Send me to the FAQY if you must, but ...

What is the correct generalization of Newtonian gravity for massive
relativistic projectiles? Assume a region of (flat) uniform
gravitational field g, and consider a relativistic projectile, rest
mass m0, altitude z. Is it still true that d^2z/dt^2 = g?

Moiraine wrote in message . ..
On 26 Jul 2003 09:06:59 -0700, (Edward Green) wrote:

Send me to the FAQY if you must, but ...

What is the correct generalization of Newtonian gravity for massive
relativistic projectiles? Assume a region of (flat) uniform
gravitational field g, and consider a relativistic projectile, rest
mass m0, altitude z. Is it still true that d^2z/dt^2 = g?

The latest gravity experiment that is close to yielding an objective
answer is Gravity Probe B. The experiment is being performed at
close to zero gravity and near the lambda point of helium to avoid
contamination of data. Should yield the most accurate compelling
evidence so far.

http://einstein.stanford.edu/gen_int...b/gpbsty1.html

Thank you for the reference. Of course experiment trumps all ... but
surely plain old GTR must make some unambiguous predictions here?

Edward Green wrote in message
om...
Moiraine wrote in message
. ..
On 26 Jul 2003 09:06:59 -0700, (Edward Green) wrote:

Send me to the FAQY if you must, but ...

What is the correct generalization of Newtonian gravity for massive
relativistic projectiles? Assume a region of (flat) uniform
gravitational field g, and consider a relativistic projectile, rest
mass m0, altitude z. Is it still true that d^2z/dt^2 = g?

The latest gravity experiment that is close to yielding an objective
answer is Gravity Probe B. The experiment is being performed at
close to zero gravity and near the lambda point of helium to avoid
contamination of data. Should yield the most accurate compelling
evidence so far.

http://einstein.stanford.edu/gen_int...b/gpbsty1.html

Thank you for the reference. Of course experiment trumps all ... but
surely plain old GTR must make some unambiguous predictions here?

Ed has it right. Use the equivalence principle. Just replace
the uniform gravitational field with a rocket ship accelerating
at g. [Old Man]

"Edward Green" wrote in message
om...
Send me to the FAQY if you must, but ...

What is the correct generalization of Newtonian gravity for massive
relativistic projectiles? Assume a region of (flat) uniform
gravitational field g, and consider a relativistic projectile, rest
mass m0, altitude z. Is it still true that d^2z/dt^2 = g?

"Kinetic Energy and the Equivalence Priciple", S. Carlip,
http://arxiv.org/abs/gr-qc/9909014

"Old Man" wrote in message news:3f23a31a_4@newsfeed...
Edward Green wrote in message
om...
....

surely plain old GTR must make some unambiguous predictions here?

Ed has it right. Use the equivalence principle. Just replace
the uniform gravitational field with a rocket ship accelerating
at g. [Old Man]

Sigh. I was afraid you might say something like that. You want me to
"think"? Ok ... girding my loins to the sticking place:

Consider the instantaneous rest frame of a rocket accelerating at 1g
in the z direction, and consider two particles at altitude z0 at t0;
one is stationary in the instantaneous rest frame of the rocket, the
other has no z component of velocity but a relativistic x component
perpendicular to z.

Well ... it is perfectly damn evident that, since in this case the
rocket is "really" doing the moving, that the two particles will
"fall" towards the floor of the rocket at exactly the same rate. The
orthogonal velocity of the second is irrelevant, it's holding at
constant z in our initial inertial frame -- and since we chose the
instantaneous rest frame, there are to first order no relativistic
corrections between rocket frame and rest frame.*

OK. That's pretty damn unambiguous. Since we may extend this
argument to arbitrary particle trajectories, we predict from the EP
that all objects accelerate at the same rate in a uniform G field,
relativistic velocities be damned.

OTOH, we know that clocks on the projectile moving past at
relativistic velocity appear to run slower by gamma; in other words,
the projectile falls past a fixed set of z-indicia in the rocket
attached frame in less proper time.
Since delta z = (1/2)gt^2, we conclude g will appear to grow by a
factor of gamma^2 on the projectile.

Any contention with any of this?

I've been thinking of that damn silly "relativistic submarine"
paradox.

A reference was provided discussing the gravitational effect of
relativistic mass: however, on this level, that doesn't seem relevant.
We have reached the conclusion that relativistically varying masses
fall exactly the same way otherwise varying masses do -- there is no
variation in the kinematics.

*Is this true? Or must we still include some correction moving back
and forth between rocket and rest clocks?

"Old Man" wrote in message news:3f249770_4@newsfeed...

Edward Green wrote in message
om...

Consider the instantaneous rest frame of a rocket accelerating at 1g
in the z direction, and consider two particles at altitude z0 at t0;
one is stationary in the instantaneous rest frame of the rocket, the
other has no z component of velocity but a relativistic x component
perpendicular to z.

Well ... it is perfectly damn evident that, since in this case the
rocket is "really" doing the moving, that the two particles will
"fall" towards the floor of the rocket at exactly the same rate. The
orthogonal velocity of the second is irrelevant, it's holding at
constant z in our initial inertial frame -- and since we chose the
instantaneous rest frame, there are to first order no relativistic
corrections between rocket frame and rest frame.*

OK. That's pretty damn unambiguous. Since we may extend this
argument to arbitrary particle trajectories, we predict from the EP
that all objects accelerate at the same rate in a uniform G field,
relativistic velocities be damned.

This sounds correct. The OTOH part, below, is too complicated
for Old Man . It's a mater of mind and time over matter. [Old Man]

No damn fair!

You have pushed me into the thorny and painful byways of "thought",
and now you must join me in the bracken field!

(I had to retype that three damn times because of software glitches,
and each time the joke grew more ornate -- so it's no my, dare I say,
damn, fault ;-).

Anyway, the snipped part below was relatively straight forward ...
just involving the time dilation in the inertial projectile's frame,
and the effect it would have on the apparent acceleration in that
frame past the ... err ... frames of the rocket. The "*" statement
worries me more, since I'm not sure if some baby-bathwater escapage
did not occur there -- needs more thought about clocks in accelerated
frames.

I did make a baby/bathwater kind of error in your orbital angular
momentum thread, BTW -- though I claim I left suitable hedges :-)
Viz., I blithly claimed a course change is a course change, and costs
us locally the same whatever we think of the global energetic orbital
implications. I still stand by this ... but the expert John Prussing
points out that your solution lies precisely in using awareness of the
global properties of the orbits to pick the optimal points to spend
your course change currency.

I am happy you have the energy to spend on sci.physics. I saw some
mock kung-fu knife fighting at a street fair Sunday, and while the
dancing and athletics were impressive, the effect was lessened by the
steely knives flapping in the wind! Your knife remains tempered,
however, and just the right foil.

(I'm trying my hand at a bit of sycophancy, and you must excuse any
clumsiness in the unfamiliar form :-).

Edward Green wrote in message
om...
"Old Man" wrote in message
news:3f249770_4@newsfeed...

Edward Green wrote in message
om...

Consider the instantaneous rest frame of a rocket accelerating at 1g
in the z direction, and consider two particles at altitude z0 at t0;
one is stationary in the instantaneous rest frame of the rocket, the
other has no z component of velocity but a relativistic x component
perpendicular to z.

Well ... it is perfectly damn evident that, since in this case the
rocket is "really" doing the moving, that the two particles will
"fall" towards the floor of the rocket at exactly the same rate. The
orthogonal velocity of the second is irrelevant, it's holding at
constant z in our initial inertial frame -- and since we chose the
instantaneous rest frame, there are to first order no relativistic
corrections between rocket frame and rest frame.*

OK. That's pretty damn unambiguous. Since we may extend this
argument to arbitrary particle trajectories, we predict from the EP
that all objects accelerate at the same rate in a uniform G field,
relativistic velocities be damned.

This sounds correct. The OTOH part, below, is too complicated
for Old Man . It's a mater of mind and time over matter. [Old Man]

No damn fair!

You have pushed me into the thorny and painful byways of "thought",
and now you must join me in the bracken field!

OK, I'll make an attempt. It's not that relativity problems aren't
interesting, but I usually screw-up.

The subject is motion in directions transverse to the relativistic
velocity vector of a massive body.

A familiar example is that of a charged particle accelerating in a
uniform electric field, wherein the particle has initial momentum
in a direction transverse to the electric field. As in Ed's problem,
we will assume that the initial transverse velocity is relativistic.
For a stationary observer, the transverse momentum is conserved,
but the transverse velocity approaches zero as the velocity parallel
to the field approaches that of light. Here is a calculation:

Consider the following problem: An observer is standing at rest at
the origin of a Cartesian coordinate system. A uniform electric field,
E, exists with direction parallel to the y - axis. A test particle with
rest
mass, M, and charge, Q, has an initial momentum, Pxo, with direction
parallel to the x - axis. The initial conditions a

Uniform electric field ( Ex, Ey, Ez ) = ( 0, E, 0 ) for all space and for
all times. Test charge momentum ( Px, Py, Pz ) = ( Pxo, 0, 0 ) at tine
T=0. M is the rest mass of the charge. The solution is obtained by
setting

dPy / dT = QE (2)

Px = Pxo = MVx / sqrt( 1- (V/C)^2 ) (3a)

Py = MVy / sqrt( 1- (V/C)^2 ) (3b)

Pz = 0 (4)

where V is the test particle speed and ( Vx, Vy, Vz ) are its vector
components. Solving these equations yields

Vy / C = QET / sqrt( (QET)^2 + (MC)^2 + Pxo^2 ) (5)

Vx / C = Pxo / sqrt( (QET)^2 + (MC)^2 + Pxo^2 ) (6)

Vz / C = 0 (7)

From equation (5), it is apparent that as time, T, increases without
bound, Vy approaches C. This is expected since the only force upon
the particle is in the y-direction. From equation (6), it is apparent that
as time, T, increases without bound, Vx approaches zero. This is not
expected since there is no force applied in the x - direction.

Now, what if we replace the electric field with a gravitational field?
That is, set QE = Mg ? Is the solution the same? For an observer
traveling on the charge, there is a definite difference: In an electric
field, the observer will feel the acceleration of the charge in the
direction of the field, and this will have an effect on his time piece,
but, in a gravitational field, the observer is in free fall and feels no
forces. This will have a different effect on his time piece. The
situation for the stationary observer is similarly complicated.
[Old Man]

In article , (Edward Green) writes:
"Old Man" wrote in message news:3f249770_4@newsfeed...

Edward Green wrote in message
om...

Consider the instantaneous rest frame of a rocket accelerating at 1g
in the z direction, and consider two particles at altitude z0 at t0;
one is stationary in the instantaneous rest frame of the rocket, the
other has no z component of velocity but a relativistic x component
perpendicular to z.

Well ... it is perfectly damn evident that, since in this case the
rocket is "really" doing the moving, that the two particles will
"fall" towards the floor of the rocket at exactly the same rate. The
orthogonal velocity of the second is irrelevant, it's holding at
constant z in our initial inertial frame -- and since we chose the
instantaneous rest frame, there are to first order no relativistic
corrections between rocket frame and rest frame.*

OK. That's pretty damn unambiguous. Since we may extend this
argument to arbitrary particle trajectories, we predict from the EP
that all objects accelerate at the same rate in a uniform G field,
relativistic velocities be damned.

This sounds correct. The OTOH part, below, is too complicated
for Old Man . It's a mater of mind and time over matter. [Old Man]

No damn fair!

You have pushed me into the thorny and painful byways of "thought",
and now you must join me in the bracken field!

(I had to retype that three damn times because of software glitches,
and each time the joke grew more ornate -- so it's no my, dare I say,
damn, fault ;-).

Anyway, the snipped part below was relatively straight forward ...
just involving the time dilation in the inertial projectile's frame,
and the effect it would have on the apparent acceleration in that
frame past the ... err ... frames of the rocket. The "*" statement
worries me more, since I'm not sure if some baby-bathwater escapage
did not occur there -- needs more thought about clocks in accelerated
frames.

I did make a baby/bathwater kind of error in your orbital angular
momentum thread, BTW -- though I claim I left suitable hedges :-)
Viz., I blithly claimed a course change is a course change, and costs
us locally the same whatever we think of the global energetic orbital
implications.

Well, you did some good thinking there, so let me add the few missing
bits.

First, yes, it is a good idea to consider things locally, it
simplifies matters enormously. Especially, it is a good idea (with
rockets) to think in terms of "short burn", one occuring fast enough
so that for all practical purposes the craft remains in the same
location on orbit during the burn. This enables you to ignore gravity
alltogether when analyzing the results.

As for "change" and "cost", you've to be a bit careful. Yes, a
momentum change is a momentum change and once you found it relative to
one reference frame, you got it. Also, the energy expended for this
change is the same relative to all reference frames. *However*, the
split of said energy into the part that's thrown away with the exhaust
and the part which stays with the craft, this *is* reference frame
dependent. And as for change in angular momentum, this is most
certainly reference frame dependent.

I still stand by this ... but the expert John Prussing
points out that your solution lies precisely in using awareness of the
global properties of the orbits to pick the optimal points to spend
your course change currency.

John knows his stuff. To put what he wrote in the most simplified
form possible, the rules of thumb a

1) For a given delta_v (meaning given amount of fuel used) you get
the maximal change in energy by performing the burn at
peri...whatever (and you should thrust along the direction of motion)

2) For a given delta_v you get the maximal change in angular momentum
by performing the burn at ap...whatever.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

In writes:

In article , (Edward Green) writes:
"Old Man" wrote in message news:3f249770_4@newsfeed...

Edward Green wrote in message
om...

Consider the instantaneous rest frame of a rocket accelerating at 1g
in the z direction, and consider two particles at altitude z0 at t0;
one is stationary in the instantaneous rest frame of the rocket, the
other has no z component of velocity but a relativistic x component
perpendicular to z.

Well ... it is perfectly damn evident that, since in this case the
rocket is "really" doing the moving, that the two particles will
"fall" towards the floor of the rocket at exactly the same rate. The
orthogonal velocity of the second is irrelevant, it's holding at
constant z in our initial inertial frame -- and since we chose the
instantaneous rest frame, there are to first order no relativistic
corrections between rocket frame and rest frame.*

OK. That's pretty damn unambiguous. Since we may extend this
argument to arbitrary particle trajectories, we predict from the EP
that all objects accelerate at the same rate in a uniform G field,
relativistic velocities be damned.

This sounds correct. The OTOH part, below, is too complicated
for Old Man . It's a mater of mind and time over matter. [Old Man]

No damn fair!

You have pushed me into the thorny and painful byways of "thought",
and now you must join me in the bracken field!

(I had to retype that three damn times because of software glitches,
and each time the joke grew more ornate -- so it's no my, dare I say,
damn, fault ;-).

Anyway, the snipped part below was relatively straight forward ...
just involving the time dilation in the inertial projectile's frame,
and the effect it would have on the apparent acceleration in that
frame past the ... err ... frames of the rocket. The "*" statement
worries me more, since I'm not sure if some baby-bathwater escapage
did not occur there -- needs more thought about clocks in accelerated
frames.

I did make a baby/bathwater kind of error in your orbital angular
momentum thread, BTW -- though I claim I left suitable hedges :-)
Viz., I blithly claimed a course change is a course change, and costs
us locally the same whatever we think of the global energetic orbital
implications.

Well, you did some good thinking there, so let me add the few missing
bits.

First, yes, it is a good idea to consider things locally, it
simplifies matters enormously. Especially, it is a good idea (with
rockets) to think in terms of "short burn", one occuring fast enough
so that for all practical purposes the craft remains in the same
location on orbit during the burn. This enables you to ignore gravity
alltogether when analyzing the results.

As for "change" and "cost", you've to be a bit careful. Yes, a
momentum change is a momentum change and once you found it relative to
one reference frame, you got it. Also, the energy expended for this
change is the same relative to all reference frames. *However*, the
split of said energy into the part that's thrown away with the exhaust
and the part which stays with the craft, this *is* reference frame
dependent. And as for change in angular momentum, this is most
certainly reference frame dependent.

I still stand by this ... but the expert John Prussing
points out that your solution lies precisely in using awareness of the
global properties of the orbits to pick the optimal points to spend
your course change currency.

John knows his stuff. To put what he wrote in the most simplified
form possible, the rules of thumb a

1) For a given delta_v (meaning given amount of fuel used) you get
the maximal change in energy by performing the burn at
peri...whatever (and you should thrust along the direction of motion)

2) For a given delta_v you get the maximal change in angular momentum
by performing the burn at ap...whatever.

And you could add (and you should thrust normal to the radius.)

You make it sound so simple! But, as you know, situations don't always
fit the ideal case. Murphy's Law seems to prevail, as usual.

Re peri...whatever, the generic term usually used is periapse, where apse
means point. In specific cases perigee, perihelion, perijove (Jupiter)
are used. And you have a choice between perilune or periselenium for the
moon, depending on whether you prefer the Latin or Greek word for moon.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"
--
John E. Prussing
University of Illinois at Urbana-Champaign
Department of Aerospace Engineering
http://www.uiuc.edu/~prussing

In article , (John E. Prussing) writes:
In writes:

In article , (Edward Green) writes:
"Old Man" wrote in message news:3f249770_4@newsfeed...

Edward Green wrote in message
om...

Consider the instantaneous rest frame of a rocket accelerating at 1g
in the z direction, and consider two particles at altitude z0 at t0;
one is stationary in the instantaneous rest frame of the rocket, the
other has no z component of velocity but a relativistic x component
perpendicular to z.

Well ... it is perfectly damn evident that, since in this case the
rocket is "really" doing the moving, that the two particles will
"fall" towards the floor of the rocket at exactly the same rate. The
orthogonal velocity of the second is irrelevant, it's holding at
constant z in our initial inertial frame -- and since we chose the
instantaneous rest frame, there are to first order no relativistic
corrections between rocket frame and rest frame.*

OK. That's pretty damn unambiguous. Since we may extend this
argument to arbitrary particle trajectories, we predict from the EP
that all objects accelerate at the same rate in a uniform G field,
relativistic velocities be damned.

This sounds correct. The OTOH part, below, is too complicated
for Old Man . It's a mater of mind and time over matter. [Old Man]

No damn fair!

You have pushed me into the thorny and painful byways of "thought",
and now you must join me in the bracken field!

(I had to retype that three damn times because of software glitches,
and each time the joke grew more ornate -- so it's no my, dare I say,
damn, fault ;-).

Anyway, the snipped part below was relatively straight forward ...
just involving the time dilation in the inertial projectile's frame,
and the effect it would have on the apparent acceleration in that
frame past the ... err ... frames of the rocket. The "*" statement
worries me more, since I'm not sure if some baby-bathwater escapage
did not occur there -- needs more thought about clocks in accelerated
frames.

I did make a baby/bathwater kind of error in your orbital angular
momentum thread, BTW -- though I claim I left suitable hedges :-)
Viz., I blithly claimed a course change is a course change, and costs
us locally the same whatever we think of the global energetic orbital
implications.

Well, you did some good thinking there, so let me add the few missing
bits.

First, yes, it is a good idea to consider things locally, it
simplifies matters enormously. Especially, it is a good idea (with
rockets) to think in terms of "short burn", one occuring fast enough
so that for all practical purposes the craft remains in the same
location on orbit during the burn. This enables you to ignore gravity
alltogether when analyzing the results.

As for "change" and "cost", you've to be a bit careful. Yes, a
momentum change is a momentum change and once you found it relative to
one reference frame, you got it. Also, the energy expended for this
change is the same relative to all reference frames. *However*, the
split of said energy into the part that's thrown away with the exhaust
and the part which stays with the craft, this *is* reference frame
dependent. And as for change in angular momentum, this is most
certainly reference frame dependent.

I still stand by this ... but the expert John Prussing
points out that your solution lies precisely in using awareness of the
global properties of the orbits to pick the optimal points to spend
your course change currency.

John knows his stuff. To put what he wrote in the most simplified
form possible, the rules of thumb a

1) For a given delta_v (meaning given amount of fuel used) you get
the maximal change in energy by performing the burn at
peri...whatever (and you should thrust along the direction of motion)

2) For a given delta_v you get the maximal change in angular momentum
by performing the burn at ap...whatever.

And you could add (and you should thrust normal to the radius.)

Yes, certainly.

You make it sound so simple! But, as you know, situations don't always
fit the ideal case. Murphy's Law seems to prevail, as usual.

That's for sure, being an experimentalist I'm painfully aware of this
fact. Murphy's Law is the one you can always count on:-)

Re peri...whatever, the generic term usually used is periapse, where apse
means point.

Oh, thanks, that's the word I was missing.

In specific cases perigee, perihelion, perijove (Jupiter)
are used. And you have a choice between perilune or periselenium for the
moon, depending on whether you prefer the Latin or Greek word for moon.

I like the sound of periselenium, but chemists may find it confusing.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"