, isn't it. If you all start at the same point, and
don't accelerate thereafter, you actually *can't* establish any angular
motion. It's all straight line motion. But if there's a change in
motion just a teeny bit after you part... just a teeny bit... Ehh, I
feel like there's something important here about singularities or
quantum mechanics, but I don't know.
| Quote: | 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?
|
I'll look over how you did this and try to get a general form for
myself. I only considered this special case, and got an answer of 150
degrees, though my confidence waxes and wanes. I'm sure it is close to
150 degrees, so your answer is in the ballpark.
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.
But you've got a general answer, and I should probably seek the same,
because I'm not at all sure I've not overlooked something, such as
propagation time.
| Quote: | How did I know the answer would be less than 180? Because the geometry of
special relativity is the same as the FRW geometry with Omega=Lambda=0, and
you problem is equivalent to adding the vertices of a triangle in that
geometry, and I know that it's hyperbolic for Omega < Omega_c.
Don't you want some units on that?
Good enough.
:-)
what point should you use to determine the center of
that coordinate system?
Er, it had better not matter, else your theory is inconsistent.
Shouldn't you use the point P to calculate the force at point P?
The choice of origin is dictated by convenience. If I'm solving the
gravitational two-body problem, the most convenient origin is the center of
mass of the system, even though neither body is ever located there. I'm free
to choose the center of mass because the answer is the same no matter what
origin I choose. You can't claim that someone's calculation is wrong just
because the point they picked isn't the one you would have picked.
F(r) = -4/3 pi G rho (r - r0)
where r0 is an arbitrary point in space. r0 is the center of a radial
gravitational field which increases without bound as you move farther from
the center.
Correct. This is essentially what Einstein calculated. To get
infinite force at point P, then, he used r=infinity.
Maybe you're right. I don't really understand what he was arguing at the end
of the paragraph.
So what's the significance of this? You could argue that the gravitational
field is not well-defined, since it depends on this arbitrary point, and you
could use this to make the case for a modification of Newton's theory
(perhaps an added exponential falloff).
... which is essentially what Einstein did?
Maybe.
But there's another way out. The
difference between any two of these fields (with different values of r0) is
a constant field, i.e. you can move r0 around by adding a constant field.
What's the added effect of the constant field? Nothing!
That is similar to my idea of taking the average field which I
mentioned to Sue.
I assume you're thinking of something like the following: to calculate the
field at P, it makes sense to take r0=P; so the field everywhere works out
to zero. Or, it makes sense to take an average over all the solutions for
all values of r0, and that average is zero everywhere.
These arguments look reasonable, but they must be wrong, for the simple
reason that F(r)=0 does not, in fact, satisfy the field equation that it was
our original goal to solve!
|
Touche'
But in fact, I've never had it as my goal to solve Einstein's Field
Equation. 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. No
units. No physical descriptions. Just a code. No questions. Just an
answer. 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.
| Quote: | You can concoct a solution to a differential
equation by any method you like, no matter how wacky (think image charges),
but only if it's actually a solution. If you're interested I can explain the
flaw in each argument, but I only have to look at the conclusion to know
that there is a flaw.
-- Ben
|
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. So our choice is important if
the number of particles is infinite. I say there is a reason to choose
point P when calculating the force on the particle at point P. (If
there is no particle at point P then there is no reason to be
calculating the force there anyway, since there is nothing for such a
force to act upon.) 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. But my argument has been that there is no
reason to calculate the force at point P from point Q, because the
force is not felt at point Q, it is felt at point P. And because there
is a reason to choose P over Q, but not Q over P, one should calculate
the force at the only point in the universe where the force on point P
has any importance, which is the point P.
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.
There's GOT to be some relationship between the effect of the
non-Lorentz-Invariant p[n]=h*n distribution and the Pauli Principle.
Well maybe there's got to be. Up until this morning, I would have just
told you it was invariant. *sigh* |
|
| Back to top |
|
|
Ben Rudiak-Gould
science forum Guru
Joined: 04 May 2005
Posts: 382
|
| Posted: Mon Jun 26, 2006 4:07 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | I worked it out with Mathematica I'm
not ready to throw the problem away just yet, but it appears you were
right--it is unbelievably close to being Lorentz Invariant, but it's
just not quite.
|
Whether it's close or not depends on h. The difference is large if h is
large, or if you compare k and k+1000000 instead of k and k+1.
| Quote: | You also asked me to find a Lorentz Invariant mass distribution. Was
that also a problem that has a solution that is not very "clean?"
|
No, that one has a nice answer. Hint: you want four-momentum (or
four-velocity) vectors that are evenly spaced on a hyperbola. To get points
evenly spaced on a hyperbola, use the hyperbolic trig functions with evenly
spaced angles.
-- Ben |
|
| Back to top |
|
|
Ben Rudiak-Gould
science forum Guru
Joined: 04 May 2005
Posts: 382
|
| Posted: Mon Jun 26, 2006 4:34 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Spoonfed wrote:
| Quote: | 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.
| Quote: | But in fact, I've never had it as my goal to solve Einstein's Field
Equation.
|
We were talking Newton, not Einstein.
| Quote: | 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.
| 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.
| 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, 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.
| 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.
| 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.
| Quote: | 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 |
|
| Back to top |
|
|
Ben Rudiak-Gould
science forum Guru
Joined: 04 May 2005
Posts: 382
|
| Posted: Mon Jun 26, 2006 8:45 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
I wrote:
| Quote: | 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 |
|
| Back to top |
|
|
Google
|
|
| Back to top |
|
|
|
|
The time now is Sat Jan 10, 2009 3:07 am | All times are GMT
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
Forum index
»
Science and Technology
»
Physics
»
Relativity
